线性规划的灵敏度分析实验报告.doc

《运筹学/线性规划》实验报告
实验室： 实验日期：
实验项目
线性规划的灵敏度分析
系 别
数学系
姓 名

学 号

班 级

指导教师

成 绩

一 实验目的
掌握用Lingo/Lindo对线性规划问题进行灵敏度分析的方法，理解解报告的内容。初步掌握对实际的线性规划问题建立数学模型，并运用计算机求解分析的一般方法。

二 实验环境
Lingo软件

三 实验内容（涉及数学模型、上机程序、实验结果、结果分析与问题解答等）
例题2-10
MODEL:
[_1] MAX= 2 * X_1 + 3 * X_2 ;
[_2] X_1 + 2 * X_2 + X_3 = 8 ;
[_3] 4 * X_1 + X_4 = 16 ;
[_4] 4 * X_2 + X_5 = 12 ;
END
编程
sets:
is/1..3/:b;
js/1..5/:c,x;
links(is,js):a;
endsets
max=@sum(js(J):c(J)*x(J));
@for(is(I):@sum(js(J):a(I,J)*x(J))=b(I));
data:
c=2 3 0 0 0;
b=8 16 12;
a=1 2 1 0 0
4 0 0 1 0
0 4 0 0 1;
end data
end

灵敏度分析

Ranges in which the basis is unchanged:

Objective Coefficient Ranges
Current Allowable Allowable
Variable Coefficient Increase Decrease
X( 1) 2.000000 INFINITY 0.5000000
X( 2) 3.000000 1.000000 3.000000
X( 3) 0.0 1.500000 INFINITY
X( 4) 0.0 0.1250000 INFINITY
X( 5) 0.0 0.7500000 0.2500000

Righthand Side Ranges
Row Current Allowable Allowable
RHS Increase Decrease
2 8.000000 2.000000 4.000000
3 16.00000 16.00000 8.000000
4 12.00000 INFINITY 4.000000
当b2在 [8,32]之间变化时 最优基不变

最优解
Global optimal solution found at iteration: 0
Objective value: 14.00000
Variable Value Reduced Cost
B( 1) 8.000000 0.000000
B( 2) 16.00000 0.000000
B( 3) 12.00000 0.000000
C( 1) 2.000000 0.000000
C( 2) 3.000000 0.000000
C( 3) 0.000000 0.000000
C( 4) 0.000000 0.000000
C( 5) 0.000000 0.000000
X( 1) 4.000000 0.000000
X( 2) 2.000000 0.000000
X( 3) 0.000000 1.500000
X( 4) 0.000000 0.1250000
X( 5) 4.000000 0.000000
A( 1, 1) 1.000000 0.000000
A( 1, 2) 2.000000 0.000000
A( 1, 3) 1.000000 0.000000
A( 1, 4) 0.000000 0.000000
A( 1, 5) 0.000000 0.000000
A( 2, 1) 4.000000 0.000000
A( 2, 2) 0.000000 0.000000
A( 2, 3) 0.000000 0.000000
A( 2, 4) 1.000000 0.000000
A( 2, 5) 0.000000 0.000000
A( 3, 1) 0.000000 0.000000
A( 3, 2) 4.000000 0.000000
A( 3, 3) 0.000000 0.000000
A( 3, 4) 0.000000 0.000000
A( 3, 5) 1.000000 0.000000

Row Slack or Surplus Dual Price
1 14.00000 1.000000
2 0.000000 1.500000
3 0.000000 0.1250000
4 0.000000 0.000000

例题2-11
模型
MAX 2 X( 1) + 3 X( 2)
SUBJECT TO
2] X( 1) + 2 X( 2) + X( 3) = 12
3] 4 X( 1) + X( 4) = 16
4] 4 X( 2) + X( 5) = 12
END

编程
sets:
is/1..3/:b;
js/1..5/:c,x;
links(is,js):a;
endsets
max=@sum(js(J):c(J)*x(J));
@for(is(I):@sum(js(J):a(I,J)*x(J))=b(I));
data:
c=2 3 0 0 0;
b=12 16 12;
a=1 2 1 0 0
4 0 0 1 0
0 4 0 0 1;
end data
end

最优解
Global optimal solution found at iteration: 2
Objective value: 17.00000
Variable Value Reduced Cost
B( 1) 12.00000 0.000000
B( 2) 16.00000 0.000000
B( 3) 12.00000 0.000000
C( 1) 2.000000 0.000000
C( 2) 3.000000 0.000000
C( 3) 0.000000 0.000000
C( 4) 0.000000 0.000000
C( 5) 0.000000 0.000000
X( 1) 4.000000 0.000000
X( 2) 3.000000 0.000000
X( 3) 2.000000 0.000000
X( 4) 0.000000 0.5000000
X( 5) 0.000000 0.7500000
A( 1, 1) 1.000000 0.000000
A( 1, 2) 2.000000 0.000000
A( 1, 3) 1.000000 0.000000
A( 1, 4) 0.000000 0.000000
A( 1, 5) 0.000000 0.000000
A( 2, 1) 4.000000 0.000000
A( 2, 2) 0.000000 0.000000
A( 2, 3) 0.000000 0.000000
A( 2, 4) 1.000000 0.000000
A( 2, 5) 0.000000 0.000000
A( 3, 1) 0.000000 0.000000
A( 3, 2) 4.000000 0.000000
A( 3, 3) 0.000000 0.000000
A( 3, 4) 0.000000 0.000000
A( 3, 5) 1.000000 0.000000

Row Slack or Surplus Dual Price
1 17.00000 1.000000
2 0.000000 0.000000
3 0.000000 0.5000000
4 0.000000 0.7500000

最优解（4,3,2,0,0）最优值z=17
分析
Ranges in which the basis is unchanged:

Objective Coefficient Ranges
Current Allowable Allowable
Variable Coefficient Increase Decrease
X( 1) 2.000000 INFINITY 2.000000
X( 2) 3.000000 INFINITY 3.000000
X( 3) 0.0 1.500000 INFINITY
X( 4) 0.0 0.5000000 INFINITY
X( 5) 0.0 0.7500000 INFINITY

Righthand Side Ranges
Row Current Allowable Allowable
RHS Increase Decrease
2 12.00000 INFINITY 2.000000
3 16.00000 8.000000 16.00000
4 12.00000 4.000000 12.00000
例题2-12
模型
MAX 2 X( 1) + 3 X( 2)
SUBJECT TO
2] X( 1) + 2 X( 2) + X( 3) = 8
3] 4 X( 1) + X( 4) = 16
4] 4 X( 2) + X( 5) = 12
END
编程
sets:
is/1..3/:b;
js/1..5/:c,x;
links(is,js):a;
endsets
max=@sum(js(J):c(J)*x(J));
@for(is(I):@sum(js(J):a(I,J)*x(J))=b(I));
data:
c=2 3 0 0 0;
b=8 16 12;
a=1 2 1 0 0
4 0 0 1 0
0 4 0 0 1;
end data
end

灵敏度分析
Ranges in which the basis is unchanged:

Objective Coefficient Ranges
Current Allowable Allowable
Variable Coefficient Increase Decrease
X( 1) 2.000000 INFINITY 0.5000000
X( 2) 3.000000 1.000000 3.000000
X( 3) 0.0 1.500000 INFINITY
X( 4) 0.0 0.1250000 INFINITY
X( 5) 0.0 0.7500000 0.2500000

Righthand Side Ranges
Row Current Allowable Allowable
RHS Increase Decrease
2 8.000000 2.000000 4.000000
3 16.00000 16.00000 8.000000
4 12.00000 INFINITY 4.000000
由灵敏度分析表知道C2在【0,4】之间变化时，最优基不变。

第六题
模型
MODEL:
[_1] MAX= 3 * X_1 + X_2 + 4 X_3 ;
[_2] 6 * X_1 + 3 * X_2 + 5 * X_3 <= 450 ;
[_3] 3 * X_1 + 4 * X_2 + 5 * X_3 <= 300 ;
END

编程
sets:
is/1..2/:b;
js/1..3/:c,x;
links(is,js):a;
endsets
max=@sum(js(J):c(J)*x(J));
@for(is(I):@sum(js(J):a(I,J)*x(J))<=b(I));
data:
c=3 1 4;
b=450 300;
a=6 3 5
3 4 5;
end data
End
最优解 Global optimal solution found.
Objective value: 270.0000
Infeasibilities: 0.000000
Total solver iterations: 2

Variable Value Reduced Cost
B( 1) 450.0000 0.000000
B( 2) 300.0000 0.000000
C( 1) 3.000000 0.000000
C( 2) 1.000000 0.000000
C( 3) 4.000000 0.000000
X( 1) 50.00000 0.000000
X( 2) 0.000000 2.000000
X( 3) 30.00000 0.000000
A( 1, 1) 6.000000 0.000000
A( 1, 2) 3.000000 0.000000
A( 1, 3) 5.000000 0.000000
A( 2, 1) 3.000000 0.000000
A( 2, 2) 4.000000 0.000000
A( 2, 3) 5.000000 0.000000

Row Slack or Surplus Dual Price
1 270.0000 1.000000
2 0.000000 0.2023000
3 0.000000 0.6000000

第一问：
A生产50 B生产0 C生产30 有最高利润270元；
第二问：
单个价值系数和右端系数变化范围的灵敏度分析结果
Ranges in which the basis is unchanged:

Objective Coefficient Ranges
Current Allowable Allowable
Variable Coefficient Increase Decrease
X( 1) 3.000000 1.800000 0.6000000
X( 2) 1.000000 2.000000 INFINITY
X( 3) 4.000000 1.000000 1.500000

Righthand Side Ranges
Row Current Allowable Allowable
RHS Increase Decrease
2 450.0000 150.0000 150.0000
3 300.0000 150.0000 75.00000
当A的利润在【2.4，4.8】之间变化时，原最优生产计划不变。
第三问：
模型
MODEL:
[_1] MAX= 3 * X_1 + X_2 + 4 * X_3 + 3 * X_4 ;
[_2] 6 * X_1 + 3 * X_2 + 5 * X_3 + 8 * X_4 <= 450 ;
[_3] 3 * X_1 + 4 * X_2 + 5 * X_3 + 2 * X_4 <= 300 ;
END
编程
sets:
is/1..2/:b;
js/1..4/:c,x;
links(is,js):a;
endsets
max=@sum(js(J):c(J)*x(J));
@for(is(I):@sum(js(J):a(I,J)*x(J))<=b(I));
data:
c=3 1 4 3;
b=450 300;
a=6 3 5 8
3 4 5 2;
end data
End

最优解
Global optimal solution found.
Objective value: 275.0000
Infeasibilities: 0.000000
Total solver iterations: 2

Variable Value Reduced Cost
B( 1) 450.0000 0.000000
B( 2) 300.0000 0.000000
C( 1) 3.000000 0.000000
C( 2) 1.000000 0.000000
C( 3) 4.000000 0.000000
C( 4) 3.000000 0.000000
X( 1) 0.000000 0.1000000
X( 2) 0.000000 1.966667
X( 3) 50.00000 0.000000
X( 4) 25.00000 0.000000
A( 1, 1) 6.000000 0.000000
A( 1, 2) 3.000000 0.000000
A( 1, 3) 5.000000 0.000000
A( 1, 4) 8.000000 0.000000
A( 2, 1) 3.000000 0.000000
A( 2, 2) 4.000000 0.000000
A( 2, 3) 5.000000 0.000000
A( 2, 4) 2.000000 0.000000

Row Slack or Surplus Dual Price
1 275.0000 1.000000
2 0.000000 0.2333333
3 0.000000 0.5666667
利润275元 值得生产。
第四问
由单个价值系数和右端系数变化范围的灵敏度分析结果
Ranges in which the basis is unchanged:
Objective Coefficient Ranges
Current Allowable Allowable
Variable Coefficient Increase Decrease
X( 1) 3.000000 1.800000 0.6000000
X( 2)