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群组: LINGO |
MODEL:
title economic pipe;
sets:
tubulation/1..9/:L,Q,A,EK;
caliber/1..5/;
LINK(tubulation,caliber):C,X,D;
ENDSETS
DATA:
!各段管长;
L=700,700,700,500,500,500,500,500,500;
!各段流量;
Q=0.135,0.060,0.015,0.015,0.045,0.015,0.015,0.015,0.015;
!节点服务水压标高;
EK=47,46,44,46,44,43,45,46,43;
!水源到节点的管段数;
A=1,2,3,2,2,3,3,3,3;
!标准管径单价;
C=42.0 55.2 64.0 83.2 103.2
38.7 42.0 55.2 64.0
25.0 30.8
25.0 30.8
30.8 38.7 42.0 55.2
25.0 30.8
25.0 30.8
25.0 30.8
25.0 30.8;
!标准管径;
D=0.25 0.30 0.35 0.40 0.50
0.20 0.25 0.30 0.35
0.1 0.15
0.1 0.15
0.15 0.20 0.25 0.30
0.1 0.15
0.1 0.15
0.1 0.15
0.1 0.15;
ENDDATA
!OBJECTIVE FUNCTION(目标函数);
[OBJ]MIN=@SUM(LINK(I,J):C(I,J)*X(I,J));
!pipe range CONSTRAINTE(管长约束);
@FOR(tubulation(I):[RANGE_CON]@SUM(caliber(J):X(I,J))=L(I););
!Stress constraint(压力约束);
@FOR(caliber(J):(@SUM(tubulation(I):(0.135^2*X(1,J)/D(1,J)^5.342))*(1.942*10^-3))<(60-47));
@FOR(CALIBER(J):(@SUM(TUBULATION(I):(0.135^2*X(1,J)/D(1,J)^5.342+0.060^2*X(2,J)/D(2,J)^5.342))*(1.942*10^-3))<(60-46));
@for(caliber(j):(@SUM(TUBULATION(I):(0.135^2*X(1,J)/D(1,J)^5.342+0.060^2*X(2,J)/D(2,J)^5.342+0.015^2*X(3,J)/D(3,J)^5.342))*(1.942*10^-3))<(60-44));
@for(caliber(j):(@SUM(TUBULATION(I):(0.135^2*X(1,J)/D(1,J)^5.342+0.015^2*X(4,J)/D(4,J)^5.342))*(1.942*10^-3))<(60-46));
@FOR(CALIBER(J):(@SUM(TUBULATION(I):(0.135^2*X(1,J)/D(1,J)^5.342+0.045^2*X(5,J)/D(5,J)^5.342))*(1.942*10^-3))<(60-44));
@for(caliber(j):(@SUM(TUBULATION(I):(0.135^2*X(1,J)/D(1,J)^5.342+0.045^2*X(5,J)/D(5,J)^5.342+0.015^2*X(6,J)/D(6,J)^5.342))*(1.942*10^-3))<(60-43));
@for(caliber(j):(@SUM(TUBULATION(I):(0.135^2*X(1,J)/D(1,J)^5.342+0.045^2*X(5,J)/D(5,J)^5.342+0.015^2*X(7,J)/D(7,J)^5.342))*(1.942*10^-3))<(60-45));
@for(caliber(j):(@SUM(TUBULATION(I):(0.135^2*X(1,J)/D(1,J)^5.342+0.060^2*X(2,J)/D(2,J)^5.342+0.015^2*X(8,J)/D(8,J)^5.342))*(1.942*10^-3))<(60-46));
@for(caliber(j):(@SUM(TUBULATION(I):(0.135^2*X(1,J)/D(1,J)^5.342+0.060^2*X(2,J)/D(2,J)^5.342+0.015^2*X(9,J)/D(9,J)^5.342))*(1.942*10^-3))<(60-43));
!速度要求;
!非负假定;
@FOR(LINK(I,J):X(I,J)>0;);
这个程序是有些问题的,求大神帮我解答!!!
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