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标题: 高手帮忙,看下这段程序!万分感激! [打印本页]

作者: ada0115    时间: 2008-5-23 13:53
标题: 高手帮忙,看下这段程序!万分感激!

我正在准备毕业论文,可是模型算法再LINGO中实现的时候一直有错误,我这是第一次用LINGO,所以正苦于找不到原因,请高手们帮帮忙!

MODEL:
Title Location Problem;
sets:
DC/1..4/:y,a,x;
vendors/1..7/:b,w;
link(DC,vendors):c,z,t;
endsets
data:
a=27.3 38.2 43.7 35.2;
b=25000 12000 6000 21000 8000 7000 3000;
c=4.5 7.5 10 8 16.8 28.7 39.6
23.6 21 23.8 28.5 18.3 9.7 16.4
29.3 27 41 32.7 13.8 8 11
15.8 19.2 19 19.3 7.8 11 17;
w=28 28 30 28 30 32 32;
t=23 24 24.5 24 26.5 30 33
32 31 31.5 33 30.5 28.5 29.5
34.5 33.5 37 35.5 30.5 29 30
29 30 30 30 27 27.5 29.5;
enddata
!Objective function(目标);
[OBJ] min=@sum(link(j,i): z(j,i)*c(j,i)*b(i))+@sum(DC(j):x(j)*a(j));
@for(link(j,i):z(j,i)*t(j,i)<w(i));
@for(vendors(i)sum(DC(j):z(j,i)*y(j))=1);
@for(DC(j)sum(DC(j):y(j))=2);
@for(DC(j)sum(vendors(i):b(i)*z(j,i))=x(j));
@for(DC(j)bin(y(j)));
@for(link(j,i)bin(z(j,i)));
END


作者: madio    时间: 2008-5-24 00:47

修改了,目前没有错误了,但是由于我的lingo是试用版不能运行这么大规模的程序,你自己试试结果对不对!

MODEL:
Title Location Problem;
sets:
DC/1..4/:y,a,x;
vendors/1..7/:b,w;
link(DC,vendors):c,z,t;
endsets
data:
a=27.3 38.2 43.7 35.2;
b=25000 12000 6000 21000 8000 7000 3000;
c=4.5 7.5 10 8 16.8 28.7 39.6
23.6 21 23.8 28.5 18.3 9.7 16.4
29.3 27 41 32.7 13.8 8 11
15.8 19.2 19 19.3 7.8 11 17;
w=28 28 30 28 30 32 32;
t=23 24 24.5 24 26.5 30 33
32 31 31.5 33 30.5 28.5 29.5
34.5 33.5 37 35.5 30.5 29 30
29 30 30 30 27 27.5 29.5;
enddata
!Objective function(目标);
[OBJ] min=@sum(link(j,i): z(j,i)*c(j,i)*b(i))+@sum(DC(j):x(j)*a(j));
@for(link(j,i):z(j,i)*t(j,i)<w(i));
@for(vendors(i)sum(DC(j):z(j,i)*y(j))=1);
@sum(DC(j):y(j))=2;
@for(DC(j)sum(vendors(i):b(i)*z(j,i))=x(j));
@for(DC(j)bin(y(j)));
@for(link(j,i)bin(z(j,i)));
END


作者: ada0115    时间: 2008-5-24 01:13

恩恩 谢谢了 我看过了 程序是没有错了

可是为什么会没有解呢?


作者: ada0115    时间: 2008-5-24 01:16
是不是约束条件不够 ,还是数据不够?
作者: madio    时间: 2008-5-24 09:05

都不是,可能是约束条件太严格了,求解困难,但是你也要找个完全版的lingo先看看结果!我这只有试用版!


作者: hubulwm    时间: 2008-5-24 14:23

Local optimal solution found.
Objective value: 3029600.
Extended solver steps: 0
Total solver iterations: 4


Model Title: Location Problem

Variable Value Reduced Cost
Y( 1) 1.000000 -2530500.
Y( 2) 1.000000 -499100.0
Y( 3) 0.000000 0.000000
Y( 4) 0.000000 0.000000
A( 1) 27.30000 0.000000
A( 2) 38.20000 0.000000
A( 3) 43.70000 0.000000
A( 4) 35.20000 0.000000
X( 1) 72000.00 0.000000
X( 2) 10000.00 0.000000
X( 3) 0.000000 0.000000
X( 4) 0.000000 0.000000
B( 1) 25000.00 0.000000
B( 2) 12000.00 0.000000
B( 3) 6000.000 0.000000
B( 4) 21000.00 0.000000
B( 5) 8000.000 0.000000
B( 6) 7000.000 0.000000
B( 7) 3000.000 0.000000
W( 1) 28.00000 0.000000
W( 2) 28.00000 0.000000
W( 3) 30.00000 0.000000
W( 4) 28.00000 0.000000
W( 5) 30.00000 0.000000
W( 6) 32.00000 0.000000
W( 7) 32.00000 0.000000
C( 1, 1) 4.500000 0.000000
C( 1, 2) 7.500000 0.000000
C( 1, 3) 10.00000 0.000000
C( 1, 4) 8.000000 0.000000
C( 1, 5) 16.80000 0.000000
C( 1, 6) 28.70000 0.000000
C( 1, 7) 39.60000 0.000000
C( 2, 1) 23.60000 0.000000
C( 2, 2) 21.00000 0.000000
C( 2, 3) 23.80000 0.000000
C( 2, 4) 28.50000 0.000000
C( 2, 5) 18.30000 0.000000
C( 2, 6) 9.700000 0.000000
C( 2, 7) 16.40000 0.000000
C( 3, 1) 29.30000 0.000000
C( 3, 2) 27.00000 0.000000
C( 3, 3) 41.00000 0.000000
C( 3, 4) 32.70000 0.000000
C( 3, 5) 13.80000 0.000000
C( 3, 6) 8.000000 0.000000
C( 3, 7) 11.00000 0.000000
C( 4, 1) 15.80000 0.000000
C( 4, 2) 19.20000 0.000000
C( 4, 3) 19.00000 0.000000
C( 4, 4) 19.30000 0.000000
C( 4, 5) 7.800000 0.000000
C( 4, 6) 11.00000 0.000000
C( 4, 7) 17.00000 0.000000
Z( 1, 1) 1.000000 0.000000
Z( 1, 2) 1.000000 0.000000
Z( 1, 3) 1.000000 0.000000
Z( 1, 4) 1.000000 0.000000
Z( 1, 5) 1.000000 0.000000
Z( 1, 6) 0.000000 56700.00
Z( 1, 7) 0.000000 36900.00
Z( 2, 1) 0.000000 750000.0
Z( 2, 2) 0.000000 292800.0
Z( 2, 3) 0.000000 148200.0
Z( 2, 4) 0.000000 659400.0
Z( 2, 5) 0.000000 99200.00
Z( 2, 6) 1.000000 0.000000
Z( 2, 7) 1.000000 0.000000
Z( 3, 1) 0.000000 1825000.
Z( 3, 2) 0.000000 848400.0
Z( 3, 3) 0.000000 508200.0
Z( 3, 4) 0.000000 1604400.
Z( 3, 5) 0.000000 460000.0
Z( 3, 6) 0.000000 361900.0
Z( 3, 7) 0.000000 164100.0
Z( 4, 1) 0.000000 1275000.
Z( 4, 2) 0.000000 652800.0
Z( 4, 3) 0.000000 325200.0
Z( 4, 4) 0.000000 1144500.
Z( 4, 5) 0.000000 344000.0
Z( 4, 6) 0.000000 323400.0
Z( 4, 7) 0.000000 156600.0
T( 1, 1) 23.00000 0.000000
T( 1, 2) 24.00000 0.000000
T( 1, 3) 24.50000 0.000000
T( 1, 4) 24.00000 0.000000
T( 1, 5) 26.50000 0.000000
T( 1, 6) 30.00000 0.000000
T( 1, 7) 33.00000 0.000000
T( 2, 1) 32.00000 0.000000
T( 2, 2) 31.00000 0.000000
T( 2, 3) 31.50000 0.000000
T( 2, 4) 33.00000 0.000000
T( 2, 5) 30.50000 0.000000
T( 2, 6) 28.50000 0.000000
T( 2, 7) 29.50000 0.000000
T( 3, 1) 34.50000 0.000000
T( 3, 2) 33.50000 0.000000
T( 3, 3) 37.00000 0.000000
T( 3, 4) 35.50000 0.000000
T( 3, 5) 30.50000 0.000000
T( 3, 6) 29.00000 0.000000
T( 3, 7) 30.00000 0.000000
T( 4, 1) 29.00000 0.000000
T( 4, 2) 30.00000 0.000000
T( 4, 3) 30.00000 0.000000
T( 4, 4) 30.00000 0.000000
T( 4, 5) 27.00000 0.000000
T( 4, 6) 27.50000 0.000000
T( 4, 7) 29.50000 0.000000

Row Slack or Surplus Dual Price
OBJ 3029600. -1.000000
2 5.000000 0.000000
3 4.000000 0.000000
4 5.500000 0.000000
5 4.000000 0.000000
6 3.500000 0.000000
7 32.00000 0.000000
8 32.00000 0.000000
9 28.00000 0.000000
10 28.00000 0.000000
11 30.00000 0.000000
12 28.00000 0.000000
13 30.00000 0.000000
14 3.500000 0.000000
15 2.500000 0.000000
16 28.00000 0.000000
17 28.00000 0.000000
18 30.00000 0.000000
19 28.00000 0.000000
20 30.00000 0.000000
21 32.00000 0.000000
22 32.00000 0.000000
23 28.00000 0.000000
24 28.00000 0.000000
25 30.00000 0.000000
26 28.00000 0.000000
27 30.00000 0.000000
28 32.00000 0.000000
29 32.00000 0.000000
30 0.000000 -795000.0
31 0.000000 -417600.0
32 0.000000 -223800.0
33 0.000000 -741300.0
34 0.000000 -352800.0
35 0.000000 -335300.0
36 0.000000 -163800.0
37 0.000000 0.000000
38 0.000000 27.30000
39 0.000000 38.20000
40 0.000000 43.70000
41 0.000000 35.20000


作者: ada0115    时间: 2008-5-25 18:57
谢谢了 万分感谢各位的帮忙!
作者: ada0115    时间: 2008-5-25 22:15
QUOTE:
以下是引用hubulwm在2008-5-24 14:23:22的发言:

Local optimal solution found.
Objective value: 3029600.
Extended solver steps: 0
Total solver iterations: 4


Model Title: Location Problem

Variable Value Reduced Cost
Y( 1) 1.000000 -2530500.
Y( 2) 1.000000 -499100.0
Y( 3) 0.000000 0.000000
Y( 4) 0.000000 0.000000
A( 1) 27.30000 0.000000
A( 2) 38.20000 0.000000
A( 3) 43.70000 0.000000
A( 4) 35.20000 0.000000
X( 1) 72000.00 0.000000
X( 2) 10000.00 0.000000
X( 3) 0.000000 0.000000
X( 4) 0.000000 0.000000
B( 1) 25000.00 0.000000
B( 2) 12000.00 0.000000
B( 3) 6000.000 0.000000
B( 4) 21000.00 0.000000
B( 5) 8000.000 0.000000
B( 6) 7000.000 0.000000
B( 7) 3000.000 0.000000
W( 1) 28.00000 0.000000
W( 2) 28.00000 0.000000
W( 3) 30.00000 0.000000
W( 4) 28.00000 0.000000
W( 5) 30.00000 0.000000
W( 6) 32.00000 0.000000
W( 7) 32.00000 0.000000
C( 1, 1) 4.500000 0.000000
C( 1, 2) 7.500000 0.000000
C( 1, 3) 10.00000 0.000000
C( 1, 4) 8.000000 0.000000
C( 1, 5) 16.80000 0.000000
C( 1, 6) 28.70000 0.000000
C( 1, 7) 39.60000 0.000000
C( 2, 1) 23.60000 0.000000
C( 2, 2) 21.00000 0.000000
C( 2, 3) 23.80000 0.000000
C( 2, 4) 28.50000 0.000000
C( 2, 5) 18.30000 0.000000
C( 2, 6) 9.700000 0.000000
C( 2, 7) 16.40000 0.000000
C( 3, 1) 29.30000 0.000000
C( 3, 2) 27.00000 0.000000
C( 3, 3) 41.00000 0.000000
C( 3, 4) 32.70000 0.000000
C( 3, 5) 13.80000 0.000000
C( 3, 6) 8.000000 0.000000
C( 3, 7) 11.00000 0.000000
C( 4, 1) 15.80000 0.000000
C( 4, 2) 19.20000 0.000000
C( 4, 3) 19.00000 0.000000
C( 4, 4) 19.30000 0.000000
C( 4, 5) 7.800000 0.000000
C( 4, 6) 11.00000 0.000000
C( 4, 7) 17.00000 0.000000
Z( 1, 1) 1.000000 0.000000
Z( 1, 2) 1.000000 0.000000
Z( 1, 3) 1.000000 0.000000
Z( 1, 4) 1.000000 0.000000
Z( 1, 5) 1.000000 0.000000
Z( 1, 6) 0.000000 56700.00
Z( 1, 7) 0.000000 36900.00
Z( 2, 1) 0.000000 750000.0
Z( 2, 2) 0.000000 292800.0
Z( 2, 3) 0.000000 148200.0
Z( 2, 4) 0.000000 659400.0
Z( 2, 5) 0.000000 99200.00
Z( 2, 6) 1.000000 0.000000
Z( 2, 7) 1.000000 0.000000
Z( 3, 1) 0.000000 1825000.
Z( 3, 2) 0.000000 848400.0
Z( 3, 3) 0.000000 508200.0
Z( 3, 4) 0.000000 1604400.
Z( 3, 5) 0.000000 460000.0
Z( 3, 6) 0.000000 361900.0
Z( 3, 7) 0.000000 164100.0
Z( 4, 1) 0.000000 1275000.
Z( 4, 2) 0.000000 652800.0
Z( 4, 3) 0.000000 325200.0
Z( 4, 4) 0.000000 1144500.
Z( 4, 5) 0.000000 344000.0
Z( 4, 6) 0.000000 323400.0
Z( 4, 7) 0.000000 156600.0
T( 1, 1) 23.00000 0.000000
T( 1, 2) 24.00000 0.000000
T( 1, 3) 24.50000 0.000000
T( 1, 4) 24.00000 0.000000
T( 1, 5) 26.50000 0.000000
T( 1, 6) 30.00000 0.000000
T( 1, 7) 33.00000 0.000000
T( 2, 1) 32.00000 0.000000
T( 2, 2) 31.00000 0.000000
T( 2, 3) 31.50000 0.000000
T( 2, 4) 33.00000 0.000000
T( 2, 5) 30.50000 0.000000
T( 2, 6) 28.50000 0.000000
T( 2, 7) 29.50000 0.000000
T( 3, 1) 34.50000 0.000000
T( 3, 2) 33.50000 0.000000
T( 3, 3) 37.00000 0.000000
T( 3, 4) 35.50000 0.000000
T( 3, 5) 30.50000 0.000000
T( 3, 6) 29.00000 0.000000
T( 3, 7) 30.00000 0.000000
T( 4, 1) 29.00000 0.000000
T( 4, 2) 30.00000 0.000000
T( 4, 3) 30.00000 0.000000
T( 4, 4) 30.00000 0.000000
T( 4, 5) 27.00000 0.000000
T( 4, 6) 27.50000 0.000000
T( 4, 7) 29.50000 0.000000

Row Slack or Surplus Dual Price
OBJ 3029600. -1.000000
2 5.000000 0.000000
3 4.000000 0.000000
4 5.500000 0.000000
5 4.000000 0.000000
6 3.500000 0.000000
7 32.00000 0.000000
8 32.00000 0.000000
9 28.00000 0.000000
10 28.00000 0.000000
11 30.00000 0.000000
12 28.00000 0.000000
13 30.00000 0.000000
14 3.500000 0.000000
15 2.500000 0.000000
16 28.00000 0.000000
17 28.00000 0.000000
18 30.00000 0.000000
19 28.00000 0.000000
20 30.00000 0.000000
21 32.00000 0.000000
22 32.00000 0.000000
23 28.00000 0.000000
24 28.00000 0.000000
25 30.00000 0.000000
26 28.00000 0.000000
27 30.00000 0.000000
28 32.00000 0.000000
29 32.00000 0.000000
30 0.000000 -795000.0
31 0.000000 -417600.0
32 0.000000 -223800.0
33 0.000000 -741300.0
34 0.000000 -352800.0
35 0.000000 -335300.0
36 0.000000 -163800.0
37 0.000000 0.000000
38 0.000000 27.30000
39 0.000000 38.20000
40 0.000000 43.70000
41 0.000000 35.20000

请问你是用什么版本的lingo运行的,能告诉我么?谢谢了 我需要下能运行的LINGO进行截图,论文使用~拜托了

QUOTE:


作者: hubulwm    时间: 2008-5-29 08:49
用LINGO8.0的Global求解程序就行了。




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