运筹学实验报告4(运输问题的编程求解

1、精选优质文档-倾情为你奉上运筹学实验报告实验课程:运筹学 实验日期: 6月9日 任课教师: 杨小康 班级：信息与计算科学 姓名：刘静 学号 一、实验名称: 线性规划模型的编程求解（2） 二、实验目的：1.掌握Lingo软件中变量定界函数，集合循环函数和IF(条件，true结果，false结果)2. 熟悉运输问题的数学模型，掌握简单运输问题数学模型的Lingo软件编程求解的方法，掌握解报告的内容3. 掌握整数规划、0-1规划和指派问题的计算机求解方法。三、实验要求: 1.给出所求解问题的数学模型；2.给出Lingo中的编程程序；3.能给出最优解和最优值；四、报告正文（文挡，数据，模型，程序，图形

2、）:1利用lingo软件求解下面运输问题的最优解。单位 销地运价产地B1B2B3B4产量A13113107A219284A3741059销量3656 程序：model:sets: AH/A1.A3/:A; BD/B1.B4/:B; link(AH,BD):C,X; endsetsdata:A=7,4,9;B=3,6,5,6;C=3 11 3 10 1 9 2 8 7 4 10 5;enddatamin=sum(link:c*x);for(AH(i):sum(BD(j):x(i,j)<=A(i);for(BD(j):sum(AH(i):x(i,j)=B(j);for(link:g

3、in(x);end结果：z=85;Global optimal solution found. Objective value: 85.00000 Objective bound: 85.00000 Infeasibilities: 0. Extended solver steps: 0 Total solver iterations: 0 Variable Value Reduced Cost A( A1) 7. 0. A( A2) 4. 0. A( A3) 9. 0. B( B1) 3. 0. B( B2) 6. 0. B( B3) 5. 0. B( B4) 6. 0. C( A1, B1

4、) 3. 0. C( A1, B2) 11.00000 0. C( A1, B3) 3. 0. C( A1, B4) 10.00000 0. C( A2, B1) 1. 0. C( A2, B2) 9. 0. C( A2, B3) 2. 0. C( A2, B4) 8. 0. C( A3, B1) 7. 0. C( A3, B2) 4. 0. C( A3, B3) 10.00000 0. C( A3, B4) 5. 0. X( A1, B1) 0. 3. X( A1, B2) 0. 11.00000 X( A1, B3) 5. 3. X( A1, B4) 2. 10.00000 X( A2,

5、B1) 3. 1. X( A2, B2) 0. 9. X( A2, B3) 0. 2. X( A2, B4) 1. 8. X( A3, B1) 0. 7. X( A3, B2) 6. 4. X( A3, B3) 0. 10.00000 X( A3, B4) 3. 5. Row Slack or Surplus Dual Price 1 85.00000 -1. 2 0. 0. 3 0. 0. 4 0. 0. 5 0. 0. 6 0. 0. 7 0. 0. 8 0. 0.2.求解下面混合整数规划min z=-3x1+4x2-2x3+5x4;4x1-x2+2x3-x4=-2;x1+x2+3x3-x

6、414;-2x1+3x2-x3+2x42;x1为整数，x2=0或1,x30,x4无约束;程序：model:sets: location1/1.4/:A,x; location2/1.3/:B; link(location2,location1):C; endsetsdata:A=-3,4,-2,5;B=-2,14,2;C=4,-1,2,-1 1,1,3,-1 -2,3,-1,2;enddatamin=sum(location1:A*x);sum(location1(j):C(1,j)*x(j)=-2;sum(location1(j):C(2,j)*x(j)<=14;sum(locatio

7、n1(j):C(3,j)*x(j)>=2;gin(x(1);bin(x(2);gin(x(3);free(x(4);end结果：z=9; Global optimal solution found. Objective value: 9. Objective bound: 9. Infeasibilities: 0. Extended solver steps: 0 Total solver iterations: 6 Model Class: MILP Total variables: 4 Nonlinear variables: 0 Integer variables: 3 Tota

8、l constraints: 4 Nonlinear constraints: 0 Total nonzeros: 16 Nonlinear nonzeros: 0 Variable Value Reduced Cost A( 1) -3. 0. A( 2) 4. 0. A( 3) -2. 0. A( 4) 5. 0. X( 1) 0. 17.00000 X( 2) 1. -1. X( 3) 0. 8. X( 4) 1. 0. B( 1) -2. 0. B( 2) 14.00000 0. B( 3) 2. 0. C( 1, 1) 4. 0. C( 1, 2) -1. 0. C( 1, 3) 2

9、. 0. C( 1, 4) -1. 0. C( 2, 1) 1. 0. C( 2, 2) 1. 0. C( 2, 3) 3. 0. C( 2, 4) -1. 0. C( 3, 1) -2. 0. C( 3, 2) 3. 0. C( 3, 3) -1. 0. C( 3, 4) 2. 0. Row Slack or Surplus Dual Price 1 9. -1. 2 0. 5. 3 14.00000 0. 4 3. 0.3. 现在要在五个工人中确定四个人来分别完成四项工作中的一项工作。由于每个工人的技术特长不同，他们完成各项工作所需的工时也不同。每个工人完成各项工作所需工时如下表所示，试找

10、出一个工作分配方案，使总工时最小。工作工人ABCD943746565475752310674程序：model:sets: worker/1.5/:A; job/1.4/:B; link(worker,job):C,X; endsetsdata:C=9 4 3 7 4 6 5 6 5 4 7 5 7 5 2 3 10 6 7 4;enddatamin=sum(link:c*x);for(worker(i):sum(job(j):x(i,j)<=1);for(job(j):sum(worker(i):x(i,j)=1);for(link:bin(x);end结果：z=3+4+3+4=14;

11、第一个人做第三个工作X( 1, 3) ，第二个人做第一个工作 X( 2, 1)，第三个人做第二个工作 X( 3, 2)，第四个人做第四个工作 X( 4, 4)； Global optimal solution found. Objective value: 14.00000 Objective bound: 14.00000 Infeasibilities: 0. Extended solver steps: 0 Total solver iterations: 0 Variable Value Reduced Cost A( 1) 0. 0. A( 2) 0. 0. A( 3) 0. 0.

12、A( 4) 0. 0. A( 5) 0. 0. B( 1) 0. 0. B( 2) 0. 0. B( 3) 0. 0. B( 4) 0. 0. C( 1, 1) 9. 0. C( 1, 2) 4. 0. C( 1, 3) 3. 0. C( 1, 4) 7. 0. C( 2, 1) 4. 0. C( 2, 2) 6. 0. C( 2, 3) 5. 0. C( 2, 4) 6. 0. C( 3, 1) 5. 0. C( 3, 2) 4. 0. C( 3, 3) 7. 0. C( 3, 4) 5. 0. C( 4, 1) 7. 0. C( 4, 2) 5. 0. C( 4, 3) 2. 0. C(

13、4, 4) 3. 0. C( 5, 1) 10.00000 0. C( 5, 2) 6. 0. C( 5, 3) 7. 0. C( 5, 4) 4. 0. X( 1, 1) 0. 9. X( 1, 2) 0. 4. X( 1, 3) 1. 3. X( 1, 4) 0. 7. X( 2, 1) 1. 4. X( 2, 2) 0. 6. X( 2, 3) 0. 5. X( 2, 4) 0. 6. X( 3, 1) 0. 5. X( 3, 2) 1. 4. X( 3, 3) 0. 7. X( 3, 4) 0. 5. X( 4, 1) 0. 7. X( 4, 2) 0. 5. X( 4, 3) 0.

14、2. X( 4, 4) 1. 3. X( 5, 1) 0. 10.00000 X( 5, 2) 0. 6. X( 5, 3) 0. 7. X( 5, 4) 0. 4. Row Slack or Surplus Dual Price 1 14.00000 -1. 2 0. 0. 3 0. 0. 4 0. 0. 5 0. 0. 6 1. 0. 7 0. 0. 8 0. 0. 9 0. 0. 10 0. 0.4 某市消防队布点问题。该市共有6个区，每个区都可以建消防站，市政府希望设置的消防站最少，但必须满足在城市任何地区发生火警时，消防车要在15min内赶到现场。据实地测定，各区之间消防车行驶的时间

15、见表，请制定一个布点最少的计划。地区1地区2地区3地区4地区5地区6地区1地区2地区3地区4地区5地区6010162827201002432171016240122721283212015252717271501420102125140满足条件的约束：x1+x2>=1;x1+x2+x6>=1;x3+x4>=1;x3+x4+x5>=1;x4+x5>=1;x2+x6>=1;程序：model:sets: place1/1.6/:x; place2/1.6/; link(place2,place1):C; endsetsdata:C=1 1 0 0 0 0 1 1

16、0 0 0 1 0 0 1 1 0 0 0 0 1 1 1 0 0 0 0 1 1 0 0 1 0 0 0 1;enddatamin=sum(place1:x);for(place2(i):sum(place1(j):C(i,j)*x(j)>=1);for(place1:bin(x);end结果：z=2；选二和四地区Global optimal solution found. Objective value: 2. Objective bound: 2. Infeasibilities: 0. Extended solver steps: 0 Total solver iteration

17、s: 0 Model Class: PILP Total variables: 6 Nonlinear variables: 0 Integer variables: 6 Total constraints: 7 Nonlinear constraints: 0 Total nonzeros: 20 Nonlinear nonzeros: 0 Variable Value Reduced Cost X( 1) 0. 1. X( 2) 1. 1. X( 3) 0. 1. X( 4) 1. 1. X( 5) 0. 1. X( 6) 0. 1. C( 1, 1) 1. 0. C( 1, 2) 1.

18、0. C( 1, 3) 0. 0. C( 1, 4) 0. 0. C( 1, 5) 0. 0. C( 1, 6) 0. 0. C( 2, 1) 1. 0. C( 2, 2) 1. 0. C( 2, 3) 0. 0. C( 2, 4) 0. 0. C( 2, 5) 0. 0. C( 2, 6) 1. 0. C( 3, 1) 0. 0. C( 3, 2) 0. 0. C( 3, 3) 1. 0. C( 3, 4) 1. 0. C( 3, 5) 0. 0. C( 3, 6) 0. 0. C( 4, 1) 0. 0. C( 4, 2) 0. 0. C( 4, 3) 1. 0. C( 4, 4) 1. 0. C( 4, 5) 1. 0. C( 4,