P中值选址问题的整数规划求解

一 、P-中值问题

p-中值选址问题是一个常见的选址问题. 问题是给定I个需求结点和J个待选设施地点, 要求选择p个地点建立设施, 使得运输成本最低. 下面是个英文的问题详细描述:

P中值选址问题的整数规划求解-LMLPHP

二、整数规划模型

p-中值选址问题的整数规划模型如下

P中值选址问题的整数规划求解-LMLPHP

P中值选址问题的整数规划求解-LMLPHP

P中值选址问题的整数规划求解-LMLPHP

三、当给定距离时候的求解

假设给定了设施-需求之间的距离矩阵,照着上面的模型即可写出Leapms模型。

//==========================================================
//The P-Median Problem
//==========================================================
min sum{j=,...,J;i=,...,I}h[i]d[i][j]y[i][j]
subject to
sum{j=,...,J}y[i][j]=|i=,...,I
y[i][j]-x[j]<=|i=,...,I;j=,...,J
sum{j=,...,J}x[j]=p
where
I,J are numbers
p is a number
h is a set
d[i][j] is a number | i=,...,I;j=,...,J
x[j] is a variable of binary|j=,...,J
y[i][j] is a variable of binary|i=,...,I;j=,...,J data
I=
J=
p=
h={3.1,2.2,4.3,5.0,7.1,8.0,9.9,3.4,4.3,3.3}
d={ }

把上面的模型保存为pemedian1.leap

在leapms中求解: 依次输入load, pmedian1, mip即可求解

P中值选址问题的整数规划求解-LMLPHP

。。。 。。。

P中值选址问题的整数规划求解-LMLPHP

四、当给定设施和需求点坐标时候的求解

与上面类似,不过设施和需求点的距离需要用data_relation 段用公式表达出来,leapms表达的模型如下

//==========================================================
//The P-Median Problem
//==========================================================
min sum{j=1,...,J;i=1,...,I}h[i]d[i][j]y[i][j] subject to sum{j=1,...,J}y[i][j]=1|i=1,...,I
y[i][j]-x[j]<=0|i=1,...,I;j=1,...,J
sum{j=1,...,J}x[j]=p where
I,J are numbers
p is a number
h is a set cdx1[i],cdy1[i] are numbers|i=1,...,I
cdx2[j],cdy2[j] are numbers|j=1,...,J d[i][j] is a number | i=1,...,I;j=1,...,J
x[j] is a variable of binary|j=1,...,J
y[i][j] is a variable of binary|i=1,...,I;j=1,...,J data_relation d[i][j]=sqrt((cdx1[i]-cdx2[j])^2+(cdy1[i]-cdy2[j])^2)|-->
i=1,...,I;j=1,...,J
data
I=10
J=10 p=6
h={3.1, 2.2, 4.3, 5.0, 7.1, 8.0, 9.9, 3.4, 4.3, 3.3} cdx1={-9.7, 8.8,8.3,7.1,-6.8, 6.7,-5.8,0.7,7 , -6 }
cdy1={-8.3,-1.9,-0.1,-3, 4 , -4.1, 4.4,2.5,4.4, 8.8} cdx2={-1.9,-8.3,2.5,0.8,-2.2, 4.1,-1.2,9.2,-0.9,1}
cdy2={-6.6,-0.9,1 ,8.2, 0.5,-8.9,-2.2,9.5,-4 ,6}

把上面的模型保存为pemedian2.leap

在leapms中求解: 依次输入load, pmedian2, mip即可求解

P中值选址问题的整数规划求解-LMLPHP

P中值选址问题的整数规划求解-LMLPHP

05-11 18:01