【分段函数】【多周期】【最大值问题】
各位零友，谁知道怎么把下面的代码改成多个周期的y值之和最小吗？比如说6个周期，自变量x(1)~x(6)始终在【200,1500】之间，求y(1)+y(2)+y(3)+y(4)+y(5)+y(6)最值。
代码如下
Model:
SETS:
Points/1..4/: b, c, y, z; ! 端点数为4，即分段数为3;
ENDSETS
DATA:
b=0 500 1000 1500;
c=0 5000 9000 12000;
y=,,,0; ! 增加的虚拟变量y(4)=0;
ENDDATA
Max= 4.8*x11 + 4.8*x21 + 5.6*x12 + 5.6*x22 - @sum(Points: c*z);
x11+x12 < x + 500;
x21+x22 < 1000;
0.5*x11 - 0.5*x21 > 0;
0.4*x12 - 0.6*x22 > 0;
@sum(Points: b*z)=x;
@for(Points(i)|i#eq#1: z(i) <= y(i));
@for(Points(i)|i#ne#1: z(i) <= y(i-1)+y(i));
@sum(Points: y)=1;
@sum(Points: z)=1;
@for(Points: @bin(y));
end
你说到了周期及自变量和目标函数，但是没有说具体的函数表达式。光看你的代码也不清楚你建的模型，你能不能仔细描述一下你的问题？ olh2008 发表于 2012-11-28 22:31 static/image/common/back.gif
你说到了周期及自变量和目标函数，但是没有说具体的函数表达式。光看你的代码也不清楚你建的模型，你能不能 ...
【问题补充】分段函数如下所示：
y=10*x 0<x<=500
y=8*x+1000 500<x<=1000
y=-x+9000 1000<x<=1500
在period（1..6）6个周期中，x>=a，a=(300,800,1300,1400,600,900)，求@sum(y)的最大值。 你第一个模型好像是个例子，你下面的模型可以通过lingo10以上的版本中的子模型来求解。
程序如下（我不知道我有没有正确的理解你的模型）
model:
sets:
ys/1..6/:a;
endsets
data:
a=300 800 1300 1400 600 900;
enddata
submodel obj1:
x>=a(1);
y=@if(x#LE#500,10*x,@if(x#LE#1000,8*x+1000,-x+9000));
max=y;
endsubmodel
submodel obj2:
x>=a(2);
y=@if(x#LE#500,10*x,@if(x#LE#1000,8*x+1000,-x+9000));
max=y;
endsubmodel
submodel obj3:
x>=a(3);
y=@if(x#LE#500,10*x,@if(x#LE#1000,8*x+1000,-x+9000));
max=y;
endsubmodel
submodel obj4:
x>=a(4);
y=@if(x#LE#500,10*x,@if(x#LE#1000,8*x+1000,-x+9000));
max=y;
endsubmodel
submodel obj5:
x>=a(5);
y=@if(x#LE#500,10*x,@if(x#LE#1000,8*x+1000,-x+9000));
max=y;
endsubmodel
submodel obj6:
x>=a(6);
y=@if(x#LE#500,10*x,@if(x#LE#1000,8*x+1000,-x+9000));
max=y;
endsubmodel
calc:
@write('a=300时的解',@newline(1));
@solve(obj1);
@write('a=800时的解',@newline(1));
@solve(obj2);
@write('a=1300时的解',@newline(1));
@solve(obj3);
@write('a=1400时的解',@newline(1));
@solve(obj4);
@write('a=600时的解',@newline(1));
@solve(obj5);
@write('a=900时的解',@newline(1));
@solve(obj6);
endcalc
end 运行结果如下：
a=300时的解
Linearization components added:
Constraints: 30
Variables: 16
Integers: 12
Global optimal solution found.
Objective value: 9000.000
Objective bound: 9000.000
Infeasibilities: 0.000000
Extended solver steps: 3
Total solver iterations: 22
Variable Value Reduced Cost
X 1000.000 0.000000
Y 9000.000 0.000000
A( 1) 300.0000 0.000000
A( 2) 800.0000 0.000000
A( 3) 1300.000 0.000000
A( 4) 1400.000 0.000000
A( 5) 600.0000 0.000000
A( 6) 900.0000 0.000000
Row Slack or Surplus Dual Price
1 700.0000 0.000000
2 0.000000 1.000000
3 9000.000 1.000000
a=800时的解
Linearization components added:
Constraints: 30
Variables: 16
Integers: 12
Global optimal solution found.
Objective value: 9000.000
Objective bound: 9000.000
Infeasibilities: 0.000000
Extended solver steps: 2
Total solver iterations: 8
Variable Value Reduced Cost
X 1000.000 0.000000
Y 9000.000 0.000000
A( 1) 300.0000 0.000000
A( 2) 800.0000 0.000000
A( 3) 1300.000 0.000000
A( 4) 1400.000 0.000000
A( 5) 600.0000 0.000000
A( 6) 900.0000 0.000000
Row Slack or Surplus Dual Price
1 200.0000 0.000000
2 0.000000 1.000000
3 9000.000 1.000000
a=1300时的解
Linearization components added:
Constraints: 30
Variables: 16
Integers: 12
Global optimal solution found.
Objective value: 7700.000
Objective bound: 7700.000
Infeasibilities: 0.000000
Extended solver steps: 0
Total solver iterations: 0
Variable Value Reduced Cost
X 1300.000 0.000000
Y 7700.000 0.000000
A( 1) 300.0000 0.000000
A( 2) 800.0000 0.000000
A( 3) 1300.000 0.000000
A( 4) 1400.000 0.000000
A( 5) 600.0000 0.000000
A( 6) 900.0000 0.000000
Row Slack or Surplus Dual Price
1 0.000000 -1.000000
2 0.000000 1.000000
3 7700.000 1.000000
a=1400时的解
Linearization components added:
Constraints: 30
Variables: 16
Integers: 12
Global optimal solution found.
Objective value: 7600.000
Objective bound: 7600.000
Infeasibilities: 0.000000
Extended solver steps: 0
Total solver iterations: 0
Variable Value Reduced Cost
X 1400.000 0.000000
Y 7600.000 0.000000
A( 1) 300.0000 0.000000
A( 2) 800.0000 0.000000
A( 3) 1300.000 0.000000
A( 4) 1400.000 0.000000
A( 5) 600.0000 0.000000
A( 6) 900.0000 0.000000
Row Slack or Surplus Dual Price
1 0.000000 -1.000000
2 0.000000 1.000000
3 7600.000 1.000000
a=600时的解
Linearization components added:
Constraints: 30
Variables: 16
Integers: 12
Global optimal solution found.
Objective value: 9000.000
Objective bound: 9000.000
Infeasibilities: 0.000000
Extended solver steps: 2
Total solver iterations: 8
Variable Value Reduced Cost
X 1000.000 0.000000
Y 9000.000 0.000000
A( 1) 300.0000 0.000000
A( 2) 800.0000 0.000000
A( 3) 1300.000 0.000000
A( 4) 1400.000 0.000000
A( 5) 600.0000 0.000000
A( 6) 900.0000 0.000000
Row Slack or Surplus Dual Price
1 400.0000 0.000000
2 0.000000 1.000000
3 9000.000 1.000000
a=900时的解
Linearization components added:
Constraints: 30
Variables: 16
Integers: 12
Global optimal solution found.
Objective value: 9000.000
Objective bound: 9000.000
Infeasibilities: 0.000000
Extended solver steps: 2
Total solver iterations: 8
Variable Value Reduced Cost
X 1000.000 0.000000
Y 9000.000 0.000000
A( 1) 300.0000 0.000000
A( 2) 800.0000 0.000000
A( 3) 1300.000 0.000000
A( 4) 1400.000 0.000000
A( 5) 600.0000 0.000000
A( 6) 900.0000 0.000000
Row Slack or Surplus Dual Price
1 100.0000 0.000000
2 0.000000 1.000000
3 9000.000 1.000000
4楼好给力啊，什么问题都解决的了，以后有问题就找你了{:soso_e113:} 强~ 学习了
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