p-median model facility location

8/10/2019 p-median model facility location

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In the p-median problem we are interested infinding the location ofpfacilities to serve demand

nodes so that the transportation cost is minimized.

The transportation cost is given by the product

of the demand at the demand node and the

distance between the demand node and the facility

that serves the demand node.

There are no capacity constraints at the

facilities.

The P-median Problem


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A B C D

E F

G IH

J LK

15 22 18

12

18

2222 25

19 21

1919

30

24

20

12

1624

25

1215

15 10 12 18

11 16

22 19 20

13

245



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As there are no capacity constraints at the facilities,

it is optimal to satisfy the demand at a demand node

from a single facility.

An optimal solution can be found by restricting the

search to the demand nodes.

Fixed costs are assumed to be equal.

The number of possible solutions is


Remarks

!!!

pnp

n


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For even moderate values of nandpthe number of possible

solutions can be very large:

For example, if n=20 andp=5, there are 15,504 solutions.

If n=50 andp=10, the problem has more than 1010possible

solutions.

If you could evaluate 1 million solutions per second, it willtake you 3 hours for total enumeration.


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Inputs:hi = demand at node i

dij= distance between demand node iand sitej

p = number of facilities

Decision variables:

otherwise0

siteatfacilityabyservedisnodeatdemandif1

otherwise0

siteatlocatedisfacilityaif1

jiY

jX

ij

j



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jX

jiY

jiXY

pX

iY

Ydh

j

ij

jij

j

j

j

ij

i j

ijiji

allfor1,0

,allfor1,0

,allfor

allfor1subject to

min



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Distance matrix for the network

Node i A B C D E F G H I J K L

A 0 15 37 55 24 60 18 33 48 40 58 67

B 15 0 22 40 38 52 33 48 42 55 61 61

C 37 22 0 18 16 30 41 28 20 58 39 39D 55 40 18 0 34 12 59 46 24 62 43 34

E 24 38 16 34 0 36 25 12 24 47 37 43

F 60 52 30 12 36 0 57 42 12 50 31 22

G 18 33 41 59 25 57 0 15 45 22 40 61

H 33 48 28 46 12 42 15 0 30 37 25 46

I 48 42 20 24 24 12 45 30 0 38 19 19

J 40 55 58 62 47 50 22 37 38 0 19 40

K 58 61 39 43 37 31 40 25 19 19 0 21

L 67 61 39 34 43 22 61 46 19 40 21 0

Site j


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Cost (demand x distance) matrix for the network

Demand

Node i A B C D E F G H I J K L

A 0 225 555 825 360 900 270 495 720 600 870 1005

B 150 0 220 400 380 520 330 480 420 550 610 610

C 444 264 0 216 192 360 492 336 240 696 468 468D 990 720 324 0 612 216 1062 828 432 1116 774 612

E 120 190 80 170 0 180 125 60 120 235 185 215

F 1440 1248 720 288 864 0 1368 1008 288 1200 744 528

G 198 363 451 649 275 627 0 165 495 242 440 671

H 528 768 448 736 192 672 240 0 480 592 400 736

I 624 546 260 312 312 156 585 390 0 494 247 247

J 880 1210 1276 1364 1034 1100 484 814 836 0 418 880

K 1102 1159 741 817 703 589 760 475 361 361 0 399

L 1340 1220 780 680 860 440 1220 920 380 800 420 0

Site j

hix dij


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Heuristic Algorithm


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A B C D E F G H I J K L

A 0 225 555 825 360 900 270 495 720 600 870 1005

B 150 0 220 400 380 520 330 480 420 550 610 610

C 444 264 0 216 192 360 492 336 240 696 468 468

D 990 720 324 0 612 216 1062 828 432 1116 774 612

E 120 190 80 170 0 180 125 60 120 235 185 215

F 1440 1248 720 288 864 0 1368 1008 288 1200 744 528G 198 363 451 649 275 627 0 165 495 242 440 671

H 528 768 448 736 192 672 240 0 480 592 400 736

I 624 546 260 312 312 156 585 390 0 494 247 247

J 880 1210 1276 1364 1034 1100 484 814 836 0 418 880

K 1102 1159 741 817 703 589 760 475 361 361 0 399

L 1340 1220 780 680 860 440 1220 920 380 800 420 0

Total 7816 7913 5855 6457 5784 5760 6936 5971 4772 6886 5576 6371

serving all the demand from a facility at A


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If we locate only one facility, it should be at site I with a

total cost of 4772.

What if we want to open a second facility?

A B ... I

A 0 225 720B 150 0 420

C 444 264 240

D 990 720 432

E 120 190 120

F 1440 1248 288

G 198 363 495

H 528 768 480

I 624 546 0

J 880 1210 836

K 1102 1159 361

L 1340 1230 380

A B

A 0 225B 150 0

C 240 240

D 432 432

E 120 120

F 288 288

G 198 363

H 480 480

I 0 0

J 836 836

K 361 361

L 380 380


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A B C D E F G H I J K L

A 0 225 555 720 360 720 270 495 720 600 720 720

B 150 0 220 400 380 420 330 420 420 420 420 420

C 240 240 0 216 192 240 240 240 240 240 240 240

D 432 432 324 0 432 216 432 432 432 432 432 432

E 120 120 80 120 0 120 120 60 120 120 120 120

F 288 288 288 288 288 0 288 288 288 288 288 288G 198 363 451 495 275 495 0 165 495 242 440 495

H 480 480 448 480 192 480 240 0 480 480 400 480

I 0 0 0 0 0 0 0 0 0 0 0 0

J 836 836 836 836 836 836 484 814 836 0 418 836

K 361 361 361 361 361 361 361 361 361 361 0 361

L 380 380 380 380 380 380 380 380 380 380 380 0

Total 3485 3725 3943 4296 3696 4268 3145 3655 4772 3563 3858 4392

It is best to open the second facility at site G.


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Node i A B C D E F G H I J K LA 0 15 37 55 24 60 18 33 48 40 58 67

B 15 0 22 40 38 52 33 48 42 55 61 61

C 37 22 0 18 16 30 41 28 20 58 39 39

D 55 40 18 0 34 12 59 46 24 62 43 34

E 24 38 16 34 0 36 25 12 24 47 37 43

F 60 52 30 12 36 0 57 42 12 50 31 22

G 18 33 41 59 25 57 0 15 45 22 40 61

H 33 48 28 46 12 42 15 0 30 37 25 46

I 48 42 20 24 24 12 45 30 0 38 19 19

J 40 55 58 62 47 50 22 37 38 0 19 40

K 58 61 39 43 37 31 40 25 19 19 0 21

L 67 61 39 34 43 22 61 46 19 40 21 0

Site j

Assign nodes A,B,E* to facility AD,F F

G,H G

C,I,K*,L I

J J


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A B C D

E F

G IH

J LK

15 22 18

12

18

22

22 25

19 21

1919

30

24

20

12

1624

25

1215

15 10 12 18

1116

22 1920

13

24

5


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Note that the optimal solution had a total cost of 1444whereas the heuristic solution provides a total cost of 1707.

This means that we can improve the solution obtained by

the heuristic. Therefore we apply an improvement

algorithm.

For each facility site, this algorithm identifies the set of

demand nodes that are served from the facility called as the

neighborhood.

Within each neighborhood, the optimal 1-median is found.


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A B C D

E F

G IH

J LK

15 22 18

12

18

22

22 25

19 21

1919

30

24

20

12

1624

25

1215

15 10 12 18

1116

2219 20

13

24

5

change=-75


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A B C D

E F

G IH

J LK

15 22 18

12

18

22

22 25

19 21

1919

30

24

20

12

1624

25

1215

15 10 12 18

1116

22 19 20

13

245

change=-60


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The new solution has a cost of 1707-135=1572.

Because of a limitation in this heuristic, some relocations

which are beneficial in a global (but not local or

neighborhood) sense are not considered.

This leads us to consider an exchange algorithm as an

improvement procedure.


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A B C D

E F

G IH

J LK

15 22 18

12

18

22

22 25

19 21

1919

30

24

20

12

1624

25

1215

15 10 12 18

1116

22 19 20

13

245


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Summary of Exchange Opportunities from NeighborhoodSearch Solution (Facilities at sitesA,F,H,IandJwith a

total cost of 1572)

Remove Replace Total ChangeWith Cost

A B 1647 75

F D 1620 48

H E 1689 117

I L 1444 -128

J K 1629 57


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A* B C D

E F*

G IH*

J* L*K

15 22 18

12

18

22

22 25

19 21

1919

30

24

20

12

1624

25

1215

15 10 12 18

11 16

22 19 20

13

245


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Summary of Exchange Opportunities from ImprovedSolution (Facilities at sitesA,F,H,JandLwith a total

cost of 1444)

Remove Replace Total ChangeWith Cost

A B 1447 3

F D 1487 43

H E 1465 21J K 1501 57

L K 1503 59

p-median model facility location

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