线性规划的灵敏度分析实验报告

1、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，理解解报告的容。 初步掌握对实际的线性规划问题建立数学模型，并利用计算机求解分析的一般方法。二 实验环境Lingo 软件三 实验容（包括数学模型、上机程序、实验结果、结果分析与问题解答等）例题 2-10MODEL:_1MAX= 2 * X_1 + 3 * X_2 ;_2X_1+2*X_2+X_3=8;_34*X_1+X_4=16;_44*X_2+X_5=12;END编程sets:is/1.3/:b;js/1

2、.5/:c,x;links(is,js):a;endsetsmax = 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 04001004001;enddataend灵敏度分析Ranges in which the basis is unchanged:Objective Coefficient RangesCurrentAllowableAllowableVariableCoefficientIncreaseDecreaseX( 1)2.000000I

3、NFINITY0.5000000X( 2)3.0000001.0000003.000000X( 3)0.01.500000INFINITYX( 4)0.00.1250000INFINITYX( 5)0.00.75000000.2500000Righthand Side RangesRowCurrentAllowableAllowableRHSIncreaseDecrease28.0000002.0000004.000000316.0000016.000008.000000412.00000INFINITY4.000000当 b2在 8,32之间变化时最优基不变最优解Global optimal

4、 solution found at iteration:0Objective value:14.00000VariableValueReduced CostB( 1)8.0000000.000000B( 2)16.000000.000000B( 3)12.000000.000000C( 1)2.0000000.000000C( 2)3.0000000.000000C( 3)0.0000000.000000C( 4)0.0000000.000000C( 5)0.0000000.000000X( 1)4.0000000.000000X( 2)2.0000000.000000X( 3)0.0000

5、001.500000X( 4)0.0000000.1250000X( 5)4.0000000.000000A( 1, 1)1.0000000.000000A( 1, 2)2.0000000.000000A( 1, 3)1.0000000.000000A( 1, 4)0.0000000.000000A( 1, 5)0.0000000.000000A( 2, 1)4.0000000.000000A( 2, 2)0.0000000.000000A( 2, 3)0.0000000.000000A( 2, 4)1.0000000.000000A( 2, 5)0.0000000.000000A( 3, 1

6、)0.0000000.000000A( 3, 2)4.0000000.000000A( 3, 3)0.0000000.000000A( 3, 4)0.0000000.000000A( 3, 5)1.0000000.000000RowSlack or Surplus Dual Price114.000001.00000020.0000001.50000030.0000000.125000040.0000000.000000例题 2-11模型MAX2X(1)+3X(2)SUBJECT TO2X(1)+2X(2)+X(3)= 1234X(1)+X(4)=1644X(2)+X(5)=12END编程se

7、ts:is/1.3/:b;js/1.5/:c,x;links(is,js):a;endsetsmax =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 04001004001;enddataend最优解Global optimal solution found at iteration:2Objective value:17.00000VariableValueReduced CostB( 1)12.000000.000000B( 2)16.0000

8、00.000000B( 3)12.000000.000000C( 1)2.0000000.000000C( 2)3.0000000.000000C( 3)0.0000000.000000C( 4)0.0000000.000000C( 5)0.0000000.000000X( 1)4.0000000.000000X( 2)3.0000000.000000X( 3)2.0000000.000000X( 4)0.0000000.5000000X( 5)0.0000000.7500000A( 1, 1)1.0000000.000000A( 1, 2)2.0000000.000000A( 1, 3)1.

9、0000000.000000A( 1, 4)0.0000000.000000A( 1, 5)0.0000000.000000A( 2, 1)4.0000000.000000A( 2, 2)0.0000000.000000A( 2, 3)0.0000000.000000A( 2, 4)1.0000000.000000A( 2, 5)0.0000000.000000A( 3, 1)0.0000000.000000A( 3, 2)4.0000000.000000A( 3, 3)0.0000000.000000A( 3, 4)0.0000000.000000A( 3, 5)1.0000000.0000

10、00RowSlack or SurplusDual Price117.000001.00000020.0000000.00000030.0000000.500000040.0000000.7500000最优解（ 4,3,2,0,0）最优值 z=17分析Ranges in which the basis is unchanged:Objective Coefficient RangesCurrentAllowableAllowableVariableCoefficientIncreaseDecreaseX( 1)2.000000INFINITY2.000000X( 2)3.000000INFIN

11、ITY3.000000X( 3)0.01.500000INFINITYX( 4)0.00.5000000INFINITYX( 5)0.00.7500000INFINITYRighthand Side RangesRowCurrentAllowableAllowableRHSIncreaseDecrease212.00000INFINITY2.000000316.000008.00000016.00000412.000004.00000012.00000例题 2-12模型MAX2X(1)+3X(2)SUBJECT TO2X(1)+2X(2)+X(3)= 834X(1)+X(4)=1644X(2)

12、+X(5)=12END编程sets:is/1.3/:b;js/1.5/:c,x;links(is,js):a;endsetsmax =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 04001004001;enddataend灵敏度分析Ranges in which the basis is unchanged:Objective Coefficient RangesCurrentAllowableAllowableVariableCoefficien

13、tIncreaseDecreaseX( 1)2.000000INFINITY0.5000000X( 2)3.0000001.0000003.000000X( 3)0.01.500000INFINITYX( 4)0.00.1250000INFINITYX( 5)0.00.75000000.2500000Righthand Side RangesRowCurrentAllowableAllowableRHSIncreaseDecrease28.0000002.0000004.000000316.0000016.000008.000000412.00000INFINITY4.000000由灵敏度分析

14、表知道C2 在【 0,4】之间变化时，最优基不变。第六题模型MODEL:_1MAX= 3 * X_1 + X_2 + 4 * X_3 ;_26*X_1+3*X_2+5*X_3=450;_33*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;endsetsmax =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 5345;enddataEnd最优解Global optimal solu

15、tion found.Objective value:270.0000Infeasibilities:0.000000Total solver iterations:2VariableValueReduced CostB( 1)450.00000.000000B( 2)300.00000.000000C( 1)3.0000000.000000C( 2)1.0000000.000000C( 3)4.0000000.000000X( 1)50.000000.000000X( 2)0.0000002.000000X( 3)30.000000.000000A( 1, 1)6.0000000.00000

16、0A( 1, 2)3.0000000.000000A( 1, 3)5.0000000.000000A( 2, 1)3.0000000.000000A( 2, 2)4.0000000.000000A( 2, 3)5.0000000.000000RowSlack or SurplusDual Price1270.00001.00000020.0000000.200000030.0000000.6000000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围的灵敏度分析结果Ranges in which the basis is unchang

17、ed:Objective Coefficient RangesCurrentAllowableAllowableVariableCoefficientIncreaseDecreaseX( 1)3.0000001.8000000.6000000X( 2)1.0000002.000000INFINITYX( 3)4.0000001.0000001.500000Righthand Side RangesRowCurrentAllowableAllowableRHSIncreaseDecrease2450.0000150.0000150.00003300.0000150.000075.00000当 A

18、 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:_1MAX=3*X_1+X_2+4*X_3+3*X_4;_26*X_1+3*X_2+5*X_3+8*X_4=450;_33*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;endsetsmax =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 83452;enddataEn

19、d最优解Global optimal solution found.Objective value:275.0000Infeasibilities:0.000000Total solver iterations:2VariableValueReduced CostB( 1)450.00000.000000B( 2)300.00000.000000C( 1)3.0000000.000000C( 2)1.0000000.000000C( 3)4.0000000.000000C( 4)3.0000000.000000X( 1)0.0000000.1000000X( 2)0.0000001.96666

20、7X( 3)50.000000.000000X( 4)25.000000.000000A( 1, 1)6.0000000.000000A( 1, 2)3.0000000.000000A( 1, 3)5.0000000.000000A( 1, 4)8.0000000.000000A( 2, 1)3.0000000.000000A( 2, 2)4.0000000.000000A( 2, 3)5.0000000.000000A( 2, 4)2.0000000.000000RowSlack or SurplusDual Price1275.00001.00000020.0000000.23333333

21、0.0000000.5666667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围的灵敏度分析结果Ranges in which the basis is unchanged:Objective Coefficient RangesCurrentAllowableAllowableVariableCoefficientIncreaseDecreaseX( 1)3.0000001.8000000.6000000X( 2)1.0000002.000000INFINITYX( 3)4.0000001.0000001.500000Righthand Side RangesRowCurre

22、ntAllowableAllowableRHSIncreaseDecrease2450.0000150.0000150.00003300.0000150.000075.00000当购买150 吨时此时可买360 元 在减去购买150 吨的进价60 元 此时可获利300 超过了原计划， 应该购买。第七题模型MODEL:_1MAX= 30 * X_1 + 20 * X_2 + 50 * X_3 ;_2 X_1 + 2 * X_2 + X_3 = 430 ;_33*X_1+2*X_3=410;_4 X_1 + 4 * X_2 = 420 ;_5 X_1 + X_2 + X_3 = 70 ;_7 X_

23、3 = 240 ;END编程sets:is/1.6/:b;js/1.3/:c,x;links(is,js):a;endsetsmax =sum (js(J):c(J)*x(J);sum (js(J):a(1,J)*x(J)=b(1);sum (js(J):a(2,J)*x(J)=b(2);sum (js(J):a(3,J)*x(J)=b(3);sum (js(J):a(4,J)*x(J)=B(5);sum (js(J):a(6,J)*x(J)=b(6);data:c=30 20 50;b=430 410 420 300 70 240;a=1 2 13 0 21 4 01 1 10 1 0001

24、;enddataend最优解Global optimal solution found.Objective value:12150.00Infeasibilities:0.000000Total solver iterations:4VariableValueReduced CostB( 1)430.00000.000000B( 2)410.00000.000000B( 3)420.00000.000000B( 4)300.00000.000000B( 5)70.000000.000000B( 6)240.00000.000000C( 1)30.000000.000000C( 2)20.000

25、000.000000C( 3)50.000000.000000X( 1)0.00000035.00000X( 2)95.000000.000000X( 3)205.00000.000000A( 1, 1)1.0000000.000000A( 1, 2)2.0000000.000000A( 1, 3)1.0000000.000000A( 2, 1)3.0000000.000000A( 2, 2)0.0000000.000000A( 2, 3)2.0000000.000000A( 3, 1)1.0000000.000000A( 3, 2)4.0000000.000000A( 3, 3)0.0000

26、000.000000A( 4, 1)1.0000000.000000A( 4, 2)1.0000000.000000A( 4, 3)1.0000000.000000A( 5, 1)0.0000000.000000A( 5, 2)1.0000000.000000A( 5, 3)0.0000000.000000A( 6, 1)0.0000000.000000A( 6, 2)0.0000000.000000A( 6, 3)1.0000000.000000RowSlack or SurplusDual Price112150.001.000000235.000000.00000030.00000015

27、.00000440.000000.00000050.00000020.00000625.000000.000000735.000000.000000最优解（ 0 95 205）最优值12150第一问模型MODEL:_1MAX= 30 * X_1 + 20 * X_2 + 60 * X_3 ;_2 X_1 + 2 * X_2 + X_3 = 430 ;_33*X_1+2*X_3=410;_4 X_1 + 4 * X_2 = 420 ;_5 X_1 + X_2 + X_3 = 70 ;_7 X_3 = 190 ;END编程sets:is/1.6/:b;js/1.3/:c,x;links(is,js

28、):a;endsetsmax =sum (js(J):c(J)*x(J);sum (js(J):a(1,J)*x(J)=b(1);sum (js(J):a(2,J)*x(J)=b(2);sum (js(J):a(3,J)*x(J)=b(3);sum (js(J):a(4,J)*x(J)=B(5);sum (js(J):a(6,J)*x(J)=b(6);data:c=30 20 60;b=430 410 420 300 70 190;a=1 2 13 0 21 4 01 1 10 1 0001;enddataend最优解Global optimal solution found.Objectiv

29、e value:13700.00Infeasibilities:0.000000Total solver iterations:4VariableValueReduced CostB( 1)430.00000.000000B( 2)410.00000.000000B( 3)420.00000.000000B( 4)300.00000.000000B( 5)70.000000.000000B( 6)190.00000.000000C( 1)30.000000.000000C( 2)20.000000.000000C( 3)60.000000.000000X( 1)10.000000.000000

30、X( 2)100.00000.000000X( 3)190.00000.000000A( 1, 1)1.0000000.000000A( 1, 2)2.0000000.000000A( 1, 3)1.0000000.000000A( 2, 1)3.0000000.000000A( 2, 2)0.0000000.000000A( 2, 3)2.0000000.000000A( 3, 1)1.0000000.000000A( 3, 2)4.0000000.000000A( 3, 3)0.0000000.000000A( 4, 1)1.0000000.000000A( 4, 2)1.0000000.

31、000000A( 4, 3)1.0000000.000000A( 5, 1)0.0000000.000000A( 5, 2)1.0000000.000000A( 5, 3)0.0000000.000000A( 6, 1)0.0000000.000000A( 6, 2)0.0000000.000000A( 6, 3)1.0000000.000000RowSlack or SurplusDual Price113700.001.000000230.000000.00000030.0000003.333333410.000000.00000050.00000020.00000630.000000.0

32、0000070.00000033.33333最优解（ 10 ， 100,190）最优值13700 ；可行。第二问由原问题的单个价值系数和右端系数变化围的灵敏度分析结果得Ranges in which the basis is unchanged:Objective Coefficient RangesCurrentAllowableAllowableVariableCoefficientIncreaseDecreaseX( 1)30.0000035.00000INFINITYX( 2)20.0000030.0000020.00000X( 3)50.00000INFINITY23.33333Ri

33、ghthand Side RangesRowCurrentAllowableAllowableRHSIncreaseDecrease2430.0000INFINITY35.000003410.000050.0000020.000004420.0000INFINITY40.000005300.000010.0000025.00000670.0000025.00000INFINITY7240.0000INFINITY35.00000当 C2增加到 310 时此时模型MODEL:_1MAX= 30 * X_1 + 20 * X_2 + 50 * X_3 ;_2 X_1 + 2 * X_2 + X_3

34、 = 430 ;_33*X_1+2*X_3=410;_4 X_1 + 4 * X_2 = 420 ;_5 X_1 + X_2 + X_3 = 70 ;_7 X_3 = 240 ;END编程sets:is/1.6/:b;js/1.3/:c,x;links(is,js):a;endsetsmax =sum (js(J):c(J)*x(J);sum (js(J):a(1,J)*x(J)=b(1);sum (js(J):a(2,J)*x(J)=b(2);sum (js(J):a(3,J)*x(J)=b(3);sum (js(J):a(4,J)*x(J)=B(5);sum (js(J):a(6,J)*x

35、(J)=b(6);data:c=30 20 50;b=430 410 420 310 70 240;a=1 2 13 0 21 4 01 1 10 1 0001;enddataend此时最优解Global optimal solution found.Objective value:12350.00Infeasibilities:0.000000Total solver iterations:4VariableValueReduced CostB( 1)430.00000.000000B( 2)410.00000.000000B( 3)420.00000.000000B( 4)310.0000

36、0.000000B( 5)70.000000.000000B( 6)240.00000.000000C( 1)30.000000.000000C( 2)20.000000.000000C( 3)50.000000.000000X( 1)0.00000050.00000X( 2)105.00000.000000X( 3)205.00000.000000A( 1, 1)1.0000000.000000A( 1, 2)2.0000000.000000A( 1, 3)1.0000000.000000A( 2, 1)3.0000000.000000A( 2, 2)0.0000000.000000A( 2

37、, 3)2.0000000.000000A( 3, 1)1.0000000.000000A( 3, 2)4.0000000.000000A( 3, 3)0.0000000.000000A( 4, 1)1.0000000.000000A( 4, 2)1.0000000.000000A( 4, 3)1.0000000.000000A( 5, 1)0.0000000.000000A( 5, 2)1.0000000.000000A( 5, 3)0.0000000.000000A( 6, 1)0.0000000.000000A( 6, 2)0.0000000.000000A( 6, 3)1.000000

38、0.000000RowSlack or Surplus Dual Price112350.001.000000215.000000.00000030.00000025.0000040.0000005.00000050.0000000.000000635.000000.000000735.000000.000000即（ 0，105,205）最优值为 12350，此时的最优值减去增加的价格150 ，得到最终的利润 12350-150=12200可行。第三问模型MODEL:_1MAX= 30 * X_1 + 20 * X_2 + 50 * X_3 ;_2 X_1 + 2 * X_2 + X_3 =

39、470 ;_33*X_1+2*X_3=450;_4 X_1 + 4 * X_2 = 420 ;_5 X_1 + X_2 + X_3 = 70 ;_7 X_3 = 240 ;END编程sets:is/1.6/:b;js/1.3/:c,x;links(is,js):a;endsetsmax =sum (js(J):c(J)*x(J);sum (js(J):a(1,J)*x(J)=b(1);sum (js(J):a(2,J)*x(J)=b(2);sum (js(J):a(3,J)*x(J)=b(3);sum (js(J):a(4,J)*x(J)=B(5);sum (js(J):a(6,J)*x(J)

40、=b(6);data:c=30 20 50;b=470 450 420 300 70 240;a=1 2 13 0 21 4 01 1 10 1 0001;enddataend最优解Global optimal solution found.Objective value:12750.00Infeasibilities:0.000000Total solver iterations:4VariableValueReduced CostB( 1)470.00000.000000B( 2)450.00000.000000B( 3)420.00000.000000B( 4)300.00000.000

41、000B( 5)70.000000.000000B( 6)240.00000.000000C( 1)30.000000.000000C( 2)20.000000.000000C( 3)50.000000.000000X( 1)0.00000035.00000X( 2)75.000000.000000X( 3)225.00000.000000A( 1, 1)1.0000000.000000A( 1, 2)2.0000000.000000A( 1, 3)1.0000000.000000A( 2, 1)3.0000000.000000A( 2, 2)0.0000000.000000A( 2, 3)2

42、.0000000.000000A( 3, 1)1.0000000.000000A( 3, 2)4.0000000.000000A( 3, 3)0.0000000.000000A( 4, 1)1.0000000.000000A( 4, 2)1.0000000.000000A( 4, 3)1.0000000.000000A( 5, 1)0.0000000.000000A( 5, 2)1.0000000.000000A( 5, 3)0.0000000.000000A( 6, 1)0.0000000.000000A( 6, 2)0.0000000.000000A( 6, 3)1.0000000.000

43、000RowSlack or Surplus Dual Price112750.001.000000295.000000.00000030.00000015.000004120.00000.00000050.00000020.0000065.0000000.000000715.000000.000000此时最优解（ 0,75,225）最优值12750 此时的最优值 12750-700=12050，即 12050便是此时的利润，不可行第四问模型MODEL:_1MAX= 30 * X_1 + 20 * X_2 + 50 * X_3 ;_2 X_1 + 2 * X_2 + X_3 = 430 ;_33*X_1+2*X_3=410;_4 X_1 + 4 * X_2 = 420 ;_5 X_1 + X_2 + X_3 = 300 ;_6 X_2 = 100 ;_7 X_3 = 240 ;END编程sets:is/1.6/:b;js/1.3/:c,x;links(is,js):a;endsetsm