EE 122: Final Review - University of California, Berkeleyinst.eecs.berkeley.edu/~ee122/fa09/notes/25-FinalReviewx4.pdf · 2 5 Routing: Persistent Oscillations Assume link cost = amount

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1

EE 122: Final Review Ion Stoica

TAs: Junda Liu, DK Moon, David Zats

http://inst.eecs.berkeley.edu/~ee122/fa09 (Materials with thanks to Vern Paxson, Jennifer Rexford,

and colleagues at UC Berkeley)

Announcements

  Project 2 due Friday, Dec 11, 11:59pm

  Final exam: Dec 17, 8-11am, 10 Evans Hall

  My office hours next week   My office hours: MW 10-11:30am   The office hours of everyone else unchanged

Final Exam

  Open book, open notes!   Crib sheets ok if you like

  Comprehensive, but greater focus on material since midterm

  Questions similar in format to the first midterm   Problem set-up descriptions + multipart fill-ins

  All answers on the exam sheets we hand out   Bring PENCIL, ERASER, no calculators needed

Outline

  Persistent Oscillations   TCP   Multicast: DVRMP   Token Bucket   Integrated & Differentiated Services   Routing in Chord

2

5

Routing: Persistent Oscillations   Assume link cost = amount of carried traffic

A

D

C

B 0 0

0 0 0 0

No traffic

A

D

C

B 1 1

0 0 0 0

B to A: 1 unit of traffic D to A: 1 unit of traffic

1 1

A

D

C

B 1 1+e

e 1 0 0

CA: e units of traffic

1 1 e

6


A

D

C

B 1 1+e

e 0

e 1 1

0 0

A

D

C

B 2+e 0

0 0 1+e 1

B to A: cost(BCDA) = 1 lower than cost(BA) = 1+e

C to A: cost(CDA) = 1 lower than cost(CBA) = 1+2*e

1 1 e

B to A: switches to BCDA

C to A: switches to CDA

7


A

D

C

B 2+e 0

0 0

e 1 1

1+e 1

B to A: cost(BA) = 0 lower than cost(BCDA) = 4+2*e C to A: cost(CBA) = 0 lower than cost(CDA) = 3+2*e D to A: cost(DCBA) = 0 lower than cost(DA) = 2+e

B to A: switches to BA

C to A: switches to CBA

D to A: switches to DCBA

A

D

C

B 0 2+e

1+e 1 0 0

1 1 e

8


B to A: cost(BCDA) = 0 lower than cost(BCDA) = 4+2*e C to A: cost(CBA) = 0 lower than cost(CDA) = 3+2*e D to A: cost(DCBA) = 0 lower than cost(DA) = 2+e

B to A: switches to BCDA

C to A: switches to CDA

D to A: switches to DA

A

D

C

B 0 2+e

1+e 1 0 0

1 1 e

A

D

C

B 2+e 0

e 0 1+e 1

1 1 e

3

Outline

  Persistent Oscillations   TCP   Wireless MAC   Multicast: DVRMP   Token Bucket   Integrated & Differentiated Services   Routing in CAN & Chord

TCP Congestion Control

  Flow control keeps one fast sender from overwhelming a slow receiver

  Congestion control keeps a set of senders from overloading the network

  Three congestion control problems:   Adjusting to bottleneck bandwidth

  Without any a priori knowledge   Could be a Gbps link; could be a modem

  Adjusting to variations in bandwidth   Sharing bandwidth between flows

11

The big picture (with timeouts)

Time

cwnd

Initially: cwnd = 1; ssthresh = infinite;

New ack received: if (cwnd < ssthresh) /* Slow Start*/ cwnd = cwnd + 1; else /* Congestion Avoidance */ cwnd = cwnd + 1/cwnd;

Timeout: /* Multiplicative decrease */ ssthresh = cwnd/2; cwnd = 1;

12


Time

cwnd

Slow Start




4

13


Time

cwnd

Timeout

Slow Start




1/2 cwnd

ssthresh

14


Time

cwnd

Timeout

Slow Start

1/2 cwnd

ssthresh




Slow Start

15


Time

cwnd

Timeout

Slow Start

AIMD

ssthresh

Slow Start

1/2 cwnd




16


Time

cwnd

Timeout

Slow Start

AIMD Timeout

ssthresh

Slow Start

1/2 cwnd




5

17


Time

cwnd

Timeout

Slow Start

AIMD Timeout

ssthresh

Slow Start

Slow Start

AIMD

1/2 cwnd




18


Time

cwnd

AIMD

Timeout

Slow Start

AIMD




AIMD

Slow Start

sstresh

19

Congestion Detection Revisited

  Wait for Retransmission Time Out (RTO)   RTO kills throughput

  In BSD TCP implementations, RTO is usually more than 500ms   The granularity of RTT estimate is 500 ms   Retransmission timeout is RTT + 4 * mean_deviation

  Solution: Don’t wait for RTO to expire

20

Fast Retransmits

  Resend a segment after 3 duplicate ACKs   Duplicate ACK means

that an out-of sequence segment was received

  Notes:   ACKs are for next

expected packet   Packet reordering can

cause duplicate ACKs   Window may be too

small to get enough duplicate ACKs

ACK 2

segment 1 cwnd = 1

cwnd = 2 segment 2 segment 3

ACK 4 cwnd = 4 segment 4

segment 5 segment 6 segment 7

ACK 3

3 duplicate ACKs

ACK 4

ACK 4

ACK 4

6

21

Fast Retransmit and Fast Recovery

  Retransmit after 3 duplicated acks   Prevent expensive timeouts

  Reduce slow starts   At steady state, cwnd oscillates around

the optimal window size

Time

cwnd

Slow Start

AIMD

Fast retransmit

22

Fast Recovery: After a Fast Retransmit   ssthresh = cwnd / 2   cwnd = ssthresh

  Instead of setting cwnd to 1, cut cwnd in half (multiplicative decrease)

  For each dup ack arrival   dupack++   Indicates packet left network, so we may be able to send more   MaxWindow = min(cwnd + dupack, AdvWin)

  Receive ack for new data (beyond initial dup ack)   dupack = 0   Exit fast recovery

  But when RTO expires still do cwnd = 1

Outline

  Persistent Oscillations   TCP   Multicast: DVRMP   Token Bucket   Integrated & Differentiated Services   Routing in CAN & Chord

Distance Vector Multicast Routing Protocol (DVRMP)

  An elegant extension to DV routing   Use shortest path DV routes to determine if link

is on the source-rooted spanning tree   Three steps in developing DVRMP

  Reverse Path Flooding   Reverse Path Broadcasting   Truncated Reverse Path Broadcasting

7

Reverse Path Flooding   What: node Y forwards

packet from node X to   all its neighbors

(except X),   iff X is the next hop

of Y to source (A)   How: just use unicast

routing tables   Why: eliminate the

loops of simple flooding protocol

4

2

10

3

6

8 9

1

5 7

A

D

B

C

E F

G

H

Unicast shortest path Multicast data packets Packets not forwarded

Reverse Path Flooding   What: node Y forwards

packet from node X to   all its neighbors

(except X),   iff X is the next hop

of Y to source (A)   How: just use unicast

routing tables   Why: eliminate the

loops of simple flooding protocol

4

2

10

3

6

8 9

1

5 7

A

D

B

C

E F

G

H

Unicast shortest path Multicast data packets Packets not forwarded

Duplicate packet

Reverse Path Broadcasting

  What: node X forwards packet to node Y iff X is next hop of Y to source (A)

  How: X infers this info from routing messages (see multicast lecture)

  Why: Avoid a router receiving duplicate packets

4

2

10

3

6

8 9

1

5 7

A

D

B

C

E F

G

H

Unicast shortest path Multicast data packets

Truncated Reverse Broadcasting

  What: don’t forward packets to non-members

  How: use prune messages

  Why: eliminate un-needed forwarding

4

2

10

3

6

8 9

1

5 7

A

D

B

C

E F

G

H

Unicast shortest path Multicast data packets Non-membership report Subscription message

8

Truncated Reverse Broadcasting

  What: don’t forward packets to non-members

  How: use prune messages

  Why: eliminate un-needed forwarding

4

2

10

3

6

8 9

1

5 7

A

D

B

C

E F

G

H

Unicast shortest path Multicast data packets Non-membership report Subscription message

Pruning Details

  Prune (Source,Group) at leaf if no members   Send Non-Membership Report (NMR) up tree

  If all children of router R send NRM, prune (S,G)   Propagate prune for (S,G) to parent R

  On timeout:   Prune dropped   Flow is reinstated   Down stream routers re-prune

  Note: a soft-state approach

Outline

  Persistent Oscillations   TCP   Multicast: DVRMP   Token Bucket   Integrated & Differentiated Services   Routing in Chord

Token Bucket and Arrival Curve

  Parameters   r – average rate, i.e., rate at which tokens fill the bucket   b – bucket depth   R – maximum link capacity or peak rate (optional parameter)

  A bit is transmitted only when there is an available token   Arrival curve – maximum number of bits transmitted within an

interval of time of size t

r bps

b bits

<= R bps

regulator time

bits

b*R/(R-r)

slope R

slope r

Arrival curve

9

Traffic Enforcement: Example   r = 100 Kbps; b = 3 Kb; R = 500 Kbps

3Kb

T = 0 : 1Kb packet arrives

(a)

2.2Kb

T = 2ms : packet transmitted b = 3Kb – 1Kb + 2ms*100Kbps = 2.2Kb

(b)

2.4Kb

T = 4ms : 3Kb packet arrives

(c)

3Kb

T = 10ms : packet needs to wait until enough tokens are in the bucket!

(d)

0.6Kb

T = 16ms : packet transmitted

(e)

Source Traffic Characterization: Arrival Curve

  Arrival curve – maximum amount of bits transmitted during any interval of time Δt

  Use token bucket to bound arrival curve

Δt

bits

Arrival curve

time

bps

Arrival Curve: Example   Arrival curve – maximum amount of bits

transmitted during any interval of time Δt   Use token bucket to bound arrival curve

bits Arrival curve

time

bps

0 1 2 3 4 5

1

2

1 2 3 4 5

1

2

3

4

(R=2,b=1,r=1)

Δt

Per-hop Reservation   End-host: specify

  the arrival rate characterized by token-bucket with parameters (b,r,R)   the maximum maximum admissible delay D

  Router: allocate bandwidth ra and buffer space Ba such that   no packet is dropped   no packet experiences a delay larger than D

bits

b*R/(R-r)

slope r Arrival curve

D Ba

slope ra

10

Outline

  Persistent Oscillations   TCP   Multicast: DVRMP   Token Bucket   Integrated and Differentiated Services   Routing in CAN & Chord

Integrated Services

  Provide service guarantees at a per-flow granularity

  Reservation Protocol   How service request gets from host to network

  Admission control algorithm   How network decides if it can accept flow

  Packet scheduling algorithms   How routers deliver service

Control Plane: Resource Reservation

Sender Receiver


Sender Receiver

Sender sends specification of traffic profile

11


Sender Receiver Path established (or perhaps admission control denies path)


Sender Receiver

The receiver accepts reservation request

Control Plane: Admission Control

Sender Receiver

Per-flow state (soft state)

Sender Receiver

Control Plane: Admission Control

Per-flow state on all routers in path

12

Data Plane

Sender Receiver

Per-flow classification on each router

Data Plane

Sender Receiver

Per-flow classification on each router

Data Plane

Sender Receiver Per-flow scheduling on each router

Differentiated Services (DiffServ)   Give some traffic better treatment than other

  App. requirements: interactive vs. bulk transfer   Economic arrangements: first-class versus coach

  What kind of better service could you give?   Fewer drops   Lower delay   Lower delay variation (jitter)

  How to know which packets get better service?   Bits in packet header

  Deals with traffic in aggregate   Provides weaker services   But much more scalable

13

Diffserv Architecture   Ingress routers - entrance to a DiffServ domain

  Police or shape traffic   Set Differentiated Service Code Point (DSCP) in IP header

  Core routers   Implement Per Hop Behavior (PHB) for each DSCP   Process packets based on DSCP

Ingress Egress

Ingress Egress

DS-1 DS-2

Edge router Core router

Differentiated Services

  Expedite Forwarding (EF)   Give packet minimal delay and loss service   P-bit denotes premium traffic

  Assured Forwarding (AS)   Packets are all serviced in order

  Makes TCP implementations perform well

  But some packets can be marked as low-drop   Think of it as priority levels for dropping

  A-bit denotes assured traffic

Comparison to Best-Effort & Intserv

Best-Effort Diffserv Intserv

Service Connectivity No isolation No guarantees

Per aggregate isolation Per aggregate guarantee

Per flow isolation Per flow guarantee

Service scope

End-to-end Domain End-to-end

Complexity No setup Long term setup Per flow steup

Scalability Highly scalable (nodes maintain only routing state)

Scalable (edge routers maintain per aggregate state; core routers per class state)

Not scalable (each router maintains per flow state)

Outline

  Persistent Oscillations   TCP   Multicast: DVRMP   Token Bucket   Integrated & Differentiated Services   Routing in CAN & Chord

14

Routing in CAN

  O(d*N1/d)   d: dimensions   N: number of nodes

  Example: route from S to D on a 2-dimensional grid   d = 2   N = m x m   Path length: 2*m =

2*N1/2

S

D

m

m

Routing in Chord using Fingers

  Finger i at node n:

80 + 20!80 + 21!

80 + 22!80 + 23!

80 + 24!

80 + 25!

(80 + 26) mod 27 = 16!0

32

45 80

20 112

96

i ft[i] 0 96 1 96 2 96 3 96 4 96

5 112 6 20

55 70

Routing in Chord using Fingers

  Route length: O(log N)   N: number of nodes

  Justification: every hop reduces the distance to destination by at least half

0

32

45 80

20 112

96

55 70 S

D

Recursive vs. Iterative Routing

0

32

45 80

20 112

96

55 70 S

D

0

32

45 80

20 112

96

55 70 S

D

Recursive Iterative

15

Final Exam

  Open book, open notes!   Crib sheets ok if you like

  Comprehensive, but greater focus on material since midterm

  Questions similar in format to the first midterm   Problem set-up descriptions + multipart fill-ins

  All answers on the exam sheets we hand out   Bring PENCIL, ERASER, no calculators needed

EE 122: Final Review - University of California, Berkeleyinst.eecs.berkeley.edu/~ee122/fa09/notes/25-FinalReviewx4.pdf · 2 5 Routing: Persistent Oscillations Assume link cost = amount

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