数学建模中选址问题(Lingo程序)

P94，例3.4 选址问题

目录

题目 (1)

第一步，旧址基础上只求运量的ＬＰ程序 (1)

第二步，旧址基础上选择新址的NLP程序 (2)

题目

6个工地的地址(坐标表示，距离单位KM)及水泥用量(单位：吨)如下表，而在P(5,1)及Q(2,7)处有两个临时料场，日储量各有20t，如何安排运输，可使总的吨公里数最小?

新料场应选何处？能节约多少吨公里数？

第一步，旧址基础上只求运量的ＬＰ程序

MODEL:

Title Location Problem;

sets:

demand/1..6/:a,b,d;

supply/1..2/:x,y,e;

link(demand,supply):c;

endsets

data:

!locations for the demand(需求点的位置);

a=1.25,8.75,0.5,5.75,3,7.25;

b=1.25,0.75,4.75,5,6.5,7.75;

!quantities of the demand and supply（供需量）;

d=3,5,4,7,6,11; e=20,20;

x,y=5,1,2,7;

enddata

init:

!initial locations for the supply（初始点）;

endinit

!Objective function（目标）;

[OBJ] min=@sum(link(i,j): c(i,j)*((x(j)-a(i))^2+(y(j)-b(i))^2)^(1/2) );

!demand constraints（需求约束）;

@for(demand(i):[DEMAND_CON] @sum(supply(j):c(i,j)) =d(i););

!supply constraints（供应约束）;

@for(supply(i):[SUPPL Y_CON] @sum(demand(j):c(j,i)) <=e(i); );

!@for(supply: @free(x);!@free(Y);!);

@for(supply: @bnd(0.5,X,8.75); @bnd(0.75,Y,7.75); );

END

运行可得到全局最优解

Global optimal solution found.

Objective value: 136.2275

Total solver iterations: 1

Model Title: Location Problem

Variable Value Reduced Cost

X( 1) 5.000000 0.000000

X( 2) 2.000000 0.000000

Y( 1) 1.000000 0.000000

Y( 2) 7.000000 0.000000

E( 1) 20.00000 0.000000

E( 2) 20.00000 0.000000

第二步，旧址基础上选择新址的NLP程序

!选新址的ＮＬＰ程序;

MODEL:

Title Location Problem;

sets:

demand/1..6/:a,b,d;

supply/1..2/:x,y,e;

link(demand,supply):c;

endsets

data:

!locations for the demand(需求点的位置);

a=1.25,8.75,0.5,5.75,3,7.25;

b=1.25,0.75,4.75,5,6.5,7.75;

!quantities of the demand and supply（供需量）;

d=3,5,4,7,6,11; e=20,20;

enddata

init:

!initial locations for the supply（初始点）;

!x,y=5,1,2,7;

endinit

!Objective function（目标）;

[OBJ] min=@sum(link(i,j): c(i,j)*((x(j)-a(i))^2+(y(j)-b(i))^2)^(1/2) );

!demand constraints（需求约束）;

@for(demand(i):[DEMAND_CON] @sum(supply(j):c(i,j)) =d(i););

!supply constraints（供应约束）;

@for(supply(i):[SUPPL Y_CON] @sum(demand(j):c(j,i)) <=e(i); );

!@for(supply: @free(x);!@free(Y);!);

@for(supply: @bnd(0.5,X,8.75); @bnd(0.75,Y,7.75); );

END

求解结果

只得到局部最优解

Local optimal solution found.

Objective value: 89.88347

Total solver iterations: 67

Model Title: Location Problem

Variable Value Reduced Cost

X( 1) 5.695966 0.000000

X( 2) 7.250000 -0.3212138E-05

Y( 1) 4.928558 0.000000

Y( 2) 7.750000 -0.1009767E-05

如果不要初始数据，可能计算时间更长，本例的结果更优：

Local optimal solution found.

Objective value: 85.26604

Total solver iterations: 29

Model Title: Location Problem

Variable Value Reduced Cost

X( 1) 3.254883 0.000000

X( 2) 7.250000 -0.2958858E-05

Y( 1) 5.652332 0.000000

Y( 2) 7.750000 -0.1114154E-05

如果想求全局最优解，结果将会出现如下错误

版本限制，但会得到一个的局部最优解，结果与不要初始数据时算出的结果一样。

Local optimal solution found.

Objective value: 85.26604

Extended solver steps: 18

Total solver iterations: 16121

Model Title: Location Problem

Variable Value Reduced Cost

X( 1) 3.254883 0.000000

X( 2) 7.250000 0.8084079E-07

Y( 1) 5.652332 0.000000

Y( 2) 7.750000 0.2675276E-06