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Lecture 26 – Problem 11-19 Page 617

Military Problem with Three CommandersSix Radar StationsEach Commander Should Be Assigned Two Radar

Stations

C1 C2 C3

S1 S2 S3 S4 S5 S6

Cost = 42 M Cost = 68 M Cost = 47 M

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Another Feasible Assignment

C1 C2 C3

S1 S3 S2 S5 S4 S6

Cost = 65 M Cost = 40 M Cost = 39 M

Total Cost For This Assignment is

65+40+39 = 144

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Cost Matrix In 000,000s

Symmetrical – See Page 617

1 2 3 4 5 6

1 - 42 65 29 31 55

2 - 20 39 40 21

3 - 68 55 22

4 - 30 39

5 - 47

6 -

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The Data File – See p 617

set N := 1 2 3 4 5 6;param cost :=1 2 421 3 651 4 291 5 311 6 552 3 202 4 392 5 402 6 213 4 683 5 553 6 224 5 304 6 395 6 47 ;

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AMPL Model – Page 1

set N; # the set of radar stationsparam cost{N,N} default 0; # cost[i,j] denotes the construction cost for # linking i and jdata data26.txt;display N;display cost;set N1;let N1 := N diff {1};display N1;

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AMPL Model – Page 2

# Since the cost matrix is symmetrical, fill in bottom half

for {i in N1} for {j in N: j < i} let cost[i,j] := cost[j,i];

display cost;

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AMPL Model - Page 3

var x{N,N} binary; # x[i,j] = 1, if i and j are a pair # = 0, otherwise

subject to OnePerRow {i in N}: sum {j in N: i<>j} x[i,j] = 1;

subject to OnePerCol {j in N}: sum {i in N: i<>j} x[i,j] = 1;

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AMPL Model – Page 4

subject to Symmetry {i in N, j in N}: x[i,j] = x[j,i];

minimize TotalCost: sum{i in N, j in N} cost[i,j]*x[i,j];

expand OnePerRow, OnePerCol, Symmetry, TotalCost;

solve;

display x;

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Output – Page 1

set N := 1 2 3 4 5 6;

cost :=1 2 421 3 651 4 291 5 311 6 552 3 202 4 392 5 402 6 213 4 683 5 553 6 224 5 304 6 395 6 47;

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Output – Page 2

set N1 := 2 3 4 5 6;

cost [*,*]: 1 2 3 4 5 6 :=1 . 42 65 29 31 552 42 . 20 39 40 213 65 20 . 68 55 224 29 39 68 . 30 395 31 40 55 30 . 476 55 21 22 39 47 .;

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Output – Page 3

s.t. OnePerRow[1]:x[1,2] + x[1,3] + x[1,4] + x[1,5] + x[1,6] = 1;

s.t. OnePerRow[2]:x[2,1] + x[2,3] + x[2,4] + x[2,5] + x[2,6] = 1;

s.t. OnePerRow[3]:x[3,1] + x[3,2] + x[3,4] + x[3,5] + x[3,6] = 1;

s.t. OnePerRow[4]:x[4,1] + x[4,2] + x[4,3] + x[4,5] + x[4,6] = 1;

s.t. OnePerRow[5]:x[5,1] + x[5,2] + x[5,3] + x[5,4] + x[5,6] = 1;

s.t. OnePerRow[6]:x[6,1] + x[6,2] + x[6,3] + x[6,4] + x[6,5] = 1;

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Output – Page 4

s.t. OnePerCol[1]:x[2,1] + x[3,1] + x[4,1] + x[5,1] + x[6,1] = 1;

s.t. OnePerCol[2]:x[1,2] + x[3,2] + x[4,2] + x[5,2] + x[6,2] = 1;

s.t. OnePerCol[3]:x[1,3] + x[2,3] + x[4,3] + x[5,3] + x[6,3] = 1;

s.t. OnePerCol[4]:x[1,4] + x[2,4] + x[3,4] + x[5,4] + x[6,4] = 1;

s.t. OnePerCol[5]:x[1,5] + x[2,5] + x[3,5] + x[4,5] + x[6,5] = 1;

s.t. OnePerCol[6]:x[1,6] + x[2,6] + x[3,6] + x[4,6] + x[5,6] = 1;

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Output – Page 5

s.t. Symmetry[1,1]:0 = 0;

s.t. Symmetry[1,2]:x[1,2] - x[2,1] = 0;

s.t. Symmetry[1,3]:x[1,3] - x[3,1] = 0;

s.t. Symmetry[1,4]:x[1,4] - x[4,1] = 0;

s.t. Symmetry[1,5]:x[1,5] - x[5,1] = 0;

s.t. Symmetry[1,6]:x[1,6] - x[6,1] = 0;

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Output – Page 6

s.t. Symmetry[6,1]:-x[1,6] + x[6,1] = 0;

s.t. Symmetry[6,2]:-x[2,6] + x[6,2] = 0;

s.t. Symmetry[6,3]:-x[3,6] + x[6,3] = 0;

s.t. Symmetry[6,4]:-x[4,6] + x[6,4] = 0;

s.t. Symmetry[6,5]:-x[5,6] + x[6,5] = 0;

s.t. Symmetry[6,6]:0 = 0;

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Output – Page 7

minimize TotalCost:

42*x[1,2] + 65*x[1,3] + 29*x[1,4] + 31*x[1,5] + 55*x[1,6] + 42*x[2,1]

+ 20*x[2,3] + 39*x[2,4] + 40*x[2,5] + 21*x[2,6] + 65*x[3,1] +

20*x[3,2] + 68*x[3,4] + 55*x[3,5] + 22*x[3,6] + 29*x[4,1] +39*x[4,2]

+ 68*x[4,3] + 30*x[4,5] + 39*x[4,6] + 31*x[5,1] + 40*x[5,2] +

55*x[5,3] + 30*x[5,4] + 47*x[5,6] + 55*x[6,1] + 21*x[6,2] + 22*x[6,3]

+ 39*x[6,4] + 47*x[6,5];

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Output – Page 8

CPLEX 8.0.0: optimal integer solution; objective 1808 MIP simplex iterations0 branch-and-bound nodesx [*,*]: 1 2 3 4 5 6 :=1 0 0 0 0 1 02 0 0 1 0 0 03 0 1 0 0 0 04 0 0 0 0 0 15 1 0 0 0 0 06 0 0 0 1 0 0;

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Optimal Assignment

C1 C2 C3

S1 S5 S2 S3 S4 S6

Cost = 31 M Cost = 20 M Cost = 39 M

Total Cost For This Assignment is

31+20+39 = 90 – Why did CPLEX give 180?