Fault Tolerant Facility Location

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Fault Tolerant Facility Location. Chaitanya SwamyDavid Shmoys Cornell University. Metric Facility Location. 2. facility. F : set of facilities . D : set of clients . Facility i has facility cost f i . c ij : distance between any i and j in D  F . - PowerPoint PPT Presentation

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Fault Tolerant Facility Location

Chaitanya Swamy David ShmoysCornell University

Theory Seminar 042002

Metric Facility Location

F set of facilitiesD set of clients

Facility i has facility cost fi

cij distance between any i and j in D F

Client j wants to be connected to rj distinct facilities

3

2

2

client

facility

Theory Seminar 042002

We want to

1) Pick a set S of facilities to open

2) Assign each client j to rj open facilities

Goal Minimize total facility cost of S + sum of distances(service cost)

3

2

2client

facility

openfacility

Theory Seminar 042002

rj=1

bullLP rounding Shmoys Tardos amp Aardal Chudak amp Shmoys Sviridenko

bullPrimal-dual algorithms Jain amp Vazirani Markakis Mahdian Saberi amp Vazirani(MMSV01) Jain Mahdian amp Saberi(JMS02)

bullLocal search Koropulu Plaxton amp Rajaraman Guha amp Khuller Charikar amp Guha

Best approx - Mahdian Ye amp Zhang (MYZ02) 152

Previous Work

Theory Seminar 042002

Previous Work (contd)

Uniform requirements rj=r

bullMarkakis et al (MMSV01) 1861

Non-uniform requirements rj

bullJain amp Vazirani O(log rmax)

bullGuha Meyerson amp Munagala 247

Our Results

bullNon-uniform rj get a 2076-approx

bullrj=r can extend JMS02 MYZ02 to

get a 152-approx

Theory Seminar 042002

LP Formulation

Primal

Min i fiyi + ji cijxij

st

i xij ge rj j

xij le yi i j

yi le 1 i

xij ge 0 yi ge 0 i j

Theory Seminar 042002

LP Formulation (contd)

Max j rjvj - i zi

st

vj le wij + cij i j

j wij le fi + zi i

vj ge 0 wij ge 0 zi ge 0 i j

Dual

Theory Seminar 042002

Complementary Slackness

Primal Slackness Conditions

bullxij gt 0 vj = wij + cij

bullyi gt 0 j wij = fi + zi

Dual Slackness Conditions

bullvj gt 0 j xij = rj

bullwij gt 0 xij = yi

bullzi gt 0 yi = 1

Theory Seminar 042002

4-approximation outline

Basic Idea vj lsquopaysrsquo for each cij stxij gt 0

Bound service cost for each copy of j by ρvj total service cost leρj

rjvj

Problem Have ndashzis in the dual

But zi gt 0 yi = 1 So can open these facilities and charge all of this cost to the LP

2

j(1) j(2)

le vj

le vj

view as rj

copiesj(c) cth copy

Theory Seminar 042002

The Algorithm

Phase 1 Clustering Ensures that each copy j(c) has a nearby open facilityIterative algorithm

S = j|rj gt 0 Fj = i|xij gt 0 in fi order

Start of iteration

1 Pick j with smallest vj

2 Cluster is M Fj with iM yi = rj

2

51

2

j

client in Sfacility in some Fj

Cluster M

Theory Seminar 042002

0

X XX

30

2

j

3 Open rj cheapest facilities in M

4 For k st Fk M connect rj copies to opened facilities Decrease rk set Fk=Fk-M

End of iteration

client in S

facility in some Fj

client not in S

X facility removed from

Fj

Cluster M

facility opened from M

Theory Seminar 042002

Analysis Phase 1

Solution is feasible each j is connected to rj distinct facilities

Lemma Facility cost lei fiyi

Proof Cost of rj cheapest facilities in M lerj (avg cost) = iM fiyi These facilities donrsquot get used again

Theory Seminar 042002

Analysis (contd)

Lemma For any j and c service cost of copy j(c) le3vj

Proof

vk le vj since k was chosen as cluster center

Service cost le vj + 2vk le 3vj

Cluster M

j(c)le vj

le vk

le vk k

Theory Seminar 042002

The Algorithm (contd)

Phase 2 Taking care of ndashzis

1 Open all (unopened) i st yi = 1

2 For any j if xij = yi = 1 disconnect a copy of j and connect it to i

j

rj = 3

X

i with yi = 1i with yi lt

1 and open

Theory Seminar 042002

Analysis Phase 2

Lemma Cost of phase 2 = fi + cij = j ljvj ndash i zi

Proof Each i with zi gt 0 is opened For iL1 all j st wij gt 0 are connected to it So

vj = (service cost) + (fi +

zi)

j ljvj = fi + cij + i zi

Let L1 = i | yi = 1

Lj = i | xij = 1 L1 and lj = |Lj|

iL1 jiLj

j|iLj

jiLjiL1

Theory Seminar 042002

Finally hellip

Theorem Total cost le 4 times the optimal cost

Proof Total cost le

i fiyi + 3j (rj ndash lj)vj + fi + cij

facility cost of phase 1 cost for

copies connected by

phase 1

cost of phase 2

lei fiyi + 3j (rj ndash lj)vj + (j ljvj ndash i

zi )

lei fiyi + 3(j rjvj ndash i zi )

le4OPT

iL1 jiLj

Theory Seminar 042002

A Randomized Algorithm

Idea Open i with probability ρyi

Expected facility cost le ρi fiyi

Hope that each copy j(c) has a nearby facility open and service cost decreases

Not quitehellip no facility may be open

Cluster facilities open ge 1 facility in each cluster

Theory Seminar 042002

Phase 1 Pruning out ndashzis

Open all i st yi = 1

For each j if xij = yi = 1 connect j to i

Let Lj = i | xij = 1 and lj = |Lj|

Cost = j ljvj ndash i zi

Lj

Fj10

rrsquoj = residual reqmt = 6

Lrsquoj

Phase 2

Open all i st frac12 le yi lt 1

For each j let Lrsquoj = i | frac12 le xij lt 1

Connect copies of j to i Lrsquoj

Lose a factor of 2

facilities opened in

phases 1 2yi = 1 frac12 le yi lt 1yi lt frac12

Set L1

Set L2

Theory Seminar 042002

Phase 3

Notation facwt(S j) = iS xij

1 Form clusters Each cluster has facwt ge frac12

2 Open facilities Open ge 1 facility in a cluster ndash used as a backup facility Open facility i with prob 2yi

3 Assign facilities to copies Each copy j(c) gets a preferred set of facilities ndash P(j(c)) with facwt ge frac12 For c d P(j(c)) P(j(d)) =

4 Connect clients Connect j(c) to the nearest i open in P(j(c)) or to a backup facility

Theory Seminar 042002

ClusteringAfter phases 1 and 2 Fj = i | xij lt frac12 Sort these by cij and distribute among the rrsquoj copies

Cj(c) = avg service cost of the cth copy denote c Cj(c) = cijxij by Cj

Initial Fj before any iterations

Cj(1)

Cj(2)

Cj(3)3

i Fj

client j

Want the following properties

Clusters to be disjoint

Each cluster have facwt ge frac12

Each j be connected to rrsquoj clusters

iFj

Theory Seminar 042002

Iterative algorithm

S = j | rrsquoj gt 0

aj = lsquoactiversquo copy of j initially = 1

Ĉj(aj) = avg distance to the first k

facilities in Fj gathering facwt ge frac12

say these facilities lsquoserversquo j

Will maintain Ĉj(aj) le Cj(aj)

X

X

X1

Fj after some iterations

X i removed from Fj

i Fjserving jĈj(3)

facilities serving j

aj = 3

4X

(aj)

(aj)

(aj)

Theory Seminar 042002

Start of iteration

1 Choose j in S with minimum vj + Ĉj(aj)

2 Form cluster M = facilities serving j Note facilities are not split

3 For k st Fk M decrease rrsquok advance ak set Fk = Fk ndash M

2j(3) Cluster M

aj = 1

4

1

XX X

aj = 4 Cluster M

aj = 2

3

client in S

facility in some Fj

X facility removed from Fj

(aj)

Theory Seminar 042002

Opening Facilities

Central facilities opened in 2 steps

1 Open exactly 1 facility in M i opened with prob qyi Acts as backup

denoted b(k ) for each k st Fk M

2 Open each i in M indep with prob (2-q)yi and independent of step 1

Non-central facilities

Cluster M

k

open with prob 2yi independent of other choices

j

(ak

)

Theory Seminar 042002

Let Sj(c) = avg dist from j to P(j(c))

= ( cijxij)facwt(P(j(c))

j)

Then c Sj(c) le 2Cj

Distributing Facilities

iP(j(c))

j

rrsquoj = 3

P(j(1))P(j(2))

P(j(3))

Copy c gets a preferred set P(j(c))Preferred sets are disjoint

Ensure facwt(P(j(c)) j) ge frac12 for all c

Possible to do so since each xij lt frac12

facility in Fj

Theory Seminar 042002

Analysis

Feasibility follows from

1 Facilities in phases 1 2 not reused

2 After clustering j is connected to rrsquoj disjoint clusters backups are distinct

3 Preferred sets are disjoint

So j connected to rj distinct facilities

Theory Seminar 042002

Facility cost

Recall L1 = i | yi = 1

Phase 2 incur a factor of 2

Phase 3 each i is opened with probability 2yi

Expected facility cost le 2 fiyifor phases 2 3

iL1

Theory Seminar 042002

Bounding backup cost denoted by B rv

D event that no i in P(j(c)) is open

Lemma E[B|D] le 2vj + Cj(c)

Proof 2 cases

Service cost I

iM Fj st cik le Ĉj(d)

Also vk + Ĉj(d) le vj + Ĉj(c) le vj + Cj(c)

k(d)

j(c)

le vj

le Ĉj(d)le vk

B

k(d)

j(c)

le vj

le vk

iM Fj cik gt Ĉj(d)

le Ĉj(d) in expectatio

n

1)

2)

backup = b(j(c))

Theory Seminar 042002

Service Cost II

Fix j c Let X(c) = service cost of j(c)

Let di = cij pi = prob i is opened = 2yi

B(c) = backup costD(c) = event that no iP(j(c)) is

openp = Pr[D(c)] = (1-p1)hellip(1-pm) le e-1

davg = weighted avg of the dis

= (i pidi)(i pi) = Sj(c)

d1

d2 dm

P(j(c)) sorted by increasing cij

j(c)

i P(j(c))

Theory Seminar 042002

Then

E[X(c)] = [p1d1 + (1-p1)p2d2 + hellip

+ (1-p1)hellip(1-pm-1)pmdm]

+ pE[B(c)|D(c)]

le (1-p)davg + p[2vj + Cj(c)]

le (1-e-1)Sj(c) + e-1[2vj +

Cj(c)]

Let X = c X(c) = service cost of j

c Sj(c) le 2Cj and c Cj(c) le 2Cj

Summing over all c = 1helliprrsquoj

E[X] le (1-e-1)2Cj + e-1(2rrsquojvj + Cj)

le 2Cj + 2e-1rrsquojvj

Theory Seminar 042002

Putting it all together

Phase 1 pay the optimal LP cost

Phases 2 3

bull Facility cost twice LP facility cost

bull Service costLose a factor of 2 for phase 2Phase 3 cost is 2(LP service cost)+2e-1(dual value)

Overall cost for le (2+2e)(LP cost) phases 2 3

Total cost le (2+2e)OPT

Theory Seminar 042002

How to improve this

bull Distribute facilities more equitably (in an expected sense) among copies - decreases prob of lsquobadrsquo event

bull Better analysis ndash maximum distance within a cluster can be bounded by 2Cj(c)

bull Balance phases 2 and 3

Theory Seminar 042002

Summary of Results

bullGive a 2076-approx algorithm for non-uniform rjs Based on LP rounding using complem slackness

bullFor rj = r extend the primal-dual algorithm of (JMS02) to get a 152-approximation

bullFault tolerant k medians with rj = r

a Primal-dual algorithm (JMS02) gives a 4-approx using Lagrangean relaxation

b LP rounding gives a factor of 8

Theory Seminar 042002

Open Questions

1 Reduce gap between rj = r non-uniform rj

2 Combinatorial algorithms for non-uniform rj primal-dual local-search

3 Constant-factor approx for fault tolerant k medians with non-uniform rjs

  • Fault Tolerant Facility Location
  • Metric Facility Location
  • PowerPoint Presentation
  • Previous Work
  • Previous Work (contd)
  • LP Formulation
  • LP Formulation (contd)
  • Complementary Slackness
  • 4-approximation outline
  • The Algorithm
  • Slide 11
  • Analysis Phase 1
  • Analysis (contd)
  • The Algorithm (contd)
  • Analysis Phase 2
  • Finally hellip
  • A Randomized Algorithm
  • Slide 18
  • Slide 19
  • Clustering
  • Slide 21
  • Slide 22
  • Opening Facilities
  • Distributing Facilities
  • Analysis
  • Facility cost
  • Service cost I
  • Service Cost II
  • Slide 29
  • Putting it all together
  • How to improve this
  • Summary of Results
  • Open Questions

    Theory Seminar 042002

    Metric Facility Location

    F set of facilitiesD set of clients

    Facility i has facility cost fi

    cij distance between any i and j in D F

    Client j wants to be connected to rj distinct facilities

    3

    2

    2

    client

    facility

    Theory Seminar 042002

    We want to

    1) Pick a set S of facilities to open

    2) Assign each client j to rj open facilities

    Goal Minimize total facility cost of S + sum of distances(service cost)

    3

    2

    2client

    facility

    openfacility

    Theory Seminar 042002

    rj=1

    bullLP rounding Shmoys Tardos amp Aardal Chudak amp Shmoys Sviridenko

    bullPrimal-dual algorithms Jain amp Vazirani Markakis Mahdian Saberi amp Vazirani(MMSV01) Jain Mahdian amp Saberi(JMS02)

    bullLocal search Koropulu Plaxton amp Rajaraman Guha amp Khuller Charikar amp Guha

    Best approx - Mahdian Ye amp Zhang (MYZ02) 152

    Previous Work

    Theory Seminar 042002

    Previous Work (contd)

    Uniform requirements rj=r

    bullMarkakis et al (MMSV01) 1861

    Non-uniform requirements rj

    bullJain amp Vazirani O(log rmax)

    bullGuha Meyerson amp Munagala 247

    Our Results

    bullNon-uniform rj get a 2076-approx

    bullrj=r can extend JMS02 MYZ02 to

    get a 152-approx

    Theory Seminar 042002

    LP Formulation

    Primal

    Min i fiyi + ji cijxij

    st

    i xij ge rj j

    xij le yi i j

    yi le 1 i

    xij ge 0 yi ge 0 i j

    Theory Seminar 042002

    LP Formulation (contd)

    Max j rjvj - i zi

    st

    vj le wij + cij i j

    j wij le fi + zi i

    vj ge 0 wij ge 0 zi ge 0 i j

    Dual

    Theory Seminar 042002

    Complementary Slackness

    Primal Slackness Conditions

    bullxij gt 0 vj = wij + cij

    bullyi gt 0 j wij = fi + zi

    Dual Slackness Conditions

    bullvj gt 0 j xij = rj

    bullwij gt 0 xij = yi

    bullzi gt 0 yi = 1

    Theory Seminar 042002

    4-approximation outline

    Basic Idea vj lsquopaysrsquo for each cij stxij gt 0

    Bound service cost for each copy of j by ρvj total service cost leρj

    rjvj

    Problem Have ndashzis in the dual

    But zi gt 0 yi = 1 So can open these facilities and charge all of this cost to the LP

    2

    j(1) j(2)

    le vj

    le vj

    view as rj

    copiesj(c) cth copy

    Theory Seminar 042002

    The Algorithm

    Phase 1 Clustering Ensures that each copy j(c) has a nearby open facilityIterative algorithm

    S = j|rj gt 0 Fj = i|xij gt 0 in fi order

    Start of iteration

    1 Pick j with smallest vj

    2 Cluster is M Fj with iM yi = rj

    2

    51

    2

    j

    client in Sfacility in some Fj

    Cluster M

    Theory Seminar 042002

    0

    X XX

    30

    2

    j

    3 Open rj cheapest facilities in M

    4 For k st Fk M connect rj copies to opened facilities Decrease rk set Fk=Fk-M

    End of iteration

    client in S

    facility in some Fj

    client not in S

    X facility removed from

    Fj

    Cluster M

    facility opened from M

    Theory Seminar 042002

    Analysis Phase 1

    Solution is feasible each j is connected to rj distinct facilities

    Lemma Facility cost lei fiyi

    Proof Cost of rj cheapest facilities in M lerj (avg cost) = iM fiyi These facilities donrsquot get used again

    Theory Seminar 042002

    Analysis (contd)

    Lemma For any j and c service cost of copy j(c) le3vj

    Proof

    vk le vj since k was chosen as cluster center

    Service cost le vj + 2vk le 3vj

    Cluster M

    j(c)le vj

    le vk

    le vk k

    Theory Seminar 042002

    The Algorithm (contd)

    Phase 2 Taking care of ndashzis

    1 Open all (unopened) i st yi = 1

    2 For any j if xij = yi = 1 disconnect a copy of j and connect it to i

    j

    rj = 3

    X

    i with yi = 1i with yi lt

    1 and open

    Theory Seminar 042002

    Analysis Phase 2

    Lemma Cost of phase 2 = fi + cij = j ljvj ndash i zi

    Proof Each i with zi gt 0 is opened For iL1 all j st wij gt 0 are connected to it So

    vj = (service cost) + (fi +

    zi)

    j ljvj = fi + cij + i zi

    Let L1 = i | yi = 1

    Lj = i | xij = 1 L1 and lj = |Lj|

    iL1 jiLj

    j|iLj

    jiLjiL1

    Theory Seminar 042002

    Finally hellip

    Theorem Total cost le 4 times the optimal cost

    Proof Total cost le

    i fiyi + 3j (rj ndash lj)vj + fi + cij

    facility cost of phase 1 cost for

    copies connected by

    phase 1

    cost of phase 2

    lei fiyi + 3j (rj ndash lj)vj + (j ljvj ndash i

    zi )

    lei fiyi + 3(j rjvj ndash i zi )

    le4OPT

    iL1 jiLj

    Theory Seminar 042002

    A Randomized Algorithm

    Idea Open i with probability ρyi

    Expected facility cost le ρi fiyi

    Hope that each copy j(c) has a nearby facility open and service cost decreases

    Not quitehellip no facility may be open

    Cluster facilities open ge 1 facility in each cluster

    Theory Seminar 042002

    Phase 1 Pruning out ndashzis

    Open all i st yi = 1

    For each j if xij = yi = 1 connect j to i

    Let Lj = i | xij = 1 and lj = |Lj|

    Cost = j ljvj ndash i zi

    Lj

    Fj10

    rrsquoj = residual reqmt = 6

    Lrsquoj

    Phase 2

    Open all i st frac12 le yi lt 1

    For each j let Lrsquoj = i | frac12 le xij lt 1

    Connect copies of j to i Lrsquoj

    Lose a factor of 2

    facilities opened in

    phases 1 2yi = 1 frac12 le yi lt 1yi lt frac12

    Set L1

    Set L2

    Theory Seminar 042002

    Phase 3

    Notation facwt(S j) = iS xij

    1 Form clusters Each cluster has facwt ge frac12

    2 Open facilities Open ge 1 facility in a cluster ndash used as a backup facility Open facility i with prob 2yi

    3 Assign facilities to copies Each copy j(c) gets a preferred set of facilities ndash P(j(c)) with facwt ge frac12 For c d P(j(c)) P(j(d)) =

    4 Connect clients Connect j(c) to the nearest i open in P(j(c)) or to a backup facility

    Theory Seminar 042002

    ClusteringAfter phases 1 and 2 Fj = i | xij lt frac12 Sort these by cij and distribute among the rrsquoj copies

    Cj(c) = avg service cost of the cth copy denote c Cj(c) = cijxij by Cj

    Initial Fj before any iterations

    Cj(1)

    Cj(2)

    Cj(3)3

    i Fj

    client j

    Want the following properties

    Clusters to be disjoint

    Each cluster have facwt ge frac12

    Each j be connected to rrsquoj clusters

    iFj

    Theory Seminar 042002

    Iterative algorithm

    S = j | rrsquoj gt 0

    aj = lsquoactiversquo copy of j initially = 1

    Ĉj(aj) = avg distance to the first k

    facilities in Fj gathering facwt ge frac12

    say these facilities lsquoserversquo j

    Will maintain Ĉj(aj) le Cj(aj)

    X

    X

    X1

    Fj after some iterations

    X i removed from Fj

    i Fjserving jĈj(3)

    facilities serving j

    aj = 3

    4X

    (aj)

    (aj)

    (aj)

    Theory Seminar 042002

    Start of iteration

    1 Choose j in S with minimum vj + Ĉj(aj)

    2 Form cluster M = facilities serving j Note facilities are not split

    3 For k st Fk M decrease rrsquok advance ak set Fk = Fk ndash M

    2j(3) Cluster M

    aj = 1

    4

    1

    XX X

    aj = 4 Cluster M

    aj = 2

    3

    client in S

    facility in some Fj

    X facility removed from Fj

    (aj)

    Theory Seminar 042002

    Opening Facilities

    Central facilities opened in 2 steps

    1 Open exactly 1 facility in M i opened with prob qyi Acts as backup

    denoted b(k ) for each k st Fk M

    2 Open each i in M indep with prob (2-q)yi and independent of step 1

    Non-central facilities

    Cluster M

    k

    open with prob 2yi independent of other choices

    j

    (ak

    )

    Theory Seminar 042002

    Let Sj(c) = avg dist from j to P(j(c))

    = ( cijxij)facwt(P(j(c))

    j)

    Then c Sj(c) le 2Cj

    Distributing Facilities

    iP(j(c))

    j

    rrsquoj = 3

    P(j(1))P(j(2))

    P(j(3))

    Copy c gets a preferred set P(j(c))Preferred sets are disjoint

    Ensure facwt(P(j(c)) j) ge frac12 for all c

    Possible to do so since each xij lt frac12

    facility in Fj

    Theory Seminar 042002

    Analysis

    Feasibility follows from

    1 Facilities in phases 1 2 not reused

    2 After clustering j is connected to rrsquoj disjoint clusters backups are distinct

    3 Preferred sets are disjoint

    So j connected to rj distinct facilities

    Theory Seminar 042002

    Facility cost

    Recall L1 = i | yi = 1

    Phase 2 incur a factor of 2

    Phase 3 each i is opened with probability 2yi

    Expected facility cost le 2 fiyifor phases 2 3

    iL1

    Theory Seminar 042002

    Bounding backup cost denoted by B rv

    D event that no i in P(j(c)) is open

    Lemma E[B|D] le 2vj + Cj(c)

    Proof 2 cases

    Service cost I

    iM Fj st cik le Ĉj(d)

    Also vk + Ĉj(d) le vj + Ĉj(c) le vj + Cj(c)

    k(d)

    j(c)

    le vj

    le Ĉj(d)le vk

    B

    k(d)

    j(c)

    le vj

    le vk

    iM Fj cik gt Ĉj(d)

    le Ĉj(d) in expectatio

    n

    1)

    2)

    backup = b(j(c))

    Theory Seminar 042002

    Service Cost II

    Fix j c Let X(c) = service cost of j(c)

    Let di = cij pi = prob i is opened = 2yi

    B(c) = backup costD(c) = event that no iP(j(c)) is

    openp = Pr[D(c)] = (1-p1)hellip(1-pm) le e-1

    davg = weighted avg of the dis

    = (i pidi)(i pi) = Sj(c)

    d1

    d2 dm

    P(j(c)) sorted by increasing cij

    j(c)

    i P(j(c))

    Theory Seminar 042002

    Then

    E[X(c)] = [p1d1 + (1-p1)p2d2 + hellip

    + (1-p1)hellip(1-pm-1)pmdm]

    + pE[B(c)|D(c)]

    le (1-p)davg + p[2vj + Cj(c)]

    le (1-e-1)Sj(c) + e-1[2vj +

    Cj(c)]

    Let X = c X(c) = service cost of j

    c Sj(c) le 2Cj and c Cj(c) le 2Cj

    Summing over all c = 1helliprrsquoj

    E[X] le (1-e-1)2Cj + e-1(2rrsquojvj + Cj)

    le 2Cj + 2e-1rrsquojvj

    Theory Seminar 042002

    Putting it all together

    Phase 1 pay the optimal LP cost

    Phases 2 3

    bull Facility cost twice LP facility cost

    bull Service costLose a factor of 2 for phase 2Phase 3 cost is 2(LP service cost)+2e-1(dual value)

    Overall cost for le (2+2e)(LP cost) phases 2 3

    Total cost le (2+2e)OPT

    Theory Seminar 042002

    How to improve this

    bull Distribute facilities more equitably (in an expected sense) among copies - decreases prob of lsquobadrsquo event

    bull Better analysis ndash maximum distance within a cluster can be bounded by 2Cj(c)

    bull Balance phases 2 and 3

    Theory Seminar 042002

    Summary of Results

    bullGive a 2076-approx algorithm for non-uniform rjs Based on LP rounding using complem slackness

    bullFor rj = r extend the primal-dual algorithm of (JMS02) to get a 152-approximation

    bullFault tolerant k medians with rj = r

    a Primal-dual algorithm (JMS02) gives a 4-approx using Lagrangean relaxation

    b LP rounding gives a factor of 8

    Theory Seminar 042002

    Open Questions

    1 Reduce gap between rj = r non-uniform rj

    2 Combinatorial algorithms for non-uniform rj primal-dual local-search

    3 Constant-factor approx for fault tolerant k medians with non-uniform rjs

    • Fault Tolerant Facility Location
    • Metric Facility Location
    • PowerPoint Presentation
    • Previous Work
    • Previous Work (contd)
    • LP Formulation
    • LP Formulation (contd)
    • Complementary Slackness
    • 4-approximation outline
    • The Algorithm
    • Slide 11
    • Analysis Phase 1
    • Analysis (contd)
    • The Algorithm (contd)
    • Analysis Phase 2
    • Finally hellip
    • A Randomized Algorithm
    • Slide 18
    • Slide 19
    • Clustering
    • Slide 21
    • Slide 22
    • Opening Facilities
    • Distributing Facilities
    • Analysis
    • Facility cost
    • Service cost I
    • Service Cost II
    • Slide 29
    • Putting it all together
    • How to improve this
    • Summary of Results
    • Open Questions

      Theory Seminar 042002

      We want to

      1) Pick a set S of facilities to open

      2) Assign each client j to rj open facilities

      Goal Minimize total facility cost of S + sum of distances(service cost)

      3

      2

      2client

      facility

      openfacility

      Theory Seminar 042002

      rj=1

      bullLP rounding Shmoys Tardos amp Aardal Chudak amp Shmoys Sviridenko

      bullPrimal-dual algorithms Jain amp Vazirani Markakis Mahdian Saberi amp Vazirani(MMSV01) Jain Mahdian amp Saberi(JMS02)

      bullLocal search Koropulu Plaxton amp Rajaraman Guha amp Khuller Charikar amp Guha

      Best approx - Mahdian Ye amp Zhang (MYZ02) 152

      Previous Work

      Theory Seminar 042002

      Previous Work (contd)

      Uniform requirements rj=r

      bullMarkakis et al (MMSV01) 1861

      Non-uniform requirements rj

      bullJain amp Vazirani O(log rmax)

      bullGuha Meyerson amp Munagala 247

      Our Results

      bullNon-uniform rj get a 2076-approx

      bullrj=r can extend JMS02 MYZ02 to

      get a 152-approx

      Theory Seminar 042002

      LP Formulation

      Primal

      Min i fiyi + ji cijxij

      st

      i xij ge rj j

      xij le yi i j

      yi le 1 i

      xij ge 0 yi ge 0 i j

      Theory Seminar 042002

      LP Formulation (contd)

      Max j rjvj - i zi

      st

      vj le wij + cij i j

      j wij le fi + zi i

      vj ge 0 wij ge 0 zi ge 0 i j

      Dual

      Theory Seminar 042002

      Complementary Slackness

      Primal Slackness Conditions

      bullxij gt 0 vj = wij + cij

      bullyi gt 0 j wij = fi + zi

      Dual Slackness Conditions

      bullvj gt 0 j xij = rj

      bullwij gt 0 xij = yi

      bullzi gt 0 yi = 1

      Theory Seminar 042002

      4-approximation outline

      Basic Idea vj lsquopaysrsquo for each cij stxij gt 0

      Bound service cost for each copy of j by ρvj total service cost leρj

      rjvj

      Problem Have ndashzis in the dual

      But zi gt 0 yi = 1 So can open these facilities and charge all of this cost to the LP

      2

      j(1) j(2)

      le vj

      le vj

      view as rj

      copiesj(c) cth copy

      Theory Seminar 042002

      The Algorithm

      Phase 1 Clustering Ensures that each copy j(c) has a nearby open facilityIterative algorithm

      S = j|rj gt 0 Fj = i|xij gt 0 in fi order

      Start of iteration

      1 Pick j with smallest vj

      2 Cluster is M Fj with iM yi = rj

      2

      51

      2

      j

      client in Sfacility in some Fj

      Cluster M

      Theory Seminar 042002

      0

      X XX

      30

      2

      j

      3 Open rj cheapest facilities in M

      4 For k st Fk M connect rj copies to opened facilities Decrease rk set Fk=Fk-M

      End of iteration

      client in S

      facility in some Fj

      client not in S

      X facility removed from

      Fj

      Cluster M

      facility opened from M

      Theory Seminar 042002

      Analysis Phase 1

      Solution is feasible each j is connected to rj distinct facilities

      Lemma Facility cost lei fiyi

      Proof Cost of rj cheapest facilities in M lerj (avg cost) = iM fiyi These facilities donrsquot get used again

      Theory Seminar 042002

      Analysis (contd)

      Lemma For any j and c service cost of copy j(c) le3vj

      Proof

      vk le vj since k was chosen as cluster center

      Service cost le vj + 2vk le 3vj

      Cluster M

      j(c)le vj

      le vk

      le vk k

      Theory Seminar 042002

      The Algorithm (contd)

      Phase 2 Taking care of ndashzis

      1 Open all (unopened) i st yi = 1

      2 For any j if xij = yi = 1 disconnect a copy of j and connect it to i

      j

      rj = 3

      X

      i with yi = 1i with yi lt

      1 and open

      Theory Seminar 042002

      Analysis Phase 2

      Lemma Cost of phase 2 = fi + cij = j ljvj ndash i zi

      Proof Each i with zi gt 0 is opened For iL1 all j st wij gt 0 are connected to it So

      vj = (service cost) + (fi +

      zi)

      j ljvj = fi + cij + i zi

      Let L1 = i | yi = 1

      Lj = i | xij = 1 L1 and lj = |Lj|

      iL1 jiLj

      j|iLj

      jiLjiL1

      Theory Seminar 042002

      Finally hellip

      Theorem Total cost le 4 times the optimal cost

      Proof Total cost le

      i fiyi + 3j (rj ndash lj)vj + fi + cij

      facility cost of phase 1 cost for

      copies connected by

      phase 1

      cost of phase 2

      lei fiyi + 3j (rj ndash lj)vj + (j ljvj ndash i

      zi )

      lei fiyi + 3(j rjvj ndash i zi )

      le4OPT

      iL1 jiLj

      Theory Seminar 042002

      A Randomized Algorithm

      Idea Open i with probability ρyi

      Expected facility cost le ρi fiyi

      Hope that each copy j(c) has a nearby facility open and service cost decreases

      Not quitehellip no facility may be open

      Cluster facilities open ge 1 facility in each cluster

      Theory Seminar 042002

      Phase 1 Pruning out ndashzis

      Open all i st yi = 1

      For each j if xij = yi = 1 connect j to i

      Let Lj = i | xij = 1 and lj = |Lj|

      Cost = j ljvj ndash i zi

      Lj

      Fj10

      rrsquoj = residual reqmt = 6

      Lrsquoj

      Phase 2

      Open all i st frac12 le yi lt 1

      For each j let Lrsquoj = i | frac12 le xij lt 1

      Connect copies of j to i Lrsquoj

      Lose a factor of 2

      facilities opened in

      phases 1 2yi = 1 frac12 le yi lt 1yi lt frac12

      Set L1

      Set L2

      Theory Seminar 042002

      Phase 3

      Notation facwt(S j) = iS xij

      1 Form clusters Each cluster has facwt ge frac12

      2 Open facilities Open ge 1 facility in a cluster ndash used as a backup facility Open facility i with prob 2yi

      3 Assign facilities to copies Each copy j(c) gets a preferred set of facilities ndash P(j(c)) with facwt ge frac12 For c d P(j(c)) P(j(d)) =

      4 Connect clients Connect j(c) to the nearest i open in P(j(c)) or to a backup facility

      Theory Seminar 042002

      ClusteringAfter phases 1 and 2 Fj = i | xij lt frac12 Sort these by cij and distribute among the rrsquoj copies

      Cj(c) = avg service cost of the cth copy denote c Cj(c) = cijxij by Cj

      Initial Fj before any iterations

      Cj(1)

      Cj(2)

      Cj(3)3

      i Fj

      client j

      Want the following properties

      Clusters to be disjoint

      Each cluster have facwt ge frac12

      Each j be connected to rrsquoj clusters

      iFj

      Theory Seminar 042002

      Iterative algorithm

      S = j | rrsquoj gt 0

      aj = lsquoactiversquo copy of j initially = 1

      Ĉj(aj) = avg distance to the first k

      facilities in Fj gathering facwt ge frac12

      say these facilities lsquoserversquo j

      Will maintain Ĉj(aj) le Cj(aj)

      X

      X

      X1

      Fj after some iterations

      X i removed from Fj

      i Fjserving jĈj(3)

      facilities serving j

      aj = 3

      4X

      (aj)

      (aj)

      (aj)

      Theory Seminar 042002

      Start of iteration

      1 Choose j in S with minimum vj + Ĉj(aj)

      2 Form cluster M = facilities serving j Note facilities are not split

      3 For k st Fk M decrease rrsquok advance ak set Fk = Fk ndash M

      2j(3) Cluster M

      aj = 1

      4

      1

      XX X

      aj = 4 Cluster M

      aj = 2

      3

      client in S

      facility in some Fj

      X facility removed from Fj

      (aj)

      Theory Seminar 042002

      Opening Facilities

      Central facilities opened in 2 steps

      1 Open exactly 1 facility in M i opened with prob qyi Acts as backup

      denoted b(k ) for each k st Fk M

      2 Open each i in M indep with prob (2-q)yi and independent of step 1

      Non-central facilities

      Cluster M

      k

      open with prob 2yi independent of other choices

      j

      (ak

      )

      Theory Seminar 042002

      Let Sj(c) = avg dist from j to P(j(c))

      = ( cijxij)facwt(P(j(c))

      j)

      Then c Sj(c) le 2Cj

      Distributing Facilities

      iP(j(c))

      j

      rrsquoj = 3

      P(j(1))P(j(2))

      P(j(3))

      Copy c gets a preferred set P(j(c))Preferred sets are disjoint

      Ensure facwt(P(j(c)) j) ge frac12 for all c

      Possible to do so since each xij lt frac12

      facility in Fj

      Theory Seminar 042002

      Analysis

      Feasibility follows from

      1 Facilities in phases 1 2 not reused

      2 After clustering j is connected to rrsquoj disjoint clusters backups are distinct

      3 Preferred sets are disjoint

      So j connected to rj distinct facilities

      Theory Seminar 042002

      Facility cost

      Recall L1 = i | yi = 1

      Phase 2 incur a factor of 2

      Phase 3 each i is opened with probability 2yi

      Expected facility cost le 2 fiyifor phases 2 3

      iL1

      Theory Seminar 042002

      Bounding backup cost denoted by B rv

      D event that no i in P(j(c)) is open

      Lemma E[B|D] le 2vj + Cj(c)

      Proof 2 cases

      Service cost I

      iM Fj st cik le Ĉj(d)

      Also vk + Ĉj(d) le vj + Ĉj(c) le vj + Cj(c)

      k(d)

      j(c)

      le vj

      le Ĉj(d)le vk

      B

      k(d)

      j(c)

      le vj

      le vk

      iM Fj cik gt Ĉj(d)

      le Ĉj(d) in expectatio

      n

      1)

      2)

      backup = b(j(c))

      Theory Seminar 042002

      Service Cost II

      Fix j c Let X(c) = service cost of j(c)

      Let di = cij pi = prob i is opened = 2yi

      B(c) = backup costD(c) = event that no iP(j(c)) is

      openp = Pr[D(c)] = (1-p1)hellip(1-pm) le e-1

      davg = weighted avg of the dis

      = (i pidi)(i pi) = Sj(c)

      d1

      d2 dm

      P(j(c)) sorted by increasing cij

      j(c)

      i P(j(c))

      Theory Seminar 042002

      Then

      E[X(c)] = [p1d1 + (1-p1)p2d2 + hellip

      + (1-p1)hellip(1-pm-1)pmdm]

      + pE[B(c)|D(c)]

      le (1-p)davg + p[2vj + Cj(c)]

      le (1-e-1)Sj(c) + e-1[2vj +

      Cj(c)]

      Let X = c X(c) = service cost of j

      c Sj(c) le 2Cj and c Cj(c) le 2Cj

      Summing over all c = 1helliprrsquoj

      E[X] le (1-e-1)2Cj + e-1(2rrsquojvj + Cj)

      le 2Cj + 2e-1rrsquojvj

      Theory Seminar 042002

      Putting it all together

      Phase 1 pay the optimal LP cost

      Phases 2 3

      bull Facility cost twice LP facility cost

      bull Service costLose a factor of 2 for phase 2Phase 3 cost is 2(LP service cost)+2e-1(dual value)

      Overall cost for le (2+2e)(LP cost) phases 2 3

      Total cost le (2+2e)OPT

      Theory Seminar 042002

      How to improve this

      bull Distribute facilities more equitably (in an expected sense) among copies - decreases prob of lsquobadrsquo event

      bull Better analysis ndash maximum distance within a cluster can be bounded by 2Cj(c)

      bull Balance phases 2 and 3

      Theory Seminar 042002

      Summary of Results

      bullGive a 2076-approx algorithm for non-uniform rjs Based on LP rounding using complem slackness

      bullFor rj = r extend the primal-dual algorithm of (JMS02) to get a 152-approximation

      bullFault tolerant k medians with rj = r

      a Primal-dual algorithm (JMS02) gives a 4-approx using Lagrangean relaxation

      b LP rounding gives a factor of 8

      Theory Seminar 042002

      Open Questions

      1 Reduce gap between rj = r non-uniform rj

      2 Combinatorial algorithms for non-uniform rj primal-dual local-search

      3 Constant-factor approx for fault tolerant k medians with non-uniform rjs

      • Fault Tolerant Facility Location
      • Metric Facility Location
      • PowerPoint Presentation
      • Previous Work
      • Previous Work (contd)
      • LP Formulation
      • LP Formulation (contd)
      • Complementary Slackness
      • 4-approximation outline
      • The Algorithm
      • Slide 11
      • Analysis Phase 1
      • Analysis (contd)
      • The Algorithm (contd)
      • Analysis Phase 2
      • Finally hellip
      • A Randomized Algorithm
      • Slide 18
      • Slide 19
      • Clustering
      • Slide 21
      • Slide 22
      • Opening Facilities
      • Distributing Facilities
      • Analysis
      • Facility cost
      • Service cost I
      • Service Cost II
      • Slide 29
      • Putting it all together
      • How to improve this
      • Summary of Results
      • Open Questions

        Theory Seminar 042002

        rj=1

        bullLP rounding Shmoys Tardos amp Aardal Chudak amp Shmoys Sviridenko

        bullPrimal-dual algorithms Jain amp Vazirani Markakis Mahdian Saberi amp Vazirani(MMSV01) Jain Mahdian amp Saberi(JMS02)

        bullLocal search Koropulu Plaxton amp Rajaraman Guha amp Khuller Charikar amp Guha

        Best approx - Mahdian Ye amp Zhang (MYZ02) 152

        Previous Work

        Theory Seminar 042002

        Previous Work (contd)

        Uniform requirements rj=r

        bullMarkakis et al (MMSV01) 1861

        Non-uniform requirements rj

        bullJain amp Vazirani O(log rmax)

        bullGuha Meyerson amp Munagala 247

        Our Results

        bullNon-uniform rj get a 2076-approx

        bullrj=r can extend JMS02 MYZ02 to

        get a 152-approx

        Theory Seminar 042002

        LP Formulation

        Primal

        Min i fiyi + ji cijxij

        st

        i xij ge rj j

        xij le yi i j

        yi le 1 i

        xij ge 0 yi ge 0 i j

        Theory Seminar 042002

        LP Formulation (contd)

        Max j rjvj - i zi

        st

        vj le wij + cij i j

        j wij le fi + zi i

        vj ge 0 wij ge 0 zi ge 0 i j

        Dual

        Theory Seminar 042002

        Complementary Slackness

        Primal Slackness Conditions

        bullxij gt 0 vj = wij + cij

        bullyi gt 0 j wij = fi + zi

        Dual Slackness Conditions

        bullvj gt 0 j xij = rj

        bullwij gt 0 xij = yi

        bullzi gt 0 yi = 1

        Theory Seminar 042002

        4-approximation outline

        Basic Idea vj lsquopaysrsquo for each cij stxij gt 0

        Bound service cost for each copy of j by ρvj total service cost leρj

        rjvj

        Problem Have ndashzis in the dual

        But zi gt 0 yi = 1 So can open these facilities and charge all of this cost to the LP

        2

        j(1) j(2)

        le vj

        le vj

        view as rj

        copiesj(c) cth copy

        Theory Seminar 042002

        The Algorithm

        Phase 1 Clustering Ensures that each copy j(c) has a nearby open facilityIterative algorithm

        S = j|rj gt 0 Fj = i|xij gt 0 in fi order

        Start of iteration

        1 Pick j with smallest vj

        2 Cluster is M Fj with iM yi = rj

        2

        51

        2

        j

        client in Sfacility in some Fj

        Cluster M

        Theory Seminar 042002

        0

        X XX

        30

        2

        j

        3 Open rj cheapest facilities in M

        4 For k st Fk M connect rj copies to opened facilities Decrease rk set Fk=Fk-M

        End of iteration

        client in S

        facility in some Fj

        client not in S

        X facility removed from

        Fj

        Cluster M

        facility opened from M

        Theory Seminar 042002

        Analysis Phase 1

        Solution is feasible each j is connected to rj distinct facilities

        Lemma Facility cost lei fiyi

        Proof Cost of rj cheapest facilities in M lerj (avg cost) = iM fiyi These facilities donrsquot get used again

        Theory Seminar 042002

        Analysis (contd)

        Lemma For any j and c service cost of copy j(c) le3vj

        Proof

        vk le vj since k was chosen as cluster center

        Service cost le vj + 2vk le 3vj

        Cluster M

        j(c)le vj

        le vk

        le vk k

        Theory Seminar 042002

        The Algorithm (contd)

        Phase 2 Taking care of ndashzis

        1 Open all (unopened) i st yi = 1

        2 For any j if xij = yi = 1 disconnect a copy of j and connect it to i

        j

        rj = 3

        X

        i with yi = 1i with yi lt

        1 and open

        Theory Seminar 042002

        Analysis Phase 2

        Lemma Cost of phase 2 = fi + cij = j ljvj ndash i zi

        Proof Each i with zi gt 0 is opened For iL1 all j st wij gt 0 are connected to it So

        vj = (service cost) + (fi +

        zi)

        j ljvj = fi + cij + i zi

        Let L1 = i | yi = 1

        Lj = i | xij = 1 L1 and lj = |Lj|

        iL1 jiLj

        j|iLj

        jiLjiL1

        Theory Seminar 042002

        Finally hellip

        Theorem Total cost le 4 times the optimal cost

        Proof Total cost le

        i fiyi + 3j (rj ndash lj)vj + fi + cij

        facility cost of phase 1 cost for

        copies connected by

        phase 1

        cost of phase 2

        lei fiyi + 3j (rj ndash lj)vj + (j ljvj ndash i

        zi )

        lei fiyi + 3(j rjvj ndash i zi )

        le4OPT

        iL1 jiLj

        Theory Seminar 042002

        A Randomized Algorithm

        Idea Open i with probability ρyi

        Expected facility cost le ρi fiyi

        Hope that each copy j(c) has a nearby facility open and service cost decreases

        Not quitehellip no facility may be open

        Cluster facilities open ge 1 facility in each cluster

        Theory Seminar 042002

        Phase 1 Pruning out ndashzis

        Open all i st yi = 1

        For each j if xij = yi = 1 connect j to i

        Let Lj = i | xij = 1 and lj = |Lj|

        Cost = j ljvj ndash i zi

        Lj

        Fj10

        rrsquoj = residual reqmt = 6

        Lrsquoj

        Phase 2

        Open all i st frac12 le yi lt 1

        For each j let Lrsquoj = i | frac12 le xij lt 1

        Connect copies of j to i Lrsquoj

        Lose a factor of 2

        facilities opened in

        phases 1 2yi = 1 frac12 le yi lt 1yi lt frac12

        Set L1

        Set L2

        Theory Seminar 042002

        Phase 3

        Notation facwt(S j) = iS xij

        1 Form clusters Each cluster has facwt ge frac12

        2 Open facilities Open ge 1 facility in a cluster ndash used as a backup facility Open facility i with prob 2yi

        3 Assign facilities to copies Each copy j(c) gets a preferred set of facilities ndash P(j(c)) with facwt ge frac12 For c d P(j(c)) P(j(d)) =

        4 Connect clients Connect j(c) to the nearest i open in P(j(c)) or to a backup facility

        Theory Seminar 042002

        ClusteringAfter phases 1 and 2 Fj = i | xij lt frac12 Sort these by cij and distribute among the rrsquoj copies

        Cj(c) = avg service cost of the cth copy denote c Cj(c) = cijxij by Cj

        Initial Fj before any iterations

        Cj(1)

        Cj(2)

        Cj(3)3

        i Fj

        client j

        Want the following properties

        Clusters to be disjoint

        Each cluster have facwt ge frac12

        Each j be connected to rrsquoj clusters

        iFj

        Theory Seminar 042002

        Iterative algorithm

        S = j | rrsquoj gt 0

        aj = lsquoactiversquo copy of j initially = 1

        Ĉj(aj) = avg distance to the first k

        facilities in Fj gathering facwt ge frac12

        say these facilities lsquoserversquo j

        Will maintain Ĉj(aj) le Cj(aj)

        X

        X

        X1

        Fj after some iterations

        X i removed from Fj

        i Fjserving jĈj(3)

        facilities serving j

        aj = 3

        4X

        (aj)

        (aj)

        (aj)

        Theory Seminar 042002

        Start of iteration

        1 Choose j in S with minimum vj + Ĉj(aj)

        2 Form cluster M = facilities serving j Note facilities are not split

        3 For k st Fk M decrease rrsquok advance ak set Fk = Fk ndash M

        2j(3) Cluster M

        aj = 1

        4

        1

        XX X

        aj = 4 Cluster M

        aj = 2

        3

        client in S

        facility in some Fj

        X facility removed from Fj

        (aj)

        Theory Seminar 042002

        Opening Facilities

        Central facilities opened in 2 steps

        1 Open exactly 1 facility in M i opened with prob qyi Acts as backup

        denoted b(k ) for each k st Fk M

        2 Open each i in M indep with prob (2-q)yi and independent of step 1

        Non-central facilities

        Cluster M

        k

        open with prob 2yi independent of other choices

        j

        (ak

        )

        Theory Seminar 042002

        Let Sj(c) = avg dist from j to P(j(c))

        = ( cijxij)facwt(P(j(c))

        j)

        Then c Sj(c) le 2Cj

        Distributing Facilities

        iP(j(c))

        j

        rrsquoj = 3

        P(j(1))P(j(2))

        P(j(3))

        Copy c gets a preferred set P(j(c))Preferred sets are disjoint

        Ensure facwt(P(j(c)) j) ge frac12 for all c

        Possible to do so since each xij lt frac12

        facility in Fj

        Theory Seminar 042002

        Analysis

        Feasibility follows from

        1 Facilities in phases 1 2 not reused

        2 After clustering j is connected to rrsquoj disjoint clusters backups are distinct

        3 Preferred sets are disjoint

        So j connected to rj distinct facilities

        Theory Seminar 042002

        Facility cost

        Recall L1 = i | yi = 1

        Phase 2 incur a factor of 2

        Phase 3 each i is opened with probability 2yi

        Expected facility cost le 2 fiyifor phases 2 3

        iL1

        Theory Seminar 042002

        Bounding backup cost denoted by B rv

        D event that no i in P(j(c)) is open

        Lemma E[B|D] le 2vj + Cj(c)

        Proof 2 cases

        Service cost I

        iM Fj st cik le Ĉj(d)

        Also vk + Ĉj(d) le vj + Ĉj(c) le vj + Cj(c)

        k(d)

        j(c)

        le vj

        le Ĉj(d)le vk

        B

        k(d)

        j(c)

        le vj

        le vk

        iM Fj cik gt Ĉj(d)

        le Ĉj(d) in expectatio

        n

        1)

        2)

        backup = b(j(c))

        Theory Seminar 042002

        Service Cost II

        Fix j c Let X(c) = service cost of j(c)

        Let di = cij pi = prob i is opened = 2yi

        B(c) = backup costD(c) = event that no iP(j(c)) is

        openp = Pr[D(c)] = (1-p1)hellip(1-pm) le e-1

        davg = weighted avg of the dis

        = (i pidi)(i pi) = Sj(c)

        d1

        d2 dm

        P(j(c)) sorted by increasing cij

        j(c)

        i P(j(c))

        Theory Seminar 042002

        Then

        E[X(c)] = [p1d1 + (1-p1)p2d2 + hellip

        + (1-p1)hellip(1-pm-1)pmdm]

        + pE[B(c)|D(c)]

        le (1-p)davg + p[2vj + Cj(c)]

        le (1-e-1)Sj(c) + e-1[2vj +

        Cj(c)]

        Let X = c X(c) = service cost of j

        c Sj(c) le 2Cj and c Cj(c) le 2Cj

        Summing over all c = 1helliprrsquoj

        E[X] le (1-e-1)2Cj + e-1(2rrsquojvj + Cj)

        le 2Cj + 2e-1rrsquojvj

        Theory Seminar 042002

        Putting it all together

        Phase 1 pay the optimal LP cost

        Phases 2 3

        bull Facility cost twice LP facility cost

        bull Service costLose a factor of 2 for phase 2Phase 3 cost is 2(LP service cost)+2e-1(dual value)

        Overall cost for le (2+2e)(LP cost) phases 2 3

        Total cost le (2+2e)OPT

        Theory Seminar 042002

        How to improve this

        bull Distribute facilities more equitably (in an expected sense) among copies - decreases prob of lsquobadrsquo event

        bull Better analysis ndash maximum distance within a cluster can be bounded by 2Cj(c)

        bull Balance phases 2 and 3

        Theory Seminar 042002

        Summary of Results

        bullGive a 2076-approx algorithm for non-uniform rjs Based on LP rounding using complem slackness

        bullFor rj = r extend the primal-dual algorithm of (JMS02) to get a 152-approximation

        bullFault tolerant k medians with rj = r

        a Primal-dual algorithm (JMS02) gives a 4-approx using Lagrangean relaxation

        b LP rounding gives a factor of 8

        Theory Seminar 042002

        Open Questions

        1 Reduce gap between rj = r non-uniform rj

        2 Combinatorial algorithms for non-uniform rj primal-dual local-search

        3 Constant-factor approx for fault tolerant k medians with non-uniform rjs

        • Fault Tolerant Facility Location
        • Metric Facility Location
        • PowerPoint Presentation
        • Previous Work
        • Previous Work (contd)
        • LP Formulation
        • LP Formulation (contd)
        • Complementary Slackness
        • 4-approximation outline
        • The Algorithm
        • Slide 11
        • Analysis Phase 1
        • Analysis (contd)
        • The Algorithm (contd)
        • Analysis Phase 2
        • Finally hellip
        • A Randomized Algorithm
        • Slide 18
        • Slide 19
        • Clustering
        • Slide 21
        • Slide 22
        • Opening Facilities
        • Distributing Facilities
        • Analysis
        • Facility cost
        • Service cost I
        • Service Cost II
        • Slide 29
        • Putting it all together
        • How to improve this
        • Summary of Results
        • Open Questions

          Theory Seminar 042002

          Previous Work (contd)

          Uniform requirements rj=r

          bullMarkakis et al (MMSV01) 1861

          Non-uniform requirements rj

          bullJain amp Vazirani O(log rmax)

          bullGuha Meyerson amp Munagala 247

          Our Results

          bullNon-uniform rj get a 2076-approx

          bullrj=r can extend JMS02 MYZ02 to

          get a 152-approx

          Theory Seminar 042002

          LP Formulation

          Primal

          Min i fiyi + ji cijxij

          st

          i xij ge rj j

          xij le yi i j

          yi le 1 i

          xij ge 0 yi ge 0 i j

          Theory Seminar 042002

          LP Formulation (contd)

          Max j rjvj - i zi

          st

          vj le wij + cij i j

          j wij le fi + zi i

          vj ge 0 wij ge 0 zi ge 0 i j

          Dual

          Theory Seminar 042002

          Complementary Slackness

          Primal Slackness Conditions

          bullxij gt 0 vj = wij + cij

          bullyi gt 0 j wij = fi + zi

          Dual Slackness Conditions

          bullvj gt 0 j xij = rj

          bullwij gt 0 xij = yi

          bullzi gt 0 yi = 1

          Theory Seminar 042002

          4-approximation outline

          Basic Idea vj lsquopaysrsquo for each cij stxij gt 0

          Bound service cost for each copy of j by ρvj total service cost leρj

          rjvj

          Problem Have ndashzis in the dual

          But zi gt 0 yi = 1 So can open these facilities and charge all of this cost to the LP

          2

          j(1) j(2)

          le vj

          le vj

          view as rj

          copiesj(c) cth copy

          Theory Seminar 042002

          The Algorithm

          Phase 1 Clustering Ensures that each copy j(c) has a nearby open facilityIterative algorithm

          S = j|rj gt 0 Fj = i|xij gt 0 in fi order

          Start of iteration

          1 Pick j with smallest vj

          2 Cluster is M Fj with iM yi = rj

          2

          51

          2

          j

          client in Sfacility in some Fj

          Cluster M

          Theory Seminar 042002

          0

          X XX

          30

          2

          j

          3 Open rj cheapest facilities in M

          4 For k st Fk M connect rj copies to opened facilities Decrease rk set Fk=Fk-M

          End of iteration

          client in S

          facility in some Fj

          client not in S

          X facility removed from

          Fj

          Cluster M

          facility opened from M

          Theory Seminar 042002

          Analysis Phase 1

          Solution is feasible each j is connected to rj distinct facilities

          Lemma Facility cost lei fiyi

          Proof Cost of rj cheapest facilities in M lerj (avg cost) = iM fiyi These facilities donrsquot get used again

          Theory Seminar 042002

          Analysis (contd)

          Lemma For any j and c service cost of copy j(c) le3vj

          Proof

          vk le vj since k was chosen as cluster center

          Service cost le vj + 2vk le 3vj

          Cluster M

          j(c)le vj

          le vk

          le vk k

          Theory Seminar 042002

          The Algorithm (contd)

          Phase 2 Taking care of ndashzis

          1 Open all (unopened) i st yi = 1

          2 For any j if xij = yi = 1 disconnect a copy of j and connect it to i

          j

          rj = 3

          X

          i with yi = 1i with yi lt

          1 and open

          Theory Seminar 042002

          Analysis Phase 2

          Lemma Cost of phase 2 = fi + cij = j ljvj ndash i zi

          Proof Each i with zi gt 0 is opened For iL1 all j st wij gt 0 are connected to it So

          vj = (service cost) + (fi +

          zi)

          j ljvj = fi + cij + i zi

          Let L1 = i | yi = 1

          Lj = i | xij = 1 L1 and lj = |Lj|

          iL1 jiLj

          j|iLj

          jiLjiL1

          Theory Seminar 042002

          Finally hellip

          Theorem Total cost le 4 times the optimal cost

          Proof Total cost le

          i fiyi + 3j (rj ndash lj)vj + fi + cij

          facility cost of phase 1 cost for

          copies connected by

          phase 1

          cost of phase 2

          lei fiyi + 3j (rj ndash lj)vj + (j ljvj ndash i

          zi )

          lei fiyi + 3(j rjvj ndash i zi )

          le4OPT

          iL1 jiLj

          Theory Seminar 042002

          A Randomized Algorithm

          Idea Open i with probability ρyi

          Expected facility cost le ρi fiyi

          Hope that each copy j(c) has a nearby facility open and service cost decreases

          Not quitehellip no facility may be open

          Cluster facilities open ge 1 facility in each cluster

          Theory Seminar 042002

          Phase 1 Pruning out ndashzis

          Open all i st yi = 1

          For each j if xij = yi = 1 connect j to i

          Let Lj = i | xij = 1 and lj = |Lj|

          Cost = j ljvj ndash i zi

          Lj

          Fj10

          rrsquoj = residual reqmt = 6

          Lrsquoj

          Phase 2

          Open all i st frac12 le yi lt 1

          For each j let Lrsquoj = i | frac12 le xij lt 1

          Connect copies of j to i Lrsquoj

          Lose a factor of 2

          facilities opened in

          phases 1 2yi = 1 frac12 le yi lt 1yi lt frac12

          Set L1

          Set L2

          Theory Seminar 042002

          Phase 3

          Notation facwt(S j) = iS xij

          1 Form clusters Each cluster has facwt ge frac12

          2 Open facilities Open ge 1 facility in a cluster ndash used as a backup facility Open facility i with prob 2yi

          3 Assign facilities to copies Each copy j(c) gets a preferred set of facilities ndash P(j(c)) with facwt ge frac12 For c d P(j(c)) P(j(d)) =

          4 Connect clients Connect j(c) to the nearest i open in P(j(c)) or to a backup facility

          Theory Seminar 042002

          ClusteringAfter phases 1 and 2 Fj = i | xij lt frac12 Sort these by cij and distribute among the rrsquoj copies

          Cj(c) = avg service cost of the cth copy denote c Cj(c) = cijxij by Cj

          Initial Fj before any iterations

          Cj(1)

          Cj(2)

          Cj(3)3

          i Fj

          client j

          Want the following properties

          Clusters to be disjoint

          Each cluster have facwt ge frac12

          Each j be connected to rrsquoj clusters

          iFj

          Theory Seminar 042002

          Iterative algorithm

          S = j | rrsquoj gt 0

          aj = lsquoactiversquo copy of j initially = 1

          Ĉj(aj) = avg distance to the first k

          facilities in Fj gathering facwt ge frac12

          say these facilities lsquoserversquo j

          Will maintain Ĉj(aj) le Cj(aj)

          X

          X

          X1

          Fj after some iterations

          X i removed from Fj

          i Fjserving jĈj(3)

          facilities serving j

          aj = 3

          4X

          (aj)

          (aj)

          (aj)

          Theory Seminar 042002

          Start of iteration

          1 Choose j in S with minimum vj + Ĉj(aj)

          2 Form cluster M = facilities serving j Note facilities are not split

          3 For k st Fk M decrease rrsquok advance ak set Fk = Fk ndash M

          2j(3) Cluster M

          aj = 1

          4

          1

          XX X

          aj = 4 Cluster M

          aj = 2

          3

          client in S

          facility in some Fj

          X facility removed from Fj

          (aj)

          Theory Seminar 042002

          Opening Facilities

          Central facilities opened in 2 steps

          1 Open exactly 1 facility in M i opened with prob qyi Acts as backup

          denoted b(k ) for each k st Fk M

          2 Open each i in M indep with prob (2-q)yi and independent of step 1

          Non-central facilities

          Cluster M

          k

          open with prob 2yi independent of other choices

          j

          (ak

          )

          Theory Seminar 042002

          Let Sj(c) = avg dist from j to P(j(c))

          = ( cijxij)facwt(P(j(c))

          j)

          Then c Sj(c) le 2Cj

          Distributing Facilities

          iP(j(c))

          j

          rrsquoj = 3

          P(j(1))P(j(2))

          P(j(3))

          Copy c gets a preferred set P(j(c))Preferred sets are disjoint

          Ensure facwt(P(j(c)) j) ge frac12 for all c

          Possible to do so since each xij lt frac12

          facility in Fj

          Theory Seminar 042002

          Analysis

          Feasibility follows from

          1 Facilities in phases 1 2 not reused

          2 After clustering j is connected to rrsquoj disjoint clusters backups are distinct

          3 Preferred sets are disjoint

          So j connected to rj distinct facilities

          Theory Seminar 042002

          Facility cost

          Recall L1 = i | yi = 1

          Phase 2 incur a factor of 2

          Phase 3 each i is opened with probability 2yi

          Expected facility cost le 2 fiyifor phases 2 3

          iL1

          Theory Seminar 042002

          Bounding backup cost denoted by B rv

          D event that no i in P(j(c)) is open

          Lemma E[B|D] le 2vj + Cj(c)

          Proof 2 cases

          Service cost I

          iM Fj st cik le Ĉj(d)

          Also vk + Ĉj(d) le vj + Ĉj(c) le vj + Cj(c)

          k(d)

          j(c)

          le vj

          le Ĉj(d)le vk

          B

          k(d)

          j(c)

          le vj

          le vk

          iM Fj cik gt Ĉj(d)

          le Ĉj(d) in expectatio

          n

          1)

          2)

          backup = b(j(c))

          Theory Seminar 042002

          Service Cost II

          Fix j c Let X(c) = service cost of j(c)

          Let di = cij pi = prob i is opened = 2yi

          B(c) = backup costD(c) = event that no iP(j(c)) is

          openp = Pr[D(c)] = (1-p1)hellip(1-pm) le e-1

          davg = weighted avg of the dis

          = (i pidi)(i pi) = Sj(c)

          d1

          d2 dm

          P(j(c)) sorted by increasing cij

          j(c)

          i P(j(c))

          Theory Seminar 042002

          Then

          E[X(c)] = [p1d1 + (1-p1)p2d2 + hellip

          + (1-p1)hellip(1-pm-1)pmdm]

          + pE[B(c)|D(c)]

          le (1-p)davg + p[2vj + Cj(c)]

          le (1-e-1)Sj(c) + e-1[2vj +

          Cj(c)]

          Let X = c X(c) = service cost of j

          c Sj(c) le 2Cj and c Cj(c) le 2Cj

          Summing over all c = 1helliprrsquoj

          E[X] le (1-e-1)2Cj + e-1(2rrsquojvj + Cj)

          le 2Cj + 2e-1rrsquojvj

          Theory Seminar 042002

          Putting it all together

          Phase 1 pay the optimal LP cost

          Phases 2 3

          bull Facility cost twice LP facility cost

          bull Service costLose a factor of 2 for phase 2Phase 3 cost is 2(LP service cost)+2e-1(dual value)

          Overall cost for le (2+2e)(LP cost) phases 2 3

          Total cost le (2+2e)OPT

          Theory Seminar 042002

          How to improve this

          bull Distribute facilities more equitably (in an expected sense) among copies - decreases prob of lsquobadrsquo event

          bull Better analysis ndash maximum distance within a cluster can be bounded by 2Cj(c)

          bull Balance phases 2 and 3

          Theory Seminar 042002

          Summary of Results

          bullGive a 2076-approx algorithm for non-uniform rjs Based on LP rounding using complem slackness

          bullFor rj = r extend the primal-dual algorithm of (JMS02) to get a 152-approximation

          bullFault tolerant k medians with rj = r

          a Primal-dual algorithm (JMS02) gives a 4-approx using Lagrangean relaxation

          b LP rounding gives a factor of 8

          Theory Seminar 042002

          Open Questions

          1 Reduce gap between rj = r non-uniform rj

          2 Combinatorial algorithms for non-uniform rj primal-dual local-search

          3 Constant-factor approx for fault tolerant k medians with non-uniform rjs

          • Fault Tolerant Facility Location
          • Metric Facility Location
          • PowerPoint Presentation
          • Previous Work
          • Previous Work (contd)
          • LP Formulation
          • LP Formulation (contd)
          • Complementary Slackness
          • 4-approximation outline
          • The Algorithm
          • Slide 11
          • Analysis Phase 1
          • Analysis (contd)
          • The Algorithm (contd)
          • Analysis Phase 2
          • Finally hellip
          • A Randomized Algorithm
          • Slide 18
          • Slide 19
          • Clustering
          • Slide 21
          • Slide 22
          • Opening Facilities
          • Distributing Facilities
          • Analysis
          • Facility cost
          • Service cost I
          • Service Cost II
          • Slide 29
          • Putting it all together
          • How to improve this
          • Summary of Results
          • Open Questions

            Theory Seminar 042002

            LP Formulation

            Primal

            Min i fiyi + ji cijxij

            st

            i xij ge rj j

            xij le yi i j

            yi le 1 i

            xij ge 0 yi ge 0 i j

            Theory Seminar 042002

            LP Formulation (contd)

            Max j rjvj - i zi

            st

            vj le wij + cij i j

            j wij le fi + zi i

            vj ge 0 wij ge 0 zi ge 0 i j

            Dual

            Theory Seminar 042002

            Complementary Slackness

            Primal Slackness Conditions

            bullxij gt 0 vj = wij + cij

            bullyi gt 0 j wij = fi + zi

            Dual Slackness Conditions

            bullvj gt 0 j xij = rj

            bullwij gt 0 xij = yi

            bullzi gt 0 yi = 1

            Theory Seminar 042002

            4-approximation outline

            Basic Idea vj lsquopaysrsquo for each cij stxij gt 0

            Bound service cost for each copy of j by ρvj total service cost leρj

            rjvj

            Problem Have ndashzis in the dual

            But zi gt 0 yi = 1 So can open these facilities and charge all of this cost to the LP

            2

            j(1) j(2)

            le vj

            le vj

            view as rj

            copiesj(c) cth copy

            Theory Seminar 042002

            The Algorithm

            Phase 1 Clustering Ensures that each copy j(c) has a nearby open facilityIterative algorithm

            S = j|rj gt 0 Fj = i|xij gt 0 in fi order

            Start of iteration

            1 Pick j with smallest vj

            2 Cluster is M Fj with iM yi = rj

            2

            51

            2

            j

            client in Sfacility in some Fj

            Cluster M

            Theory Seminar 042002

            0

            X XX

            30

            2

            j

            3 Open rj cheapest facilities in M

            4 For k st Fk M connect rj copies to opened facilities Decrease rk set Fk=Fk-M

            End of iteration

            client in S

            facility in some Fj

            client not in S

            X facility removed from

            Fj

            Cluster M

            facility opened from M

            Theory Seminar 042002

            Analysis Phase 1

            Solution is feasible each j is connected to rj distinct facilities

            Lemma Facility cost lei fiyi

            Proof Cost of rj cheapest facilities in M lerj (avg cost) = iM fiyi These facilities donrsquot get used again

            Theory Seminar 042002

            Analysis (contd)

            Lemma For any j and c service cost of copy j(c) le3vj

            Proof

            vk le vj since k was chosen as cluster center

            Service cost le vj + 2vk le 3vj

            Cluster M

            j(c)le vj

            le vk

            le vk k

            Theory Seminar 042002

            The Algorithm (contd)

            Phase 2 Taking care of ndashzis

            1 Open all (unopened) i st yi = 1

            2 For any j if xij = yi = 1 disconnect a copy of j and connect it to i

            j

            rj = 3

            X

            i with yi = 1i with yi lt

            1 and open

            Theory Seminar 042002

            Analysis Phase 2

            Lemma Cost of phase 2 = fi + cij = j ljvj ndash i zi

            Proof Each i with zi gt 0 is opened For iL1 all j st wij gt 0 are connected to it So

            vj = (service cost) + (fi +

            zi)

            j ljvj = fi + cij + i zi

            Let L1 = i | yi = 1

            Lj = i | xij = 1 L1 and lj = |Lj|

            iL1 jiLj

            j|iLj

            jiLjiL1

            Theory Seminar 042002

            Finally hellip

            Theorem Total cost le 4 times the optimal cost

            Proof Total cost le

            i fiyi + 3j (rj ndash lj)vj + fi + cij

            facility cost of phase 1 cost for

            copies connected by

            phase 1

            cost of phase 2

            lei fiyi + 3j (rj ndash lj)vj + (j ljvj ndash i

            zi )

            lei fiyi + 3(j rjvj ndash i zi )

            le4OPT

            iL1 jiLj

            Theory Seminar 042002

            A Randomized Algorithm

            Idea Open i with probability ρyi

            Expected facility cost le ρi fiyi

            Hope that each copy j(c) has a nearby facility open and service cost decreases

            Not quitehellip no facility may be open

            Cluster facilities open ge 1 facility in each cluster

            Theory Seminar 042002

            Phase 1 Pruning out ndashzis

            Open all i st yi = 1

            For each j if xij = yi = 1 connect j to i

            Let Lj = i | xij = 1 and lj = |Lj|

            Cost = j ljvj ndash i zi

            Lj

            Fj10

            rrsquoj = residual reqmt = 6

            Lrsquoj

            Phase 2

            Open all i st frac12 le yi lt 1

            For each j let Lrsquoj = i | frac12 le xij lt 1

            Connect copies of j to i Lrsquoj

            Lose a factor of 2

            facilities opened in

            phases 1 2yi = 1 frac12 le yi lt 1yi lt frac12

            Set L1

            Set L2

            Theory Seminar 042002

            Phase 3

            Notation facwt(S j) = iS xij

            1 Form clusters Each cluster has facwt ge frac12

            2 Open facilities Open ge 1 facility in a cluster ndash used as a backup facility Open facility i with prob 2yi

            3 Assign facilities to copies Each copy j(c) gets a preferred set of facilities ndash P(j(c)) with facwt ge frac12 For c d P(j(c)) P(j(d)) =

            4 Connect clients Connect j(c) to the nearest i open in P(j(c)) or to a backup facility

            Theory Seminar 042002

            ClusteringAfter phases 1 and 2 Fj = i | xij lt frac12 Sort these by cij and distribute among the rrsquoj copies

            Cj(c) = avg service cost of the cth copy denote c Cj(c) = cijxij by Cj

            Initial Fj before any iterations

            Cj(1)

            Cj(2)

            Cj(3)3

            i Fj

            client j

            Want the following properties

            Clusters to be disjoint

            Each cluster have facwt ge frac12

            Each j be connected to rrsquoj clusters

            iFj

            Theory Seminar 042002

            Iterative algorithm

            S = j | rrsquoj gt 0

            aj = lsquoactiversquo copy of j initially = 1

            Ĉj(aj) = avg distance to the first k

            facilities in Fj gathering facwt ge frac12

            say these facilities lsquoserversquo j

            Will maintain Ĉj(aj) le Cj(aj)

            X

            X

            X1

            Fj after some iterations

            X i removed from Fj

            i Fjserving jĈj(3)

            facilities serving j

            aj = 3

            4X

            (aj)

            (aj)

            (aj)

            Theory Seminar 042002

            Start of iteration

            1 Choose j in S with minimum vj + Ĉj(aj)

            2 Form cluster M = facilities serving j Note facilities are not split

            3 For k st Fk M decrease rrsquok advance ak set Fk = Fk ndash M

            2j(3) Cluster M

            aj = 1

            4

            1

            XX X

            aj = 4 Cluster M

            aj = 2

            3

            client in S

            facility in some Fj

            X facility removed from Fj

            (aj)

            Theory Seminar 042002

            Opening Facilities

            Central facilities opened in 2 steps

            1 Open exactly 1 facility in M i opened with prob qyi Acts as backup

            denoted b(k ) for each k st Fk M

            2 Open each i in M indep with prob (2-q)yi and independent of step 1

            Non-central facilities

            Cluster M

            k

            open with prob 2yi independent of other choices

            j

            (ak

            )

            Theory Seminar 042002

            Let Sj(c) = avg dist from j to P(j(c))

            = ( cijxij)facwt(P(j(c))

            j)

            Then c Sj(c) le 2Cj

            Distributing Facilities

            iP(j(c))

            j

            rrsquoj = 3

            P(j(1))P(j(2))

            P(j(3))

            Copy c gets a preferred set P(j(c))Preferred sets are disjoint

            Ensure facwt(P(j(c)) j) ge frac12 for all c

            Possible to do so since each xij lt frac12

            facility in Fj

            Theory Seminar 042002

            Analysis

            Feasibility follows from

            1 Facilities in phases 1 2 not reused

            2 After clustering j is connected to rrsquoj disjoint clusters backups are distinct

            3 Preferred sets are disjoint

            So j connected to rj distinct facilities

            Theory Seminar 042002

            Facility cost

            Recall L1 = i | yi = 1

            Phase 2 incur a factor of 2

            Phase 3 each i is opened with probability 2yi

            Expected facility cost le 2 fiyifor phases 2 3

            iL1

            Theory Seminar 042002

            Bounding backup cost denoted by B rv

            D event that no i in P(j(c)) is open

            Lemma E[B|D] le 2vj + Cj(c)

            Proof 2 cases

            Service cost I

            iM Fj st cik le Ĉj(d)

            Also vk + Ĉj(d) le vj + Ĉj(c) le vj + Cj(c)

            k(d)

            j(c)

            le vj

            le Ĉj(d)le vk

            B

            k(d)

            j(c)

            le vj

            le vk

            iM Fj cik gt Ĉj(d)

            le Ĉj(d) in expectatio

            n

            1)

            2)

            backup = b(j(c))

            Theory Seminar 042002

            Service Cost II

            Fix j c Let X(c) = service cost of j(c)

            Let di = cij pi = prob i is opened = 2yi

            B(c) = backup costD(c) = event that no iP(j(c)) is

            openp = Pr[D(c)] = (1-p1)hellip(1-pm) le e-1

            davg = weighted avg of the dis

            = (i pidi)(i pi) = Sj(c)

            d1

            d2 dm

            P(j(c)) sorted by increasing cij

            j(c)

            i P(j(c))

            Theory Seminar 042002

            Then

            E[X(c)] = [p1d1 + (1-p1)p2d2 + hellip

            + (1-p1)hellip(1-pm-1)pmdm]

            + pE[B(c)|D(c)]

            le (1-p)davg + p[2vj + Cj(c)]

            le (1-e-1)Sj(c) + e-1[2vj +

            Cj(c)]

            Let X = c X(c) = service cost of j

            c Sj(c) le 2Cj and c Cj(c) le 2Cj

            Summing over all c = 1helliprrsquoj

            E[X] le (1-e-1)2Cj + e-1(2rrsquojvj + Cj)

            le 2Cj + 2e-1rrsquojvj

            Theory Seminar 042002

            Putting it all together

            Phase 1 pay the optimal LP cost

            Phases 2 3

            bull Facility cost twice LP facility cost

            bull Service costLose a factor of 2 for phase 2Phase 3 cost is 2(LP service cost)+2e-1(dual value)

            Overall cost for le (2+2e)(LP cost) phases 2 3

            Total cost le (2+2e)OPT

            Theory Seminar 042002

            How to improve this

            bull Distribute facilities more equitably (in an expected sense) among copies - decreases prob of lsquobadrsquo event

            bull Better analysis ndash maximum distance within a cluster can be bounded by 2Cj(c)

            bull Balance phases 2 and 3

            Theory Seminar 042002

            Summary of Results

            bullGive a 2076-approx algorithm for non-uniform rjs Based on LP rounding using complem slackness

            bullFor rj = r extend the primal-dual algorithm of (JMS02) to get a 152-approximation

            bullFault tolerant k medians with rj = r

            a Primal-dual algorithm (JMS02) gives a 4-approx using Lagrangean relaxation

            b LP rounding gives a factor of 8

            Theory Seminar 042002

            Open Questions

            1 Reduce gap between rj = r non-uniform rj

            2 Combinatorial algorithms for non-uniform rj primal-dual local-search

            3 Constant-factor approx for fault tolerant k medians with non-uniform rjs

            • Fault Tolerant Facility Location
            • Metric Facility Location
            • PowerPoint Presentation
            • Previous Work
            • Previous Work (contd)
            • LP Formulation
            • LP Formulation (contd)
            • Complementary Slackness
            • 4-approximation outline
            • The Algorithm
            • Slide 11
            • Analysis Phase 1
            • Analysis (contd)
            • The Algorithm (contd)
            • Analysis Phase 2
            • Finally hellip
            • A Randomized Algorithm
            • Slide 18
            • Slide 19
            • Clustering
            • Slide 21
            • Slide 22
            • Opening Facilities
            • Distributing Facilities
            • Analysis
            • Facility cost
            • Service cost I
            • Service Cost II
            • Slide 29
            • Putting it all together
            • How to improve this
            • Summary of Results
            • Open Questions

              Theory Seminar 042002

              LP Formulation (contd)

              Max j rjvj - i zi

              st

              vj le wij + cij i j

              j wij le fi + zi i

              vj ge 0 wij ge 0 zi ge 0 i j

              Dual

              Theory Seminar 042002

              Complementary Slackness

              Primal Slackness Conditions

              bullxij gt 0 vj = wij + cij

              bullyi gt 0 j wij = fi + zi

              Dual Slackness Conditions

              bullvj gt 0 j xij = rj

              bullwij gt 0 xij = yi

              bullzi gt 0 yi = 1

              Theory Seminar 042002

              4-approximation outline

              Basic Idea vj lsquopaysrsquo for each cij stxij gt 0

              Bound service cost for each copy of j by ρvj total service cost leρj

              rjvj

              Problem Have ndashzis in the dual

              But zi gt 0 yi = 1 So can open these facilities and charge all of this cost to the LP

              2

              j(1) j(2)

              le vj

              le vj

              view as rj

              copiesj(c) cth copy

              Theory Seminar 042002

              The Algorithm

              Phase 1 Clustering Ensures that each copy j(c) has a nearby open facilityIterative algorithm

              S = j|rj gt 0 Fj = i|xij gt 0 in fi order

              Start of iteration

              1 Pick j with smallest vj

              2 Cluster is M Fj with iM yi = rj

              2

              51

              2

              j

              client in Sfacility in some Fj

              Cluster M

              Theory Seminar 042002

              0

              X XX

              30

              2

              j

              3 Open rj cheapest facilities in M

              4 For k st Fk M connect rj copies to opened facilities Decrease rk set Fk=Fk-M

              End of iteration

              client in S

              facility in some Fj

              client not in S

              X facility removed from

              Fj

              Cluster M

              facility opened from M

              Theory Seminar 042002

              Analysis Phase 1

              Solution is feasible each j is connected to rj distinct facilities

              Lemma Facility cost lei fiyi

              Proof Cost of rj cheapest facilities in M lerj (avg cost) = iM fiyi These facilities donrsquot get used again

              Theory Seminar 042002

              Analysis (contd)

              Lemma For any j and c service cost of copy j(c) le3vj

              Proof

              vk le vj since k was chosen as cluster center

              Service cost le vj + 2vk le 3vj

              Cluster M

              j(c)le vj

              le vk

              le vk k

              Theory Seminar 042002

              The Algorithm (contd)

              Phase 2 Taking care of ndashzis

              1 Open all (unopened) i st yi = 1

              2 For any j if xij = yi = 1 disconnect a copy of j and connect it to i

              j

              rj = 3

              X

              i with yi = 1i with yi lt

              1 and open

              Theory Seminar 042002

              Analysis Phase 2

              Lemma Cost of phase 2 = fi + cij = j ljvj ndash i zi

              Proof Each i with zi gt 0 is opened For iL1 all j st wij gt 0 are connected to it So

              vj = (service cost) + (fi +

              zi)

              j ljvj = fi + cij + i zi

              Let L1 = i | yi = 1

              Lj = i | xij = 1 L1 and lj = |Lj|

              iL1 jiLj

              j|iLj

              jiLjiL1

              Theory Seminar 042002

              Finally hellip

              Theorem Total cost le 4 times the optimal cost

              Proof Total cost le

              i fiyi + 3j (rj ndash lj)vj + fi + cij

              facility cost of phase 1 cost for

              copies connected by

              phase 1

              cost of phase 2

              lei fiyi + 3j (rj ndash lj)vj + (j ljvj ndash i

              zi )

              lei fiyi + 3(j rjvj ndash i zi )

              le4OPT

              iL1 jiLj

              Theory Seminar 042002

              A Randomized Algorithm

              Idea Open i with probability ρyi

              Expected facility cost le ρi fiyi

              Hope that each copy j(c) has a nearby facility open and service cost decreases

              Not quitehellip no facility may be open

              Cluster facilities open ge 1 facility in each cluster

              Theory Seminar 042002

              Phase 1 Pruning out ndashzis

              Open all i st yi = 1

              For each j if xij = yi = 1 connect j to i

              Let Lj = i | xij = 1 and lj = |Lj|

              Cost = j ljvj ndash i zi

              Lj

              Fj10

              rrsquoj = residual reqmt = 6

              Lrsquoj

              Phase 2

              Open all i st frac12 le yi lt 1

              For each j let Lrsquoj = i | frac12 le xij lt 1

              Connect copies of j to i Lrsquoj

              Lose a factor of 2

              facilities opened in

              phases 1 2yi = 1 frac12 le yi lt 1yi lt frac12

              Set L1

              Set L2

              Theory Seminar 042002

              Phase 3

              Notation facwt(S j) = iS xij

              1 Form clusters Each cluster has facwt ge frac12

              2 Open facilities Open ge 1 facility in a cluster ndash used as a backup facility Open facility i with prob 2yi

              3 Assign facilities to copies Each copy j(c) gets a preferred set of facilities ndash P(j(c)) with facwt ge frac12 For c d P(j(c)) P(j(d)) =

              4 Connect clients Connect j(c) to the nearest i open in P(j(c)) or to a backup facility

              Theory Seminar 042002

              ClusteringAfter phases 1 and 2 Fj = i | xij lt frac12 Sort these by cij and distribute among the rrsquoj copies

              Cj(c) = avg service cost of the cth copy denote c Cj(c) = cijxij by Cj

              Initial Fj before any iterations

              Cj(1)

              Cj(2)

              Cj(3)3

              i Fj

              client j

              Want the following properties

              Clusters to be disjoint

              Each cluster have facwt ge frac12

              Each j be connected to rrsquoj clusters

              iFj

              Theory Seminar 042002

              Iterative algorithm

              S = j | rrsquoj gt 0

              aj = lsquoactiversquo copy of j initially = 1

              Ĉj(aj) = avg distance to the first k

              facilities in Fj gathering facwt ge frac12

              say these facilities lsquoserversquo j

              Will maintain Ĉj(aj) le Cj(aj)

              X

              X

              X1

              Fj after some iterations

              X i removed from Fj

              i Fjserving jĈj(3)

              facilities serving j

              aj = 3

              4X

              (aj)

              (aj)

              (aj)

              Theory Seminar 042002

              Start of iteration

              1 Choose j in S with minimum vj + Ĉj(aj)

              2 Form cluster M = facilities serving j Note facilities are not split

              3 For k st Fk M decrease rrsquok advance ak set Fk = Fk ndash M

              2j(3) Cluster M

              aj = 1

              4

              1

              XX X

              aj = 4 Cluster M

              aj = 2

              3

              client in S

              facility in some Fj

              X facility removed from Fj

              (aj)

              Theory Seminar 042002

              Opening Facilities

              Central facilities opened in 2 steps

              1 Open exactly 1 facility in M i opened with prob qyi Acts as backup

              denoted b(k ) for each k st Fk M

              2 Open each i in M indep with prob (2-q)yi and independent of step 1

              Non-central facilities

              Cluster M

              k

              open with prob 2yi independent of other choices

              j

              (ak

              )

              Theory Seminar 042002

              Let Sj(c) = avg dist from j to P(j(c))

              = ( cijxij)facwt(P(j(c))

              j)

              Then c Sj(c) le 2Cj

              Distributing Facilities

              iP(j(c))

              j

              rrsquoj = 3

              P(j(1))P(j(2))

              P(j(3))

              Copy c gets a preferred set P(j(c))Preferred sets are disjoint

              Ensure facwt(P(j(c)) j) ge frac12 for all c

              Possible to do so since each xij lt frac12

              facility in Fj

              Theory Seminar 042002

              Analysis

              Feasibility follows from

              1 Facilities in phases 1 2 not reused

              2 After clustering j is connected to rrsquoj disjoint clusters backups are distinct

              3 Preferred sets are disjoint

              So j connected to rj distinct facilities

              Theory Seminar 042002

              Facility cost

              Recall L1 = i | yi = 1

              Phase 2 incur a factor of 2

              Phase 3 each i is opened with probability 2yi

              Expected facility cost le 2 fiyifor phases 2 3

              iL1

              Theory Seminar 042002

              Bounding backup cost denoted by B rv

              D event that no i in P(j(c)) is open

              Lemma E[B|D] le 2vj + Cj(c)

              Proof 2 cases

              Service cost I

              iM Fj st cik le Ĉj(d)

              Also vk + Ĉj(d) le vj + Ĉj(c) le vj + Cj(c)

              k(d)

              j(c)

              le vj

              le Ĉj(d)le vk

              B

              k(d)

              j(c)

              le vj

              le vk

              iM Fj cik gt Ĉj(d)

              le Ĉj(d) in expectatio

              n

              1)

              2)

              backup = b(j(c))

              Theory Seminar 042002

              Service Cost II

              Fix j c Let X(c) = service cost of j(c)

              Let di = cij pi = prob i is opened = 2yi

              B(c) = backup costD(c) = event that no iP(j(c)) is

              openp = Pr[D(c)] = (1-p1)hellip(1-pm) le e-1

              davg = weighted avg of the dis

              = (i pidi)(i pi) = Sj(c)

              d1

              d2 dm

              P(j(c)) sorted by increasing cij

              j(c)

              i P(j(c))

              Theory Seminar 042002

              Then

              E[X(c)] = [p1d1 + (1-p1)p2d2 + hellip

              + (1-p1)hellip(1-pm-1)pmdm]

              + pE[B(c)|D(c)]

              le (1-p)davg + p[2vj + Cj(c)]

              le (1-e-1)Sj(c) + e-1[2vj +

              Cj(c)]

              Let X = c X(c) = service cost of j

              c Sj(c) le 2Cj and c Cj(c) le 2Cj

              Summing over all c = 1helliprrsquoj

              E[X] le (1-e-1)2Cj + e-1(2rrsquojvj + Cj)

              le 2Cj + 2e-1rrsquojvj

              Theory Seminar 042002

              Putting it all together

              Phase 1 pay the optimal LP cost

              Phases 2 3

              bull Facility cost twice LP facility cost

              bull Service costLose a factor of 2 for phase 2Phase 3 cost is 2(LP service cost)+2e-1(dual value)

              Overall cost for le (2+2e)(LP cost) phases 2 3

              Total cost le (2+2e)OPT

              Theory Seminar 042002

              How to improve this

              bull Distribute facilities more equitably (in an expected sense) among copies - decreases prob of lsquobadrsquo event

              bull Better analysis ndash maximum distance within a cluster can be bounded by 2Cj(c)

              bull Balance phases 2 and 3

              Theory Seminar 042002

              Summary of Results

              bullGive a 2076-approx algorithm for non-uniform rjs Based on LP rounding using complem slackness

              bullFor rj = r extend the primal-dual algorithm of (JMS02) to get a 152-approximation

              bullFault tolerant k medians with rj = r

              a Primal-dual algorithm (JMS02) gives a 4-approx using Lagrangean relaxation

              b LP rounding gives a factor of 8

              Theory Seminar 042002

              Open Questions

              1 Reduce gap between rj = r non-uniform rj

              2 Combinatorial algorithms for non-uniform rj primal-dual local-search

              3 Constant-factor approx for fault tolerant k medians with non-uniform rjs

              • Fault Tolerant Facility Location
              • Metric Facility Location
              • PowerPoint Presentation
              • Previous Work
              • Previous Work (contd)
              • LP Formulation
              • LP Formulation (contd)
              • Complementary Slackness
              • 4-approximation outline
              • The Algorithm
              • Slide 11
              • Analysis Phase 1
              • Analysis (contd)
              • The Algorithm (contd)
              • Analysis Phase 2
              • Finally hellip
              • A Randomized Algorithm
              • Slide 18
              • Slide 19
              • Clustering
              • Slide 21
              • Slide 22
              • Opening Facilities
              • Distributing Facilities
              • Analysis
              • Facility cost
              • Service cost I
              • Service Cost II
              • Slide 29
              • Putting it all together
              • How to improve this
              • Summary of Results
              • Open Questions

                Theory Seminar 042002

                Complementary Slackness

                Primal Slackness Conditions

                bullxij gt 0 vj = wij + cij

                bullyi gt 0 j wij = fi + zi

                Dual Slackness Conditions

                bullvj gt 0 j xij = rj

                bullwij gt 0 xij = yi

                bullzi gt 0 yi = 1

                Theory Seminar 042002

                4-approximation outline

                Basic Idea vj lsquopaysrsquo for each cij stxij gt 0

                Bound service cost for each copy of j by ρvj total service cost leρj

                rjvj

                Problem Have ndashzis in the dual

                But zi gt 0 yi = 1 So can open these facilities and charge all of this cost to the LP

                2

                j(1) j(2)

                le vj

                le vj

                view as rj

                copiesj(c) cth copy

                Theory Seminar 042002

                The Algorithm

                Phase 1 Clustering Ensures that each copy j(c) has a nearby open facilityIterative algorithm

                S = j|rj gt 0 Fj = i|xij gt 0 in fi order

                Start of iteration

                1 Pick j with smallest vj

                2 Cluster is M Fj with iM yi = rj

                2

                51

                2

                j

                client in Sfacility in some Fj

                Cluster M

                Theory Seminar 042002

                0

                X XX

                30

                2

                j

                3 Open rj cheapest facilities in M

                4 For k st Fk M connect rj copies to opened facilities Decrease rk set Fk=Fk-M

                End of iteration

                client in S

                facility in some Fj

                client not in S

                X facility removed from

                Fj

                Cluster M

                facility opened from M

                Theory Seminar 042002

                Analysis Phase 1

                Solution is feasible each j is connected to rj distinct facilities

                Lemma Facility cost lei fiyi

                Proof Cost of rj cheapest facilities in M lerj (avg cost) = iM fiyi These facilities donrsquot get used again

                Theory Seminar 042002

                Analysis (contd)

                Lemma For any j and c service cost of copy j(c) le3vj

                Proof

                vk le vj since k was chosen as cluster center

                Service cost le vj + 2vk le 3vj

                Cluster M

                j(c)le vj

                le vk

                le vk k

                Theory Seminar 042002

                The Algorithm (contd)

                Phase 2 Taking care of ndashzis

                1 Open all (unopened) i st yi = 1

                2 For any j if xij = yi = 1 disconnect a copy of j and connect it to i

                j

                rj = 3

                X

                i with yi = 1i with yi lt

                1 and open

                Theory Seminar 042002

                Analysis Phase 2

                Lemma Cost of phase 2 = fi + cij = j ljvj ndash i zi

                Proof Each i with zi gt 0 is opened For iL1 all j st wij gt 0 are connected to it So

                vj = (service cost) + (fi +

                zi)

                j ljvj = fi + cij + i zi

                Let L1 = i | yi = 1

                Lj = i | xij = 1 L1 and lj = |Lj|

                iL1 jiLj

                j|iLj

                jiLjiL1

                Theory Seminar 042002

                Finally hellip

                Theorem Total cost le 4 times the optimal cost

                Proof Total cost le

                i fiyi + 3j (rj ndash lj)vj + fi + cij

                facility cost of phase 1 cost for

                copies connected by

                phase 1

                cost of phase 2

                lei fiyi + 3j (rj ndash lj)vj + (j ljvj ndash i

                zi )

                lei fiyi + 3(j rjvj ndash i zi )

                le4OPT

                iL1 jiLj

                Theory Seminar 042002

                A Randomized Algorithm

                Idea Open i with probability ρyi

                Expected facility cost le ρi fiyi

                Hope that each copy j(c) has a nearby facility open and service cost decreases

                Not quitehellip no facility may be open

                Cluster facilities open ge 1 facility in each cluster

                Theory Seminar 042002

                Phase 1 Pruning out ndashzis

                Open all i st yi = 1

                For each j if xij = yi = 1 connect j to i

                Let Lj = i | xij = 1 and lj = |Lj|

                Cost = j ljvj ndash i zi

                Lj

                Fj10

                rrsquoj = residual reqmt = 6

                Lrsquoj

                Phase 2

                Open all i st frac12 le yi lt 1

                For each j let Lrsquoj = i | frac12 le xij lt 1

                Connect copies of j to i Lrsquoj

                Lose a factor of 2

                facilities opened in

                phases 1 2yi = 1 frac12 le yi lt 1yi lt frac12

                Set L1

                Set L2

                Theory Seminar 042002

                Phase 3

                Notation facwt(S j) = iS xij

                1 Form clusters Each cluster has facwt ge frac12

                2 Open facilities Open ge 1 facility in a cluster ndash used as a backup facility Open facility i with prob 2yi

                3 Assign facilities to copies Each copy j(c) gets a preferred set of facilities ndash P(j(c)) with facwt ge frac12 For c d P(j(c)) P(j(d)) =

                4 Connect clients Connect j(c) to the nearest i open in P(j(c)) or to a backup facility

                Theory Seminar 042002

                ClusteringAfter phases 1 and 2 Fj = i | xij lt frac12 Sort these by cij and distribute among the rrsquoj copies

                Cj(c) = avg service cost of the cth copy denote c Cj(c) = cijxij by Cj

                Initial Fj before any iterations

                Cj(1)

                Cj(2)

                Cj(3)3

                i Fj

                client j

                Want the following properties

                Clusters to be disjoint

                Each cluster have facwt ge frac12

                Each j be connected to rrsquoj clusters

                iFj

                Theory Seminar 042002

                Iterative algorithm

                S = j | rrsquoj gt 0

                aj = lsquoactiversquo copy of j initially = 1

                Ĉj(aj) = avg distance to the first k

                facilities in Fj gathering facwt ge frac12

                say these facilities lsquoserversquo j

                Will maintain Ĉj(aj) le Cj(aj)

                X

                X

                X1

                Fj after some iterations

                X i removed from Fj

                i Fjserving jĈj(3)

                facilities serving j

                aj = 3

                4X

                (aj)

                (aj)

                (aj)

                Theory Seminar 042002

                Start of iteration

                1 Choose j in S with minimum vj + Ĉj(aj)

                2 Form cluster M = facilities serving j Note facilities are not split

                3 For k st Fk M decrease rrsquok advance ak set Fk = Fk ndash M

                2j(3) Cluster M

                aj = 1

                4

                1

                XX X

                aj = 4 Cluster M

                aj = 2

                3

                client in S

                facility in some Fj

                X facility removed from Fj

                (aj)

                Theory Seminar 042002

                Opening Facilities

                Central facilities opened in 2 steps

                1 Open exactly 1 facility in M i opened with prob qyi Acts as backup

                denoted b(k ) for each k st Fk M

                2 Open each i in M indep with prob (2-q)yi and independent of step 1

                Non-central facilities

                Cluster M

                k

                open with prob 2yi independent of other choices

                j

                (ak

                )

                Theory Seminar 042002

                Let Sj(c) = avg dist from j to P(j(c))

                = ( cijxij)facwt(P(j(c))

                j)

                Then c Sj(c) le 2Cj

                Distributing Facilities

                iP(j(c))

                j

                rrsquoj = 3

                P(j(1))P(j(2))

                P(j(3))

                Copy c gets a preferred set P(j(c))Preferred sets are disjoint

                Ensure facwt(P(j(c)) j) ge frac12 for all c

                Possible to do so since each xij lt frac12

                facility in Fj

                Theory Seminar 042002

                Analysis

                Feasibility follows from

                1 Facilities in phases 1 2 not reused

                2 After clustering j is connected to rrsquoj disjoint clusters backups are distinct

                3 Preferred sets are disjoint

                So j connected to rj distinct facilities

                Theory Seminar 042002

                Facility cost

                Recall L1 = i | yi = 1

                Phase 2 incur a factor of 2

                Phase 3 each i is opened with probability 2yi

                Expected facility cost le 2 fiyifor phases 2 3

                iL1

                Theory Seminar 042002

                Bounding backup cost denoted by B rv

                D event that no i in P(j(c)) is open

                Lemma E[B|D] le 2vj + Cj(c)

                Proof 2 cases

                Service cost I

                iM Fj st cik le Ĉj(d)

                Also vk + Ĉj(d) le vj + Ĉj(c) le vj + Cj(c)

                k(d)

                j(c)

                le vj

                le Ĉj(d)le vk

                B

                k(d)

                j(c)

                le vj

                le vk

                iM Fj cik gt Ĉj(d)

                le Ĉj(d) in expectatio

                n

                1)

                2)

                backup = b(j(c))

                Theory Seminar 042002

                Service Cost II

                Fix j c Let X(c) = service cost of j(c)

                Let di = cij pi = prob i is opened = 2yi

                B(c) = backup costD(c) = event that no iP(j(c)) is

                openp = Pr[D(c)] = (1-p1)hellip(1-pm) le e-1

                davg = weighted avg of the dis

                = (i pidi)(i pi) = Sj(c)

                d1

                d2 dm

                P(j(c)) sorted by increasing cij

                j(c)

                i P(j(c))

                Theory Seminar 042002

                Then

                E[X(c)] = [p1d1 + (1-p1)p2d2 + hellip

                + (1-p1)hellip(1-pm-1)pmdm]

                + pE[B(c)|D(c)]

                le (1-p)davg + p[2vj + Cj(c)]

                le (1-e-1)Sj(c) + e-1[2vj +

                Cj(c)]

                Let X = c X(c) = service cost of j

                c Sj(c) le 2Cj and c Cj(c) le 2Cj

                Summing over all c = 1helliprrsquoj

                E[X] le (1-e-1)2Cj + e-1(2rrsquojvj + Cj)

                le 2Cj + 2e-1rrsquojvj

                Theory Seminar 042002

                Putting it all together

                Phase 1 pay the optimal LP cost

                Phases 2 3

                bull Facility cost twice LP facility cost

                bull Service costLose a factor of 2 for phase 2Phase 3 cost is 2(LP service cost)+2e-1(dual value)

                Overall cost for le (2+2e)(LP cost) phases 2 3

                Total cost le (2+2e)OPT

                Theory Seminar 042002

                How to improve this

                bull Distribute facilities more equitably (in an expected sense) among copies - decreases prob of lsquobadrsquo event

                bull Better analysis ndash maximum distance within a cluster can be bounded by 2Cj(c)

                bull Balance phases 2 and 3

                Theory Seminar 042002

                Summary of Results

                bullGive a 2076-approx algorithm for non-uniform rjs Based on LP rounding using complem slackness

                bullFor rj = r extend the primal-dual algorithm of (JMS02) to get a 152-approximation

                bullFault tolerant k medians with rj = r

                a Primal-dual algorithm (JMS02) gives a 4-approx using Lagrangean relaxation

                b LP rounding gives a factor of 8

                Theory Seminar 042002

                Open Questions

                1 Reduce gap between rj = r non-uniform rj

                2 Combinatorial algorithms for non-uniform rj primal-dual local-search

                3 Constant-factor approx for fault tolerant k medians with non-uniform rjs

                • Fault Tolerant Facility Location
                • Metric Facility Location
                • PowerPoint Presentation
                • Previous Work
                • Previous Work (contd)
                • LP Formulation
                • LP Formulation (contd)
                • Complementary Slackness
                • 4-approximation outline
                • The Algorithm
                • Slide 11
                • Analysis Phase 1
                • Analysis (contd)
                • The Algorithm (contd)
                • Analysis Phase 2
                • Finally hellip
                • A Randomized Algorithm
                • Slide 18
                • Slide 19
                • Clustering
                • Slide 21
                • Slide 22
                • Opening Facilities
                • Distributing Facilities
                • Analysis
                • Facility cost
                • Service cost I
                • Service Cost II
                • Slide 29
                • Putting it all together
                • How to improve this
                • Summary of Results
                • Open Questions

                  Theory Seminar 042002

                  4-approximation outline

                  Basic Idea vj lsquopaysrsquo for each cij stxij gt 0

                  Bound service cost for each copy of j by ρvj total service cost leρj

                  rjvj

                  Problem Have ndashzis in the dual

                  But zi gt 0 yi = 1 So can open these facilities and charge all of this cost to the LP

                  2

                  j(1) j(2)

                  le vj

                  le vj

                  view as rj

                  copiesj(c) cth copy

                  Theory Seminar 042002

                  The Algorithm

                  Phase 1 Clustering Ensures that each copy j(c) has a nearby open facilityIterative algorithm

                  S = j|rj gt 0 Fj = i|xij gt 0 in fi order

                  Start of iteration

                  1 Pick j with smallest vj

                  2 Cluster is M Fj with iM yi = rj

                  2

                  51

                  2

                  j

                  client in Sfacility in some Fj

                  Cluster M

                  Theory Seminar 042002

                  0

                  X XX

                  30

                  2

                  j

                  3 Open rj cheapest facilities in M

                  4 For k st Fk M connect rj copies to opened facilities Decrease rk set Fk=Fk-M

                  End of iteration

                  client in S

                  facility in some Fj

                  client not in S

                  X facility removed from

                  Fj

                  Cluster M

                  facility opened from M

                  Theory Seminar 042002

                  Analysis Phase 1

                  Solution is feasible each j is connected to rj distinct facilities

                  Lemma Facility cost lei fiyi

                  Proof Cost of rj cheapest facilities in M lerj (avg cost) = iM fiyi These facilities donrsquot get used again

                  Theory Seminar 042002

                  Analysis (contd)

                  Lemma For any j and c service cost of copy j(c) le3vj

                  Proof

                  vk le vj since k was chosen as cluster center

                  Service cost le vj + 2vk le 3vj

                  Cluster M

                  j(c)le vj

                  le vk

                  le vk k

                  Theory Seminar 042002

                  The Algorithm (contd)

                  Phase 2 Taking care of ndashzis

                  1 Open all (unopened) i st yi = 1

                  2 For any j if xij = yi = 1 disconnect a copy of j and connect it to i

                  j

                  rj = 3

                  X

                  i with yi = 1i with yi lt

                  1 and open

                  Theory Seminar 042002

                  Analysis Phase 2

                  Lemma Cost of phase 2 = fi + cij = j ljvj ndash i zi

                  Proof Each i with zi gt 0 is opened For iL1 all j st wij gt 0 are connected to it So

                  vj = (service cost) + (fi +

                  zi)

                  j ljvj = fi + cij + i zi

                  Let L1 = i | yi = 1

                  Lj = i | xij = 1 L1 and lj = |Lj|

                  iL1 jiLj

                  j|iLj

                  jiLjiL1

                  Theory Seminar 042002

                  Finally hellip

                  Theorem Total cost le 4 times the optimal cost

                  Proof Total cost le

                  i fiyi + 3j (rj ndash lj)vj + fi + cij

                  facility cost of phase 1 cost for

                  copies connected by

                  phase 1

                  cost of phase 2

                  lei fiyi + 3j (rj ndash lj)vj + (j ljvj ndash i

                  zi )

                  lei fiyi + 3(j rjvj ndash i zi )

                  le4OPT

                  iL1 jiLj

                  Theory Seminar 042002

                  A Randomized Algorithm

                  Idea Open i with probability ρyi

                  Expected facility cost le ρi fiyi

                  Hope that each copy j(c) has a nearby facility open and service cost decreases

                  Not quitehellip no facility may be open

                  Cluster facilities open ge 1 facility in each cluster

                  Theory Seminar 042002

                  Phase 1 Pruning out ndashzis

                  Open all i st yi = 1

                  For each j if xij = yi = 1 connect j to i

                  Let Lj = i | xij = 1 and lj = |Lj|

                  Cost = j ljvj ndash i zi

                  Lj

                  Fj10

                  rrsquoj = residual reqmt = 6

                  Lrsquoj

                  Phase 2

                  Open all i st frac12 le yi lt 1

                  For each j let Lrsquoj = i | frac12 le xij lt 1

                  Connect copies of j to i Lrsquoj

                  Lose a factor of 2

                  facilities opened in

                  phases 1 2yi = 1 frac12 le yi lt 1yi lt frac12

                  Set L1

                  Set L2

                  Theory Seminar 042002

                  Phase 3

                  Notation facwt(S j) = iS xij

                  1 Form clusters Each cluster has facwt ge frac12

                  2 Open facilities Open ge 1 facility in a cluster ndash used as a backup facility Open facility i with prob 2yi

                  3 Assign facilities to copies Each copy j(c) gets a preferred set of facilities ndash P(j(c)) with facwt ge frac12 For c d P(j(c)) P(j(d)) =

                  4 Connect clients Connect j(c) to the nearest i open in P(j(c)) or to a backup facility

                  Theory Seminar 042002

                  ClusteringAfter phases 1 and 2 Fj = i | xij lt frac12 Sort these by cij and distribute among the rrsquoj copies

                  Cj(c) = avg service cost of the cth copy denote c Cj(c) = cijxij by Cj

                  Initial Fj before any iterations

                  Cj(1)

                  Cj(2)

                  Cj(3)3

                  i Fj

                  client j

                  Want the following properties

                  Clusters to be disjoint

                  Each cluster have facwt ge frac12

                  Each j be connected to rrsquoj clusters

                  iFj

                  Theory Seminar 042002

                  Iterative algorithm

                  S = j | rrsquoj gt 0

                  aj = lsquoactiversquo copy of j initially = 1

                  Ĉj(aj) = avg distance to the first k

                  facilities in Fj gathering facwt ge frac12

                  say these facilities lsquoserversquo j

                  Will maintain Ĉj(aj) le Cj(aj)

                  X

                  X

                  X1

                  Fj after some iterations

                  X i removed from Fj

                  i Fjserving jĈj(3)

                  facilities serving j

                  aj = 3

                  4X

                  (aj)

                  (aj)

                  (aj)

                  Theory Seminar 042002

                  Start of iteration

                  1 Choose j in S with minimum vj + Ĉj(aj)

                  2 Form cluster M = facilities serving j Note facilities are not split

                  3 For k st Fk M decrease rrsquok advance ak set Fk = Fk ndash M

                  2j(3) Cluster M

                  aj = 1

                  4

                  1

                  XX X

                  aj = 4 Cluster M

                  aj = 2

                  3

                  client in S

                  facility in some Fj

                  X facility removed from Fj

                  (aj)

                  Theory Seminar 042002

                  Opening Facilities

                  Central facilities opened in 2 steps

                  1 Open exactly 1 facility in M i opened with prob qyi Acts as backup

                  denoted b(k ) for each k st Fk M

                  2 Open each i in M indep with prob (2-q)yi and independent of step 1

                  Non-central facilities

                  Cluster M

                  k

                  open with prob 2yi independent of other choices

                  j

                  (ak

                  )

                  Theory Seminar 042002

                  Let Sj(c) = avg dist from j to P(j(c))

                  = ( cijxij)facwt(P(j(c))

                  j)

                  Then c Sj(c) le 2Cj

                  Distributing Facilities

                  iP(j(c))

                  j

                  rrsquoj = 3

                  P(j(1))P(j(2))

                  P(j(3))

                  Copy c gets a preferred set P(j(c))Preferred sets are disjoint

                  Ensure facwt(P(j(c)) j) ge frac12 for all c

                  Possible to do so since each xij lt frac12

                  facility in Fj

                  Theory Seminar 042002

                  Analysis

                  Feasibility follows from

                  1 Facilities in phases 1 2 not reused

                  2 After clustering j is connected to rrsquoj disjoint clusters backups are distinct

                  3 Preferred sets are disjoint

                  So j connected to rj distinct facilities

                  Theory Seminar 042002

                  Facility cost

                  Recall L1 = i | yi = 1

                  Phase 2 incur a factor of 2

                  Phase 3 each i is opened with probability 2yi

                  Expected facility cost le 2 fiyifor phases 2 3

                  iL1

                  Theory Seminar 042002

                  Bounding backup cost denoted by B rv

                  D event that no i in P(j(c)) is open

                  Lemma E[B|D] le 2vj + Cj(c)

                  Proof 2 cases

                  Service cost I

                  iM Fj st cik le Ĉj(d)

                  Also vk + Ĉj(d) le vj + Ĉj(c) le vj + Cj(c)

                  k(d)

                  j(c)

                  le vj

                  le Ĉj(d)le vk

                  B

                  k(d)

                  j(c)

                  le vj

                  le vk

                  iM Fj cik gt Ĉj(d)

                  le Ĉj(d) in expectatio

                  n

                  1)

                  2)

                  backup = b(j(c))

                  Theory Seminar 042002

                  Service Cost II

                  Fix j c Let X(c) = service cost of j(c)

                  Let di = cij pi = prob i is opened = 2yi

                  B(c) = backup costD(c) = event that no iP(j(c)) is

                  openp = Pr[D(c)] = (1-p1)hellip(1-pm) le e-1

                  davg = weighted avg of the dis

                  = (i pidi)(i pi) = Sj(c)

                  d1

                  d2 dm

                  P(j(c)) sorted by increasing cij

                  j(c)

                  i P(j(c))

                  Theory Seminar 042002

                  Then

                  E[X(c)] = [p1d1 + (1-p1)p2d2 + hellip

                  + (1-p1)hellip(1-pm-1)pmdm]

                  + pE[B(c)|D(c)]

                  le (1-p)davg + p[2vj + Cj(c)]

                  le (1-e-1)Sj(c) + e-1[2vj +

                  Cj(c)]

                  Let X = c X(c) = service cost of j

                  c Sj(c) le 2Cj and c Cj(c) le 2Cj

                  Summing over all c = 1helliprrsquoj

                  E[X] le (1-e-1)2Cj + e-1(2rrsquojvj + Cj)

                  le 2Cj + 2e-1rrsquojvj

                  Theory Seminar 042002

                  Putting it all together

                  Phase 1 pay the optimal LP cost

                  Phases 2 3

                  bull Facility cost twice LP facility cost

                  bull Service costLose a factor of 2 for phase 2Phase 3 cost is 2(LP service cost)+2e-1(dual value)

                  Overall cost for le (2+2e)(LP cost) phases 2 3

                  Total cost le (2+2e)OPT

                  Theory Seminar 042002

                  How to improve this

                  bull Distribute facilities more equitably (in an expected sense) among copies - decreases prob of lsquobadrsquo event

                  bull Better analysis ndash maximum distance within a cluster can be bounded by 2Cj(c)

                  bull Balance phases 2 and 3

                  Theory Seminar 042002

                  Summary of Results

                  bullGive a 2076-approx algorithm for non-uniform rjs Based on LP rounding using complem slackness

                  bullFor rj = r extend the primal-dual algorithm of (JMS02) to get a 152-approximation

                  bullFault tolerant k medians with rj = r

                  a Primal-dual algorithm (JMS02) gives a 4-approx using Lagrangean relaxation

                  b LP rounding gives a factor of 8

                  Theory Seminar 042002

                  Open Questions

                  1 Reduce gap between rj = r non-uniform rj

                  2 Combinatorial algorithms for non-uniform rj primal-dual local-search

                  3 Constant-factor approx for fault tolerant k medians with non-uniform rjs

                  • Fault Tolerant Facility Location
                  • Metric Facility Location
                  • PowerPoint Presentation
                  • Previous Work
                  • Previous Work (contd)
                  • LP Formulation
                  • LP Formulation (contd)
                  • Complementary Slackness
                  • 4-approximation outline
                  • The Algorithm
                  • Slide 11
                  • Analysis Phase 1
                  • Analysis (contd)
                  • The Algorithm (contd)
                  • Analysis Phase 2
                  • Finally hellip
                  • A Randomized Algorithm
                  • Slide 18
                  • Slide 19
                  • Clustering
                  • Slide 21
                  • Slide 22
                  • Opening Facilities
                  • Distributing Facilities
                  • Analysis
                  • Facility cost
                  • Service cost I
                  • Service Cost II
                  • Slide 29
                  • Putting it all together
                  • How to improve this
                  • Summary of Results
                  • Open Questions

                    Theory Seminar 042002

                    The Algorithm

                    Phase 1 Clustering Ensures that each copy j(c) has a nearby open facilityIterative algorithm

                    S = j|rj gt 0 Fj = i|xij gt 0 in fi order

                    Start of iteration

                    1 Pick j with smallest vj

                    2 Cluster is M Fj with iM yi = rj

                    2

                    51

                    2

                    j

                    client in Sfacility in some Fj

                    Cluster M

                    Theory Seminar 042002

                    0

                    X XX

                    30

                    2

                    j

                    3 Open rj cheapest facilities in M

                    4 For k st Fk M connect rj copies to opened facilities Decrease rk set Fk=Fk-M

                    End of iteration

                    client in S

                    facility in some Fj

                    client not in S

                    X facility removed from

                    Fj

                    Cluster M

                    facility opened from M

                    Theory Seminar 042002

                    Analysis Phase 1

                    Solution is feasible each j is connected to rj distinct facilities

                    Lemma Facility cost lei fiyi

                    Proof Cost of rj cheapest facilities in M lerj (avg cost) = iM fiyi These facilities donrsquot get used again

                    Theory Seminar 042002

                    Analysis (contd)

                    Lemma For any j and c service cost of copy j(c) le3vj

                    Proof

                    vk le vj since k was chosen as cluster center

                    Service cost le vj + 2vk le 3vj

                    Cluster M

                    j(c)le vj

                    le vk

                    le vk k

                    Theory Seminar 042002

                    The Algorithm (contd)

                    Phase 2 Taking care of ndashzis

                    1 Open all (unopened) i st yi = 1

                    2 For any j if xij = yi = 1 disconnect a copy of j and connect it to i

                    j

                    rj = 3

                    X

                    i with yi = 1i with yi lt

                    1 and open

                    Theory Seminar 042002

                    Analysis Phase 2

                    Lemma Cost of phase 2 = fi + cij = j ljvj ndash i zi

                    Proof Each i with zi gt 0 is opened For iL1 all j st wij gt 0 are connected to it So

                    vj = (service cost) + (fi +

                    zi)

                    j ljvj = fi + cij + i zi

                    Let L1 = i | yi = 1

                    Lj = i | xij = 1 L1 and lj = |Lj|

                    iL1 jiLj

                    j|iLj

                    jiLjiL1

                    Theory Seminar 042002

                    Finally hellip

                    Theorem Total cost le 4 times the optimal cost

                    Proof Total cost le

                    i fiyi + 3j (rj ndash lj)vj + fi + cij

                    facility cost of phase 1 cost for

                    copies connected by

                    phase 1

                    cost of phase 2

                    lei fiyi + 3j (rj ndash lj)vj + (j ljvj ndash i

                    zi )

                    lei fiyi + 3(j rjvj ndash i zi )

                    le4OPT

                    iL1 jiLj

                    Theory Seminar 042002

                    A Randomized Algorithm

                    Idea Open i with probability ρyi

                    Expected facility cost le ρi fiyi

                    Hope that each copy j(c) has a nearby facility open and service cost decreases

                    Not quitehellip no facility may be open

                    Cluster facilities open ge 1 facility in each cluster

                    Theory Seminar 042002

                    Phase 1 Pruning out ndashzis

                    Open all i st yi = 1

                    For each j if xij = yi = 1 connect j to i

                    Let Lj = i | xij = 1 and lj = |Lj|

                    Cost = j ljvj ndash i zi

                    Lj

                    Fj10

                    rrsquoj = residual reqmt = 6

                    Lrsquoj

                    Phase 2

                    Open all i st frac12 le yi lt 1

                    For each j let Lrsquoj = i | frac12 le xij lt 1

                    Connect copies of j to i Lrsquoj

                    Lose a factor of 2

                    facilities opened in

                    phases 1 2yi = 1 frac12 le yi lt 1yi lt frac12

                    Set L1

                    Set L2

                    Theory Seminar 042002

                    Phase 3

                    Notation facwt(S j) = iS xij

                    1 Form clusters Each cluster has facwt ge frac12

                    2 Open facilities Open ge 1 facility in a cluster ndash used as a backup facility Open facility i with prob 2yi

                    3 Assign facilities to copies Each copy j(c) gets a preferred set of facilities ndash P(j(c)) with facwt ge frac12 For c d P(j(c)) P(j(d)) =

                    4 Connect clients Connect j(c) to the nearest i open in P(j(c)) or to a backup facility

                    Theory Seminar 042002

                    ClusteringAfter phases 1 and 2 Fj = i | xij lt frac12 Sort these by cij and distribute among the rrsquoj copies

                    Cj(c) = avg service cost of the cth copy denote c Cj(c) = cijxij by Cj

                    Initial Fj before any iterations

                    Cj(1)

                    Cj(2)

                    Cj(3)3

                    i Fj

                    client j

                    Want the following properties

                    Clusters to be disjoint

                    Each cluster have facwt ge frac12

                    Each j be connected to rrsquoj clusters

                    iFj

                    Theory Seminar 042002

                    Iterative algorithm

                    S = j | rrsquoj gt 0

                    aj = lsquoactiversquo copy of j initially = 1

                    Ĉj(aj) = avg distance to the first k

                    facilities in Fj gathering facwt ge frac12

                    say these facilities lsquoserversquo j

                    Will maintain Ĉj(aj) le Cj(aj)

                    X

                    X

                    X1

                    Fj after some iterations

                    X i removed from Fj

                    i Fjserving jĈj(3)

                    facilities serving j

                    aj = 3

                    4X

                    (aj)

                    (aj)

                    (aj)

                    Theory Seminar 042002

                    Start of iteration

                    1 Choose j in S with minimum vj + Ĉj(aj)

                    2 Form cluster M = facilities serving j Note facilities are not split

                    3 For k st Fk M decrease rrsquok advance ak set Fk = Fk ndash M

                    2j(3) Cluster M

                    aj = 1

                    4

                    1

                    XX X

                    aj = 4 Cluster M

                    aj = 2

                    3

                    client in S

                    facility in some Fj

                    X facility removed from Fj

                    (aj)

                    Theory Seminar 042002

                    Opening Facilities

                    Central facilities opened in 2 steps

                    1 Open exactly 1 facility in M i opened with prob qyi Acts as backup

                    denoted b(k ) for each k st Fk M

                    2 Open each i in M indep with prob (2-q)yi and independent of step 1

                    Non-central facilities

                    Cluster M

                    k

                    open with prob 2yi independent of other choices

                    j

                    (ak

                    )

                    Theory Seminar 042002

                    Let Sj(c) = avg dist from j to P(j(c))

                    = ( cijxij)facwt(P(j(c))

                    j)

                    Then c Sj(c) le 2Cj

                    Distributing Facilities

                    iP(j(c))

                    j

                    rrsquoj = 3

                    P(j(1))P(j(2))

                    P(j(3))

                    Copy c gets a preferred set P(j(c))Preferred sets are disjoint

                    Ensure facwt(P(j(c)) j) ge frac12 for all c

                    Possible to do so since each xij lt frac12

                    facility in Fj

                    Theory Seminar 042002

                    Analysis

                    Feasibility follows from

                    1 Facilities in phases 1 2 not reused

                    2 After clustering j is connected to rrsquoj disjoint clusters backups are distinct

                    3 Preferred sets are disjoint

                    So j connected to rj distinct facilities

                    Theory Seminar 042002

                    Facility cost

                    Recall L1 = i | yi = 1

                    Phase 2 incur a factor of 2

                    Phase 3 each i is opened with probability 2yi

                    Expected facility cost le 2 fiyifor phases 2 3

                    iL1

                    Theory Seminar 042002

                    Bounding backup cost denoted by B rv

                    D event that no i in P(j(c)) is open

                    Lemma E[B|D] le 2vj + Cj(c)

                    Proof 2 cases

                    Service cost I

                    iM Fj st cik le Ĉj(d)

                    Also vk + Ĉj(d) le vj + Ĉj(c) le vj + Cj(c)

                    k(d)

                    j(c)

                    le vj

                    le Ĉj(d)le vk

                    B

                    k(d)

                    j(c)

                    le vj

                    le vk

                    iM Fj cik gt Ĉj(d)

                    le Ĉj(d) in expectatio

                    n

                    1)

                    2)

                    backup = b(j(c))

                    Theory Seminar 042002

                    Service Cost II

                    Fix j c Let X(c) = service cost of j(c)

                    Let di = cij pi = prob i is opened = 2yi

                    B(c) = backup costD(c) = event that no iP(j(c)) is

                    openp = Pr[D(c)] = (1-p1)hellip(1-pm) le e-1

                    davg = weighted avg of the dis

                    = (i pidi)(i pi) = Sj(c)

                    d1

                    d2 dm

                    P(j(c)) sorted by increasing cij

                    j(c)

                    i P(j(c))

                    Theory Seminar 042002

                    Then

                    E[X(c)] = [p1d1 + (1-p1)p2d2 + hellip

                    + (1-p1)hellip(1-pm-1)pmdm]

                    + pE[B(c)|D(c)]

                    le (1-p)davg + p[2vj + Cj(c)]

                    le (1-e-1)Sj(c) + e-1[2vj +

                    Cj(c)]

                    Let X = c X(c) = service cost of j

                    c Sj(c) le 2Cj and c Cj(c) le 2Cj

                    Summing over all c = 1helliprrsquoj

                    E[X] le (1-e-1)2Cj + e-1(2rrsquojvj + Cj)

                    le 2Cj + 2e-1rrsquojvj

                    Theory Seminar 042002

                    Putting it all together

                    Phase 1 pay the optimal LP cost

                    Phases 2 3

                    bull Facility cost twice LP facility cost

                    bull Service costLose a factor of 2 for phase 2Phase 3 cost is 2(LP service cost)+2e-1(dual value)

                    Overall cost for le (2+2e)(LP cost) phases 2 3

                    Total cost le (2+2e)OPT

                    Theory Seminar 042002

                    How to improve this

                    bull Distribute facilities more equitably (in an expected sense) among copies - decreases prob of lsquobadrsquo event

                    bull Better analysis ndash maximum distance within a cluster can be bounded by 2Cj(c)

                    bull Balance phases 2 and 3

                    Theory Seminar 042002

                    Summary of Results

                    bullGive a 2076-approx algorithm for non-uniform rjs Based on LP rounding using complem slackness

                    bullFor rj = r extend the primal-dual algorithm of (JMS02) to get a 152-approximation

                    bullFault tolerant k medians with rj = r

                    a Primal-dual algorithm (JMS02) gives a 4-approx using Lagrangean relaxation

                    b LP rounding gives a factor of 8

                    Theory Seminar 042002

                    Open Questions

                    1 Reduce gap between rj = r non-uniform rj

                    2 Combinatorial algorithms for non-uniform rj primal-dual local-search

                    3 Constant-factor approx for fault tolerant k medians with non-uniform rjs

                    • Fault Tolerant Facility Location
                    • Metric Facility Location
                    • PowerPoint Presentation
                    • Previous Work
                    • Previous Work (contd)
                    • LP Formulation
                    • LP Formulation (contd)
                    • Complementary Slackness
                    • 4-approximation outline
                    • The Algorithm
                    • Slide 11
                    • Analysis Phase 1
                    • Analysis (contd)
                    • The Algorithm (contd)
                    • Analysis Phase 2
                    • Finally hellip
                    • A Randomized Algorithm
                    • Slide 18
                    • Slide 19
                    • Clustering
                    • Slide 21
                    • Slide 22
                    • Opening Facilities
                    • Distributing Facilities
                    • Analysis
                    • Facility cost
                    • Service cost I
                    • Service Cost II
                    • Slide 29
                    • Putting it all together
                    • How to improve this
                    • Summary of Results
                    • Open Questions

                      Theory Seminar 042002

                      0

                      X XX

                      30

                      2

                      j

                      3 Open rj cheapest facilities in M

                      4 For k st Fk M connect rj copies to opened facilities Decrease rk set Fk=Fk-M

                      End of iteration

                      client in S

                      facility in some Fj

                      client not in S

                      X facility removed from

                      Fj

                      Cluster M

                      facility opened from M

                      Theory Seminar 042002

                      Analysis Phase 1

                      Solution is feasible each j is connected to rj distinct facilities

                      Lemma Facility cost lei fiyi

                      Proof Cost of rj cheapest facilities in M lerj (avg cost) = iM fiyi These facilities donrsquot get used again

                      Theory Seminar 042002

                      Analysis (contd)

                      Lemma For any j and c service cost of copy j(c) le3vj

                      Proof

                      vk le vj since k was chosen as cluster center

                      Service cost le vj + 2vk le 3vj

                      Cluster M

                      j(c)le vj

                      le vk

                      le vk k

                      Theory Seminar 042002

                      The Algorithm (contd)

                      Phase 2 Taking care of ndashzis

                      1 Open all (unopened) i st yi = 1

                      2 For any j if xij = yi = 1 disconnect a copy of j and connect it to i

                      j

                      rj = 3

                      X

                      i with yi = 1i with yi lt

                      1 and open

                      Theory Seminar 042002

                      Analysis Phase 2

                      Lemma Cost of phase 2 = fi + cij = j ljvj ndash i zi

                      Proof Each i with zi gt 0 is opened For iL1 all j st wij gt 0 are connected to it So

                      vj = (service cost) + (fi +

                      zi)

                      j ljvj = fi + cij + i zi

                      Let L1 = i | yi = 1

                      Lj = i | xij = 1 L1 and lj = |Lj|

                      iL1 jiLj

                      j|iLj

                      jiLjiL1

                      Theory Seminar 042002

                      Finally hellip

                      Theorem Total cost le 4 times the optimal cost

                      Proof Total cost le

                      i fiyi + 3j (rj ndash lj)vj + fi + cij

                      facility cost of phase 1 cost for

                      copies connected by

                      phase 1

                      cost of phase 2

                      lei fiyi + 3j (rj ndash lj)vj + (j ljvj ndash i

                      zi )

                      lei fiyi + 3(j rjvj ndash i zi )

                      le4OPT

                      iL1 jiLj

                      Theory Seminar 042002

                      A Randomized Algorithm

                      Idea Open i with probability ρyi

                      Expected facility cost le ρi fiyi

                      Hope that each copy j(c) has a nearby facility open and service cost decreases

                      Not quitehellip no facility may be open

                      Cluster facilities open ge 1 facility in each cluster

                      Theory Seminar 042002

                      Phase 1 Pruning out ndashzis

                      Open all i st yi = 1

                      For each j if xij = yi = 1 connect j to i

                      Let Lj = i | xij = 1 and lj = |Lj|

                      Cost = j ljvj ndash i zi

                      Lj

                      Fj10

                      rrsquoj = residual reqmt = 6

                      Lrsquoj

                      Phase 2

                      Open all i st frac12 le yi lt 1

                      For each j let Lrsquoj = i | frac12 le xij lt 1

                      Connect copies of j to i Lrsquoj

                      Lose a factor of 2

                      facilities opened in

                      phases 1 2yi = 1 frac12 le yi lt 1yi lt frac12

                      Set L1

                      Set L2

                      Theory Seminar 042002

                      Phase 3

                      Notation facwt(S j) = iS xij

                      1 Form clusters Each cluster has facwt ge frac12

                      2 Open facilities Open ge 1 facility in a cluster ndash used as a backup facility Open facility i with prob 2yi

                      3 Assign facilities to copies Each copy j(c) gets a preferred set of facilities ndash P(j(c)) with facwt ge frac12 For c d P(j(c)) P(j(d)) =

                      4 Connect clients Connect j(c) to the nearest i open in P(j(c)) or to a backup facility

                      Theory Seminar 042002

                      ClusteringAfter phases 1 and 2 Fj = i | xij lt frac12 Sort these by cij and distribute among the rrsquoj copies

                      Cj(c) = avg service cost of the cth copy denote c Cj(c) = cijxij by Cj

                      Initial Fj before any iterations

                      Cj(1)

                      Cj(2)

                      Cj(3)3

                      i Fj

                      client j

                      Want the following properties

                      Clusters to be disjoint

                      Each cluster have facwt ge frac12

                      Each j be connected to rrsquoj clusters

                      iFj

                      Theory Seminar 042002

                      Iterative algorithm

                      S = j | rrsquoj gt 0

                      aj = lsquoactiversquo copy of j initially = 1

                      Ĉj(aj) = avg distance to the first k

                      facilities in Fj gathering facwt ge frac12

                      say these facilities lsquoserversquo j

                      Will maintain Ĉj(aj) le Cj(aj)

                      X

                      X

                      X1

                      Fj after some iterations

                      X i removed from Fj

                      i Fjserving jĈj(3)

                      facilities serving j

                      aj = 3

                      4X

                      (aj)

                      (aj)

                      (aj)

                      Theory Seminar 042002

                      Start of iteration

                      1 Choose j in S with minimum vj + Ĉj(aj)

                      2 Form cluster M = facilities serving j Note facilities are not split

                      3 For k st Fk M decrease rrsquok advance ak set Fk = Fk ndash M

                      2j(3) Cluster M

                      aj = 1

                      4

                      1

                      XX X

                      aj = 4 Cluster M

                      aj = 2

                      3

                      client in S

                      facility in some Fj

                      X facility removed from Fj

                      (aj)

                      Theory Seminar 042002

                      Opening Facilities

                      Central facilities opened in 2 steps

                      1 Open exactly 1 facility in M i opened with prob qyi Acts as backup

                      denoted b(k ) for each k st Fk M

                      2 Open each i in M indep with prob (2-q)yi and independent of step 1

                      Non-central facilities

                      Cluster M

                      k

                      open with prob 2yi independent of other choices

                      j

                      (ak

                      )

                      Theory Seminar 042002

                      Let Sj(c) = avg dist from j to P(j(c))

                      = ( cijxij)facwt(P(j(c))

                      j)

                      Then c Sj(c) le 2Cj

                      Distributing Facilities

                      iP(j(c))

                      j

                      rrsquoj = 3

                      P(j(1))P(j(2))

                      P(j(3))

                      Copy c gets a preferred set P(j(c))Preferred sets are disjoint

                      Ensure facwt(P(j(c)) j) ge frac12 for all c

                      Possible to do so since each xij lt frac12

                      facility in Fj

                      Theory Seminar 042002

                      Analysis

                      Feasibility follows from

                      1 Facilities in phases 1 2 not reused

                      2 After clustering j is connected to rrsquoj disjoint clusters backups are distinct

                      3 Preferred sets are disjoint

                      So j connected to rj distinct facilities

                      Theory Seminar 042002

                      Facility cost

                      Recall L1 = i | yi = 1

                      Phase 2 incur a factor of 2

                      Phase 3 each i is opened with probability 2yi

                      Expected facility cost le 2 fiyifor phases 2 3

                      iL1

                      Theory Seminar 042002

                      Bounding backup cost denoted by B rv

                      D event that no i in P(j(c)) is open

                      Lemma E[B|D] le 2vj + Cj(c)

                      Proof 2 cases

                      Service cost I

                      iM Fj st cik le Ĉj(d)

                      Also vk + Ĉj(d) le vj + Ĉj(c) le vj + Cj(c)

                      k(d)

                      j(c)

                      le vj

                      le Ĉj(d)le vk

                      B

                      k(d)

                      j(c)

                      le vj

                      le vk

                      iM Fj cik gt Ĉj(d)

                      le Ĉj(d) in expectatio

                      n

                      1)

                      2)

                      backup = b(j(c))

                      Theory Seminar 042002

                      Service Cost II

                      Fix j c Let X(c) = service cost of j(c)

                      Let di = cij pi = prob i is opened = 2yi

                      B(c) = backup costD(c) = event that no iP(j(c)) is

                      openp = Pr[D(c)] = (1-p1)hellip(1-pm) le e-1

                      davg = weighted avg of the dis

                      = (i pidi)(i pi) = Sj(c)

                      d1

                      d2 dm

                      P(j(c)) sorted by increasing cij

                      j(c)

                      i P(j(c))

                      Theory Seminar 042002

                      Then

                      E[X(c)] = [p1d1 + (1-p1)p2d2 + hellip

                      + (1-p1)hellip(1-pm-1)pmdm]

                      + pE[B(c)|D(c)]

                      le (1-p)davg + p[2vj + Cj(c)]

                      le (1-e-1)Sj(c) + e-1[2vj +

                      Cj(c)]

                      Let X = c X(c) = service cost of j

                      c Sj(c) le 2Cj and c Cj(c) le 2Cj

                      Summing over all c = 1helliprrsquoj

                      E[X] le (1-e-1)2Cj + e-1(2rrsquojvj + Cj)

                      le 2Cj + 2e-1rrsquojvj

                      Theory Seminar 042002

                      Putting it all together

                      Phase 1 pay the optimal LP cost

                      Phases 2 3

                      bull Facility cost twice LP facility cost

                      bull Service costLose a factor of 2 for phase 2Phase 3 cost is 2(LP service cost)+2e-1(dual value)

                      Overall cost for le (2+2e)(LP cost) phases 2 3

                      Total cost le (2+2e)OPT

                      Theory Seminar 042002

                      How to improve this

                      bull Distribute facilities more equitably (in an expected sense) among copies - decreases prob of lsquobadrsquo event

                      bull Better analysis ndash maximum distance within a cluster can be bounded by 2Cj(c)

                      bull Balance phases 2 and 3

                      Theory Seminar 042002

                      Summary of Results

                      bullGive a 2076-approx algorithm for non-uniform rjs Based on LP rounding using complem slackness

                      bullFor rj = r extend the primal-dual algorithm of (JMS02) to get a 152-approximation

                      bullFault tolerant k medians with rj = r

                      a Primal-dual algorithm (JMS02) gives a 4-approx using Lagrangean relaxation

                      b LP rounding gives a factor of 8

                      Theory Seminar 042002

                      Open Questions

                      1 Reduce gap between rj = r non-uniform rj

                      2 Combinatorial algorithms for non-uniform rj primal-dual local-search

                      3 Constant-factor approx for fault tolerant k medians with non-uniform rjs

                      • Fault Tolerant Facility Location
                      • Metric Facility Location
                      • PowerPoint Presentation
                      • Previous Work
                      • Previous Work (contd)
                      • LP Formulation
                      • LP Formulation (contd)
                      • Complementary Slackness
                      • 4-approximation outline
                      • The Algorithm
                      • Slide 11
                      • Analysis Phase 1
                      • Analysis (contd)
                      • The Algorithm (contd)
                      • Analysis Phase 2
                      • Finally hellip
                      • A Randomized Algorithm
                      • Slide 18
                      • Slide 19
                      • Clustering
                      • Slide 21
                      • Slide 22
                      • Opening Facilities
                      • Distributing Facilities
                      • Analysis
                      • Facility cost
                      • Service cost I
                      • Service Cost II
                      • Slide 29
                      • Putting it all together
                      • How to improve this
                      • Summary of Results
                      • Open Questions

                        Theory Seminar 042002

                        Analysis Phase 1

                        Solution is feasible each j is connected to rj distinct facilities

                        Lemma Facility cost lei fiyi

                        Proof Cost of rj cheapest facilities in M lerj (avg cost) = iM fiyi These facilities donrsquot get used again

                        Theory Seminar 042002

                        Analysis (contd)

                        Lemma For any j and c service cost of copy j(c) le3vj

                        Proof

                        vk le vj since k was chosen as cluster center

                        Service cost le vj + 2vk le 3vj

                        Cluster M

                        j(c)le vj

                        le vk

                        le vk k

                        Theory Seminar 042002

                        The Algorithm (contd)

                        Phase 2 Taking care of ndashzis

                        1 Open all (unopened) i st yi = 1

                        2 For any j if xij = yi = 1 disconnect a copy of j and connect it to i

                        j

                        rj = 3

                        X

                        i with yi = 1i with yi lt

                        1 and open

                        Theory Seminar 042002

                        Analysis Phase 2

                        Lemma Cost of phase 2 = fi + cij = j ljvj ndash i zi

                        Proof Each i with zi gt 0 is opened For iL1 all j st wij gt 0 are connected to it So

                        vj = (service cost) + (fi +

                        zi)

                        j ljvj = fi + cij + i zi

                        Let L1 = i | yi = 1

                        Lj = i | xij = 1 L1 and lj = |Lj|

                        iL1 jiLj

                        j|iLj

                        jiLjiL1

                        Theory Seminar 042002

                        Finally hellip

                        Theorem Total cost le 4 times the optimal cost

                        Proof Total cost le

                        i fiyi + 3j (rj ndash lj)vj + fi + cij

                        facility cost of phase 1 cost for

                        copies connected by

                        phase 1

                        cost of phase 2

                        lei fiyi + 3j (rj ndash lj)vj + (j ljvj ndash i

                        zi )

                        lei fiyi + 3(j rjvj ndash i zi )

                        le4OPT

                        iL1 jiLj

                        Theory Seminar 042002

                        A Randomized Algorithm

                        Idea Open i with probability ρyi

                        Expected facility cost le ρi fiyi

                        Hope that each copy j(c) has a nearby facility open and service cost decreases

                        Not quitehellip no facility may be open

                        Cluster facilities open ge 1 facility in each cluster

                        Theory Seminar 042002

                        Phase 1 Pruning out ndashzis

                        Open all i st yi = 1

                        For each j if xij = yi = 1 connect j to i

                        Let Lj = i | xij = 1 and lj = |Lj|

                        Cost = j ljvj ndash i zi

                        Lj

                        Fj10

                        rrsquoj = residual reqmt = 6

                        Lrsquoj

                        Phase 2

                        Open all i st frac12 le yi lt 1

                        For each j let Lrsquoj = i | frac12 le xij lt 1

                        Connect copies of j to i Lrsquoj

                        Lose a factor of 2

                        facilities opened in

                        phases 1 2yi = 1 frac12 le yi lt 1yi lt frac12

                        Set L1

                        Set L2

                        Theory Seminar 042002

                        Phase 3

                        Notation facwt(S j) = iS xij

                        1 Form clusters Each cluster has facwt ge frac12

                        2 Open facilities Open ge 1 facility in a cluster ndash used as a backup facility Open facility i with prob 2yi

                        3 Assign facilities to copies Each copy j(c) gets a preferred set of facilities ndash P(j(c)) with facwt ge frac12 For c d P(j(c)) P(j(d)) =

                        4 Connect clients Connect j(c) to the nearest i open in P(j(c)) or to a backup facility

                        Theory Seminar 042002

                        ClusteringAfter phases 1 and 2 Fj = i | xij lt frac12 Sort these by cij and distribute among the rrsquoj copies

                        Cj(c) = avg service cost of the cth copy denote c Cj(c) = cijxij by Cj

                        Initial Fj before any iterations

                        Cj(1)

                        Cj(2)

                        Cj(3)3

                        i Fj

                        client j

                        Want the following properties

                        Clusters to be disjoint

                        Each cluster have facwt ge frac12

                        Each j be connected to rrsquoj clusters

                        iFj

                        Theory Seminar 042002

                        Iterative algorithm

                        S = j | rrsquoj gt 0

                        aj = lsquoactiversquo copy of j initially = 1

                        Ĉj(aj) = avg distance to the first k

                        facilities in Fj gathering facwt ge frac12

                        say these facilities lsquoserversquo j

                        Will maintain Ĉj(aj) le Cj(aj)

                        X

                        X

                        X1

                        Fj after some iterations

                        X i removed from Fj

                        i Fjserving jĈj(3)

                        facilities serving j

                        aj = 3

                        4X

                        (aj)

                        (aj)

                        (aj)

                        Theory Seminar 042002

                        Start of iteration

                        1 Choose j in S with minimum vj + Ĉj(aj)

                        2 Form cluster M = facilities serving j Note facilities are not split

                        3 For k st Fk M decrease rrsquok advance ak set Fk = Fk ndash M

                        2j(3) Cluster M

                        aj = 1

                        4

                        1

                        XX X

                        aj = 4 Cluster M

                        aj = 2

                        3

                        client in S

                        facility in some Fj

                        X facility removed from Fj

                        (aj)

                        Theory Seminar 042002

                        Opening Facilities

                        Central facilities opened in 2 steps

                        1 Open exactly 1 facility in M i opened with prob qyi Acts as backup

                        denoted b(k ) for each k st Fk M

                        2 Open each i in M indep with prob (2-q)yi and independent of step 1

                        Non-central facilities

                        Cluster M

                        k

                        open with prob 2yi independent of other choices

                        j

                        (ak

                        )

                        Theory Seminar 042002

                        Let Sj(c) = avg dist from j to P(j(c))

                        = ( cijxij)facwt(P(j(c))

                        j)

                        Then c Sj(c) le 2Cj

                        Distributing Facilities

                        iP(j(c))

                        j

                        rrsquoj = 3

                        P(j(1))P(j(2))

                        P(j(3))

                        Copy c gets a preferred set P(j(c))Preferred sets are disjoint

                        Ensure facwt(P(j(c)) j) ge frac12 for all c

                        Possible to do so since each xij lt frac12

                        facility in Fj

                        Theory Seminar 042002

                        Analysis

                        Feasibility follows from

                        1 Facilities in phases 1 2 not reused

                        2 After clustering j is connected to rrsquoj disjoint clusters backups are distinct

                        3 Preferred sets are disjoint

                        So j connected to rj distinct facilities

                        Theory Seminar 042002

                        Facility cost

                        Recall L1 = i | yi = 1

                        Phase 2 incur a factor of 2

                        Phase 3 each i is opened with probability 2yi

                        Expected facility cost le 2 fiyifor phases 2 3

                        iL1

                        Theory Seminar 042002

                        Bounding backup cost denoted by B rv

                        D event that no i in P(j(c)) is open

                        Lemma E[B|D] le 2vj + Cj(c)

                        Proof 2 cases

                        Service cost I

                        iM Fj st cik le Ĉj(d)

                        Also vk + Ĉj(d) le vj + Ĉj(c) le vj + Cj(c)

                        k(d)

                        j(c)

                        le vj

                        le Ĉj(d)le vk

                        B

                        k(d)

                        j(c)

                        le vj

                        le vk

                        iM Fj cik gt Ĉj(d)

                        le Ĉj(d) in expectatio

                        n

                        1)

                        2)

                        backup = b(j(c))

                        Theory Seminar 042002

                        Service Cost II

                        Fix j c Let X(c) = service cost of j(c)

                        Let di = cij pi = prob i is opened = 2yi

                        B(c) = backup costD(c) = event that no iP(j(c)) is

                        openp = Pr[D(c)] = (1-p1)hellip(1-pm) le e-1

                        davg = weighted avg of the dis

                        = (i pidi)(i pi) = Sj(c)

                        d1

                        d2 dm

                        P(j(c)) sorted by increasing cij

                        j(c)

                        i P(j(c))

                        Theory Seminar 042002

                        Then

                        E[X(c)] = [p1d1 + (1-p1)p2d2 + hellip

                        + (1-p1)hellip(1-pm-1)pmdm]

                        + pE[B(c)|D(c)]

                        le (1-p)davg + p[2vj + Cj(c)]

                        le (1-e-1)Sj(c) + e-1[2vj +

                        Cj(c)]

                        Let X = c X(c) = service cost of j

                        c Sj(c) le 2Cj and c Cj(c) le 2Cj

                        Summing over all c = 1helliprrsquoj

                        E[X] le (1-e-1)2Cj + e-1(2rrsquojvj + Cj)

                        le 2Cj + 2e-1rrsquojvj

                        Theory Seminar 042002

                        Putting it all together

                        Phase 1 pay the optimal LP cost

                        Phases 2 3

                        bull Facility cost twice LP facility cost

                        bull Service costLose a factor of 2 for phase 2Phase 3 cost is 2(LP service cost)+2e-1(dual value)

                        Overall cost for le (2+2e)(LP cost) phases 2 3

                        Total cost le (2+2e)OPT

                        Theory Seminar 042002

                        How to improve this

                        bull Distribute facilities more equitably (in an expected sense) among copies - decreases prob of lsquobadrsquo event

                        bull Better analysis ndash maximum distance within a cluster can be bounded by 2Cj(c)

                        bull Balance phases 2 and 3

                        Theory Seminar 042002

                        Summary of Results

                        bullGive a 2076-approx algorithm for non-uniform rjs Based on LP rounding using complem slackness

                        bullFor rj = r extend the primal-dual algorithm of (JMS02) to get a 152-approximation

                        bullFault tolerant k medians with rj = r

                        a Primal-dual algorithm (JMS02) gives a 4-approx using Lagrangean relaxation

                        b LP rounding gives a factor of 8

                        Theory Seminar 042002

                        Open Questions

                        1 Reduce gap between rj = r non-uniform rj

                        2 Combinatorial algorithms for non-uniform rj primal-dual local-search

                        3 Constant-factor approx for fault tolerant k medians with non-uniform rjs

                        • Fault Tolerant Facility Location
                        • Metric Facility Location
                        • PowerPoint Presentation
                        • Previous Work
                        • Previous Work (contd)
                        • LP Formulation
                        • LP Formulation (contd)
                        • Complementary Slackness
                        • 4-approximation outline
                        • The Algorithm
                        • Slide 11
                        • Analysis Phase 1
                        • Analysis (contd)
                        • The Algorithm (contd)
                        • Analysis Phase 2
                        • Finally hellip
                        • A Randomized Algorithm
                        • Slide 18
                        • Slide 19
                        • Clustering
                        • Slide 21
                        • Slide 22
                        • Opening Facilities
                        • Distributing Facilities
                        • Analysis
                        • Facility cost
                        • Service cost I
                        • Service Cost II
                        • Slide 29
                        • Putting it all together
                        • How to improve this
                        • Summary of Results
                        • Open Questions

                          Theory Seminar 042002

                          Analysis (contd)

                          Lemma For any j and c service cost of copy j(c) le3vj

                          Proof

                          vk le vj since k was chosen as cluster center

                          Service cost le vj + 2vk le 3vj

                          Cluster M

                          j(c)le vj

                          le vk

                          le vk k

                          Theory Seminar 042002

                          The Algorithm (contd)

                          Phase 2 Taking care of ndashzis

                          1 Open all (unopened) i st yi = 1

                          2 For any j if xij = yi = 1 disconnect a copy of j and connect it to i

                          j

                          rj = 3

                          X

                          i with yi = 1i with yi lt

                          1 and open

                          Theory Seminar 042002

                          Analysis Phase 2

                          Lemma Cost of phase 2 = fi + cij = j ljvj ndash i zi

                          Proof Each i with zi gt 0 is opened For iL1 all j st wij gt 0 are connected to it So

                          vj = (service cost) + (fi +

                          zi)

                          j ljvj = fi + cij + i zi

                          Let L1 = i | yi = 1

                          Lj = i | xij = 1 L1 and lj = |Lj|

                          iL1 jiLj

                          j|iLj

                          jiLjiL1

                          Theory Seminar 042002

                          Finally hellip

                          Theorem Total cost le 4 times the optimal cost

                          Proof Total cost le

                          i fiyi + 3j (rj ndash lj)vj + fi + cij

                          facility cost of phase 1 cost for

                          copies connected by

                          phase 1

                          cost of phase 2

                          lei fiyi + 3j (rj ndash lj)vj + (j ljvj ndash i

                          zi )

                          lei fiyi + 3(j rjvj ndash i zi )

                          le4OPT

                          iL1 jiLj

                          Theory Seminar 042002

                          A Randomized Algorithm

                          Idea Open i with probability ρyi

                          Expected facility cost le ρi fiyi

                          Hope that each copy j(c) has a nearby facility open and service cost decreases

                          Not quitehellip no facility may be open

                          Cluster facilities open ge 1 facility in each cluster

                          Theory Seminar 042002

                          Phase 1 Pruning out ndashzis

                          Open all i st yi = 1

                          For each j if xij = yi = 1 connect j to i

                          Let Lj = i | xij = 1 and lj = |Lj|

                          Cost = j ljvj ndash i zi

                          Lj

                          Fj10

                          rrsquoj = residual reqmt = 6

                          Lrsquoj

                          Phase 2

                          Open all i st frac12 le yi lt 1

                          For each j let Lrsquoj = i | frac12 le xij lt 1

                          Connect copies of j to i Lrsquoj

                          Lose a factor of 2

                          facilities opened in

                          phases 1 2yi = 1 frac12 le yi lt 1yi lt frac12

                          Set L1

                          Set L2

                          Theory Seminar 042002

                          Phase 3

                          Notation facwt(S j) = iS xij

                          1 Form clusters Each cluster has facwt ge frac12

                          2 Open facilities Open ge 1 facility in a cluster ndash used as a backup facility Open facility i with prob 2yi

                          3 Assign facilities to copies Each copy j(c) gets a preferred set of facilities ndash P(j(c)) with facwt ge frac12 For c d P(j(c)) P(j(d)) =

                          4 Connect clients Connect j(c) to the nearest i open in P(j(c)) or to a backup facility

                          Theory Seminar 042002

                          ClusteringAfter phases 1 and 2 Fj = i | xij lt frac12 Sort these by cij and distribute among the rrsquoj copies

                          Cj(c) = avg service cost of the cth copy denote c Cj(c) = cijxij by Cj

                          Initial Fj before any iterations

                          Cj(1)

                          Cj(2)

                          Cj(3)3

                          i Fj

                          client j

                          Want the following properties

                          Clusters to be disjoint

                          Each cluster have facwt ge frac12

                          Each j be connected to rrsquoj clusters

                          iFj

                          Theory Seminar 042002

                          Iterative algorithm

                          S = j | rrsquoj gt 0

                          aj = lsquoactiversquo copy of j initially = 1

                          Ĉj(aj) = avg distance to the first k

                          facilities in Fj gathering facwt ge frac12

                          say these facilities lsquoserversquo j

                          Will maintain Ĉj(aj) le Cj(aj)

                          X

                          X

                          X1

                          Fj after some iterations

                          X i removed from Fj

                          i Fjserving jĈj(3)

                          facilities serving j

                          aj = 3

                          4X

                          (aj)

                          (aj)

                          (aj)

                          Theory Seminar 042002

                          Start of iteration

                          1 Choose j in S with minimum vj + Ĉj(aj)

                          2 Form cluster M = facilities serving j Note facilities are not split

                          3 For k st Fk M decrease rrsquok advance ak set Fk = Fk ndash M

                          2j(3) Cluster M

                          aj = 1

                          4

                          1

                          XX X

                          aj = 4 Cluster M

                          aj = 2

                          3

                          client in S

                          facility in some Fj

                          X facility removed from Fj

                          (aj)

                          Theory Seminar 042002

                          Opening Facilities

                          Central facilities opened in 2 steps

                          1 Open exactly 1 facility in M i opened with prob qyi Acts as backup

                          denoted b(k ) for each k st Fk M

                          2 Open each i in M indep with prob (2-q)yi and independent of step 1

                          Non-central facilities

                          Cluster M

                          k

                          open with prob 2yi independent of other choices

                          j

                          (ak

                          )

                          Theory Seminar 042002

                          Let Sj(c) = avg dist from j to P(j(c))

                          = ( cijxij)facwt(P(j(c))

                          j)

                          Then c Sj(c) le 2Cj

                          Distributing Facilities

                          iP(j(c))

                          j

                          rrsquoj = 3

                          P(j(1))P(j(2))

                          P(j(3))

                          Copy c gets a preferred set P(j(c))Preferred sets are disjoint

                          Ensure facwt(P(j(c)) j) ge frac12 for all c

                          Possible to do so since each xij lt frac12

                          facility in Fj

                          Theory Seminar 042002

                          Analysis

                          Feasibility follows from

                          1 Facilities in phases 1 2 not reused

                          2 After clustering j is connected to rrsquoj disjoint clusters backups are distinct

                          3 Preferred sets are disjoint

                          So j connected to rj distinct facilities

                          Theory Seminar 042002

                          Facility cost

                          Recall L1 = i | yi = 1

                          Phase 2 incur a factor of 2

                          Phase 3 each i is opened with probability 2yi

                          Expected facility cost le 2 fiyifor phases 2 3

                          iL1

                          Theory Seminar 042002

                          Bounding backup cost denoted by B rv

                          D event that no i in P(j(c)) is open

                          Lemma E[B|D] le 2vj + Cj(c)

                          Proof 2 cases

                          Service cost I

                          iM Fj st cik le Ĉj(d)

                          Also vk + Ĉj(d) le vj + Ĉj(c) le vj + Cj(c)

                          k(d)

                          j(c)

                          le vj

                          le Ĉj(d)le vk

                          B

                          k(d)

                          j(c)

                          le vj

                          le vk

                          iM Fj cik gt Ĉj(d)

                          le Ĉj(d) in expectatio

                          n

                          1)

                          2)

                          backup = b(j(c))

                          Theory Seminar 042002

                          Service Cost II

                          Fix j c Let X(c) = service cost of j(c)

                          Let di = cij pi = prob i is opened = 2yi

                          B(c) = backup costD(c) = event that no iP(j(c)) is

                          openp = Pr[D(c)] = (1-p1)hellip(1-pm) le e-1

                          davg = weighted avg of the dis

                          = (i pidi)(i pi) = Sj(c)

                          d1

                          d2 dm

                          P(j(c)) sorted by increasing cij

                          j(c)

                          i P(j(c))

                          Theory Seminar 042002

                          Then

                          E[X(c)] = [p1d1 + (1-p1)p2d2 + hellip

                          + (1-p1)hellip(1-pm-1)pmdm]

                          + pE[B(c)|D(c)]

                          le (1-p)davg + p[2vj + Cj(c)]

                          le (1-e-1)Sj(c) + e-1[2vj +

                          Cj(c)]

                          Let X = c X(c) = service cost of j

                          c Sj(c) le 2Cj and c Cj(c) le 2Cj

                          Summing over all c = 1helliprrsquoj

                          E[X] le (1-e-1)2Cj + e-1(2rrsquojvj + Cj)

                          le 2Cj + 2e-1rrsquojvj

                          Theory Seminar 042002

                          Putting it all together

                          Phase 1 pay the optimal LP cost

                          Phases 2 3

                          bull Facility cost twice LP facility cost

                          bull Service costLose a factor of 2 for phase 2Phase 3 cost is 2(LP service cost)+2e-1(dual value)

                          Overall cost for le (2+2e)(LP cost) phases 2 3

                          Total cost le (2+2e)OPT

                          Theory Seminar 042002

                          How to improve this

                          bull Distribute facilities more equitably (in an expected sense) among copies - decreases prob of lsquobadrsquo event

                          bull Better analysis ndash maximum distance within a cluster can be bounded by 2Cj(c)

                          bull Balance phases 2 and 3

                          Theory Seminar 042002

                          Summary of Results

                          bullGive a 2076-approx algorithm for non-uniform rjs Based on LP rounding using complem slackness

                          bullFor rj = r extend the primal-dual algorithm of (JMS02) to get a 152-approximation

                          bullFault tolerant k medians with rj = r

                          a Primal-dual algorithm (JMS02) gives a 4-approx using Lagrangean relaxation

                          b LP rounding gives a factor of 8

                          Theory Seminar 042002

                          Open Questions

                          1 Reduce gap between rj = r non-uniform rj

                          2 Combinatorial algorithms for non-uniform rj primal-dual local-search

                          3 Constant-factor approx for fault tolerant k medians with non-uniform rjs

                          • Fault Tolerant Facility Location
                          • Metric Facility Location
                          • PowerPoint Presentation
                          • Previous Work
                          • Previous Work (contd)
                          • LP Formulation
                          • LP Formulation (contd)
                          • Complementary Slackness
                          • 4-approximation outline
                          • The Algorithm
                          • Slide 11
                          • Analysis Phase 1
                          • Analysis (contd)
                          • The Algorithm (contd)
                          • Analysis Phase 2
                          • Finally hellip
                          • A Randomized Algorithm
                          • Slide 18
                          • Slide 19
                          • Clustering
                          • Slide 21
                          • Slide 22
                          • Opening Facilities
                          • Distributing Facilities
                          • Analysis
                          • Facility cost
                          • Service cost I
                          • Service Cost II
                          • Slide 29
                          • Putting it all together
                          • How to improve this
                          • Summary of Results
                          • Open Questions

                            Theory Seminar 042002

                            The Algorithm (contd)

                            Phase 2 Taking care of ndashzis

                            1 Open all (unopened) i st yi = 1

                            2 For any j if xij = yi = 1 disconnect a copy of j and connect it to i

                            j

                            rj = 3

                            X

                            i with yi = 1i with yi lt

                            1 and open

                            Theory Seminar 042002

                            Analysis Phase 2

                            Lemma Cost of phase 2 = fi + cij = j ljvj ndash i zi

                            Proof Each i with zi gt 0 is opened For iL1 all j st wij gt 0 are connected to it So

                            vj = (service cost) + (fi +

                            zi)

                            j ljvj = fi + cij + i zi

                            Let L1 = i | yi = 1

                            Lj = i | xij = 1 L1 and lj = |Lj|

                            iL1 jiLj

                            j|iLj

                            jiLjiL1

                            Theory Seminar 042002

                            Finally hellip

                            Theorem Total cost le 4 times the optimal cost

                            Proof Total cost le

                            i fiyi + 3j (rj ndash lj)vj + fi + cij

                            facility cost of phase 1 cost for

                            copies connected by

                            phase 1

                            cost of phase 2

                            lei fiyi + 3j (rj ndash lj)vj + (j ljvj ndash i

                            zi )

                            lei fiyi + 3(j rjvj ndash i zi )

                            le4OPT

                            iL1 jiLj

                            Theory Seminar 042002

                            A Randomized Algorithm

                            Idea Open i with probability ρyi

                            Expected facility cost le ρi fiyi

                            Hope that each copy j(c) has a nearby facility open and service cost decreases

                            Not quitehellip no facility may be open

                            Cluster facilities open ge 1 facility in each cluster

                            Theory Seminar 042002

                            Phase 1 Pruning out ndashzis

                            Open all i st yi = 1

                            For each j if xij = yi = 1 connect j to i

                            Let Lj = i | xij = 1 and lj = |Lj|

                            Cost = j ljvj ndash i zi

                            Lj

                            Fj10

                            rrsquoj = residual reqmt = 6

                            Lrsquoj

                            Phase 2

                            Open all i st frac12 le yi lt 1

                            For each j let Lrsquoj = i | frac12 le xij lt 1

                            Connect copies of j to i Lrsquoj

                            Lose a factor of 2

                            facilities opened in

                            phases 1 2yi = 1 frac12 le yi lt 1yi lt frac12

                            Set L1

                            Set L2

                            Theory Seminar 042002

                            Phase 3

                            Notation facwt(S j) = iS xij

                            1 Form clusters Each cluster has facwt ge frac12

                            2 Open facilities Open ge 1 facility in a cluster ndash used as a backup facility Open facility i with prob 2yi

                            3 Assign facilities to copies Each copy j(c) gets a preferred set of facilities ndash P(j(c)) with facwt ge frac12 For c d P(j(c)) P(j(d)) =

                            4 Connect clients Connect j(c) to the nearest i open in P(j(c)) or to a backup facility

                            Theory Seminar 042002

                            ClusteringAfter phases 1 and 2 Fj = i | xij lt frac12 Sort these by cij and distribute among the rrsquoj copies

                            Cj(c) = avg service cost of the cth copy denote c Cj(c) = cijxij by Cj

                            Initial Fj before any iterations

                            Cj(1)

                            Cj(2)

                            Cj(3)3

                            i Fj

                            client j

                            Want the following properties

                            Clusters to be disjoint

                            Each cluster have facwt ge frac12

                            Each j be connected to rrsquoj clusters

                            iFj

                            Theory Seminar 042002

                            Iterative algorithm

                            S = j | rrsquoj gt 0

                            aj = lsquoactiversquo copy of j initially = 1

                            Ĉj(aj) = avg distance to the first k

                            facilities in Fj gathering facwt ge frac12

                            say these facilities lsquoserversquo j

                            Will maintain Ĉj(aj) le Cj(aj)

                            X

                            X

                            X1

                            Fj after some iterations

                            X i removed from Fj

                            i Fjserving jĈj(3)

                            facilities serving j

                            aj = 3

                            4X

                            (aj)

                            (aj)

                            (aj)

                            Theory Seminar 042002

                            Start of iteration

                            1 Choose j in S with minimum vj + Ĉj(aj)

                            2 Form cluster M = facilities serving j Note facilities are not split

                            3 For k st Fk M decrease rrsquok advance ak set Fk = Fk ndash M

                            2j(3) Cluster M

                            aj = 1

                            4

                            1

                            XX X

                            aj = 4 Cluster M

                            aj = 2

                            3

                            client in S

                            facility in some Fj

                            X facility removed from Fj

                            (aj)

                            Theory Seminar 042002

                            Opening Facilities

                            Central facilities opened in 2 steps

                            1 Open exactly 1 facility in M i opened with prob qyi Acts as backup

                            denoted b(k ) for each k st Fk M

                            2 Open each i in M indep with prob (2-q)yi and independent of step 1

                            Non-central facilities

                            Cluster M

                            k

                            open with prob 2yi independent of other choices

                            j

                            (ak

                            )

                            Theory Seminar 042002

                            Let Sj(c) = avg dist from j to P(j(c))

                            = ( cijxij)facwt(P(j(c))

                            j)

                            Then c Sj(c) le 2Cj

                            Distributing Facilities

                            iP(j(c))

                            j

                            rrsquoj = 3

                            P(j(1))P(j(2))

                            P(j(3))

                            Copy c gets a preferred set P(j(c))Preferred sets are disjoint

                            Ensure facwt(P(j(c)) j) ge frac12 for all c

                            Possible to do so since each xij lt frac12

                            facility in Fj

                            Theory Seminar 042002

                            Analysis

                            Feasibility follows from

                            1 Facilities in phases 1 2 not reused

                            2 After clustering j is connected to rrsquoj disjoint clusters backups are distinct

                            3 Preferred sets are disjoint

                            So j connected to rj distinct facilities

                            Theory Seminar 042002

                            Facility cost

                            Recall L1 = i | yi = 1

                            Phase 2 incur a factor of 2

                            Phase 3 each i is opened with probability 2yi

                            Expected facility cost le 2 fiyifor phases 2 3

                            iL1

                            Theory Seminar 042002

                            Bounding backup cost denoted by B rv

                            D event that no i in P(j(c)) is open

                            Lemma E[B|D] le 2vj + Cj(c)

                            Proof 2 cases

                            Service cost I

                            iM Fj st cik le Ĉj(d)

                            Also vk + Ĉj(d) le vj + Ĉj(c) le vj + Cj(c)

                            k(d)

                            j(c)

                            le vj

                            le Ĉj(d)le vk

                            B

                            k(d)

                            j(c)

                            le vj

                            le vk

                            iM Fj cik gt Ĉj(d)

                            le Ĉj(d) in expectatio

                            n

                            1)

                            2)

                            backup = b(j(c))

                            Theory Seminar 042002

                            Service Cost II

                            Fix j c Let X(c) = service cost of j(c)

                            Let di = cij pi = prob i is opened = 2yi

                            B(c) = backup costD(c) = event that no iP(j(c)) is

                            openp = Pr[D(c)] = (1-p1)hellip(1-pm) le e-1

                            davg = weighted avg of the dis

                            = (i pidi)(i pi) = Sj(c)

                            d1

                            d2 dm

                            P(j(c)) sorted by increasing cij

                            j(c)

                            i P(j(c))

                            Theory Seminar 042002

                            Then

                            E[X(c)] = [p1d1 + (1-p1)p2d2 + hellip

                            + (1-p1)hellip(1-pm-1)pmdm]

                            + pE[B(c)|D(c)]

                            le (1-p)davg + p[2vj + Cj(c)]

                            le (1-e-1)Sj(c) + e-1[2vj +

                            Cj(c)]

                            Let X = c X(c) = service cost of j

                            c Sj(c) le 2Cj and c Cj(c) le 2Cj

                            Summing over all c = 1helliprrsquoj

                            E[X] le (1-e-1)2Cj + e-1(2rrsquojvj + Cj)

                            le 2Cj + 2e-1rrsquojvj

                            Theory Seminar 042002

                            Putting it all together

                            Phase 1 pay the optimal LP cost

                            Phases 2 3

                            bull Facility cost twice LP facility cost

                            bull Service costLose a factor of 2 for phase 2Phase 3 cost is 2(LP service cost)+2e-1(dual value)

                            Overall cost for le (2+2e)(LP cost) phases 2 3

                            Total cost le (2+2e)OPT

                            Theory Seminar 042002

                            How to improve this

                            bull Distribute facilities more equitably (in an expected sense) among copies - decreases prob of lsquobadrsquo event

                            bull Better analysis ndash maximum distance within a cluster can be bounded by 2Cj(c)

                            bull Balance phases 2 and 3

                            Theory Seminar 042002

                            Summary of Results

                            bullGive a 2076-approx algorithm for non-uniform rjs Based on LP rounding using complem slackness

                            bullFor rj = r extend the primal-dual algorithm of (JMS02) to get a 152-approximation

                            bullFault tolerant k medians with rj = r

                            a Primal-dual algorithm (JMS02) gives a 4-approx using Lagrangean relaxation

                            b LP rounding gives a factor of 8

                            Theory Seminar 042002

                            Open Questions

                            1 Reduce gap between rj = r non-uniform rj

                            2 Combinatorial algorithms for non-uniform rj primal-dual local-search

                            3 Constant-factor approx for fault tolerant k medians with non-uniform rjs

                            • Fault Tolerant Facility Location
                            • Metric Facility Location
                            • PowerPoint Presentation
                            • Previous Work
                            • Previous Work (contd)
                            • LP Formulation
                            • LP Formulation (contd)
                            • Complementary Slackness
                            • 4-approximation outline
                            • The Algorithm
                            • Slide 11
                            • Analysis Phase 1
                            • Analysis (contd)
                            • The Algorithm (contd)
                            • Analysis Phase 2
                            • Finally hellip
                            • A Randomized Algorithm
                            • Slide 18
                            • Slide 19
                            • Clustering
                            • Slide 21
                            • Slide 22
                            • Opening Facilities
                            • Distributing Facilities
                            • Analysis
                            • Facility cost
                            • Service cost I
                            • Service Cost II
                            • Slide 29
                            • Putting it all together
                            • How to improve this
                            • Summary of Results
                            • Open Questions

                              Theory Seminar 042002

                              Analysis Phase 2

                              Lemma Cost of phase 2 = fi + cij = j ljvj ndash i zi

                              Proof Each i with zi gt 0 is opened For iL1 all j st wij gt 0 are connected to it So

                              vj = (service cost) + (fi +

                              zi)

                              j ljvj = fi + cij + i zi

                              Let L1 = i | yi = 1

                              Lj = i | xij = 1 L1 and lj = |Lj|

                              iL1 jiLj

                              j|iLj

                              jiLjiL1

                              Theory Seminar 042002

                              Finally hellip

                              Theorem Total cost le 4 times the optimal cost

                              Proof Total cost le

                              i fiyi + 3j (rj ndash lj)vj + fi + cij

                              facility cost of phase 1 cost for

                              copies connected by

                              phase 1

                              cost of phase 2

                              lei fiyi + 3j (rj ndash lj)vj + (j ljvj ndash i

                              zi )

                              lei fiyi + 3(j rjvj ndash i zi )

                              le4OPT

                              iL1 jiLj

                              Theory Seminar 042002

                              A Randomized Algorithm

                              Idea Open i with probability ρyi

                              Expected facility cost le ρi fiyi

                              Hope that each copy j(c) has a nearby facility open and service cost decreases

                              Not quitehellip no facility may be open

                              Cluster facilities open ge 1 facility in each cluster

                              Theory Seminar 042002

                              Phase 1 Pruning out ndashzis

                              Open all i st yi = 1

                              For each j if xij = yi = 1 connect j to i

                              Let Lj = i | xij = 1 and lj = |Lj|

                              Cost = j ljvj ndash i zi

                              Lj

                              Fj10

                              rrsquoj = residual reqmt = 6

                              Lrsquoj

                              Phase 2

                              Open all i st frac12 le yi lt 1

                              For each j let Lrsquoj = i | frac12 le xij lt 1

                              Connect copies of j to i Lrsquoj

                              Lose a factor of 2

                              facilities opened in

                              phases 1 2yi = 1 frac12 le yi lt 1yi lt frac12

                              Set L1

                              Set L2

                              Theory Seminar 042002

                              Phase 3

                              Notation facwt(S j) = iS xij

                              1 Form clusters Each cluster has facwt ge frac12

                              2 Open facilities Open ge 1 facility in a cluster ndash used as a backup facility Open facility i with prob 2yi

                              3 Assign facilities to copies Each copy j(c) gets a preferred set of facilities ndash P(j(c)) with facwt ge frac12 For c d P(j(c)) P(j(d)) =

                              4 Connect clients Connect j(c) to the nearest i open in P(j(c)) or to a backup facility

                              Theory Seminar 042002

                              ClusteringAfter phases 1 and 2 Fj = i | xij lt frac12 Sort these by cij and distribute among the rrsquoj copies

                              Cj(c) = avg service cost of the cth copy denote c Cj(c) = cijxij by Cj

                              Initial Fj before any iterations

                              Cj(1)

                              Cj(2)

                              Cj(3)3

                              i Fj

                              client j

                              Want the following properties

                              Clusters to be disjoint

                              Each cluster have facwt ge frac12

                              Each j be connected to rrsquoj clusters

                              iFj

                              Theory Seminar 042002

                              Iterative algorithm

                              S = j | rrsquoj gt 0

                              aj = lsquoactiversquo copy of j initially = 1

                              Ĉj(aj) = avg distance to the first k

                              facilities in Fj gathering facwt ge frac12

                              say these facilities lsquoserversquo j

                              Will maintain Ĉj(aj) le Cj(aj)

                              X

                              X

                              X1

                              Fj after some iterations

                              X i removed from Fj

                              i Fjserving jĈj(3)

                              facilities serving j

                              aj = 3

                              4X

                              (aj)

                              (aj)

                              (aj)

                              Theory Seminar 042002

                              Start of iteration

                              1 Choose j in S with minimum vj + Ĉj(aj)

                              2 Form cluster M = facilities serving j Note facilities are not split

                              3 For k st Fk M decrease rrsquok advance ak set Fk = Fk ndash M

                              2j(3) Cluster M

                              aj = 1

                              4

                              1

                              XX X

                              aj = 4 Cluster M

                              aj = 2

                              3

                              client in S

                              facility in some Fj

                              X facility removed from Fj

                              (aj)

                              Theory Seminar 042002

                              Opening Facilities

                              Central facilities opened in 2 steps

                              1 Open exactly 1 facility in M i opened with prob qyi Acts as backup

                              denoted b(k ) for each k st Fk M

                              2 Open each i in M indep with prob (2-q)yi and independent of step 1

                              Non-central facilities

                              Cluster M

                              k

                              open with prob 2yi independent of other choices

                              j

                              (ak

                              )

                              Theory Seminar 042002

                              Let Sj(c) = avg dist from j to P(j(c))

                              = ( cijxij)facwt(P(j(c))

                              j)

                              Then c Sj(c) le 2Cj

                              Distributing Facilities

                              iP(j(c))

                              j

                              rrsquoj = 3

                              P(j(1))P(j(2))

                              P(j(3))

                              Copy c gets a preferred set P(j(c))Preferred sets are disjoint

                              Ensure facwt(P(j(c)) j) ge frac12 for all c

                              Possible to do so since each xij lt frac12

                              facility in Fj

                              Theory Seminar 042002

                              Analysis

                              Feasibility follows from

                              1 Facilities in phases 1 2 not reused

                              2 After clustering j is connected to rrsquoj disjoint clusters backups are distinct

                              3 Preferred sets are disjoint

                              So j connected to rj distinct facilities

                              Theory Seminar 042002

                              Facility cost

                              Recall L1 = i | yi = 1

                              Phase 2 incur a factor of 2

                              Phase 3 each i is opened with probability 2yi

                              Expected facility cost le 2 fiyifor phases 2 3

                              iL1

                              Theory Seminar 042002

                              Bounding backup cost denoted by B rv

                              D event that no i in P(j(c)) is open

                              Lemma E[B|D] le 2vj + Cj(c)

                              Proof 2 cases

                              Service cost I

                              iM Fj st cik le Ĉj(d)

                              Also vk + Ĉj(d) le vj + Ĉj(c) le vj + Cj(c)

                              k(d)

                              j(c)

                              le vj

                              le Ĉj(d)le vk

                              B

                              k(d)

                              j(c)

                              le vj

                              le vk

                              iM Fj cik gt Ĉj(d)

                              le Ĉj(d) in expectatio

                              n

                              1)

                              2)

                              backup = b(j(c))

                              Theory Seminar 042002

                              Service Cost II

                              Fix j c Let X(c) = service cost of j(c)

                              Let di = cij pi = prob i is opened = 2yi

                              B(c) = backup costD(c) = event that no iP(j(c)) is

                              openp = Pr[D(c)] = (1-p1)hellip(1-pm) le e-1

                              davg = weighted avg of the dis

                              = (i pidi)(i pi) = Sj(c)

                              d1

                              d2 dm

                              P(j(c)) sorted by increasing cij

                              j(c)

                              i P(j(c))

                              Theory Seminar 042002

                              Then

                              E[X(c)] = [p1d1 + (1-p1)p2d2 + hellip

                              + (1-p1)hellip(1-pm-1)pmdm]

                              + pE[B(c)|D(c)]

                              le (1-p)davg + p[2vj + Cj(c)]

                              le (1-e-1)Sj(c) + e-1[2vj +

                              Cj(c)]

                              Let X = c X(c) = service cost of j

                              c Sj(c) le 2Cj and c Cj(c) le 2Cj

                              Summing over all c = 1helliprrsquoj

                              E[X] le (1-e-1)2Cj + e-1(2rrsquojvj + Cj)

                              le 2Cj + 2e-1rrsquojvj

                              Theory Seminar 042002

                              Putting it all together

                              Phase 1 pay the optimal LP cost

                              Phases 2 3

                              bull Facility cost twice LP facility cost

                              bull Service costLose a factor of 2 for phase 2Phase 3 cost is 2(LP service cost)+2e-1(dual value)

                              Overall cost for le (2+2e)(LP cost) phases 2 3

                              Total cost le (2+2e)OPT

                              Theory Seminar 042002

                              How to improve this

                              bull Distribute facilities more equitably (in an expected sense) among copies - decreases prob of lsquobadrsquo event

                              bull Better analysis ndash maximum distance within a cluster can be bounded by 2Cj(c)

                              bull Balance phases 2 and 3

                              Theory Seminar 042002

                              Summary of Results

                              bullGive a 2076-approx algorithm for non-uniform rjs Based on LP rounding using complem slackness

                              bullFor rj = r extend the primal-dual algorithm of (JMS02) to get a 152-approximation

                              bullFault tolerant k medians with rj = r

                              a Primal-dual algorithm (JMS02) gives a 4-approx using Lagrangean relaxation

                              b LP rounding gives a factor of 8

                              Theory Seminar 042002

                              Open Questions

                              1 Reduce gap between rj = r non-uniform rj

                              2 Combinatorial algorithms for non-uniform rj primal-dual local-search

                              3 Constant-factor approx for fault tolerant k medians with non-uniform rjs

                              • Fault Tolerant Facility Location
                              • Metric Facility Location
                              • PowerPoint Presentation
                              • Previous Work
                              • Previous Work (contd)
                              • LP Formulation
                              • LP Formulation (contd)
                              • Complementary Slackness
                              • 4-approximation outline
                              • The Algorithm
                              • Slide 11
                              • Analysis Phase 1
                              • Analysis (contd)
                              • The Algorithm (contd)
                              • Analysis Phase 2
                              • Finally hellip
                              • A Randomized Algorithm
                              • Slide 18
                              • Slide 19
                              • Clustering
                              • Slide 21
                              • Slide 22
                              • Opening Facilities
                              • Distributing Facilities
                              • Analysis
                              • Facility cost
                              • Service cost I
                              • Service Cost II
                              • Slide 29
                              • Putting it all together
                              • How to improve this
                              • Summary of Results
                              • Open Questions

                                Theory Seminar 042002

                                Finally hellip

                                Theorem Total cost le 4 times the optimal cost

                                Proof Total cost le

                                i fiyi + 3j (rj ndash lj)vj + fi + cij

                                facility cost of phase 1 cost for

                                copies connected by

                                phase 1

                                cost of phase 2

                                lei fiyi + 3j (rj ndash lj)vj + (j ljvj ndash i

                                zi )

                                lei fiyi + 3(j rjvj ndash i zi )

                                le4OPT

                                iL1 jiLj

                                Theory Seminar 042002

                                A Randomized Algorithm

                                Idea Open i with probability ρyi

                                Expected facility cost le ρi fiyi

                                Hope that each copy j(c) has a nearby facility open and service cost decreases

                                Not quitehellip no facility may be open

                                Cluster facilities open ge 1 facility in each cluster

                                Theory Seminar 042002

                                Phase 1 Pruning out ndashzis

                                Open all i st yi = 1

                                For each j if xij = yi = 1 connect j to i

                                Let Lj = i | xij = 1 and lj = |Lj|

                                Cost = j ljvj ndash i zi

                                Lj

                                Fj10

                                rrsquoj = residual reqmt = 6

                                Lrsquoj

                                Phase 2

                                Open all i st frac12 le yi lt 1

                                For each j let Lrsquoj = i | frac12 le xij lt 1

                                Connect copies of j to i Lrsquoj

                                Lose a factor of 2

                                facilities opened in

                                phases 1 2yi = 1 frac12 le yi lt 1yi lt frac12

                                Set L1

                                Set L2

                                Theory Seminar 042002

                                Phase 3

                                Notation facwt(S j) = iS xij

                                1 Form clusters Each cluster has facwt ge frac12

                                2 Open facilities Open ge 1 facility in a cluster ndash used as a backup facility Open facility i with prob 2yi

                                3 Assign facilities to copies Each copy j(c) gets a preferred set of facilities ndash P(j(c)) with facwt ge frac12 For c d P(j(c)) P(j(d)) =

                                4 Connect clients Connect j(c) to the nearest i open in P(j(c)) or to a backup facility

                                Theory Seminar 042002

                                ClusteringAfter phases 1 and 2 Fj = i | xij lt frac12 Sort these by cij and distribute among the rrsquoj copies

                                Cj(c) = avg service cost of the cth copy denote c Cj(c) = cijxij by Cj

                                Initial Fj before any iterations

                                Cj(1)

                                Cj(2)

                                Cj(3)3

                                i Fj

                                client j

                                Want the following properties

                                Clusters to be disjoint

                                Each cluster have facwt ge frac12

                                Each j be connected to rrsquoj clusters

                                iFj

                                Theory Seminar 042002

                                Iterative algorithm

                                S = j | rrsquoj gt 0

                                aj = lsquoactiversquo copy of j initially = 1

                                Ĉj(aj) = avg distance to the first k

                                facilities in Fj gathering facwt ge frac12

                                say these facilities lsquoserversquo j

                                Will maintain Ĉj(aj) le Cj(aj)

                                X

                                X

                                X1

                                Fj after some iterations

                                X i removed from Fj

                                i Fjserving jĈj(3)

                                facilities serving j

                                aj = 3

                                4X

                                (aj)

                                (aj)

                                (aj)

                                Theory Seminar 042002

                                Start of iteration

                                1 Choose j in S with minimum vj + Ĉj(aj)

                                2 Form cluster M = facilities serving j Note facilities are not split

                                3 For k st Fk M decrease rrsquok advance ak set Fk = Fk ndash M

                                2j(3) Cluster M

                                aj = 1

                                4

                                1

                                XX X

                                aj = 4 Cluster M

                                aj = 2

                                3

                                client in S

                                facility in some Fj

                                X facility removed from Fj

                                (aj)

                                Theory Seminar 042002

                                Opening Facilities

                                Central facilities opened in 2 steps

                                1 Open exactly 1 facility in M i opened with prob qyi Acts as backup

                                denoted b(k ) for each k st Fk M

                                2 Open each i in M indep with prob (2-q)yi and independent of step 1

                                Non-central facilities

                                Cluster M

                                k

                                open with prob 2yi independent of other choices

                                j

                                (ak

                                )

                                Theory Seminar 042002

                                Let Sj(c) = avg dist from j to P(j(c))

                                = ( cijxij)facwt(P(j(c))

                                j)

                                Then c Sj(c) le 2Cj

                                Distributing Facilities

                                iP(j(c))

                                j

                                rrsquoj = 3

                                P(j(1))P(j(2))

                                P(j(3))

                                Copy c gets a preferred set P(j(c))Preferred sets are disjoint

                                Ensure facwt(P(j(c)) j) ge frac12 for all c

                                Possible to do so since each xij lt frac12

                                facility in Fj

                                Theory Seminar 042002

                                Analysis

                                Feasibility follows from

                                1 Facilities in phases 1 2 not reused

                                2 After clustering j is connected to rrsquoj disjoint clusters backups are distinct

                                3 Preferred sets are disjoint

                                So j connected to rj distinct facilities

                                Theory Seminar 042002

                                Facility cost

                                Recall L1 = i | yi = 1

                                Phase 2 incur a factor of 2

                                Phase 3 each i is opened with probability 2yi

                                Expected facility cost le 2 fiyifor phases 2 3

                                iL1

                                Theory Seminar 042002

                                Bounding backup cost denoted by B rv

                                D event that no i in P(j(c)) is open

                                Lemma E[B|D] le 2vj + Cj(c)

                                Proof 2 cases

                                Service cost I

                                iM Fj st cik le Ĉj(d)

                                Also vk + Ĉj(d) le vj + Ĉj(c) le vj + Cj(c)

                                k(d)

                                j(c)

                                le vj

                                le Ĉj(d)le vk

                                B

                                k(d)

                                j(c)

                                le vj

                                le vk

                                iM Fj cik gt Ĉj(d)

                                le Ĉj(d) in expectatio

                                n

                                1)

                                2)

                                backup = b(j(c))

                                Theory Seminar 042002

                                Service Cost II

                                Fix j c Let X(c) = service cost of j(c)

                                Let di = cij pi = prob i is opened = 2yi

                                B(c) = backup costD(c) = event that no iP(j(c)) is

                                openp = Pr[D(c)] = (1-p1)hellip(1-pm) le e-1

                                davg = weighted avg of the dis

                                = (i pidi)(i pi) = Sj(c)

                                d1

                                d2 dm

                                P(j(c)) sorted by increasing cij

                                j(c)

                                i P(j(c))

                                Theory Seminar 042002

                                Then

                                E[X(c)] = [p1d1 + (1-p1)p2d2 + hellip

                                + (1-p1)hellip(1-pm-1)pmdm]

                                + pE[B(c)|D(c)]

                                le (1-p)davg + p[2vj + Cj(c)]

                                le (1-e-1)Sj(c) + e-1[2vj +

                                Cj(c)]

                                Let X = c X(c) = service cost of j

                                c Sj(c) le 2Cj and c Cj(c) le 2Cj

                                Summing over all c = 1helliprrsquoj

                                E[X] le (1-e-1)2Cj + e-1(2rrsquojvj + Cj)

                                le 2Cj + 2e-1rrsquojvj

                                Theory Seminar 042002

                                Putting it all together

                                Phase 1 pay the optimal LP cost

                                Phases 2 3

                                bull Facility cost twice LP facility cost

                                bull Service costLose a factor of 2 for phase 2Phase 3 cost is 2(LP service cost)+2e-1(dual value)

                                Overall cost for le (2+2e)(LP cost) phases 2 3

                                Total cost le (2+2e)OPT

                                Theory Seminar 042002

                                How to improve this

                                bull Distribute facilities more equitably (in an expected sense) among copies - decreases prob of lsquobadrsquo event

                                bull Better analysis ndash maximum distance within a cluster can be bounded by 2Cj(c)

                                bull Balance phases 2 and 3

                                Theory Seminar 042002

                                Summary of Results

                                bullGive a 2076-approx algorithm for non-uniform rjs Based on LP rounding using complem slackness

                                bullFor rj = r extend the primal-dual algorithm of (JMS02) to get a 152-approximation

                                bullFault tolerant k medians with rj = r

                                a Primal-dual algorithm (JMS02) gives a 4-approx using Lagrangean relaxation

                                b LP rounding gives a factor of 8

                                Theory Seminar 042002

                                Open Questions

                                1 Reduce gap between rj = r non-uniform rj

                                2 Combinatorial algorithms for non-uniform rj primal-dual local-search

                                3 Constant-factor approx for fault tolerant k medians with non-uniform rjs

                                • Fault Tolerant Facility Location
                                • Metric Facility Location
                                • PowerPoint Presentation
                                • Previous Work
                                • Previous Work (contd)
                                • LP Formulation
                                • LP Formulation (contd)
                                • Complementary Slackness
                                • 4-approximation outline
                                • The Algorithm
                                • Slide 11
                                • Analysis Phase 1
                                • Analysis (contd)
                                • The Algorithm (contd)
                                • Analysis Phase 2
                                • Finally hellip
                                • A Randomized Algorithm
                                • Slide 18
                                • Slide 19
                                • Clustering
                                • Slide 21
                                • Slide 22
                                • Opening Facilities
                                • Distributing Facilities
                                • Analysis
                                • Facility cost
                                • Service cost I
                                • Service Cost II
                                • Slide 29
                                • Putting it all together
                                • How to improve this
                                • Summary of Results
                                • Open Questions

                                  Theory Seminar 042002

                                  A Randomized Algorithm

                                  Idea Open i with probability ρyi

                                  Expected facility cost le ρi fiyi

                                  Hope that each copy j(c) has a nearby facility open and service cost decreases

                                  Not quitehellip no facility may be open

                                  Cluster facilities open ge 1 facility in each cluster

                                  Theory Seminar 042002

                                  Phase 1 Pruning out ndashzis

                                  Open all i st yi = 1

                                  For each j if xij = yi = 1 connect j to i

                                  Let Lj = i | xij = 1 and lj = |Lj|

                                  Cost = j ljvj ndash i zi

                                  Lj

                                  Fj10

                                  rrsquoj = residual reqmt = 6

                                  Lrsquoj

                                  Phase 2

                                  Open all i st frac12 le yi lt 1

                                  For each j let Lrsquoj = i | frac12 le xij lt 1

                                  Connect copies of j to i Lrsquoj

                                  Lose a factor of 2

                                  facilities opened in

                                  phases 1 2yi = 1 frac12 le yi lt 1yi lt frac12

                                  Set L1

                                  Set L2

                                  Theory Seminar 042002

                                  Phase 3

                                  Notation facwt(S j) = iS xij

                                  1 Form clusters Each cluster has facwt ge frac12

                                  2 Open facilities Open ge 1 facility in a cluster ndash used as a backup facility Open facility i with prob 2yi

                                  3 Assign facilities to copies Each copy j(c) gets a preferred set of facilities ndash P(j(c)) with facwt ge frac12 For c d P(j(c)) P(j(d)) =

                                  4 Connect clients Connect j(c) to the nearest i open in P(j(c)) or to a backup facility

                                  Theory Seminar 042002

                                  ClusteringAfter phases 1 and 2 Fj = i | xij lt frac12 Sort these by cij and distribute among the rrsquoj copies

                                  Cj(c) = avg service cost of the cth copy denote c Cj(c) = cijxij by Cj

                                  Initial Fj before any iterations

                                  Cj(1)

                                  Cj(2)

                                  Cj(3)3

                                  i Fj

                                  client j

                                  Want the following properties

                                  Clusters to be disjoint

                                  Each cluster have facwt ge frac12

                                  Each j be connected to rrsquoj clusters

                                  iFj

                                  Theory Seminar 042002

                                  Iterative algorithm

                                  S = j | rrsquoj gt 0

                                  aj = lsquoactiversquo copy of j initially = 1

                                  Ĉj(aj) = avg distance to the first k

                                  facilities in Fj gathering facwt ge frac12

                                  say these facilities lsquoserversquo j

                                  Will maintain Ĉj(aj) le Cj(aj)

                                  X

                                  X

                                  X1

                                  Fj after some iterations

                                  X i removed from Fj

                                  i Fjserving jĈj(3)

                                  facilities serving j

                                  aj = 3

                                  4X

                                  (aj)

                                  (aj)

                                  (aj)

                                  Theory Seminar 042002

                                  Start of iteration

                                  1 Choose j in S with minimum vj + Ĉj(aj)

                                  2 Form cluster M = facilities serving j Note facilities are not split

                                  3 For k st Fk M decrease rrsquok advance ak set Fk = Fk ndash M

                                  2j(3) Cluster M

                                  aj = 1

                                  4

                                  1

                                  XX X

                                  aj = 4 Cluster M

                                  aj = 2

                                  3

                                  client in S

                                  facility in some Fj

                                  X facility removed from Fj

                                  (aj)

                                  Theory Seminar 042002

                                  Opening Facilities

                                  Central facilities opened in 2 steps

                                  1 Open exactly 1 facility in M i opened with prob qyi Acts as backup

                                  denoted b(k ) for each k st Fk M

                                  2 Open each i in M indep with prob (2-q)yi and independent of step 1

                                  Non-central facilities

                                  Cluster M

                                  k

                                  open with prob 2yi independent of other choices

                                  j

                                  (ak

                                  )

                                  Theory Seminar 042002

                                  Let Sj(c) = avg dist from j to P(j(c))

                                  = ( cijxij)facwt(P(j(c))

                                  j)

                                  Then c Sj(c) le 2Cj

                                  Distributing Facilities

                                  iP(j(c))

                                  j

                                  rrsquoj = 3

                                  P(j(1))P(j(2))

                                  P(j(3))

                                  Copy c gets a preferred set P(j(c))Preferred sets are disjoint

                                  Ensure facwt(P(j(c)) j) ge frac12 for all c

                                  Possible to do so since each xij lt frac12

                                  facility in Fj

                                  Theory Seminar 042002

                                  Analysis

                                  Feasibility follows from

                                  1 Facilities in phases 1 2 not reused

                                  2 After clustering j is connected to rrsquoj disjoint clusters backups are distinct

                                  3 Preferred sets are disjoint

                                  So j connected to rj distinct facilities

                                  Theory Seminar 042002

                                  Facility cost

                                  Recall L1 = i | yi = 1

                                  Phase 2 incur a factor of 2

                                  Phase 3 each i is opened with probability 2yi

                                  Expected facility cost le 2 fiyifor phases 2 3

                                  iL1

                                  Theory Seminar 042002

                                  Bounding backup cost denoted by B rv

                                  D event that no i in P(j(c)) is open

                                  Lemma E[B|D] le 2vj + Cj(c)

                                  Proof 2 cases

                                  Service cost I

                                  iM Fj st cik le Ĉj(d)

                                  Also vk + Ĉj(d) le vj + Ĉj(c) le vj + Cj(c)

                                  k(d)

                                  j(c)

                                  le vj

                                  le Ĉj(d)le vk

                                  B

                                  k(d)

                                  j(c)

                                  le vj

                                  le vk

                                  iM Fj cik gt Ĉj(d)

                                  le Ĉj(d) in expectatio

                                  n

                                  1)

                                  2)

                                  backup = b(j(c))

                                  Theory Seminar 042002

                                  Service Cost II

                                  Fix j c Let X(c) = service cost of j(c)

                                  Let di = cij pi = prob i is opened = 2yi

                                  B(c) = backup costD(c) = event that no iP(j(c)) is

                                  openp = Pr[D(c)] = (1-p1)hellip(1-pm) le e-1

                                  davg = weighted avg of the dis

                                  = (i pidi)(i pi) = Sj(c)

                                  d1

                                  d2 dm

                                  P(j(c)) sorted by increasing cij

                                  j(c)

                                  i P(j(c))

                                  Theory Seminar 042002

                                  Then

                                  E[X(c)] = [p1d1 + (1-p1)p2d2 + hellip

                                  + (1-p1)hellip(1-pm-1)pmdm]

                                  + pE[B(c)|D(c)]

                                  le (1-p)davg + p[2vj + Cj(c)]

                                  le (1-e-1)Sj(c) + e-1[2vj +

                                  Cj(c)]

                                  Let X = c X(c) = service cost of j

                                  c Sj(c) le 2Cj and c Cj(c) le 2Cj

                                  Summing over all c = 1helliprrsquoj

                                  E[X] le (1-e-1)2Cj + e-1(2rrsquojvj + Cj)

                                  le 2Cj + 2e-1rrsquojvj

                                  Theory Seminar 042002

                                  Putting it all together

                                  Phase 1 pay the optimal LP cost

                                  Phases 2 3

                                  bull Facility cost twice LP facility cost

                                  bull Service costLose a factor of 2 for phase 2Phase 3 cost is 2(LP service cost)+2e-1(dual value)

                                  Overall cost for le (2+2e)(LP cost) phases 2 3

                                  Total cost le (2+2e)OPT

                                  Theory Seminar 042002

                                  How to improve this

                                  bull Distribute facilities more equitably (in an expected sense) among copies - decreases prob of lsquobadrsquo event

                                  bull Better analysis ndash maximum distance within a cluster can be bounded by 2Cj(c)

                                  bull Balance phases 2 and 3

                                  Theory Seminar 042002

                                  Summary of Results

                                  bullGive a 2076-approx algorithm for non-uniform rjs Based on LP rounding using complem slackness

                                  bullFor rj = r extend the primal-dual algorithm of (JMS02) to get a 152-approximation

                                  bullFault tolerant k medians with rj = r

                                  a Primal-dual algorithm (JMS02) gives a 4-approx using Lagrangean relaxation

                                  b LP rounding gives a factor of 8

                                  Theory Seminar 042002

                                  Open Questions

                                  1 Reduce gap between rj = r non-uniform rj

                                  2 Combinatorial algorithms for non-uniform rj primal-dual local-search

                                  3 Constant-factor approx for fault tolerant k medians with non-uniform rjs

                                  • Fault Tolerant Facility Location
                                  • Metric Facility Location
                                  • PowerPoint Presentation
                                  • Previous Work
                                  • Previous Work (contd)
                                  • LP Formulation
                                  • LP Formulation (contd)
                                  • Complementary Slackness
                                  • 4-approximation outline
                                  • The Algorithm
                                  • Slide 11
                                  • Analysis Phase 1
                                  • Analysis (contd)
                                  • The Algorithm (contd)
                                  • Analysis Phase 2
                                  • Finally hellip
                                  • A Randomized Algorithm
                                  • Slide 18
                                  • Slide 19
                                  • Clustering
                                  • Slide 21
                                  • Slide 22
                                  • Opening Facilities
                                  • Distributing Facilities
                                  • Analysis
                                  • Facility cost
                                  • Service cost I
                                  • Service Cost II
                                  • Slide 29
                                  • Putting it all together
                                  • How to improve this
                                  • Summary of Results
                                  • Open Questions

                                    Theory Seminar 042002

                                    Phase 1 Pruning out ndashzis

                                    Open all i st yi = 1

                                    For each j if xij = yi = 1 connect j to i

                                    Let Lj = i | xij = 1 and lj = |Lj|

                                    Cost = j ljvj ndash i zi

                                    Lj

                                    Fj10

                                    rrsquoj = residual reqmt = 6

                                    Lrsquoj

                                    Phase 2

                                    Open all i st frac12 le yi lt 1

                                    For each j let Lrsquoj = i | frac12 le xij lt 1

                                    Connect copies of j to i Lrsquoj

                                    Lose a factor of 2

                                    facilities opened in

                                    phases 1 2yi = 1 frac12 le yi lt 1yi lt frac12

                                    Set L1

                                    Set L2

                                    Theory Seminar 042002

                                    Phase 3

                                    Notation facwt(S j) = iS xij

                                    1 Form clusters Each cluster has facwt ge frac12

                                    2 Open facilities Open ge 1 facility in a cluster ndash used as a backup facility Open facility i with prob 2yi

                                    3 Assign facilities to copies Each copy j(c) gets a preferred set of facilities ndash P(j(c)) with facwt ge frac12 For c d P(j(c)) P(j(d)) =

                                    4 Connect clients Connect j(c) to the nearest i open in P(j(c)) or to a backup facility

                                    Theory Seminar 042002

                                    ClusteringAfter phases 1 and 2 Fj = i | xij lt frac12 Sort these by cij and distribute among the rrsquoj copies

                                    Cj(c) = avg service cost of the cth copy denote c Cj(c) = cijxij by Cj

                                    Initial Fj before any iterations

                                    Cj(1)

                                    Cj(2)

                                    Cj(3)3

                                    i Fj

                                    client j

                                    Want the following properties

                                    Clusters to be disjoint

                                    Each cluster have facwt ge frac12

                                    Each j be connected to rrsquoj clusters

                                    iFj

                                    Theory Seminar 042002

                                    Iterative algorithm

                                    S = j | rrsquoj gt 0

                                    aj = lsquoactiversquo copy of j initially = 1

                                    Ĉj(aj) = avg distance to the first k

                                    facilities in Fj gathering facwt ge frac12

                                    say these facilities lsquoserversquo j

                                    Will maintain Ĉj(aj) le Cj(aj)

                                    X

                                    X

                                    X1

                                    Fj after some iterations

                                    X i removed from Fj

                                    i Fjserving jĈj(3)

                                    facilities serving j

                                    aj = 3

                                    4X

                                    (aj)

                                    (aj)

                                    (aj)

                                    Theory Seminar 042002

                                    Start of iteration

                                    1 Choose j in S with minimum vj + Ĉj(aj)

                                    2 Form cluster M = facilities serving j Note facilities are not split

                                    3 For k st Fk M decrease rrsquok advance ak set Fk = Fk ndash M

                                    2j(3) Cluster M

                                    aj = 1

                                    4

                                    1

                                    XX X

                                    aj = 4 Cluster M

                                    aj = 2

                                    3

                                    client in S

                                    facility in some Fj

                                    X facility removed from Fj

                                    (aj)

                                    Theory Seminar 042002

                                    Opening Facilities

                                    Central facilities opened in 2 steps

                                    1 Open exactly 1 facility in M i opened with prob qyi Acts as backup

                                    denoted b(k ) for each k st Fk M

                                    2 Open each i in M indep with prob (2-q)yi and independent of step 1

                                    Non-central facilities

                                    Cluster M

                                    k

                                    open with prob 2yi independent of other choices

                                    j

                                    (ak

                                    )

                                    Theory Seminar 042002

                                    Let Sj(c) = avg dist from j to P(j(c))

                                    = ( cijxij)facwt(P(j(c))

                                    j)

                                    Then c Sj(c) le 2Cj

                                    Distributing Facilities

                                    iP(j(c))

                                    j

                                    rrsquoj = 3

                                    P(j(1))P(j(2))

                                    P(j(3))

                                    Copy c gets a preferred set P(j(c))Preferred sets are disjoint

                                    Ensure facwt(P(j(c)) j) ge frac12 for all c

                                    Possible to do so since each xij lt frac12

                                    facility in Fj

                                    Theory Seminar 042002

                                    Analysis

                                    Feasibility follows from

                                    1 Facilities in phases 1 2 not reused

                                    2 After clustering j is connected to rrsquoj disjoint clusters backups are distinct

                                    3 Preferred sets are disjoint

                                    So j connected to rj distinct facilities

                                    Theory Seminar 042002

                                    Facility cost

                                    Recall L1 = i | yi = 1

                                    Phase 2 incur a factor of 2

                                    Phase 3 each i is opened with probability 2yi

                                    Expected facility cost le 2 fiyifor phases 2 3

                                    iL1

                                    Theory Seminar 042002

                                    Bounding backup cost denoted by B rv

                                    D event that no i in P(j(c)) is open

                                    Lemma E[B|D] le 2vj + Cj(c)

                                    Proof 2 cases

                                    Service cost I

                                    iM Fj st cik le Ĉj(d)

                                    Also vk + Ĉj(d) le vj + Ĉj(c) le vj + Cj(c)

                                    k(d)

                                    j(c)

                                    le vj

                                    le Ĉj(d)le vk

                                    B

                                    k(d)

                                    j(c)

                                    le vj

                                    le vk

                                    iM Fj cik gt Ĉj(d)

                                    le Ĉj(d) in expectatio

                                    n

                                    1)

                                    2)

                                    backup = b(j(c))

                                    Theory Seminar 042002

                                    Service Cost II

                                    Fix j c Let X(c) = service cost of j(c)

                                    Let di = cij pi = prob i is opened = 2yi

                                    B(c) = backup costD(c) = event that no iP(j(c)) is

                                    openp = Pr[D(c)] = (1-p1)hellip(1-pm) le e-1

                                    davg = weighted avg of the dis

                                    = (i pidi)(i pi) = Sj(c)

                                    d1

                                    d2 dm

                                    P(j(c)) sorted by increasing cij

                                    j(c)

                                    i P(j(c))

                                    Theory Seminar 042002

                                    Then

                                    E[X(c)] = [p1d1 + (1-p1)p2d2 + hellip

                                    + (1-p1)hellip(1-pm-1)pmdm]

                                    + pE[B(c)|D(c)]

                                    le (1-p)davg + p[2vj + Cj(c)]

                                    le (1-e-1)Sj(c) + e-1[2vj +

                                    Cj(c)]

                                    Let X = c X(c) = service cost of j

                                    c Sj(c) le 2Cj and c Cj(c) le 2Cj

                                    Summing over all c = 1helliprrsquoj

                                    E[X] le (1-e-1)2Cj + e-1(2rrsquojvj + Cj)

                                    le 2Cj + 2e-1rrsquojvj

                                    Theory Seminar 042002

                                    Putting it all together

                                    Phase 1 pay the optimal LP cost

                                    Phases 2 3

                                    bull Facility cost twice LP facility cost

                                    bull Service costLose a factor of 2 for phase 2Phase 3 cost is 2(LP service cost)+2e-1(dual value)

                                    Overall cost for le (2+2e)(LP cost) phases 2 3

                                    Total cost le (2+2e)OPT

                                    Theory Seminar 042002

                                    How to improve this

                                    bull Distribute facilities more equitably (in an expected sense) among copies - decreases prob of lsquobadrsquo event

                                    bull Better analysis ndash maximum distance within a cluster can be bounded by 2Cj(c)

                                    bull Balance phases 2 and 3

                                    Theory Seminar 042002

                                    Summary of Results

                                    bullGive a 2076-approx algorithm for non-uniform rjs Based on LP rounding using complem slackness

                                    bullFor rj = r extend the primal-dual algorithm of (JMS02) to get a 152-approximation

                                    bullFault tolerant k medians with rj = r

                                    a Primal-dual algorithm (JMS02) gives a 4-approx using Lagrangean relaxation

                                    b LP rounding gives a factor of 8

                                    Theory Seminar 042002

                                    Open Questions

                                    1 Reduce gap between rj = r non-uniform rj

                                    2 Combinatorial algorithms for non-uniform rj primal-dual local-search

                                    3 Constant-factor approx for fault tolerant k medians with non-uniform rjs

                                    • Fault Tolerant Facility Location
                                    • Metric Facility Location
                                    • PowerPoint Presentation
                                    • Previous Work
                                    • Previous Work (contd)
                                    • LP Formulation
                                    • LP Formulation (contd)
                                    • Complementary Slackness
                                    • 4-approximation outline
                                    • The Algorithm
                                    • Slide 11
                                    • Analysis Phase 1
                                    • Analysis (contd)
                                    • The Algorithm (contd)
                                    • Analysis Phase 2
                                    • Finally hellip
                                    • A Randomized Algorithm
                                    • Slide 18
                                    • Slide 19
                                    • Clustering
                                    • Slide 21
                                    • Slide 22
                                    • Opening Facilities
                                    • Distributing Facilities
                                    • Analysis
                                    • Facility cost
                                    • Service cost I
                                    • Service Cost II
                                    • Slide 29
                                    • Putting it all together
                                    • How to improve this
                                    • Summary of Results
                                    • Open Questions

                                      Theory Seminar 042002

                                      Phase 3

                                      Notation facwt(S j) = iS xij

                                      1 Form clusters Each cluster has facwt ge frac12

                                      2 Open facilities Open ge 1 facility in a cluster ndash used as a backup facility Open facility i with prob 2yi

                                      3 Assign facilities to copies Each copy j(c) gets a preferred set of facilities ndash P(j(c)) with facwt ge frac12 For c d P(j(c)) P(j(d)) =

                                      4 Connect clients Connect j(c) to the nearest i open in P(j(c)) or to a backup facility

                                      Theory Seminar 042002

                                      ClusteringAfter phases 1 and 2 Fj = i | xij lt frac12 Sort these by cij and distribute among the rrsquoj copies

                                      Cj(c) = avg service cost of the cth copy denote c Cj(c) = cijxij by Cj

                                      Initial Fj before any iterations

                                      Cj(1)

                                      Cj(2)

                                      Cj(3)3

                                      i Fj

                                      client j

                                      Want the following properties

                                      Clusters to be disjoint

                                      Each cluster have facwt ge frac12

                                      Each j be connected to rrsquoj clusters

                                      iFj

                                      Theory Seminar 042002

                                      Iterative algorithm

                                      S = j | rrsquoj gt 0

                                      aj = lsquoactiversquo copy of j initially = 1

                                      Ĉj(aj) = avg distance to the first k

                                      facilities in Fj gathering facwt ge frac12

                                      say these facilities lsquoserversquo j

                                      Will maintain Ĉj(aj) le Cj(aj)

                                      X

                                      X

                                      X1

                                      Fj after some iterations

                                      X i removed from Fj

                                      i Fjserving jĈj(3)

                                      facilities serving j

                                      aj = 3

                                      4X

                                      (aj)

                                      (aj)

                                      (aj)

                                      Theory Seminar 042002

                                      Start of iteration

                                      1 Choose j in S with minimum vj + Ĉj(aj)

                                      2 Form cluster M = facilities serving j Note facilities are not split

                                      3 For k st Fk M decrease rrsquok advance ak set Fk = Fk ndash M

                                      2j(3) Cluster M

                                      aj = 1

                                      4

                                      1

                                      XX X

                                      aj = 4 Cluster M

                                      aj = 2

                                      3

                                      client in S

                                      facility in some Fj

                                      X facility removed from Fj

                                      (aj)

                                      Theory Seminar 042002

                                      Opening Facilities

                                      Central facilities opened in 2 steps

                                      1 Open exactly 1 facility in M i opened with prob qyi Acts as backup

                                      denoted b(k ) for each k st Fk M

                                      2 Open each i in M indep with prob (2-q)yi and independent of step 1

                                      Non-central facilities

                                      Cluster M

                                      k

                                      open with prob 2yi independent of other choices

                                      j

                                      (ak

                                      )

                                      Theory Seminar 042002

                                      Let Sj(c) = avg dist from j to P(j(c))

                                      = ( cijxij)facwt(P(j(c))

                                      j)

                                      Then c Sj(c) le 2Cj

                                      Distributing Facilities

                                      iP(j(c))

                                      j

                                      rrsquoj = 3

                                      P(j(1))P(j(2))

                                      P(j(3))

                                      Copy c gets a preferred set P(j(c))Preferred sets are disjoint

                                      Ensure facwt(P(j(c)) j) ge frac12 for all c

                                      Possible to do so since each xij lt frac12

                                      facility in Fj

                                      Theory Seminar 042002

                                      Analysis

                                      Feasibility follows from

                                      1 Facilities in phases 1 2 not reused

                                      2 After clustering j is connected to rrsquoj disjoint clusters backups are distinct

                                      3 Preferred sets are disjoint

                                      So j connected to rj distinct facilities

                                      Theory Seminar 042002

                                      Facility cost

                                      Recall L1 = i | yi = 1

                                      Phase 2 incur a factor of 2

                                      Phase 3 each i is opened with probability 2yi

                                      Expected facility cost le 2 fiyifor phases 2 3

                                      iL1

                                      Theory Seminar 042002

                                      Bounding backup cost denoted by B rv

                                      D event that no i in P(j(c)) is open

                                      Lemma E[B|D] le 2vj + Cj(c)

                                      Proof 2 cases

                                      Service cost I

                                      iM Fj st cik le Ĉj(d)

                                      Also vk + Ĉj(d) le vj + Ĉj(c) le vj + Cj(c)

                                      k(d)

                                      j(c)

                                      le vj

                                      le Ĉj(d)le vk

                                      B

                                      k(d)

                                      j(c)

                                      le vj

                                      le vk

                                      iM Fj cik gt Ĉj(d)

                                      le Ĉj(d) in expectatio

                                      n

                                      1)

                                      2)

                                      backup = b(j(c))

                                      Theory Seminar 042002

                                      Service Cost II

                                      Fix j c Let X(c) = service cost of j(c)

                                      Let di = cij pi = prob i is opened = 2yi

                                      B(c) = backup costD(c) = event that no iP(j(c)) is

                                      openp = Pr[D(c)] = (1-p1)hellip(1-pm) le e-1

                                      davg = weighted avg of the dis

                                      = (i pidi)(i pi) = Sj(c)

                                      d1

                                      d2 dm

                                      P(j(c)) sorted by increasing cij

                                      j(c)

                                      i P(j(c))

                                      Theory Seminar 042002

                                      Then

                                      E[X(c)] = [p1d1 + (1-p1)p2d2 + hellip

                                      + (1-p1)hellip(1-pm-1)pmdm]

                                      + pE[B(c)|D(c)]

                                      le (1-p)davg + p[2vj + Cj(c)]

                                      le (1-e-1)Sj(c) + e-1[2vj +

                                      Cj(c)]

                                      Let X = c X(c) = service cost of j

                                      c Sj(c) le 2Cj and c Cj(c) le 2Cj

                                      Summing over all c = 1helliprrsquoj

                                      E[X] le (1-e-1)2Cj + e-1(2rrsquojvj + Cj)

                                      le 2Cj + 2e-1rrsquojvj

                                      Theory Seminar 042002

                                      Putting it all together

                                      Phase 1 pay the optimal LP cost

                                      Phases 2 3

                                      bull Facility cost twice LP facility cost

                                      bull Service costLose a factor of 2 for phase 2Phase 3 cost is 2(LP service cost)+2e-1(dual value)

                                      Overall cost for le (2+2e)(LP cost) phases 2 3

                                      Total cost le (2+2e)OPT

                                      Theory Seminar 042002

                                      How to improve this

                                      bull Distribute facilities more equitably (in an expected sense) among copies - decreases prob of lsquobadrsquo event

                                      bull Better analysis ndash maximum distance within a cluster can be bounded by 2Cj(c)

                                      bull Balance phases 2 and 3

                                      Theory Seminar 042002

                                      Summary of Results

                                      bullGive a 2076-approx algorithm for non-uniform rjs Based on LP rounding using complem slackness

                                      bullFor rj = r extend the primal-dual algorithm of (JMS02) to get a 152-approximation

                                      bullFault tolerant k medians with rj = r

                                      a Primal-dual algorithm (JMS02) gives a 4-approx using Lagrangean relaxation

                                      b LP rounding gives a factor of 8

                                      Theory Seminar 042002

                                      Open Questions

                                      1 Reduce gap between rj = r non-uniform rj

                                      2 Combinatorial algorithms for non-uniform rj primal-dual local-search

                                      3 Constant-factor approx for fault tolerant k medians with non-uniform rjs

                                      • Fault Tolerant Facility Location
                                      • Metric Facility Location
                                      • PowerPoint Presentation
                                      • Previous Work
                                      • Previous Work (contd)
                                      • LP Formulation
                                      • LP Formulation (contd)
                                      • Complementary Slackness
                                      • 4-approximation outline
                                      • The Algorithm
                                      • Slide 11
                                      • Analysis Phase 1
                                      • Analysis (contd)
                                      • The Algorithm (contd)
                                      • Analysis Phase 2
                                      • Finally hellip
                                      • A Randomized Algorithm
                                      • Slide 18
                                      • Slide 19
                                      • Clustering
                                      • Slide 21
                                      • Slide 22
                                      • Opening Facilities
                                      • Distributing Facilities
                                      • Analysis
                                      • Facility cost
                                      • Service cost I
                                      • Service Cost II
                                      • Slide 29
                                      • Putting it all together
                                      • How to improve this
                                      • Summary of Results
                                      • Open Questions

                                        Theory Seminar 042002

                                        ClusteringAfter phases 1 and 2 Fj = i | xij lt frac12 Sort these by cij and distribute among the rrsquoj copies

                                        Cj(c) = avg service cost of the cth copy denote c Cj(c) = cijxij by Cj

                                        Initial Fj before any iterations

                                        Cj(1)

                                        Cj(2)

                                        Cj(3)3

                                        i Fj

                                        client j

                                        Want the following properties

                                        Clusters to be disjoint

                                        Each cluster have facwt ge frac12

                                        Each j be connected to rrsquoj clusters

                                        iFj

                                        Theory Seminar 042002

                                        Iterative algorithm

                                        S = j | rrsquoj gt 0

                                        aj = lsquoactiversquo copy of j initially = 1

                                        Ĉj(aj) = avg distance to the first k

                                        facilities in Fj gathering facwt ge frac12

                                        say these facilities lsquoserversquo j

                                        Will maintain Ĉj(aj) le Cj(aj)

                                        X

                                        X

                                        X1

                                        Fj after some iterations

                                        X i removed from Fj

                                        i Fjserving jĈj(3)

                                        facilities serving j

                                        aj = 3

                                        4X

                                        (aj)

                                        (aj)

                                        (aj)

                                        Theory Seminar 042002

                                        Start of iteration

                                        1 Choose j in S with minimum vj + Ĉj(aj)

                                        2 Form cluster M = facilities serving j Note facilities are not split

                                        3 For k st Fk M decrease rrsquok advance ak set Fk = Fk ndash M

                                        2j(3) Cluster M

                                        aj = 1

                                        4

                                        1

                                        XX X

                                        aj = 4 Cluster M

                                        aj = 2

                                        3

                                        client in S

                                        facility in some Fj

                                        X facility removed from Fj

                                        (aj)

                                        Theory Seminar 042002

                                        Opening Facilities

                                        Central facilities opened in 2 steps

                                        1 Open exactly 1 facility in M i opened with prob qyi Acts as backup

                                        denoted b(k ) for each k st Fk M

                                        2 Open each i in M indep with prob (2-q)yi and independent of step 1

                                        Non-central facilities

                                        Cluster M

                                        k

                                        open with prob 2yi independent of other choices

                                        j

                                        (ak

                                        )

                                        Theory Seminar 042002

                                        Let Sj(c) = avg dist from j to P(j(c))

                                        = ( cijxij)facwt(P(j(c))

                                        j)

                                        Then c Sj(c) le 2Cj

                                        Distributing Facilities

                                        iP(j(c))

                                        j

                                        rrsquoj = 3

                                        P(j(1))P(j(2))

                                        P(j(3))

                                        Copy c gets a preferred set P(j(c))Preferred sets are disjoint

                                        Ensure facwt(P(j(c)) j) ge frac12 for all c

                                        Possible to do so since each xij lt frac12

                                        facility in Fj

                                        Theory Seminar 042002

                                        Analysis

                                        Feasibility follows from

                                        1 Facilities in phases 1 2 not reused

                                        2 After clustering j is connected to rrsquoj disjoint clusters backups are distinct

                                        3 Preferred sets are disjoint

                                        So j connected to rj distinct facilities

                                        Theory Seminar 042002

                                        Facility cost

                                        Recall L1 = i | yi = 1

                                        Phase 2 incur a factor of 2

                                        Phase 3 each i is opened with probability 2yi

                                        Expected facility cost le 2 fiyifor phases 2 3

                                        iL1

                                        Theory Seminar 042002

                                        Bounding backup cost denoted by B rv

                                        D event that no i in P(j(c)) is open

                                        Lemma E[B|D] le 2vj + Cj(c)

                                        Proof 2 cases

                                        Service cost I

                                        iM Fj st cik le Ĉj(d)

                                        Also vk + Ĉj(d) le vj + Ĉj(c) le vj + Cj(c)

                                        k(d)

                                        j(c)

                                        le vj

                                        le Ĉj(d)le vk

                                        B

                                        k(d)

                                        j(c)

                                        le vj

                                        le vk

                                        iM Fj cik gt Ĉj(d)

                                        le Ĉj(d) in expectatio

                                        n

                                        1)

                                        2)

                                        backup = b(j(c))

                                        Theory Seminar 042002

                                        Service Cost II

                                        Fix j c Let X(c) = service cost of j(c)

                                        Let di = cij pi = prob i is opened = 2yi

                                        B(c) = backup costD(c) = event that no iP(j(c)) is

                                        openp = Pr[D(c)] = (1-p1)hellip(1-pm) le e-1

                                        davg = weighted avg of the dis

                                        = (i pidi)(i pi) = Sj(c)

                                        d1

                                        d2 dm

                                        P(j(c)) sorted by increasing cij

                                        j(c)

                                        i P(j(c))

                                        Theory Seminar 042002

                                        Then

                                        E[X(c)] = [p1d1 + (1-p1)p2d2 + hellip

                                        + (1-p1)hellip(1-pm-1)pmdm]

                                        + pE[B(c)|D(c)]

                                        le (1-p)davg + p[2vj + Cj(c)]

                                        le (1-e-1)Sj(c) + e-1[2vj +

                                        Cj(c)]

                                        Let X = c X(c) = service cost of j

                                        c Sj(c) le 2Cj and c Cj(c) le 2Cj

                                        Summing over all c = 1helliprrsquoj

                                        E[X] le (1-e-1)2Cj + e-1(2rrsquojvj + Cj)

                                        le 2Cj + 2e-1rrsquojvj

                                        Theory Seminar 042002

                                        Putting it all together

                                        Phase 1 pay the optimal LP cost

                                        Phases 2 3

                                        bull Facility cost twice LP facility cost

                                        bull Service costLose a factor of 2 for phase 2Phase 3 cost is 2(LP service cost)+2e-1(dual value)

                                        Overall cost for le (2+2e)(LP cost) phases 2 3

                                        Total cost le (2+2e)OPT

                                        Theory Seminar 042002

                                        How to improve this

                                        bull Distribute facilities more equitably (in an expected sense) among copies - decreases prob of lsquobadrsquo event

                                        bull Better analysis ndash maximum distance within a cluster can be bounded by 2Cj(c)

                                        bull Balance phases 2 and 3

                                        Theory Seminar 042002

                                        Summary of Results

                                        bullGive a 2076-approx algorithm for non-uniform rjs Based on LP rounding using complem slackness

                                        bullFor rj = r extend the primal-dual algorithm of (JMS02) to get a 152-approximation

                                        bullFault tolerant k medians with rj = r

                                        a Primal-dual algorithm (JMS02) gives a 4-approx using Lagrangean relaxation

                                        b LP rounding gives a factor of 8

                                        Theory Seminar 042002

                                        Open Questions

                                        1 Reduce gap between rj = r non-uniform rj

                                        2 Combinatorial algorithms for non-uniform rj primal-dual local-search

                                        3 Constant-factor approx for fault tolerant k medians with non-uniform rjs

                                        • Fault Tolerant Facility Location
                                        • Metric Facility Location
                                        • PowerPoint Presentation
                                        • Previous Work
                                        • Previous Work (contd)
                                        • LP Formulation
                                        • LP Formulation (contd)
                                        • Complementary Slackness
                                        • 4-approximation outline
                                        • The Algorithm
                                        • Slide 11
                                        • Analysis Phase 1
                                        • Analysis (contd)
                                        • The Algorithm (contd)
                                        • Analysis Phase 2
                                        • Finally hellip
                                        • A Randomized Algorithm
                                        • Slide 18
                                        • Slide 19
                                        • Clustering
                                        • Slide 21
                                        • Slide 22
                                        • Opening Facilities
                                        • Distributing Facilities
                                        • Analysis
                                        • Facility cost
                                        • Service cost I
                                        • Service Cost II
                                        • Slide 29
                                        • Putting it all together
                                        • How to improve this
                                        • Summary of Results
                                        • Open Questions

                                          Theory Seminar 042002

                                          Iterative algorithm

                                          S = j | rrsquoj gt 0

                                          aj = lsquoactiversquo copy of j initially = 1

                                          Ĉj(aj) = avg distance to the first k

                                          facilities in Fj gathering facwt ge frac12

                                          say these facilities lsquoserversquo j

                                          Will maintain Ĉj(aj) le Cj(aj)

                                          X

                                          X

                                          X1

                                          Fj after some iterations

                                          X i removed from Fj

                                          i Fjserving jĈj(3)

                                          facilities serving j

                                          aj = 3

                                          4X

                                          (aj)

                                          (aj)

                                          (aj)

                                          Theory Seminar 042002

                                          Start of iteration

                                          1 Choose j in S with minimum vj + Ĉj(aj)

                                          2 Form cluster M = facilities serving j Note facilities are not split

                                          3 For k st Fk M decrease rrsquok advance ak set Fk = Fk ndash M

                                          2j(3) Cluster M

                                          aj = 1

                                          4

                                          1

                                          XX X

                                          aj = 4 Cluster M

                                          aj = 2

                                          3

                                          client in S

                                          facility in some Fj

                                          X facility removed from Fj

                                          (aj)

                                          Theory Seminar 042002

                                          Opening Facilities

                                          Central facilities opened in 2 steps

                                          1 Open exactly 1 facility in M i opened with prob qyi Acts as backup

                                          denoted b(k ) for each k st Fk M

                                          2 Open each i in M indep with prob (2-q)yi and independent of step 1

                                          Non-central facilities

                                          Cluster M

                                          k

                                          open with prob 2yi independent of other choices

                                          j

                                          (ak

                                          )

                                          Theory Seminar 042002

                                          Let Sj(c) = avg dist from j to P(j(c))

                                          = ( cijxij)facwt(P(j(c))

                                          j)

                                          Then c Sj(c) le 2Cj

                                          Distributing Facilities

                                          iP(j(c))

                                          j

                                          rrsquoj = 3

                                          P(j(1))P(j(2))

                                          P(j(3))

                                          Copy c gets a preferred set P(j(c))Preferred sets are disjoint

                                          Ensure facwt(P(j(c)) j) ge frac12 for all c

                                          Possible to do so since each xij lt frac12

                                          facility in Fj

                                          Theory Seminar 042002

                                          Analysis

                                          Feasibility follows from

                                          1 Facilities in phases 1 2 not reused

                                          2 After clustering j is connected to rrsquoj disjoint clusters backups are distinct

                                          3 Preferred sets are disjoint

                                          So j connected to rj distinct facilities

                                          Theory Seminar 042002

                                          Facility cost

                                          Recall L1 = i | yi = 1

                                          Phase 2 incur a factor of 2

                                          Phase 3 each i is opened with probability 2yi

                                          Expected facility cost le 2 fiyifor phases 2 3

                                          iL1

                                          Theory Seminar 042002

                                          Bounding backup cost denoted by B rv

                                          D event that no i in P(j(c)) is open

                                          Lemma E[B|D] le 2vj + Cj(c)

                                          Proof 2 cases

                                          Service cost I

                                          iM Fj st cik le Ĉj(d)

                                          Also vk + Ĉj(d) le vj + Ĉj(c) le vj + Cj(c)

                                          k(d)

                                          j(c)

                                          le vj

                                          le Ĉj(d)le vk

                                          B

                                          k(d)

                                          j(c)

                                          le vj

                                          le vk

                                          iM Fj cik gt Ĉj(d)

                                          le Ĉj(d) in expectatio

                                          n

                                          1)

                                          2)

                                          backup = b(j(c))

                                          Theory Seminar 042002

                                          Service Cost II

                                          Fix j c Let X(c) = service cost of j(c)

                                          Let di = cij pi = prob i is opened = 2yi

                                          B(c) = backup costD(c) = event that no iP(j(c)) is

                                          openp = Pr[D(c)] = (1-p1)hellip(1-pm) le e-1

                                          davg = weighted avg of the dis

                                          = (i pidi)(i pi) = Sj(c)

                                          d1

                                          d2 dm

                                          P(j(c)) sorted by increasing cij

                                          j(c)

                                          i P(j(c))

                                          Theory Seminar 042002

                                          Then

                                          E[X(c)] = [p1d1 + (1-p1)p2d2 + hellip

                                          + (1-p1)hellip(1-pm-1)pmdm]

                                          + pE[B(c)|D(c)]

                                          le (1-p)davg + p[2vj + Cj(c)]

                                          le (1-e-1)Sj(c) + e-1[2vj +

                                          Cj(c)]

                                          Let X = c X(c) = service cost of j

                                          c Sj(c) le 2Cj and c Cj(c) le 2Cj

                                          Summing over all c = 1helliprrsquoj

                                          E[X] le (1-e-1)2Cj + e-1(2rrsquojvj + Cj)

                                          le 2Cj + 2e-1rrsquojvj

                                          Theory Seminar 042002

                                          Putting it all together

                                          Phase 1 pay the optimal LP cost

                                          Phases 2 3

                                          bull Facility cost twice LP facility cost

                                          bull Service costLose a factor of 2 for phase 2Phase 3 cost is 2(LP service cost)+2e-1(dual value)

                                          Overall cost for le (2+2e)(LP cost) phases 2 3

                                          Total cost le (2+2e)OPT

                                          Theory Seminar 042002

                                          How to improve this

                                          bull Distribute facilities more equitably (in an expected sense) among copies - decreases prob of lsquobadrsquo event

                                          bull Better analysis ndash maximum distance within a cluster can be bounded by 2Cj(c)

                                          bull Balance phases 2 and 3

                                          Theory Seminar 042002

                                          Summary of Results

                                          bullGive a 2076-approx algorithm for non-uniform rjs Based on LP rounding using complem slackness

                                          bullFor rj = r extend the primal-dual algorithm of (JMS02) to get a 152-approximation

                                          bullFault tolerant k medians with rj = r

                                          a Primal-dual algorithm (JMS02) gives a 4-approx using Lagrangean relaxation

                                          b LP rounding gives a factor of 8

                                          Theory Seminar 042002

                                          Open Questions

                                          1 Reduce gap between rj = r non-uniform rj

                                          2 Combinatorial algorithms for non-uniform rj primal-dual local-search

                                          3 Constant-factor approx for fault tolerant k medians with non-uniform rjs

                                          • Fault Tolerant Facility Location
                                          • Metric Facility Location
                                          • PowerPoint Presentation
                                          • Previous Work
                                          • Previous Work (contd)
                                          • LP Formulation
                                          • LP Formulation (contd)
                                          • Complementary Slackness
                                          • 4-approximation outline
                                          • The Algorithm
                                          • Slide 11
                                          • Analysis Phase 1
                                          • Analysis (contd)
                                          • The Algorithm (contd)
                                          • Analysis Phase 2
                                          • Finally hellip
                                          • A Randomized Algorithm
                                          • Slide 18
                                          • Slide 19
                                          • Clustering
                                          • Slide 21
                                          • Slide 22
                                          • Opening Facilities
                                          • Distributing Facilities
                                          • Analysis
                                          • Facility cost
                                          • Service cost I
                                          • Service Cost II
                                          • Slide 29
                                          • Putting it all together
                                          • How to improve this
                                          • Summary of Results
                                          • Open Questions

                                            Theory Seminar 042002

                                            Start of iteration

                                            1 Choose j in S with minimum vj + Ĉj(aj)

                                            2 Form cluster M = facilities serving j Note facilities are not split

                                            3 For k st Fk M decrease rrsquok advance ak set Fk = Fk ndash M

                                            2j(3) Cluster M

                                            aj = 1

                                            4

                                            1

                                            XX X

                                            aj = 4 Cluster M

                                            aj = 2

                                            3

                                            client in S

                                            facility in some Fj

                                            X facility removed from Fj

                                            (aj)

                                            Theory Seminar 042002

                                            Opening Facilities

                                            Central facilities opened in 2 steps

                                            1 Open exactly 1 facility in M i opened with prob qyi Acts as backup

                                            denoted b(k ) for each k st Fk M

                                            2 Open each i in M indep with prob (2-q)yi and independent of step 1

                                            Non-central facilities

                                            Cluster M

                                            k

                                            open with prob 2yi independent of other choices

                                            j

                                            (ak

                                            )

                                            Theory Seminar 042002

                                            Let Sj(c) = avg dist from j to P(j(c))

                                            = ( cijxij)facwt(P(j(c))

                                            j)

                                            Then c Sj(c) le 2Cj

                                            Distributing Facilities

                                            iP(j(c))

                                            j

                                            rrsquoj = 3

                                            P(j(1))P(j(2))

                                            P(j(3))

                                            Copy c gets a preferred set P(j(c))Preferred sets are disjoint

                                            Ensure facwt(P(j(c)) j) ge frac12 for all c

                                            Possible to do so since each xij lt frac12

                                            facility in Fj

                                            Theory Seminar 042002

                                            Analysis

                                            Feasibility follows from

                                            1 Facilities in phases 1 2 not reused

                                            2 After clustering j is connected to rrsquoj disjoint clusters backups are distinct

                                            3 Preferred sets are disjoint

                                            So j connected to rj distinct facilities

                                            Theory Seminar 042002

                                            Facility cost

                                            Recall L1 = i | yi = 1

                                            Phase 2 incur a factor of 2

                                            Phase 3 each i is opened with probability 2yi

                                            Expected facility cost le 2 fiyifor phases 2 3

                                            iL1

                                            Theory Seminar 042002

                                            Bounding backup cost denoted by B rv

                                            D event that no i in P(j(c)) is open

                                            Lemma E[B|D] le 2vj + Cj(c)

                                            Proof 2 cases

                                            Service cost I

                                            iM Fj st cik le Ĉj(d)

                                            Also vk + Ĉj(d) le vj + Ĉj(c) le vj + Cj(c)

                                            k(d)

                                            j(c)

                                            le vj

                                            le Ĉj(d)le vk

                                            B

                                            k(d)

                                            j(c)

                                            le vj

                                            le vk

                                            iM Fj cik gt Ĉj(d)

                                            le Ĉj(d) in expectatio

                                            n

                                            1)

                                            2)

                                            backup = b(j(c))

                                            Theory Seminar 042002

                                            Service Cost II

                                            Fix j c Let X(c) = service cost of j(c)

                                            Let di = cij pi = prob i is opened = 2yi

                                            B(c) = backup costD(c) = event that no iP(j(c)) is

                                            openp = Pr[D(c)] = (1-p1)hellip(1-pm) le e-1

                                            davg = weighted avg of the dis

                                            = (i pidi)(i pi) = Sj(c)

                                            d1

                                            d2 dm

                                            P(j(c)) sorted by increasing cij

                                            j(c)

                                            i P(j(c))

                                            Theory Seminar 042002

                                            Then

                                            E[X(c)] = [p1d1 + (1-p1)p2d2 + hellip

                                            + (1-p1)hellip(1-pm-1)pmdm]

                                            + pE[B(c)|D(c)]

                                            le (1-p)davg + p[2vj + Cj(c)]

                                            le (1-e-1)Sj(c) + e-1[2vj +

                                            Cj(c)]

                                            Let X = c X(c) = service cost of j

                                            c Sj(c) le 2Cj and c Cj(c) le 2Cj

                                            Summing over all c = 1helliprrsquoj

                                            E[X] le (1-e-1)2Cj + e-1(2rrsquojvj + Cj)

                                            le 2Cj + 2e-1rrsquojvj

                                            Theory Seminar 042002

                                            Putting it all together

                                            Phase 1 pay the optimal LP cost

                                            Phases 2 3

                                            bull Facility cost twice LP facility cost

                                            bull Service costLose a factor of 2 for phase 2Phase 3 cost is 2(LP service cost)+2e-1(dual value)

                                            Overall cost for le (2+2e)(LP cost) phases 2 3

                                            Total cost le (2+2e)OPT

                                            Theory Seminar 042002

                                            How to improve this

                                            bull Distribute facilities more equitably (in an expected sense) among copies - decreases prob of lsquobadrsquo event

                                            bull Better analysis ndash maximum distance within a cluster can be bounded by 2Cj(c)

                                            bull Balance phases 2 and 3

                                            Theory Seminar 042002

                                            Summary of Results

                                            bullGive a 2076-approx algorithm for non-uniform rjs Based on LP rounding using complem slackness

                                            bullFor rj = r extend the primal-dual algorithm of (JMS02) to get a 152-approximation

                                            bullFault tolerant k medians with rj = r

                                            a Primal-dual algorithm (JMS02) gives a 4-approx using Lagrangean relaxation

                                            b LP rounding gives a factor of 8

                                            Theory Seminar 042002

                                            Open Questions

                                            1 Reduce gap between rj = r non-uniform rj

                                            2 Combinatorial algorithms for non-uniform rj primal-dual local-search

                                            3 Constant-factor approx for fault tolerant k medians with non-uniform rjs

                                            • Fault Tolerant Facility Location
                                            • Metric Facility Location
                                            • PowerPoint Presentation
                                            • Previous Work
                                            • Previous Work (contd)
                                            • LP Formulation
                                            • LP Formulation (contd)
                                            • Complementary Slackness
                                            • 4-approximation outline
                                            • The Algorithm
                                            • Slide 11
                                            • Analysis Phase 1
                                            • Analysis (contd)
                                            • The Algorithm (contd)
                                            • Analysis Phase 2
                                            • Finally hellip
                                            • A Randomized Algorithm
                                            • Slide 18
                                            • Slide 19
                                            • Clustering
                                            • Slide 21
                                            • Slide 22
                                            • Opening Facilities
                                            • Distributing Facilities
                                            • Analysis
                                            • Facility cost
                                            • Service cost I
                                            • Service Cost II
                                            • Slide 29
                                            • Putting it all together
                                            • How to improve this
                                            • Summary of Results
                                            • Open Questions

                                              Theory Seminar 042002

                                              Opening Facilities

                                              Central facilities opened in 2 steps

                                              1 Open exactly 1 facility in M i opened with prob qyi Acts as backup

                                              denoted b(k ) for each k st Fk M

                                              2 Open each i in M indep with prob (2-q)yi and independent of step 1

                                              Non-central facilities

                                              Cluster M

                                              k

                                              open with prob 2yi independent of other choices

                                              j

                                              (ak

                                              )

                                              Theory Seminar 042002

                                              Let Sj(c) = avg dist from j to P(j(c))

                                              = ( cijxij)facwt(P(j(c))

                                              j)

                                              Then c Sj(c) le 2Cj

                                              Distributing Facilities

                                              iP(j(c))

                                              j

                                              rrsquoj = 3

                                              P(j(1))P(j(2))

                                              P(j(3))

                                              Copy c gets a preferred set P(j(c))Preferred sets are disjoint

                                              Ensure facwt(P(j(c)) j) ge frac12 for all c

                                              Possible to do so since each xij lt frac12

                                              facility in Fj

                                              Theory Seminar 042002

                                              Analysis

                                              Feasibility follows from

                                              1 Facilities in phases 1 2 not reused

                                              2 After clustering j is connected to rrsquoj disjoint clusters backups are distinct

                                              3 Preferred sets are disjoint

                                              So j connected to rj distinct facilities

                                              Theory Seminar 042002

                                              Facility cost

                                              Recall L1 = i | yi = 1

                                              Phase 2 incur a factor of 2

                                              Phase 3 each i is opened with probability 2yi

                                              Expected facility cost le 2 fiyifor phases 2 3

                                              iL1

                                              Theory Seminar 042002

                                              Bounding backup cost denoted by B rv

                                              D event that no i in P(j(c)) is open

                                              Lemma E[B|D] le 2vj + Cj(c)

                                              Proof 2 cases

                                              Service cost I

                                              iM Fj st cik le Ĉj(d)

                                              Also vk + Ĉj(d) le vj + Ĉj(c) le vj + Cj(c)

                                              k(d)

                                              j(c)

                                              le vj

                                              le Ĉj(d)le vk

                                              B

                                              k(d)

                                              j(c)

                                              le vj

                                              le vk

                                              iM Fj cik gt Ĉj(d)

                                              le Ĉj(d) in expectatio

                                              n

                                              1)

                                              2)

                                              backup = b(j(c))

                                              Theory Seminar 042002

                                              Service Cost II

                                              Fix j c Let X(c) = service cost of j(c)

                                              Let di = cij pi = prob i is opened = 2yi

                                              B(c) = backup costD(c) = event that no iP(j(c)) is

                                              openp = Pr[D(c)] = (1-p1)hellip(1-pm) le e-1

                                              davg = weighted avg of the dis

                                              = (i pidi)(i pi) = Sj(c)

                                              d1

                                              d2 dm

                                              P(j(c)) sorted by increasing cij

                                              j(c)

                                              i P(j(c))

                                              Theory Seminar 042002

                                              Then

                                              E[X(c)] = [p1d1 + (1-p1)p2d2 + hellip

                                              + (1-p1)hellip(1-pm-1)pmdm]

                                              + pE[B(c)|D(c)]

                                              le (1-p)davg + p[2vj + Cj(c)]

                                              le (1-e-1)Sj(c) + e-1[2vj +

                                              Cj(c)]

                                              Let X = c X(c) = service cost of j

                                              c Sj(c) le 2Cj and c Cj(c) le 2Cj

                                              Summing over all c = 1helliprrsquoj

                                              E[X] le (1-e-1)2Cj + e-1(2rrsquojvj + Cj)

                                              le 2Cj + 2e-1rrsquojvj

                                              Theory Seminar 042002

                                              Putting it all together

                                              Phase 1 pay the optimal LP cost

                                              Phases 2 3

                                              bull Facility cost twice LP facility cost

                                              bull Service costLose a factor of 2 for phase 2Phase 3 cost is 2(LP service cost)+2e-1(dual value)

                                              Overall cost for le (2+2e)(LP cost) phases 2 3

                                              Total cost le (2+2e)OPT

                                              Theory Seminar 042002

                                              How to improve this

                                              bull Distribute facilities more equitably (in an expected sense) among copies - decreases prob of lsquobadrsquo event

                                              bull Better analysis ndash maximum distance within a cluster can be bounded by 2Cj(c)

                                              bull Balance phases 2 and 3

                                              Theory Seminar 042002

                                              Summary of Results

                                              bullGive a 2076-approx algorithm for non-uniform rjs Based on LP rounding using complem slackness

                                              bullFor rj = r extend the primal-dual algorithm of (JMS02) to get a 152-approximation

                                              bullFault tolerant k medians with rj = r

                                              a Primal-dual algorithm (JMS02) gives a 4-approx using Lagrangean relaxation

                                              b LP rounding gives a factor of 8

                                              Theory Seminar 042002

                                              Open Questions

                                              1 Reduce gap between rj = r non-uniform rj

                                              2 Combinatorial algorithms for non-uniform rj primal-dual local-search

                                              3 Constant-factor approx for fault tolerant k medians with non-uniform rjs

                                              • Fault Tolerant Facility Location
                                              • Metric Facility Location
                                              • PowerPoint Presentation
                                              • Previous Work
                                              • Previous Work (contd)
                                              • LP Formulation
                                              • LP Formulation (contd)
                                              • Complementary Slackness
                                              • 4-approximation outline
                                              • The Algorithm
                                              • Slide 11
                                              • Analysis Phase 1
                                              • Analysis (contd)
                                              • The Algorithm (contd)
                                              • Analysis Phase 2
                                              • Finally hellip
                                              • A Randomized Algorithm
                                              • Slide 18
                                              • Slide 19
                                              • Clustering
                                              • Slide 21
                                              • Slide 22
                                              • Opening Facilities
                                              • Distributing Facilities
                                              • Analysis
                                              • Facility cost
                                              • Service cost I
                                              • Service Cost II
                                              • Slide 29
                                              • Putting it all together
                                              • How to improve this
                                              • Summary of Results
                                              • Open Questions

                                                Theory Seminar 042002

                                                Let Sj(c) = avg dist from j to P(j(c))

                                                = ( cijxij)facwt(P(j(c))

                                                j)

                                                Then c Sj(c) le 2Cj

                                                Distributing Facilities

                                                iP(j(c))

                                                j

                                                rrsquoj = 3

                                                P(j(1))P(j(2))

                                                P(j(3))

                                                Copy c gets a preferred set P(j(c))Preferred sets are disjoint

                                                Ensure facwt(P(j(c)) j) ge frac12 for all c

                                                Possible to do so since each xij lt frac12

                                                facility in Fj

                                                Theory Seminar 042002

                                                Analysis

                                                Feasibility follows from

                                                1 Facilities in phases 1 2 not reused

                                                2 After clustering j is connected to rrsquoj disjoint clusters backups are distinct

                                                3 Preferred sets are disjoint

                                                So j connected to rj distinct facilities

                                                Theory Seminar 042002

                                                Facility cost

                                                Recall L1 = i | yi = 1

                                                Phase 2 incur a factor of 2

                                                Phase 3 each i is opened with probability 2yi

                                                Expected facility cost le 2 fiyifor phases 2 3

                                                iL1

                                                Theory Seminar 042002

                                                Bounding backup cost denoted by B rv

                                                D event that no i in P(j(c)) is open

                                                Lemma E[B|D] le 2vj + Cj(c)

                                                Proof 2 cases

                                                Service cost I

                                                iM Fj st cik le Ĉj(d)

                                                Also vk + Ĉj(d) le vj + Ĉj(c) le vj + Cj(c)

                                                k(d)

                                                j(c)

                                                le vj

                                                le Ĉj(d)le vk

                                                B

                                                k(d)

                                                j(c)

                                                le vj

                                                le vk

                                                iM Fj cik gt Ĉj(d)

                                                le Ĉj(d) in expectatio

                                                n

                                                1)

                                                2)

                                                backup = b(j(c))

                                                Theory Seminar 042002

                                                Service Cost II

                                                Fix j c Let X(c) = service cost of j(c)

                                                Let di = cij pi = prob i is opened = 2yi

                                                B(c) = backup costD(c) = event that no iP(j(c)) is

                                                openp = Pr[D(c)] = (1-p1)hellip(1-pm) le e-1

                                                davg = weighted avg of the dis

                                                = (i pidi)(i pi) = Sj(c)

                                                d1

                                                d2 dm

                                                P(j(c)) sorted by increasing cij

                                                j(c)

                                                i P(j(c))

                                                Theory Seminar 042002

                                                Then

                                                E[X(c)] = [p1d1 + (1-p1)p2d2 + hellip

                                                + (1-p1)hellip(1-pm-1)pmdm]

                                                + pE[B(c)|D(c)]

                                                le (1-p)davg + p[2vj + Cj(c)]

                                                le (1-e-1)Sj(c) + e-1[2vj +

                                                Cj(c)]

                                                Let X = c X(c) = service cost of j

                                                c Sj(c) le 2Cj and c Cj(c) le 2Cj

                                                Summing over all c = 1helliprrsquoj

                                                E[X] le (1-e-1)2Cj + e-1(2rrsquojvj + Cj)

                                                le 2Cj + 2e-1rrsquojvj

                                                Theory Seminar 042002

                                                Putting it all together

                                                Phase 1 pay the optimal LP cost

                                                Phases 2 3

                                                bull Facility cost twice LP facility cost

                                                bull Service costLose a factor of 2 for phase 2Phase 3 cost is 2(LP service cost)+2e-1(dual value)

                                                Overall cost for le (2+2e)(LP cost) phases 2 3

                                                Total cost le (2+2e)OPT

                                                Theory Seminar 042002

                                                How to improve this

                                                bull Distribute facilities more equitably (in an expected sense) among copies - decreases prob of lsquobadrsquo event

                                                bull Better analysis ndash maximum distance within a cluster can be bounded by 2Cj(c)

                                                bull Balance phases 2 and 3

                                                Theory Seminar 042002

                                                Summary of Results

                                                bullGive a 2076-approx algorithm for non-uniform rjs Based on LP rounding using complem slackness

                                                bullFor rj = r extend the primal-dual algorithm of (JMS02) to get a 152-approximation

                                                bullFault tolerant k medians with rj = r

                                                a Primal-dual algorithm (JMS02) gives a 4-approx using Lagrangean relaxation

                                                b LP rounding gives a factor of 8

                                                Theory Seminar 042002

                                                Open Questions

                                                1 Reduce gap between rj = r non-uniform rj

                                                2 Combinatorial algorithms for non-uniform rj primal-dual local-search

                                                3 Constant-factor approx for fault tolerant k medians with non-uniform rjs

                                                • Fault Tolerant Facility Location
                                                • Metric Facility Location
                                                • PowerPoint Presentation
                                                • Previous Work
                                                • Previous Work (contd)
                                                • LP Formulation
                                                • LP Formulation (contd)
                                                • Complementary Slackness
                                                • 4-approximation outline
                                                • The Algorithm
                                                • Slide 11
                                                • Analysis Phase 1
                                                • Analysis (contd)
                                                • The Algorithm (contd)
                                                • Analysis Phase 2
                                                • Finally hellip
                                                • A Randomized Algorithm
                                                • Slide 18
                                                • Slide 19
                                                • Clustering
                                                • Slide 21
                                                • Slide 22
                                                • Opening Facilities
                                                • Distributing Facilities
                                                • Analysis
                                                • Facility cost
                                                • Service cost I
                                                • Service Cost II
                                                • Slide 29
                                                • Putting it all together
                                                • How to improve this
                                                • Summary of Results
                                                • Open Questions

                                                  Theory Seminar 042002

                                                  Analysis

                                                  Feasibility follows from

                                                  1 Facilities in phases 1 2 not reused

                                                  2 After clustering j is connected to rrsquoj disjoint clusters backups are distinct

                                                  3 Preferred sets are disjoint

                                                  So j connected to rj distinct facilities

                                                  Theory Seminar 042002

                                                  Facility cost

                                                  Recall L1 = i | yi = 1

                                                  Phase 2 incur a factor of 2

                                                  Phase 3 each i is opened with probability 2yi

                                                  Expected facility cost le 2 fiyifor phases 2 3

                                                  iL1

                                                  Theory Seminar 042002

                                                  Bounding backup cost denoted by B rv

                                                  D event that no i in P(j(c)) is open

                                                  Lemma E[B|D] le 2vj + Cj(c)

                                                  Proof 2 cases

                                                  Service cost I

                                                  iM Fj st cik le Ĉj(d)

                                                  Also vk + Ĉj(d) le vj + Ĉj(c) le vj + Cj(c)

                                                  k(d)

                                                  j(c)

                                                  le vj

                                                  le Ĉj(d)le vk

                                                  B

                                                  k(d)

                                                  j(c)

                                                  le vj

                                                  le vk

                                                  iM Fj cik gt Ĉj(d)

                                                  le Ĉj(d) in expectatio

                                                  n

                                                  1)

                                                  2)

                                                  backup = b(j(c))

                                                  Theory Seminar 042002

                                                  Service Cost II

                                                  Fix j c Let X(c) = service cost of j(c)

                                                  Let di = cij pi = prob i is opened = 2yi

                                                  B(c) = backup costD(c) = event that no iP(j(c)) is

                                                  openp = Pr[D(c)] = (1-p1)hellip(1-pm) le e-1

                                                  davg = weighted avg of the dis

                                                  = (i pidi)(i pi) = Sj(c)

                                                  d1

                                                  d2 dm

                                                  P(j(c)) sorted by increasing cij

                                                  j(c)

                                                  i P(j(c))

                                                  Theory Seminar 042002

                                                  Then

                                                  E[X(c)] = [p1d1 + (1-p1)p2d2 + hellip

                                                  + (1-p1)hellip(1-pm-1)pmdm]

                                                  + pE[B(c)|D(c)]

                                                  le (1-p)davg + p[2vj + Cj(c)]

                                                  le (1-e-1)Sj(c) + e-1[2vj +

                                                  Cj(c)]

                                                  Let X = c X(c) = service cost of j

                                                  c Sj(c) le 2Cj and c Cj(c) le 2Cj

                                                  Summing over all c = 1helliprrsquoj

                                                  E[X] le (1-e-1)2Cj + e-1(2rrsquojvj + Cj)

                                                  le 2Cj + 2e-1rrsquojvj

                                                  Theory Seminar 042002

                                                  Putting it all together

                                                  Phase 1 pay the optimal LP cost

                                                  Phases 2 3

                                                  bull Facility cost twice LP facility cost

                                                  bull Service costLose a factor of 2 for phase 2Phase 3 cost is 2(LP service cost)+2e-1(dual value)

                                                  Overall cost for le (2+2e)(LP cost) phases 2 3

                                                  Total cost le (2+2e)OPT

                                                  Theory Seminar 042002

                                                  How to improve this

                                                  bull Distribute facilities more equitably (in an expected sense) among copies - decreases prob of lsquobadrsquo event

                                                  bull Better analysis ndash maximum distance within a cluster can be bounded by 2Cj(c)

                                                  bull Balance phases 2 and 3

                                                  Theory Seminar 042002

                                                  Summary of Results

                                                  bullGive a 2076-approx algorithm for non-uniform rjs Based on LP rounding using complem slackness

                                                  bullFor rj = r extend the primal-dual algorithm of (JMS02) to get a 152-approximation

                                                  bullFault tolerant k medians with rj = r

                                                  a Primal-dual algorithm (JMS02) gives a 4-approx using Lagrangean relaxation

                                                  b LP rounding gives a factor of 8

                                                  Theory Seminar 042002

                                                  Open Questions

                                                  1 Reduce gap between rj = r non-uniform rj

                                                  2 Combinatorial algorithms for non-uniform rj primal-dual local-search

                                                  3 Constant-factor approx for fault tolerant k medians with non-uniform rjs

                                                  • Fault Tolerant Facility Location
                                                  • Metric Facility Location
                                                  • PowerPoint Presentation
                                                  • Previous Work
                                                  • Previous Work (contd)
                                                  • LP Formulation
                                                  • LP Formulation (contd)
                                                  • Complementary Slackness
                                                  • 4-approximation outline
                                                  • The Algorithm
                                                  • Slide 11
                                                  • Analysis Phase 1
                                                  • Analysis (contd)
                                                  • The Algorithm (contd)
                                                  • Analysis Phase 2
                                                  • Finally hellip
                                                  • A Randomized Algorithm
                                                  • Slide 18
                                                  • Slide 19
                                                  • Clustering
                                                  • Slide 21
                                                  • Slide 22
                                                  • Opening Facilities
                                                  • Distributing Facilities
                                                  • Analysis
                                                  • Facility cost
                                                  • Service cost I
                                                  • Service Cost II
                                                  • Slide 29
                                                  • Putting it all together
                                                  • How to improve this
                                                  • Summary of Results
                                                  • Open Questions

                                                    Theory Seminar 042002

                                                    Facility cost

                                                    Recall L1 = i | yi = 1

                                                    Phase 2 incur a factor of 2

                                                    Phase 3 each i is opened with probability 2yi

                                                    Expected facility cost le 2 fiyifor phases 2 3

                                                    iL1

                                                    Theory Seminar 042002

                                                    Bounding backup cost denoted by B rv

                                                    D event that no i in P(j(c)) is open

                                                    Lemma E[B|D] le 2vj + Cj(c)

                                                    Proof 2 cases

                                                    Service cost I

                                                    iM Fj st cik le Ĉj(d)

                                                    Also vk + Ĉj(d) le vj + Ĉj(c) le vj + Cj(c)

                                                    k(d)

                                                    j(c)

                                                    le vj

                                                    le Ĉj(d)le vk

                                                    B

                                                    k(d)

                                                    j(c)

                                                    le vj

                                                    le vk

                                                    iM Fj cik gt Ĉj(d)

                                                    le Ĉj(d) in expectatio

                                                    n

                                                    1)

                                                    2)

                                                    backup = b(j(c))

                                                    Theory Seminar 042002

                                                    Service Cost II

                                                    Fix j c Let X(c) = service cost of j(c)

                                                    Let di = cij pi = prob i is opened = 2yi

                                                    B(c) = backup costD(c) = event that no iP(j(c)) is

                                                    openp = Pr[D(c)] = (1-p1)hellip(1-pm) le e-1

                                                    davg = weighted avg of the dis

                                                    = (i pidi)(i pi) = Sj(c)

                                                    d1

                                                    d2 dm

                                                    P(j(c)) sorted by increasing cij

                                                    j(c)

                                                    i P(j(c))

                                                    Theory Seminar 042002

                                                    Then

                                                    E[X(c)] = [p1d1 + (1-p1)p2d2 + hellip

                                                    + (1-p1)hellip(1-pm-1)pmdm]

                                                    + pE[B(c)|D(c)]

                                                    le (1-p)davg + p[2vj + Cj(c)]

                                                    le (1-e-1)Sj(c) + e-1[2vj +

                                                    Cj(c)]

                                                    Let X = c X(c) = service cost of j

                                                    c Sj(c) le 2Cj and c Cj(c) le 2Cj

                                                    Summing over all c = 1helliprrsquoj

                                                    E[X] le (1-e-1)2Cj + e-1(2rrsquojvj + Cj)

                                                    le 2Cj + 2e-1rrsquojvj

                                                    Theory Seminar 042002

                                                    Putting it all together

                                                    Phase 1 pay the optimal LP cost

                                                    Phases 2 3

                                                    bull Facility cost twice LP facility cost

                                                    bull Service costLose a factor of 2 for phase 2Phase 3 cost is 2(LP service cost)+2e-1(dual value)

                                                    Overall cost for le (2+2e)(LP cost) phases 2 3

                                                    Total cost le (2+2e)OPT

                                                    Theory Seminar 042002

                                                    How to improve this

                                                    bull Distribute facilities more equitably (in an expected sense) among copies - decreases prob of lsquobadrsquo event

                                                    bull Better analysis ndash maximum distance within a cluster can be bounded by 2Cj(c)

                                                    bull Balance phases 2 and 3

                                                    Theory Seminar 042002

                                                    Summary of Results

                                                    bullGive a 2076-approx algorithm for non-uniform rjs Based on LP rounding using complem slackness

                                                    bullFor rj = r extend the primal-dual algorithm of (JMS02) to get a 152-approximation

                                                    bullFault tolerant k medians with rj = r

                                                    a Primal-dual algorithm (JMS02) gives a 4-approx using Lagrangean relaxation

                                                    b LP rounding gives a factor of 8

                                                    Theory Seminar 042002

                                                    Open Questions

                                                    1 Reduce gap between rj = r non-uniform rj

                                                    2 Combinatorial algorithms for non-uniform rj primal-dual local-search

                                                    3 Constant-factor approx for fault tolerant k medians with non-uniform rjs

                                                    • Fault Tolerant Facility Location
                                                    • Metric Facility Location
                                                    • PowerPoint Presentation
                                                    • Previous Work
                                                    • Previous Work (contd)
                                                    • LP Formulation
                                                    • LP Formulation (contd)
                                                    • Complementary Slackness
                                                    • 4-approximation outline
                                                    • The Algorithm
                                                    • Slide 11
                                                    • Analysis Phase 1
                                                    • Analysis (contd)
                                                    • The Algorithm (contd)
                                                    • Analysis Phase 2
                                                    • Finally hellip
                                                    • A Randomized Algorithm
                                                    • Slide 18
                                                    • Slide 19
                                                    • Clustering
                                                    • Slide 21
                                                    • Slide 22
                                                    • Opening Facilities
                                                    • Distributing Facilities
                                                    • Analysis
                                                    • Facility cost
                                                    • Service cost I
                                                    • Service Cost II
                                                    • Slide 29
                                                    • Putting it all together
                                                    • How to improve this
                                                    • Summary of Results
                                                    • Open Questions

                                                      Theory Seminar 042002

                                                      Bounding backup cost denoted by B rv

                                                      D event that no i in P(j(c)) is open

                                                      Lemma E[B|D] le 2vj + Cj(c)

                                                      Proof 2 cases

                                                      Service cost I

                                                      iM Fj st cik le Ĉj(d)

                                                      Also vk + Ĉj(d) le vj + Ĉj(c) le vj + Cj(c)

                                                      k(d)

                                                      j(c)

                                                      le vj

                                                      le Ĉj(d)le vk

                                                      B

                                                      k(d)

                                                      j(c)

                                                      le vj

                                                      le vk

                                                      iM Fj cik gt Ĉj(d)

                                                      le Ĉj(d) in expectatio

                                                      n

                                                      1)

                                                      2)

                                                      backup = b(j(c))

                                                      Theory Seminar 042002

                                                      Service Cost II

                                                      Fix j c Let X(c) = service cost of j(c)

                                                      Let di = cij pi = prob i is opened = 2yi

                                                      B(c) = backup costD(c) = event that no iP(j(c)) is

                                                      openp = Pr[D(c)] = (1-p1)hellip(1-pm) le e-1

                                                      davg = weighted avg of the dis

                                                      = (i pidi)(i pi) = Sj(c)

                                                      d1

                                                      d2 dm

                                                      P(j(c)) sorted by increasing cij

                                                      j(c)

                                                      i P(j(c))

                                                      Theory Seminar 042002

                                                      Then

                                                      E[X(c)] = [p1d1 + (1-p1)p2d2 + hellip

                                                      + (1-p1)hellip(1-pm-1)pmdm]

                                                      + pE[B(c)|D(c)]

                                                      le (1-p)davg + p[2vj + Cj(c)]

                                                      le (1-e-1)Sj(c) + e-1[2vj +

                                                      Cj(c)]

                                                      Let X = c X(c) = service cost of j

                                                      c Sj(c) le 2Cj and c Cj(c) le 2Cj

                                                      Summing over all c = 1helliprrsquoj

                                                      E[X] le (1-e-1)2Cj + e-1(2rrsquojvj + Cj)

                                                      le 2Cj + 2e-1rrsquojvj

                                                      Theory Seminar 042002

                                                      Putting it all together

                                                      Phase 1 pay the optimal LP cost

                                                      Phases 2 3

                                                      bull Facility cost twice LP facility cost

                                                      bull Service costLose a factor of 2 for phase 2Phase 3 cost is 2(LP service cost)+2e-1(dual value)

                                                      Overall cost for le (2+2e)(LP cost) phases 2 3

                                                      Total cost le (2+2e)OPT

                                                      Theory Seminar 042002

                                                      How to improve this

                                                      bull Distribute facilities more equitably (in an expected sense) among copies - decreases prob of lsquobadrsquo event

                                                      bull Better analysis ndash maximum distance within a cluster can be bounded by 2Cj(c)

                                                      bull Balance phases 2 and 3

                                                      Theory Seminar 042002

                                                      Summary of Results

                                                      bullGive a 2076-approx algorithm for non-uniform rjs Based on LP rounding using complem slackness

                                                      bullFor rj = r extend the primal-dual algorithm of (JMS02) to get a 152-approximation

                                                      bullFault tolerant k medians with rj = r

                                                      a Primal-dual algorithm (JMS02) gives a 4-approx using Lagrangean relaxation

                                                      b LP rounding gives a factor of 8

                                                      Theory Seminar 042002

                                                      Open Questions

                                                      1 Reduce gap between rj = r non-uniform rj

                                                      2 Combinatorial algorithms for non-uniform rj primal-dual local-search

                                                      3 Constant-factor approx for fault tolerant k medians with non-uniform rjs

                                                      • Fault Tolerant Facility Location
                                                      • Metric Facility Location
                                                      • PowerPoint Presentation
                                                      • Previous Work
                                                      • Previous Work (contd)
                                                      • LP Formulation
                                                      • LP Formulation (contd)
                                                      • Complementary Slackness
                                                      • 4-approximation outline
                                                      • The Algorithm
                                                      • Slide 11
                                                      • Analysis Phase 1
                                                      • Analysis (contd)
                                                      • The Algorithm (contd)
                                                      • Analysis Phase 2
                                                      • Finally hellip
                                                      • A Randomized Algorithm
                                                      • Slide 18
                                                      • Slide 19
                                                      • Clustering
                                                      • Slide 21
                                                      • Slide 22
                                                      • Opening Facilities
                                                      • Distributing Facilities
                                                      • Analysis
                                                      • Facility cost
                                                      • Service cost I
                                                      • Service Cost II
                                                      • Slide 29
                                                      • Putting it all together
                                                      • How to improve this
                                                      • Summary of Results
                                                      • Open Questions

                                                        Theory Seminar 042002

                                                        Service Cost II

                                                        Fix j c Let X(c) = service cost of j(c)

                                                        Let di = cij pi = prob i is opened = 2yi

                                                        B(c) = backup costD(c) = event that no iP(j(c)) is

                                                        openp = Pr[D(c)] = (1-p1)hellip(1-pm) le e-1

                                                        davg = weighted avg of the dis

                                                        = (i pidi)(i pi) = Sj(c)

                                                        d1

                                                        d2 dm

                                                        P(j(c)) sorted by increasing cij

                                                        j(c)

                                                        i P(j(c))

                                                        Theory Seminar 042002

                                                        Then

                                                        E[X(c)] = [p1d1 + (1-p1)p2d2 + hellip

                                                        + (1-p1)hellip(1-pm-1)pmdm]

                                                        + pE[B(c)|D(c)]

                                                        le (1-p)davg + p[2vj + Cj(c)]

                                                        le (1-e-1)Sj(c) + e-1[2vj +

                                                        Cj(c)]

                                                        Let X = c X(c) = service cost of j

                                                        c Sj(c) le 2Cj and c Cj(c) le 2Cj

                                                        Summing over all c = 1helliprrsquoj

                                                        E[X] le (1-e-1)2Cj + e-1(2rrsquojvj + Cj)

                                                        le 2Cj + 2e-1rrsquojvj

                                                        Theory Seminar 042002

                                                        Putting it all together

                                                        Phase 1 pay the optimal LP cost

                                                        Phases 2 3

                                                        bull Facility cost twice LP facility cost

                                                        bull Service costLose a factor of 2 for phase 2Phase 3 cost is 2(LP service cost)+2e-1(dual value)

                                                        Overall cost for le (2+2e)(LP cost) phases 2 3

                                                        Total cost le (2+2e)OPT

                                                        Theory Seminar 042002

                                                        How to improve this

                                                        bull Distribute facilities more equitably (in an expected sense) among copies - decreases prob of lsquobadrsquo event

                                                        bull Better analysis ndash maximum distance within a cluster can be bounded by 2Cj(c)

                                                        bull Balance phases 2 and 3

                                                        Theory Seminar 042002

                                                        Summary of Results

                                                        bullGive a 2076-approx algorithm for non-uniform rjs Based on LP rounding using complem slackness

                                                        bullFor rj = r extend the primal-dual algorithm of (JMS02) to get a 152-approximation

                                                        bullFault tolerant k medians with rj = r

                                                        a Primal-dual algorithm (JMS02) gives a 4-approx using Lagrangean relaxation

                                                        b LP rounding gives a factor of 8

                                                        Theory Seminar 042002

                                                        Open Questions

                                                        1 Reduce gap between rj = r non-uniform rj

                                                        2 Combinatorial algorithms for non-uniform rj primal-dual local-search

                                                        3 Constant-factor approx for fault tolerant k medians with non-uniform rjs

                                                        • Fault Tolerant Facility Location
                                                        • Metric Facility Location
                                                        • PowerPoint Presentation
                                                        • Previous Work
                                                        • Previous Work (contd)
                                                        • LP Formulation
                                                        • LP Formulation (contd)
                                                        • Complementary Slackness
                                                        • 4-approximation outline
                                                        • The Algorithm
                                                        • Slide 11
                                                        • Analysis Phase 1
                                                        • Analysis (contd)
                                                        • The Algorithm (contd)
                                                        • Analysis Phase 2
                                                        • Finally hellip
                                                        • A Randomized Algorithm
                                                        • Slide 18
                                                        • Slide 19
                                                        • Clustering
                                                        • Slide 21
                                                        • Slide 22
                                                        • Opening Facilities
                                                        • Distributing Facilities
                                                        • Analysis
                                                        • Facility cost
                                                        • Service cost I
                                                        • Service Cost II
                                                        • Slide 29
                                                        • Putting it all together
                                                        • How to improve this
                                                        • Summary of Results
                                                        • Open Questions

                                                          Theory Seminar 042002

                                                          Then

                                                          E[X(c)] = [p1d1 + (1-p1)p2d2 + hellip

                                                          + (1-p1)hellip(1-pm-1)pmdm]

                                                          + pE[B(c)|D(c)]

                                                          le (1-p)davg + p[2vj + Cj(c)]

                                                          le (1-e-1)Sj(c) + e-1[2vj +

                                                          Cj(c)]

                                                          Let X = c X(c) = service cost of j

                                                          c Sj(c) le 2Cj and c Cj(c) le 2Cj

                                                          Summing over all c = 1helliprrsquoj

                                                          E[X] le (1-e-1)2Cj + e-1(2rrsquojvj + Cj)

                                                          le 2Cj + 2e-1rrsquojvj

                                                          Theory Seminar 042002

                                                          Putting it all together

                                                          Phase 1 pay the optimal LP cost

                                                          Phases 2 3

                                                          bull Facility cost twice LP facility cost

                                                          bull Service costLose a factor of 2 for phase 2Phase 3 cost is 2(LP service cost)+2e-1(dual value)

                                                          Overall cost for le (2+2e)(LP cost) phases 2 3

                                                          Total cost le (2+2e)OPT

                                                          Theory Seminar 042002

                                                          How to improve this

                                                          bull Distribute facilities more equitably (in an expected sense) among copies - decreases prob of lsquobadrsquo event

                                                          bull Better analysis ndash maximum distance within a cluster can be bounded by 2Cj(c)

                                                          bull Balance phases 2 and 3

                                                          Theory Seminar 042002

                                                          Summary of Results

                                                          bullGive a 2076-approx algorithm for non-uniform rjs Based on LP rounding using complem slackness

                                                          bullFor rj = r extend the primal-dual algorithm of (JMS02) to get a 152-approximation

                                                          bullFault tolerant k medians with rj = r

                                                          a Primal-dual algorithm (JMS02) gives a 4-approx using Lagrangean relaxation

                                                          b LP rounding gives a factor of 8

                                                          Theory Seminar 042002

                                                          Open Questions

                                                          1 Reduce gap between rj = r non-uniform rj

                                                          2 Combinatorial algorithms for non-uniform rj primal-dual local-search

                                                          3 Constant-factor approx for fault tolerant k medians with non-uniform rjs

                                                          • Fault Tolerant Facility Location
                                                          • Metric Facility Location
                                                          • PowerPoint Presentation
                                                          • Previous Work
                                                          • Previous Work (contd)
                                                          • LP Formulation
                                                          • LP Formulation (contd)
                                                          • Complementary Slackness
                                                          • 4-approximation outline
                                                          • The Algorithm
                                                          • Slide 11
                                                          • Analysis Phase 1
                                                          • Analysis (contd)
                                                          • The Algorithm (contd)
                                                          • Analysis Phase 2
                                                          • Finally hellip
                                                          • A Randomized Algorithm
                                                          • Slide 18
                                                          • Slide 19
                                                          • Clustering
                                                          • Slide 21
                                                          • Slide 22
                                                          • Opening Facilities
                                                          • Distributing Facilities
                                                          • Analysis
                                                          • Facility cost
                                                          • Service cost I
                                                          • Service Cost II
                                                          • Slide 29
                                                          • Putting it all together
                                                          • How to improve this
                                                          • Summary of Results
                                                          • Open Questions

                                                            Theory Seminar 042002

                                                            Putting it all together

                                                            Phase 1 pay the optimal LP cost

                                                            Phases 2 3

                                                            bull Facility cost twice LP facility cost

                                                            bull Service costLose a factor of 2 for phase 2Phase 3 cost is 2(LP service cost)+2e-1(dual value)

                                                            Overall cost for le (2+2e)(LP cost) phases 2 3

                                                            Total cost le (2+2e)OPT

                                                            Theory Seminar 042002

                                                            How to improve this

                                                            bull Distribute facilities more equitably (in an expected sense) among copies - decreases prob of lsquobadrsquo event

                                                            bull Better analysis ndash maximum distance within a cluster can be bounded by 2Cj(c)

                                                            bull Balance phases 2 and 3

                                                            Theory Seminar 042002

                                                            Summary of Results

                                                            bullGive a 2076-approx algorithm for non-uniform rjs Based on LP rounding using complem slackness

                                                            bullFor rj = r extend the primal-dual algorithm of (JMS02) to get a 152-approximation

                                                            bullFault tolerant k medians with rj = r

                                                            a Primal-dual algorithm (JMS02) gives a 4-approx using Lagrangean relaxation

                                                            b LP rounding gives a factor of 8

                                                            Theory Seminar 042002

                                                            Open Questions

                                                            1 Reduce gap between rj = r non-uniform rj

                                                            2 Combinatorial algorithms for non-uniform rj primal-dual local-search

                                                            3 Constant-factor approx for fault tolerant k medians with non-uniform rjs

                                                            • Fault Tolerant Facility Location
                                                            • Metric Facility Location
                                                            • PowerPoint Presentation
                                                            • Previous Work
                                                            • Previous Work (contd)
                                                            • LP Formulation
                                                            • LP Formulation (contd)
                                                            • Complementary Slackness
                                                            • 4-approximation outline
                                                            • The Algorithm
                                                            • Slide 11
                                                            • Analysis Phase 1
                                                            • Analysis (contd)
                                                            • The Algorithm (contd)
                                                            • Analysis Phase 2
                                                            • Finally hellip
                                                            • A Randomized Algorithm
                                                            • Slide 18
                                                            • Slide 19
                                                            • Clustering
                                                            • Slide 21
                                                            • Slide 22
                                                            • Opening Facilities
                                                            • Distributing Facilities
                                                            • Analysis
                                                            • Facility cost
                                                            • Service cost I
                                                            • Service Cost II
                                                            • Slide 29
                                                            • Putting it all together
                                                            • How to improve this
                                                            • Summary of Results
                                                            • Open Questions

                                                              Theory Seminar 042002

                                                              How to improve this

                                                              bull Distribute facilities more equitably (in an expected sense) among copies - decreases prob of lsquobadrsquo event

                                                              bull Better analysis ndash maximum distance within a cluster can be bounded by 2Cj(c)

                                                              bull Balance phases 2 and 3

                                                              Theory Seminar 042002

                                                              Summary of Results

                                                              bullGive a 2076-approx algorithm for non-uniform rjs Based on LP rounding using complem slackness

                                                              bullFor rj = r extend the primal-dual algorithm of (JMS02) to get a 152-approximation

                                                              bullFault tolerant k medians with rj = r

                                                              a Primal-dual algorithm (JMS02) gives a 4-approx using Lagrangean relaxation

                                                              b LP rounding gives a factor of 8

                                                              Theory Seminar 042002

                                                              Open Questions

                                                              1 Reduce gap between rj = r non-uniform rj

                                                              2 Combinatorial algorithms for non-uniform rj primal-dual local-search

                                                              3 Constant-factor approx for fault tolerant k medians with non-uniform rjs

                                                              • Fault Tolerant Facility Location
                                                              • Metric Facility Location
                                                              • PowerPoint Presentation
                                                              • Previous Work
                                                              • Previous Work (contd)
                                                              • LP Formulation
                                                              • LP Formulation (contd)
                                                              • Complementary Slackness
                                                              • 4-approximation outline
                                                              • The Algorithm
                                                              • Slide 11
                                                              • Analysis Phase 1
                                                              • Analysis (contd)
                                                              • The Algorithm (contd)
                                                              • Analysis Phase 2
                                                              • Finally hellip
                                                              • A Randomized Algorithm
                                                              • Slide 18
                                                              • Slide 19
                                                              • Clustering
                                                              • Slide 21
                                                              • Slide 22
                                                              • Opening Facilities
                                                              • Distributing Facilities
                                                              • Analysis
                                                              • Facility cost
                                                              • Service cost I
                                                              • Service Cost II
                                                              • Slide 29
                                                              • Putting it all together
                                                              • How to improve this
                                                              • Summary of Results
                                                              • Open Questions

                                                                Theory Seminar 042002

                                                                Summary of Results

                                                                bullGive a 2076-approx algorithm for non-uniform rjs Based on LP rounding using complem slackness

                                                                bullFor rj = r extend the primal-dual algorithm of (JMS02) to get a 152-approximation

                                                                bullFault tolerant k medians with rj = r

                                                                a Primal-dual algorithm (JMS02) gives a 4-approx using Lagrangean relaxation

                                                                b LP rounding gives a factor of 8

                                                                Theory Seminar 042002

                                                                Open Questions

                                                                1 Reduce gap between rj = r non-uniform rj

                                                                2 Combinatorial algorithms for non-uniform rj primal-dual local-search

                                                                3 Constant-factor approx for fault tolerant k medians with non-uniform rjs

                                                                • Fault Tolerant Facility Location
                                                                • Metric Facility Location
                                                                • PowerPoint Presentation
                                                                • Previous Work
                                                                • Previous Work (contd)
                                                                • LP Formulation
                                                                • LP Formulation (contd)
                                                                • Complementary Slackness
                                                                • 4-approximation outline
                                                                • The Algorithm
                                                                • Slide 11
                                                                • Analysis Phase 1
                                                                • Analysis (contd)
                                                                • The Algorithm (contd)
                                                                • Analysis Phase 2
                                                                • Finally hellip
                                                                • A Randomized Algorithm
                                                                • Slide 18
                                                                • Slide 19
                                                                • Clustering
                                                                • Slide 21
                                                                • Slide 22
                                                                • Opening Facilities
                                                                • Distributing Facilities
                                                                • Analysis
                                                                • Facility cost
                                                                • Service cost I
                                                                • Service Cost II
                                                                • Slide 29
                                                                • Putting it all together
                                                                • How to improve this
                                                                • Summary of Results
                                                                • Open Questions

                                                                  Theory Seminar 042002

                                                                  Open Questions

                                                                  1 Reduce gap between rj = r non-uniform rj

                                                                  2 Combinatorial algorithms for non-uniform rj primal-dual local-search

                                                                  3 Constant-factor approx for fault tolerant k medians with non-uniform rjs

                                                                  • Fault Tolerant Facility Location
                                                                  • Metric Facility Location
                                                                  • PowerPoint Presentation
                                                                  • Previous Work
                                                                  • Previous Work (contd)
                                                                  • LP Formulation
                                                                  • LP Formulation (contd)
                                                                  • Complementary Slackness
                                                                  • 4-approximation outline
                                                                  • The Algorithm
                                                                  • Slide 11
                                                                  • Analysis Phase 1
                                                                  • Analysis (contd)
                                                                  • The Algorithm (contd)
                                                                  • Analysis Phase 2
                                                                  • Finally hellip
                                                                  • A Randomized Algorithm
                                                                  • Slide 18
                                                                  • Slide 19
                                                                  • Clustering
                                                                  • Slide 21
                                                                  • Slide 22
                                                                  • Opening Facilities
                                                                  • Distributing Facilities
                                                                  • Analysis
                                                                  • Facility cost
                                                                  • Service cost I
                                                                  • Service Cost II
                                                                  • Slide 29
                                                                  • Putting it all together
                                                                  • How to improve this
                                                                  • Summary of Results
                                                                  • Open Questions

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