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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，理解解报告的内容。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软件三实验内容（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: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se
2、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=8 16 12;a=1 2 1 0 04 0 0 1 00 4 0 0 1;end dataend灵敏度分析Ranges in which the basis is unchanged:Objective Coefficient RangesCurrentAllowableAllowableVariableCoefficientI
3、ncreaseDecreaseX( 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当b2在8,32
4、 之间变化时 最优基不变最优解Global optimal solution foundat iteration:0Objective value:14.00000VariableValueReducedCos tB( 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
5、)2.0000000.000000X( 3)0.0000001.500000X( 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(
6、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)1.0000000.000000RowSlackor SurplusDual Price114.000001.00000020.0000001.50000030.0000000.125000040.0000000.000000例题2-11模型MAX2 X(1)+3X(2)SUBJECT TO2X( 1) +2X(2)+X(3)=1234 X( 1)
7、+X(4)=1644 X( 2)+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=12 16 12;a=1 2 1 0 04 0 0 1 00 4 0 0 1;end dataend最优解Global optimal solution found at iteration:2Objective value:17.00000VariableValueR
8、educed CostB( 1)12.000000.000000B( 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)3.0000000.000000X( 3)2.0000000.000000X( 4)0.0000000.5000000X( 5)0.0000000.7500000A( 1, 1)1.0000
9、000.000000A( 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)0.0000000.000000A(3,2)4.0000000.000000A(3,3)0.0000000.000000A(3,4)0.0000000
10、.000000A(3,5)1.0000000.000000RowSlack or SurplusDual Price117.000001.00000020.0000000.00000030.0000000500000040.00000007500000最优解(4,3,2,0,0)最优值z = 17分析Ranges in which theoasis is unchanged:ObjectiveCoefficientRangesCurrentAllowableAllowableVariableCoefficientIncreaseDecreaseX(1)2.000000INFINITY2.000
11、000X(2)3.000000INFINITY3.000000X(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MAX2 X(1)+3X(2)SUBJECT TO2X( 1) +2X(2)+X(3)=834
12、 X( 1)+X(4)=1644 X( 2)+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=23000;b=816 12;a=121004001004001;end dataend灵敏度分析Ranges in which the basis is unchanged:ObjectiveCoefficientRangesCurrentAllowableAllowableVariableCoefficientIncreaseDecrease
