线性规划的灵敏度分析实验报告

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 0 4 0 0 1 0 0 4 0 0 1;end dataend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 0.2500000Righthand 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: 14.00000Variable 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 0.1250000X( 5) 4.000000 0.000000A( 1, 1) 1.000000 0.000000A( 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 0.000000Row 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;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 0 4 0 0 1 0 0 4 0 0 1;end dataend最优解 Global optimal solution found at iteration: 2 Objective value: 17.00000Variable 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 0.000000A( 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 0.000000A( 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;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 0 4 0 0 1 0 0 4 0 0 1;end dataend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 0.2500000Righthand 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;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 5 3 4 5;end dataEnd最优解 Global optimal solution found. Objective value: 270.0000 Infeasibilities: 0.000000 Total solver iterations: 2Variable 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 0.2000000 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;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 8 3 4 5 2;end dataEnd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) 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: _1 MAX= 30 * X_1 + 20 * X_2 + 50 * X_3 ; _2 X_1 + 2 * X_2 + X_3 = 430 ; _3 3 * 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(J)=b(6);data:c=30 20 50;b=430 410 420 300 70 240;a=1 2 1 3 0 2 1 4 0 1 1 1 0 1 0 0 0 1;end dataend最优解 Global optimal solution found. Objective value: 12150.00 Infeasibilities: 0.000000 Total solver iterations: 4 Variable Value Reduced Cost B( 1) 430.0000 0.000000 B( 2) 410.0000 0.000000 B( 3) 420.0000 0.000000 B( 4) 300.0000 0.000000 B( 5) 70.00000 0.000000 B( 6) 240.0000 0.000000 C( 1) 30.00000 0.000000 C( 2) 20.00000 0.000000 C( 3) 50.00000 0.000000 X( 1) 0.000000 35.00000 X( 2) 95.00000 0.000000 X( 3) 205.0000 0.000000 A( 1, 1) 1.000000 0.000000 A( 1, 2) 2.000000 0.000000 A( 1, 3) 1.000000 0.000000 A( 2, 1) 3.000000 0.000000 A( 2, 2) 0.000000 0.000000 A( 2, 3) 2.000000 0.000000 A( 3, 1) 1.000000 0.000000 A( 3, 2) 4.000000 0.000000 A( 3, 3) 0.000000 0.000000 A( 4, 1) 1.000000 0.000000 A( 4, 2) 1.000000 0.000000 A( 4, 3) 1.000000 0.000000 A( 5, 1)