物资紧急调运问题(完稿1)(5)

2025-07-22

http://www.docin.com/p-487106051.html

附录

附录1:

Floyd算法代码:

function [d,r]=floyd(w) n=length(w); for i=1:n

for j=1:n;

d(i,j)=w(i,j); r(i,j)=j; end end

for k=1:n for i=1:n for j=1:n

if d(i,k)+d(k,j)

附录2:

求问题一中第一阶段的调运程序的代码及结果: MODEL: sets:

num_i/1..5/:a1,a2,a3,a4; num_j/1..2/:f;

link(num_i,num_j):cost,x; endsets

[OBJ]min=@sum(link(i,j):cost(i,j)*x(i,j)); @for(num_j(j):@sum(link(i,j):x(i,j))=f(j));

@for(num_i(i):a1(i)-@sum(link(i,j):x(i,j))>=a3(i)); @for(link(i,j):x(i,j)>=0); @for(link(i,j):@gin(x(i,j));); data:

cost=120,321.6, 157.6,177.6, 200.4,146.4,

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288,210, 110.4,152.4; f=1000,700;

a1=100000,100000,1000000,450,800; a3=0,0,0,200,300; enddata

END(源代码)

Global optimal solution found.

Objective value: 217680.0 Objective bound: 217680.0 Infeasibilities: 0.000000 Extended solver steps: 0 Total solver iterations: 0

Variable Value Reduced Cost A1( 1) 100000.0 0.000000 A1( 2) 100000.0 0.000000 A1( 3) 1000000. 0.000000 A1( 4) 450.0000 0.000000 A1( 5) 800.0000 0.000000 A2( 1) 0.000000 0.000000 A2( 2) 0.000000 0.000000 A2( 3) 0.000000 0.000000 A2( 4) 0.000000 0.000000 A2( 5) 0.000000 0.000000 A3( 1) 0.000000 0.000000 A3( 2) 0.000000 0.000000 A3( 3) 0.000000 0.000000 A3( 4) 200.0000 0.000000 A3( 5) 300.0000 0.000000 A4( 1) 0.000000 0.000000 A4( 2) 0.000000 0.000000 A4( 3) 0.000000 0.000000 A4( 4) 0.000000 0.000000 A4( 5) 0.000000 0.000000 F( 1) 1000.000 0.000000 F( 2) 700.0000 0.000000 COST( 1, 1) 120.0000 0.000000 COST( 1, 2) 321.6000 0.000000 COST( 2, 1) 157.6000 0.000000 COST( 2, 2) 177.6000 0.000000

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COST( 3, 1) 200.4000 0.000000 COST( 3, 2) 146.4000 0.000000 COST( 4, 1) 288.0000 0.000000 COST( 4, 2) 210.0000 0.000000 COST( 5, 1) 110.4000 0.000000 COST( 5, 2) 152.4000 0.000000 X( 1, 1) 500.0000 120.0000 X( 1, 2) 0.000000 321.6000 X( 2, 1) 0.000000 157.6000 X( 2, 2) 0.000000 177.6000 X( 3, 1) 0.000000 200.4000 X( 3, 2) 700.0000 146.4000 X( 4, 1) 0.000000 288.0000 X( 4, 2) 0.000000 210.0000 X( 5, 1) 500.0000 110.4000 X( 5, 2) 0.000000 152.4000

Row Slack or Surplus Dual Price OBJ 217680.0 -1.000000 2 0.000000 0.000000 3 0.000000 0.000000 4 99500.00 0.000000 5 100000.0 0.000000 6 999300.0 0.000000 7 250.0000 0.000000 8 0.000000 0.000000 9 500.0000 0.000000 10 0.000000 0.000000 11 0.000000 0.000000 12 0.000000 0.000000 13 0.000000 0.000000 14 700.0000 0.000000 15 0.000000 0.000000 16 0.000000 0.000000 17 500.0000 0.000000 18 0.000000 0.000000

求问题一中第二阶段的调运程序的代码及结果: MODEL: sets:

num_i/1..4/:a1,a2,a3,a4; num_j/1..7/:f;

link(num_i,num_j):cost,x; endsets

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[OBJ]min=@sum(link(i,j):cost(i,j)*x(i,j)); @for(num_j(j):@sum(link(i,j):x(i,j))=f(j));

@for(num_i(i):a1(i)-@sum(link(i,j):x(i,j))>=a3(i)); @for(link(i,j):x(i,j)>=0); @for(link(i,j):@gin(x(i,j));); data:

cost=184.8,150,230.4,156,344.4,256.8,372, 69.6,188.4,189.6,247.2,303.6,141.6,331.2, 268.8,398.4,90,404.4,174,196.8,111.6, 356.4,486,177.6,492,321.6,284.4,199.2; f=300,330,100,120,170,110,100; a1=100000,100000,100000,450; a3=0,0,0,300; enddata

END(源代码)

Global optimal solution found.

Objective value: 154416.0 Objective bound: 154416.0 Infeasibilities: 0.000000 Extended solver steps: 0 Total solver iterations: 0

Variable Value Reduced Cost A1( 1) 100000.0 0.000000 A1( 2) 100000.0 0.000000 A1( 3) 100000.0 0.000000 A1( 4) 450.0000 0.000000 A2( 1) 0.000000 0.000000 A2( 2) 0.000000 0.000000 A2( 3) 0.000000 0.000000 A2( 4) 0.000000 0.000000 A3( 1) 0.000000 0.000000 A3( 2) 0.000000 0.000000 A3( 3) 0.000000 0.000000 A3( 4) 300.0000 0.000000 A4( 1) 0.000000 0.000000 A4( 2) 0.000000 0.000000 A4( 3) 0.000000 0.000000 A4( 4) 0.000000 0.000000 F( 1) 300.0000 0.000000 F( 2) 330.0000 0.000000

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F( 3) 100.0000 0.000000 F( 4) 120.0000 0.000000 F( 5) 170.0000 0.000000 F( 6) 110.0000 0.000000 F( 7) 100.0000 0.000000 COST( 1, 1) 184.8000 0.000000 COST( 1, 2) 150.0000 0.000000 COST( 1, 3) 230.4000 0.000000 COST( 1, 4) 156.0000 0.000000 COST( 1, 5) 344.4000 0.000000 COST( 1, 6) 256.8000 0.000000 COST( 1, 7) 372.0000 0.000000 COST( 2, 1) 69.60000 0.000000 COST( 2, 2) 188.4000 0.000000 COST( 2, 3) 189.6000 0.000000 COST( 2, 4) 247.2000 0.000000 COST( 2, 5) 303.6000 0.000000 COST( 2, 6) 141.6000 0.000000 COST( 2, 7) 331.2000 0.000000 COST( 3, 1) 268.8000 0.000000 COST( 3, 2) 398.4000 0.000000 COST( 3, 3) 90.00000 0.000000 COST( 3, 4) 404.4000 0.000000 COST( 3, 5) 174.0000 0.000000 COST( 3, 6) 196.8000 0.000000 COST( 3, 7) 111.6000 0.000000 COST( 4, 1) 356.4000 0.000000 COST( 4, 2) 486.0000 0.000000 COST( 4, 3) 177.6000 0.000000 COST( 4, 4) 492.0000 0.000000 COST( 4, 5) 321.6000 0.000000 COST( 4, 6) 284.4000 0.000000 COST( 4, 7) 199.2000 0.000000 X( 1, 1) 0.000000 184.8000 X( 1, 2) 330.0000 150.0000 X( 1, 3) 0.000000 230.4000 X( 1, 4) 120.0000 156.0000 X( 1, 5) 0.000000 344.4000 X( 1, 6) 0.000000 256.8000 X( 1, 7) 0.000000 372.0000 X( 2, 1) 300.0000 69.60000 X( 2, 2) 0.000000 188.4000 X( 2, 3) 0.000000 189.6000 X( 2, 4) 0.000000 247.2000

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