LINGO8.0 for windows软件及应用（编译）

kaiyin · 发表于 2004-9-13 21:11:39

<

>太好了，谢谢您的资料啊，有什么不懂的还望不吝赐教哦

zyhwan · 发表于 2004-9-14 07:59:31

<

>回第三楼：<

>问题很简单，修改如下：<

>model:
sets:
tou/1..4/:s,r,p,u,x,y,z;
endsets
data:
r=0.28 0.21 0.23 0.25;
p=0.01 0.02 0.045 0.065;
u=103 198 52 40;
enddata
max <a href="mailt=@sum" target="_blank" >=@sum</A>( tou(i):s(i)*r(i)-x(i)*(p(i)*u(i)+(s(i)-u(i))*p(i)*y(i)));
@sum(tou(i):s(i))<10000;
@for(tou(i):x(i)=@if(s(i) #gt#0,1,0));
@for(tou(i):x(i)=@if(s(i) #eq#0,0,1));
@for(tou:z=u-s);
@for(tou(i):y(i)=@if(z(i) #gt# 0,0,1));
@for(tou(i):y(i)=@if(z(i) #le# 0,1,0));
endLocal optimal solution found at iteration: 4
 Objective value: 82.45000
 Variable Value Reduced Cost
 S( 1) 103.0000 0.000000
 S( 2) 198.0000 0.000000
 S( 3) 52.00000 0.000000
 S( 4) 40.00000 0.000000
 R( 1) 0.2800000 0.000000
 R( 2) 0.2100000 0.000000
 R( 3) 0.2300000 0.000000
 R( 4) 0.2500000 0.000000
 P( 1) 0.1000000E-01 0.000000
 P( 2) 0.2000000E-01 0.000000
 P( 3) 0.4500000E-01 0.000000
 P( 4) 0.6500000E-01 0.000000
 U( 1) 103.0000 0.000000
 U( 2) 198.0000 0.000000
 U( 3) 52.00000 0.000000
 U( 4) 40.00000 0.000000
 X( 1) 1.000000 0.000000
 X( 2) 1.000000 0.000000
 X( 3) 1.000000 0.000000
 X( 4) 1.000000 0.000000
 Y( 1) 1.000000 -0.1350000E-07
 Y( 2) 1.000000 -0.9500000E-08
 Y( 3) 1.000000 -0.9250000E-08
 Y( 4) 1.000000 -0.9250000E-08
 Z( 1) 0.000000 0.000000
 Z( 2) 0.000000 0.000000
 Z( 3) 0.000000 0.000000
 Z( 4) 0.000000 0.000000

ho000000 · 发表于 2004-9-14 17:56:16

godanddog29 · 发表于 2004-9-14 18:54:36

呵呵，不错不错，谢谢搂住

都市女杀手 · 发表于 2004-9-15 23:53:19

<

>高手请问画线的部分为什么错了？<

>model:
sets:
 days/mon..sun/:reuired,start;
endsets
data:
 !每天所需的最少职员数；
 reuired=20 16 13 16 19 14 12;
 @text(':\out.txt')=days'至少需要的职员数为'start;
enddata
!最小化每周所需职员数；
<a href="mailtmin=@sum(days:start" target="_blank" >min=@sum(days:start</A>);
@for(days(J):
 @sum(days(I)|I #le# 5:
 start(@wrap(J+I+2,7)))>= required(J));
end

wanbaocheng · 发表于 2004-9-16 04:03:42

<

>model:
sets:
 days/mon..sun/:required,start;
endsets
data:
 !每天所需的最少职员数；
 reuired=20 16 13 16 19 14 12;
 @text(':\out.txt')=days'至少需要的职员数为'start;
enddata
!最小化每周所需职员数；
<a href="mailtmin=@sum(days:start" target="_blank" >min=@sum(days:start</A>);
@for(days(J):
 @sum(days(I)|I #le# 5:
 start(@wrap(J+I+2,7)))>= required(J));
end<

>落了一个q，已用下划线标出（红字）

zyhwan · 发表于 2004-9-17 07:19:31

<

>model:
sets:
days/mon..sun/:required,start;
endsets
data:
!每天所需的最少职员数;
required=20 16 13 16 19 14 12;
@text('d:\out.txt')=days'至少需要的职员数为'start;
enddata
!最小化每周所需职员数;
<a href="mailtmin=@sum(days:start" target="_blank" >min=@sum(days:start</A>);
@for(days(J):
@sum(days(I)|I #le# 5:start(@wrap(J-I+1,7)))>= required(J)
);
end<

> Global optimal solution found at iteration: 0
Objective value: 22.00000<

> Variable Value Reduced Cost
 REQUIRED( MON) 20.00000 0.000000
 REQUIRED( TUE) 16.00000 0.000000
 REQUIRED( WED) 13.00000 0.000000
 REQUIRED( THU) 16.00000 0.000000
 REQUIRED( FRI) 19.00000 0.000000
 REQUIRED( SAT) 14.00000 0.000000
 REQUIRED( SUN) 12.00000 0.000000
 START( MON) 8.000000 0.000000
 START( TUE) 2.000000 0.000000
 START( WED) 0.000000 0.3333333
 START( THU) 6.000000 0.000000
 START( FRI) 3.000000 0.000000
 START( SAT) 3.000000 0.000000
 START( SUN) 0.000000 0.000000 Row Slack or Surplus Dual Price
 1 22.00000 -1.000000
 2 0.000000 -0.3333333
 3 0.000000 -0.3333333
 4 0.000000 0.000000
 5 0.000000 -0.3333333
 6 0.000000 0.000000
 7 0.000000 -0.3333333
 8 0.000000 0.000000

southboy · 发表于 2004-9-18 07:34:22

<

>我急需Lingo教程，请问哪位大侠帮忙<

>,谢谢<

><a href="mailtweimingtan@email.jlu.edu.cn" target="_blank" >weimingtan@email.jlu.edu.cn</A>

shirley2004 · 发表于 2004-9-18 22:40:05

<

>哪位高手帮我修改一下这段小程序啊<

>model:
sets:
xialiao/fangshi,liaoshu/:b
days/1..4/:c,x;
nu/1..2/:n;
endsets
data:
b=1 2 3 4
 0 0 1 4
 3 4 5 6
 7 5 6 7
c=4 3 1 2;
n=22 55 33 44;
enddata
min =@sum( cai(i):c(i)*x(i));
@sum(cai(i):x(i))<=400;

@for(nu(i):
@sum(cai(i):xialiao(i,j)*x(i))=n(i));
end<

>谢谢

zyhwan · 发表于 2004-9-20 13:57:10

<

>修改及答案<

>model:<

>sets:days/1..4/:c,x;nu/1..4/:n;xialiao(nu,days):b;endsetsdata:b=1 2 3 4
 0 0 1 4
 3 4 5 6
 7 5 6 7;c=4 3 1 2;n=22 55 33 44;enddatamin <a href="mailt=@sum" target="_blank" >=@sum</A>( days:c*x);@sum(days:x)<=400; @for(nu(i): @sum(days(j):b(i,j)*x(i))=n(i) );endGlobal optimal solution found at iteration: 0
 Objective value: 47.15333
 Variable Value Reduced Cost
 C( 1) 4.000000 0.000000
 C( 2) 3.000000 0.000000
 C( 3) 1.000000 0.000000
 C( 4) 2.000000 0.000000
 X( 1) 2.200000 0.000000
 X( 2) 11.00000 0.000000
 X( 3) 1.833333 0.000000
 X( 4) 1.760000 0.000000
 N( 1) 22.00000 0.000000
 N( 2) 55.00000 0.000000
 N( 3) 33.00000 0.000000
 N( 4) 44.00000 0.000000
 B( 1, 1) 1.000000 0.000000
 B( 1, 2) 2.000000 0.000000
 B( 1, 3) 3.000000 0.000000
 B( 1, 4) 4.000000 0.000000
 B( 2, 1) 0.000000 0.000000
 B( 2, 2) 0.000000 0.000000
 B( 2, 3) 1.000000 0.000000
 B( 2, 4) 4.000000 0.000000
 B( 3, 1) 3.000000 0.000000
 B( 3, 2) 4.000000 0.000000
 B( 3, 3) 5.000000 0.000000
 B( 3, 4) 6.000000 0.000000
 B( 4, 1) 7.000000 0.000000
 B( 4, 2) 5.000000 0.000000
 B( 4, 3) 6.000000 0.000000
 B( 4, 4) 7.000000 0.000000 Row Slack or Surplus Dual Price
 1 47.15333 -1.000000
 2 383.2067 0.000000
 3 0.000000 -0.4000000
 4 0.000000 -0.6000000
 5 0.000000 -0.5555556E-01
 6 0.000000 -0.8000000E-01

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