Chapter 1 Strategic Problems: Location. QEM - Chapter 1 Location problems Full truck loadtransportation … CW 1 CW 2 Central warehouse FTL or tourstransportation.

Chapter 1

Strategic Problems: Location

QEM - Chapter 1

Location problems

Full truck loadtransportation

…CW1 CW2Central warehouse

FTL or tourstransportation

…DC1 DC2 DC3 DC4Distribution centers

tourstransportation

…C1 C2 C3 C4customers

…PS1 PS2 PS3 PS4Production site

(c) Prof. Richard F. Hartl Kapitel 3 / 2

QEM - Chapter 1

More levels possible (regional warehouses)

Can be delegated to logistics service providers

Decision problems Number and types of warehouses Location of warehouses Transportation problem (assignment of customers)


QEM - Chapter 1

Median Problem

Simplest location problem Represent in complete graph. Nodes i are customers with weights bi

Choose one node as location of warehouse Minimize total weighted distance from warehouse

Definition: Median directed graph (one way streets…):

σ(i) = ∑dijbj → min.

undirected graph:σout(i) = ∑dijbj → min … out median

σin(i) = ∑djibj → min … in median


QEM - Chapter 1

Example: fromDomschke und Drexl (Logistik: Standorte, 1990, Kapitel 3.3.1)

0 12 2 10 6 12

2 0 3 3 7 5

12 10 0 8 4 10

4 2 5 0 5 2

8 6 9 4 0 6

11 9 12 7 3 0

2/0

1/4

3/2

4/3

5/1

6/2

22 2

2

3

3

3

4

4

5

D=

4

0

2

3

1

2

b=


Distance betweenlocations

weight

QEM - Chapter 1

Example: Median

i\j 1 2 3 4 5 6 i/j 1 2 3 4 5 6

OUT e.g: emergency delivery of goods IN e.g.: collection of hazardous waste

4*0 0*12 2*2 3*10 6*1 2*12 64

4*12 0*10 2*0 3*8 4*1 2*10 96

4*4 0*2 2*5 3*0 5*1 2*2 35

4*8 0*6 2*9 3*4 0*1 2*6 74

4*11 0*9 2*12 3*7 3*1 2*0 92

4*2 0*0 2*3 3*3 7*1 2*5 40

66

2*12

3*4

1*8

2*11

0*2

4*0

98

2*10

3*2

1*6

2*9

0*0

4*12

56

2*0

3*5

1*9

2*12

0*3

4*2

74

2*8

3*0

1*4

2*7

0*3

4*10

53

8

15

0

6

0

24

80

20

6

6

0

0

481

3

4

5

6

2

σout(i)

1

3

4

5

6

2

σin(i)City 4 is median since 35+74 = 109 minimal


QEM - Chapter 1

Related Problem: Center

Median Node with min total weighted distance

→ min.

Center Node with min Maximum (weighted) Distance

→ min.

j

jijbd (i)

jijj

bdmax (i)


QEM - Chapter 1

24

8

15

0

6

0

24

Solution

i\j 1 2 3 4 5 6 i/j 1 2 3 4 5 6

4*0 0*12 2*2 3*10 6*1 2*12 30

4*12 0*10 2*0 3*8 4*1 2*10 48

4*4 0*2 2*5 3*0 5*1 2*2 16

4*8 0*6 2*9 3*4 0*1 2*6 32

4*11 0*9 2*12 3*7 3*1 2*0 44

4*2 0*0 2*3 3*3 7*1 2*5 10

24

2*12

3*4

1*8

2*11

0*2

4*0

48

2*10

3*2

1*6

2*9

0*0

4*12

24

2*0

3*5

1*9

2*12

0*3

4*2

40

2*8

3*0

1*4

2*7

0*3

4*10

48

20

6

6

0

0

481

3

4

5

6

2

out(i)

1

3

4

5

6

2

in(i)City1 is center since 30+24 = 54 minimal


QEM - Chapter 1

Uncapacitated (single-stage) Warehouse Location Problem – LP-Formulation

single-stage WLP: warehouse

customer:

W1 W2 m

C1 C2 C3 C4 n

Deliver goods to n customers each customer has given demand Exist: m potential warehouse locations Warhouse in location i causes fixed costs fi

Transportation costs i j are cij if total demand of j comes from i.


QEM - Chapter 1

Problem:

How many warehouses?(many/few high/low fixed costs, low/high transportation costs

Where?

Goal:

Satisfy all demand

minimize total cost (fixed + transportation)

transportation to warehouses is ignored


QEM - Chapter 1

Example: from Domschke & Drexl (Logistik: Standorte, 1990, Kapitel 3.3.1)

Solution 1: all warehouses

i\j 1 2 3 4 5 6 7 fi

1 1 2 10 9 6 7 3 5

2 2 9 0 7 3 6 10 7

3 7 6 1 5 3 10 5 5

4 6 5 10 2 6 3 6 6

5 6 4 6 3 7 2 6 5

i\j 1 2 3 4 5 6 7 fi

1 1 2 10 9 6 7 3 5

3 7 6 1 5 3 10 5 5

Fixed costs = 5+7+5+6+5 = 28 high

Transp. costs = 1+2+0+2+3+2+3 = 13

Total costs = 28 + 13 = 41

Fixed costs = 5+5 = 10

Transp. costs = 1+2+1+5+3+7+3 = 22

Total costs = 10 + 22 = 32

Solution 2: just warehouses 1 and 3


QEM - Chapter 1

Formulation as LP (MIP) yi … Binary variable for i = 1, …, m:

yi = 1 if location i is chosen for warehouse

0 otherwise

when locations are decided: transportation cost easy (closest location) Problem: 2m-1 possibilities (exp…)

xij … real „assignment“ oder transportation variable für i = 1, …,m and j = 1, …, n:

xij = fraction of demand of customer j devivered from location i.


QEM - Chapter 1

MIP for WLP

min ),(11 1

m

i

ii

m

i

n

j

ijij yfxcyxZ

miijx11

transportation cost+ fixed cost

Delivery only from locations i that are built

Satisfy total demand of customer j

yi is binary

xij non negative

xij ≤ yi

i = 1, …, m

j = 1, …,n

i = 1, …, mFor all i and j0

}1,0{

ij

i

x

y

j = 1, …,n

m

i

ijx1

1


QEM - Chapter 1

Problem:

m*n real Variablen und m binary → for a few 100 potential locations exact solution difficult → Heuristics

Heuristics: Construction or Start heuristics (find initial feasible solution)

Add Drop

Improvement heuristics (improve starting or incumbent solution)


QEM - Chapter 1

ADD for WLP

Notation:I:={1,…,m} set of all potential locations

I0 set of (finally) forbidden locations (y i = 0 fixed)

Iovl set of preliminary forbidden locations (y i = 0 tentaitively)

I1 set of included (built, realized) locations (y i =1 fixed)

reduction in transportation cost, if location i is built in addition to current loc.

Z total cost (objective)

i


QEM - Chapter 1

Initialzation:

Determine, which location to build if just one location is built:

row sum of cost matrix ci := ∑cij … transportation cost

choose location k with minimal cost ck + fk

set I1 = {k}, Iovl = I – {k} und Z = ck + fk … incumbent solution

compute savings of transportation cost ωij = max {ckj – cij, 0} for all locations i from Io

vl and all customers j as well as row sum ωi … choose maximum ωi

Example: first location k=5 with Z:= c5 + f5 = 39, I1 = {5}, Iovl = {1,2,3,4}


QEM - Chapter 1

i\j 1 2 3 4 5 6 7 fi ci fi + ci

1 1 2 10 9 6 7 3 5 38 43

2 2 9 0 7 3 6 10 7 37 44

3 7 6 1 5 3 10 5 5 37 42

4 6 5 10 2 6 3 6 6 38 44

5 6 4 6 3 7 2 6 5 34 39

i\j 1 2 3 4 5 6 7 ωi fi

1

2

3

4

ωij is saving in transportation cost when delivering to custonmer j, by opening additional location i.

→ row sum ωi is total saving in transportation cost when opening additional location i.

11

5 2 1 3

45 1

4 6 4

2 6

10 5

14 7

11 5


QEM - Chapter 1

Iteration:

in each iteration fix as built the location from Iovl, with the largest total saving:

},{11 kII Iovl = Io

vl – {k} and Z = Z – ωk + fk

Fild potential location k from Iovl, where saving in transportation cost minus

additional fixed cost ωk – fk is maximum.

Also, forbid all locations (finally) where saving in transportation cost are smaller than additional fixed costs

Update the savings in transpotrtation cost for all locations Iovl and all customers j :

ωij = max {ωij - ωkj, 0}

vlIi 0For all with ωi ≤ fi : }{ }{ 0000 iIIandiII vlvl


QEM - Chapter 1

Stiopping criterion: Stiop if no more cost saving are possible by additional locations from Io

vl

Build locationd from set I1.

Total cost Z assignment: xij = 1 iff

} { min 1 Ihcc hjij

i\j 1 2 3 4 5 6 7 ωi fi

1

2

3

4

Fix k = 2

Forbid i = 4

Beispiel: Iteration 1

Because of ω4 < f4 location 4 is forbidden finally. Location k=2 is built.

Now Z = 39 – 7 = 32 and Iovl = {1,3}, I1 = {2,5}, Io = {4}.

Update savings ωij.

4 6 4

1 1

5 2 1 3

5 4 1

11 5

14 7

10 5

2 6


QEM - Chapter 1

Iteration 2:

Location 3 is forbidden, location k = 1 is finally built.

i\j 1 2 3 4 5 6 7 ωi fi

1

3

Fix k = 1

Forbid i = 3

1 2 3

1

6 5

1 5

Ergebnis:

Final solution I1 = {1,2,5}, Io = {3,4} and Z = 32 – 1 = 31.

Build locations 1, 2 and 5 Customers {1,2,7} are delivered from location 1, {3,5} from location 2, and {4,6}

from location 5. Total cost Z = 31.


QEM - Chapter 1

DROP for WLP

Die Set Iovl is replaced by I1

vl.

I1vl set of preliminarily built locations (yi =1 tentatively)

DROP works the other way round compared to ADD, i.e. start with all locations temporarily built; in each iteration remove one location…

Initialisation: I1vl = I, I0

= I1 = { }

Iteration In each Iteration delete that location from I1

vl (finally), which reduces total cost most.

If deleting would let total cost increase, fix this location as built


QEM - Chapter 1

Expand matrix C: Row m+1 (row m+2) contains smallest ch1j (second smallest ch2j)

only consider locations not finally deleted → 0 Ii

i\j 1 2 3 4 5 6 7 δi fi

1 1 2 10 9 6 7 3 5

2 2 9 0 7 3 6 10 7

3 7 6 1 5 3 10 5 5

4 6 5 10 2 6 3 6 6

5 6 4 6 3 7 2 6 5

ch1j 6

ch2j 7

h1 8

h2 9

build

delete

Example: Initialisation and Iteration 1: I1vl ={1,2,3,4,5}

Row m+3 (row m+4) contains row number h1 (and h2) where smallest (second smallest) cost elemet occurs. If location h1 (from I1

vl) is dropped, transportation cost for customer Kunden j increase by ch2j - ch1j

5

1

0

1

1

2

1

4

5

0

2

1

3

2

4

3

5

3

2

3

3

2

5

3

4

3

1

5

32

2

1

1


QEM - Chapter 1

For all i from I1vl compute increase in transportation cost δi if I is finally dropped. δi

is sum of differences between smallest and second smallest cost element in rows where i = h1 contains the smallest element.

Iteration 2: I1

vl = {3,4,5}, I1 = {1}, I0

= {2}

Omit row 2 because finally dropped. Update remaining 4 rows, where changes are only possible where smallest or second smallest element occurred

Keep row 1 since I1 = {1}, but 1 is no candidate for dropping. Hence do not

compute δi there.

2 examples:

If fixed costs savings fi exceed additional transportation cost δi, finally drop i. In Iteration 1 location 1 is finally built.

δ1 = (c21 – c11) + (c52 – c12) + (c37 – c17) = 5 δ2 = (c33 – c23) + (c35 – c25) = 1


QEM - Chapter 1

i\j 1 2 3 4 5 6 7

1 1 2 10 9 6 7 3

3 7 6 1 5 3 10 5

4 6 5 10 2 6 3 6

5 6 4 6 3 7 2 6

δi fi

-

5

6

5

build

forbid

Location 3 is finally built, location 4 finally dropped.

ch1j

ch2j

h1

h2

2

1

4

5

1

3

6

5

2

4

3

5

3

3

6

1

2

5

3

4

3

1

5

34

6

1

1

-

8

1

1


QEM - Chapter 1

Iteration 3: I1

vl = {5}, I1 = {1,3}, I0 = {2,4}

Location 5 is finally built

i\j 1 2 3 4 5 6 7

1 1 2 10 9 6 7 3

3 7 6 1 5 3 10 5

5 6 4 6 3 7 2 6

δi fi

-

-

5 build

ch1j

ch2j

h1

h2

2

1

4

5

1

3

6

5

3

5

5

3

3

3

6

1

2

5

7

1

3

1

5

35

6

1

1

-

-

7


QEM - Chapter 1

Result:

Build locations I1 = {1,3,5}

Deliver customers {1,2,7} from 1, customers {3,5} from 3, and customers {4,6} from 5.

Total cost Z = 30 (slightly better than ADD – can be the other way round)


QEM - Chapter 1

Improvement for WLP

In each iteration you can do:

Replace a built location (from I1) by a forbidden location (from I0). Choose first improvement of best improvement

Using rules of DROP-Algorithm delete 1 or more locations, so that cost decrease most (or increase least) and then apply ADD as long as cost savings are possible.

Using rules of ADD-Algorithm add 1 or more locations, so that cost decrease most (or increase least) and then apply DROP as long as cost savings are possible.


P-Median

Number of facilities is fixed … p Typically fixed costs are not needed (but can be

considered if not uniform)

QEM - Chapter 1 28(c) Prof. Richard F. Hartl

QEM - Chapter 1

MIP for p-Median

min ),(1 1

m

i

n

j

ijijxcyxZ

miijx11

transportation cost+ fixed cost

Delivery only from locations i that are built

Satisfy total demand of customer j

yi is binary

xij non negative

xij ≤ yi

i = 1, …, m

j = 1, …,n

i = 1, …, mFor all i and j0

}1,0{

ij

i

x

y

j = 1, …,n

m

i

ijx1

1


Exactly p facilities

m

ii py

1

Chapter 1 Strategic Problems: Location. QEM - Chapter 1 Location problems Full truck loadtransportation … CW 1 CW 2 Central warehouse FTL or tourstransportation.

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