运筹学实验报告lingo软件的使用习题代码.docx

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运筹学实验报告lingo软件的使用习题代码.docx

运筹学实验报告lingo软件的使用习题代码

 

运筹学

实验报告

 

姓名:

学号:

班级:

相关问题说明:

一、实验性质和教学目的

本实验是运筹学课内安排的上机操作实验。

目的在于了解、熟悉计算机Lingo软件在运筹学模型求解中的作用,激发学习兴趣,提高学习效果,增强自身的动手能力,提高实际应用能力。

二、实验基本要求

要求学生:

1.实验前认真做好理论准备,仔细阅读实验指导书;

2.遵从教师指导,认真完成实验任务,按时按质提交实验报告。

三、主要参考资料

1.LINGO软件

2.LINGO8.0及其在环境系统优化中的应用,天津大学出版社,2005

3.优化建模与LINDO/LINGO软件,清华大学出版社,2005

4.运筹学编写组主编,运筹学(修订版),清华大学出版社,1990

5.蓝伯雄主编,管理数学(下)—运筹学,清华大学出版社,1997

6.胡运权主编,运筹学习题集(修订版),清华大学出版社,1995

7.胡运权主编,运筹学教程(第二版),清华大学出版社,2003

 

实验内容

1、线性规划问题:

(1)给出原始代码;

(2)计算结果(包括灵敏度分析,求解结果粘贴);

(3)回答下列问题(手写):

a)最优解及最优目标函数值是多少;

b)资源的对偶价格各为多少,并说明对偶价格的含义;

c)为了使目标函数值增加最多,让你选择一个约束条件,将它的常数项增加一个单位,你将选择哪一个约束条件?

这时目标函数值将是多少?

d)对x2的目标函数系数进行灵敏度分析;

e)对第2个约束的约束右端项进行灵敏度分析;

f)结合本题的结果解释“ReducedCost”的含义。

对偶价格就是说约束方程右端变量增加1对目标函数值的影响

答案:

(1)代码

max=8*x1+6*x2;

9*x1+8*x2<=12;

7*x1+11*x2<=24;

9*x1+11*x2<=13;

x1>=0;

x2>=0;

(2)计算结果

Globaloptimalsolutionfound.

Objectivevalue:

10.66667

Totalsolveriterations:

2

VariableValueReducedCost

X11.3333330.000000

X20.0000001.111111

RowSlackorSurplusDualPrice

110.666671.000000

20.0000000.8888889

314.666670.000000

41.0000000.000000

51.3333330.000000

60.0000000.000000

Rangesinwhichthebasisisunchanged:

ObjectiveCoefficientRanges

CurrentAllowableAllowable

VariableCoefficientIncreaseDecrease

X18.000000INFINITY1.250000

X26.0000001.111111INFINITY

RighthandSideRanges

RowCurrentAllowableAllowable

RHSIncreaseDecrease

212.000001.00000012.00000

324.00000INFINITY14.66667

413.00000INFINITY1.000000

50.01.333333INFINITY

60.00.0INFINITY

(3)a)

b)

c)

d)

e)

f)

2、运输问题:

已知6个发点8个收点的最小费用运输问题。

产销量及单位运价如下表。

销地

cij

产地

B1

B2

B3

B4

B5

B6

B7

B8

产量

A1

6

2

9

7

4

2

5

9

55

A2

4

5

5

3

8

5

3

2

47

A3

5

2

1

3

7

4

8

3

42

A4

7

6

7

9

9

2

7

1

52

A5

2

3

6

5

7

2

6

5

41

A6

5

9

2

2

8

1

4

3

32

销量

60

55

51

43

41

52

43

38

 

(1)给出原始代码;

(2)计算结果(决策变量求解结果粘贴)

MinZ=CijXij

Xij<=bj(j=1...8)销量约束

Xij=ai(i=1...6)产量约束

Xij≥0(i=1...6;j=1...8)

代码:

model:

!

6发点8model:

!

6发点8收点运输问题;

sets:

warehouses/wh1..wh6/:

capacity;

vendors/v1..v8/:

demand;

links(warehouses,vendors):

cost,volume;

endsets

min=@sum(links:

cost*volume);!

目标函数;

@for(vendors(J):

@sum(warehouses(I):

volume(I,J))<=demand(J));!

需求约束;

@for(warehouses(I):

@sum(vendors(J):

volume(I,J))=capacity(I));!

产量约束;

!

这里是数据;

data:

capacity=554742524132;

demand=6055514341524338;

cost=62974259

45538532

52137483

76799271

23657265

59228143;

enddata

end

 

答案

Globaloptimalsolutionfound.

Objectivevalue:

473.0000

Infeasibilities:

0.000000

Totalsolveriterations:

9

ModelClass:

LP

Totalvariables:

48

Nonlinearvariables:

0

Integervariables:

0

Totalconstraints:

15

Nonlinearconstraints:

0

Totalnonzeros:

144

Nonlinearnonzeros:

0

VariableValueReducedCost

CAPACITY(WH1)55.000000.000000

CAPACITY(WH2)47.000000.000000

CAPACITY(WH3)42.000000.000000

CAPACITY(WH4)52.000000.000000

CAPACITY(WH5)41.000000.000000

CAPACITY(WH6)32.000000.000000

DEMAND(V1)60.000000.000000

DEMAND(V2)55.000000.000000

DEMAND(V3)51.000000.000000

DEMAND(V4)43.000000.000000

DEMAND(V5)41.000000.000000

DEMAND(V6)52.000000.000000

DEMAND(V7)43.000000.000000

DEMAND(V8)38.000000.000000

COST(WH1,V1)6.0000000.000000

COST(WH1,V2)2.0000000.000000

COST(WH1,V3)9.0000000.000000

COST(WH1,V4)7.0000000.000000

COST(WH1,V5)4.0000000.000000

COST(WH1,V6)2.0000000.000000

COST(WH1,V7)5.0000000.000000

COST(WH1,V8)9.0000000.000000

COST(WH2,V1)4.0000000.000000

COST(WH2,V2)5.0000000.000000

COST(WH2,V3)5.0000000.000000

COST(WH2,V4)3.0000000.000000

COST(WH2,V5)8.0000000.000000

COST(WH2,V6)5.0000000.000000

COST(WH2,V7)3.0000000.000000

COST(WH2,V8)2.0000000.000000

COST(WH3,V1)5.0000000.000000

COST(WH3,V2)2.0000000.000000

COST(WH3,V3)1.0000000.000000

COST(WH3,V4)3.0000000.000000

COST(WH3,V5)7.0000000.000000

COST(WH3,V6)4.0000000.000000

COST(WH3,V7)8.0000000.000000

COST(WH3,V8)3.0000000.000000

COST(WH4,V1)7.0000000.000000

COST(WH4,V2)6.0000000.000000

COST(WH4,V3)7.0000000.000000

COST(WH4,V4)9.0000000.000000

COST(WH4,V5)9.0000000.000000

COST(WH4,V6)2.0000000.000000

COST(WH4,V7)7.0000000.000000

COST(WH4,V8)1.0000000.000000

COST(WH5,V1)2.0000000.000000

COST(WH5,V2)3.0000000.000000

COST(WH5,V3)6.0000000.000000

COST(WH5,V4)5.0000000.000000

COST(WH5,V5)7.0000000.000000

COST(WH5,V6)2.0000000.000000

COST(WH5,V7)6.0000000.000000

COST(WH5,V8)5.0000000.000000

COST(WH6,V1)5.0000000.000000

COST(WH6,V2)9.0000000.000000

COST(WH6,V3)2.0000000.000000

COST(WH6,V4)2.0000000.000000

COST(WH6,V5)8.0000000.000000

COST(WH6,V6)1.0000000.000000

COST(WH6,V7)4.0000000.000000

COST(WH6,V8)3.0000000.000000

VOLUME(WH1,V1)0.0000004.000000

VOLUME(WH1,V2)55.000000.000000

VOLUME(WH1,V3)0.0000007.000000

VOLUME(WH1,V4)0.0000005.000000

VOLUME(WH1,V5)0.0000002.000000

VOLUME(WH1,V6)0.0000000.000000

VOLUME(WH1,V7)0.0000003.000000

VOLUME(WH1,V8)0.0000008.000000

VOLUME(WH2,V1)0.0000001.000000

VOLUME(WH2,V2)0.0000002.000000

VOLUME(WH2,V3)0.0000002.000000

VOLUME(WH2,V4)43.000000.000000

VOLUME(WH2,V5)0.0000005.000000

VOLUME(WH2,V6)0.0000002.000000

VOLUME(WH2,V7)4.0000000.000000

VOLUME(WH2,V8)0.0000000.000000

VOLUME(WH3,V1)0.0000004.000000

VOLUME(WH3,V2)0.0000001.000000

VOLUME(WH3,V3)42.000000.000000

VOLUME(WH3,V4)0.0000002.000000

VOLUME(WH3,V5)0.0000006.000000

VOLUME(WH3,V6)0.0000003.000000

VOLUME(WH3,V7)0.0000007.000000

VOLUME(WH3,V8)0.0000003.000000

VOLUME(WH4,V1)0.0000005.000000

VOLUME(WH4,V2)0.0000004.000000

VOLUME(WH4,V3)0.0000005.000000

VOLUME(WH4,V4)0.0000007.000000

VOLUME(WH4,V5)0.0000007.000000

VOLUME(WH4,V6)14.000000.000000

VOLUME(WH4,V7)0.0000005.000000

VOLUME(WH4,V8)38.000000.000000

VOLUME(WH5,V1)41.000000.000000

VOLUME(WH5,V2)0.0000001.000000

VOLUME(WH5,V3)0.0000004.000000

VOLUME(WH5,V4)0.0000003.000000

VOLUME(WH5,V5)0.0000005.000000

VOLUME(WH5,V6)0.0000000.000000

VOLUME(WH5,V7)0.0000004.000000

VOLUME(WH5,V8)0.0000004.000000

VOLUME(WH6,V1)0.0000004.000000

VOLUME(WH6,V2)0.0000008.000000

VOLUME(WH6,V3)0.0000001.000000

VOLUME(WH6,V4)0.0000001.000000

VOLUME(WH6,V5)0.0000007.000000

VOLUME(WH6,V6)32.000000.000000

VOLUME(WH6,V7)0.0000003.000000

VOLUME(WH6,V8)0.0000003.000000

RowSlackorSurplusDualPrice

1473.0000-1.000000

219.000000.000000

30.0000000.000000

49.0000000.000000

50.0000000.000000

641.000000.000000

76.0000000.000000

839.000000.000000

90.0000001.000000

100.000000-2.000000

110.000000-3.000000

120.000000-1.000000

130.000000-2.000000

140.000000-2.000000

150.000000-1.000000

3、一般整数规划问题:

某服务部门各时段(每2h为一时段)需要的服务员人数见下表。

按规定,服务员连续工作8h(即四个时段)为一班。

现要求安排服务员的工作时间,使服务部门服务员总数最少。

时段

1

2

3

4

5

6

7

8

服务员最少数目

10

8

9

11

13

8

5

3

(1)给出原始代码;

(2)计算结果(决策变量求解结果粘贴)

model:

sets:

time/x1..x8/:

required,start;

endsets

data:

!

每天所需的最少职员数;

required=10891113853;

enddata

!

最小化每周所需职员数;

min=@sum(time:

start);

@for(time(J):

@sum(time(I)|I#le#4:

start(@wrap(J+I+2,8)))>=required(J));

end

结果

Globaloptimalsolutionfound.

Objectivevalue:

23.00000

Totalsolveriterations:

3

 

VariableValueReducedCost

REQUIRED(X1)10.000000.000000

REQUIRED(X2)8.0000000.000000

REQUIRED(X3)9.0000000.000000

REQUIRED(X4)11.000000.000000

REQUIRED(X5)13.000000.000000

REQUIRED(X6)8.0000000.000000

REQUIRED(X7)5.0000000.000000

REQUIRED(X8)3.0000000.000000

START(X1)13.000000.000000

START(X2)0.0000000.000000

START(X3)0.0000000.000000

START(X4)2.0000000.000000

START(X5)8.0000000.000000

START(X6)0.0000000.000000

START(X7)0.0000000.000000

START(X8)0.0000000.000000

RowSlackorSurplusDualPrice

123.00000-1.000000

20.000000-1.000000

30.0000000.000000

44.0000000.000000

52.0000000.000000

60.000000-1.000000

77.0000000.000000

85.0000000.000000

97.0000000.000000

4、指派问题:

已知如下效率矩阵,求极大化指派问题。

 

B1

B2

B3

B4

B5

A1

4

8

7

15

12

A2

7

9

17

14

10

A3

6

9

12

8

7

A4

6

7

14

6

10

A5

6

9

12

10

6

(1)给出原始代码;

(2)计算结果(决策变量求解结果粘贴)

model:

!

5个工人,5个工作的分配问题;

sets:

workers/w1..w5/;

jobs/j1..j5/;

links(workers,jobs):

cost,volume;

endsets

!

目标函数;

min=@sum(links:

cost*volume);

!

每个工人只能有一份工作;

@for(workers(I):

@sum(jobs(J):

volume(I,J))=1;

);

!

每份工作只能有一个工人;

@for(jobs(J):

@sum(workers(I):

volume(I,J))=1;

);

data:

cost=4871512

79171410

691287

6714610

6912106;

enddata

end

答案

Globaloptimalsolutionfound.

Objectivevalue:

34.00000

Totalsolveriterations:

10

 

VariableValueReducedCost

COST(W1,J1)4.0000000.000000

COST(W1,J2)8.0000000.000000

COST(W1,J3)7.0000000.000000

COST(W1,J4)15.000000.000000

COST(W1,J5)12.000000.000000

COST(W2,J1)7.0000000.000000

COST(W2,J2)9.0000000.000000

COST(W2,J3)17.000000.000000

COST(W2,J4)14.000000.000000

COST(W2,J5)10.000000.000000

COST(W3,J1)6.0000000.000000

COST(W3,J2)9.0000000.000000

COST(W3,J3)12.000000.000000

COST(W3,J4)8.0000000.000000

COST(W3,J5)7.0000000.000000

COST(W4,J1)6.0000000.000000

COST(W4,J

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