《LINGO实习报告》word版.doc

Lingo实习报告姓名：张永桥班级：信息计算101801班学号：201018030131LINGO实习报告1. 直接用LINGO来解如下二次规划问题： 程序： max=98*x1+277*x2-x12-2*x22-0.3*x1*x2; x1+x2=100;x1=2*x2;gin(x1);gin(x2); 2. 例 SAILCO公司需要决定下四个季度的帆船生产量。下四个季度的帆船需求量分别是40条，60条，75条，25条，这些需求必须按时满足。每个季度正常的生产能力是40条帆船，每条船的生产费用为400美元。如果加班生产，每条船的生产费用为450美元。每个季度末，每条船的库存费用为20美元。假定生产提前期为0，初始库存为10条船。如何安排生产可使总费用最小？程序：sets:quarters/1,2,3,4/:dem,rp,op,inv;endsetsmin=sum(quarters:400*rp+450*op+20*inv);for(quarters(i):rp(i)40);for(quarters(i)|i#gt#1: inv(i)=inv(i-1)+rp(i)+op(i)-dem(i););inv(1)=10+rp(1)+op(1)+-dem(1);data:dem=40,60,75,25;enddata运行结果： Global optimal solution found. Objective value: 145750.0 Infeasibilities: 0.000000 Total solver iterations: 4 Variable Value Reduced Cost XQ( 1) 40.00000 0.000000 XQ( 2) 60.00000 0.000000 XQ( 3) 75.00000 0.000000 XQ( 4) 25.00000 0.000000 XQ( 5) 30.00000 0.000000 XQ( 6) 65.00000 0.000000 XQ( 7) 50.00000 0.000000 XQ( 8) 20.00000 0.000000 ZC( 1) 40.00000 0.000000 ZC( 2) 40.00000 0.000000 ZC( 3) 40.00000 0.000000 ZC( 4) 40.00000 0.000000 ZC( 5) 40.00000 0.000000 ZC( 6) 40.00000 0.000000 ZC( 7) 40.00000 0.000000 ZC( 8) 20.00000 0.000000 JB( 1) 0.000000 20.00000 JB( 2) 10.00000 0.000000 JB( 3) 35.00000 0.000000 JB( 4) 0.000000 40.00000 JB( 5) 0.000000 20.00000 JB( 6) 0.000000 0.000000 JB( 7) 10.00000 0.000000 JB( 8) 0.000000 50.00000 KC( 1) 10.00000 0.000000 KC( 2) 0.000000 20.00000 KC( 3) 0.000000 60.00000 KC( 4) 15.00000 0.000000 KC( 5) 25.00000 0.000000 KC( 6) 0.000000 20.00000 KC( 7) 0.000000 70.00000 KC( 8) 0.000000 420.0000 Row Slack or Surplus Dual Price 1 145750.0 -1.000000 2 0.000000 30.00000 3 0.000000 50.00000 4 0.000000 50.00000 5 0.000000 10.00000 6 0.000000 30.00000 7 0.000000 50.00000 8 0.000000 50.00000 9 20.00000 0.000000 10 0.000000 430.0000 11 0.000000 450.0000 12 0.000000 450.0000 13 0.000000 410.0000 14 0.000000 430.0000 15 0.000000 450.0000 16 0.000000 450.0000 17 0.000000 400.00003.例3.4 建筑工地的位置(用平面坐标a, b表示，距离单位：公里)及水泥日用量d(吨)下表给出。有两个临时料场位于P (5,1), Q (2, 7),日储量各有20吨。从A, B两料场分别向各工地运送多少吨水泥，使总的吨公里数最小。两个新的料场应建在何处，节省的吨公里数有多大？a1.258.750.55.7537.25b1.250.754.7556.57.75d3547611程序：model: sets:gdjh/1.6/:a,b,d;lcjh/1,2/:x,y,e;gdlcjh(gdjh,lcjh):c; endsets data:a=1.25,8.75,0.5,5.75,3,7.25;b=1.25,0.75,4.75,5,6.5,7.75;d=3,5,4,7,6,11;x,y=5,1,2,7;e=20,20;enddata min=sum(gdlcjh(i,j):c(i,j)*(x(j)-a(i)2+(y(j)-b(i)2)0.5); for(gdjh(i):sum(lcjh(j):c(i,j)=d(i); for(lcjh(j):sum(gdjh(i):c(i,j)=e(j);end 运行结果： Global optimal solution found. Objective value: 136.2275 Infeasibilities: 0.000000 Total solver iterations: 1 Variable Value Reduced Cost A( 1) 1.250000 0.000000 A( 2) 8.750000 0.000000 A( 3) 0.5000000 0.000000 A( 4) 5.750000 0.000000 A( 5) 3.000000 0.000000 A( 6) 7.250000 0.000000 B( 1) 1.250000 0.000000 B( 2) 0.7500000 0.000000 B( 3) 4.750000 0.000000 B( 4) 5.000000 0.000000 B( 5) 6.500000 0.000000 B( 6) 7.750000 0.000000 D( 1) 3.000000 0.000000 D( 2) 5.000000 0.000000 D( 3) 4.000000 0.000000 D( 4) 7.000000 0.000000 D( 5) 6.000000 0.000000 D( 6) 11.00000 0.000000 X( 1) 5.000000 0.000000 X( 2) 2.000000 0.000000 Y( 1) 1.000000 0.000000 Y( 2) 7.000000 0.000000 E( 1) 20.00000 0.000000 E( 2) 20.00000 0.000000 C( 1, 1) 3.000000 0.000000 C( 1, 2) 0.000000 3.852207 C( 2, 1) 5.000000 0.000000 C( 2, 2) 0.000000 7.252685 C( 3, 1) 0.000000 1.341700 C( 3, 2) 4.000000 0.000000 C( 4, 1) 7.000000 0.000000 C( 4, 2) 0.000000 1.992119 C( 5, 1) 0.000000 2.922492 C( 5, 2) 6.000000 0.000000 C( 6, 1) 1.000000 0.000000 C( 6, 2) 10.00000 0.000000 Row Slack or Surplus Dual Price 1 136.2275 -1.000000 2 0.000000 -3.758324 3 0.000000 -3.758324 4 0.000000 -4.515987 5 0.000000 -4.069705 6 0.000000 -2.929858 7 0.000000 -7.115125 8 4.000000 0.000000 9 0.000000 1.8118244.例 (最短路问题) 在纵横交错的公路网中，货车司机希望找到一条从一个城市到另一个城市的最短路. 下图表示的是公路网, 节点表示货车可以停靠的城市,弧上的权表示两个城市之间的距离(百公里). 那么,货车从城市S出发到达城市T,如何选择行驶路线,使所经过的路程最短?STA1 A2 A3 B1 B2 C1 C2 633665874678956程序：model: sets:city/s,a1,a2,a3,b1,b2,c1,c2,t/:l;gljh(city,city)/s,a1 s,a2 s,a3 a1,b1 a1,b2 a2,b1 a2,b2 a3,b1 a3,b2 b1,c1 b1,c2 b2,c1 b2,c2 c1,t c2,t/:d; endsets data:l=0,;d=6,3,3,6,5,8,6,7,4,6,7,8,9,5,6;enddata for(city(i)|i#gt#1:l(i)=min(gljh(j,i):l(j)+d(j,i);end 运行结果：Feasible solution found. Total solver iterations: 0 Variable Value L( S) 0.000000 L( A1) 6.000000 L( A2) 3.000000 L( A3) 3.000000 L( B1) 10.00000 L( B2) 7.000000 L( C1) 15.00000 L( C2) 16.00000 L( T) 20.00000 D( S, A1) 6.000000 D( S, A2) 3.000000 D( S, A3) 3.000000 D( A1, B1) 6.000000 D( A1, B2) 5.000000 D( A2, B1) 8.000000 D( A2, B2) 6.000000 D( A3, B1) 7.000000 D( A3, B2) 4.000000 D( B1, C1) 6.000000 D( B1, C2) 7.000000 D( B2, C1) 8.000000 D( B2, C2) 9.000000 D( C1, T) 5.000000 D( C2, T) 6.000000 Row Slack or Surplus 1 0.000000 2 0.000000 3 0.000000 4 0.000000 5 0.000000 6 0.000000 7 0.000000 8 0.0000005.例 某班8名同学准备分成4个调查队(每队两人)前往4个地区进行社会调查。这8名同学两两之间组队的效率如下表所示(由于对称性，只列出了严格上三角部分)，问如何组队可以使总效率最高？学生S1S2S3S4S5S6S7S8S1-9342156S2-173521S3-44292S4-1552S5-876S6-23S7-4程序：model: sets:xsjh/1.8/;xljh(xsjh,xsjh)|&2#gt#&1:xl,y; endsets data:xl=9,3,4,2,1,5,6,1,7,3,5,2,1,4,4,2,9,2,1,5,5,2,8,7,6,2,3,4;enddata max=sum(xljh(i,j):xl(i,j)*y(i,j);for(xsjh(k): sum(xljh(i,j)|(i#eq#k)#or#(j#eq#k):y(i,j)=1);for(xljh(i,j):bin(y(i,j);end 运行程序：Global optimal solution found. Objective value: 30.00000 Objective bound: 30.00000 Infeasibilities: 0.000000 Extended solver steps: 0 Total solver iterations: 0 Variable Value Reduced Cost XL( 1, 2) 9.000000 0.000000 XL( 1, 3) 3.000000 0.000000 XL( 1, 4) 4.000000 0.000000 XL( 1, 5) 2.000000 0.000000 XL( 1, 6) 1.000000 0.000000 XL( 1, 7) 5.000000 0.000000 XL( 1, 8) 6.000000 0.000000 XL( 2, 3) 1.000000 0.000000 XL( 2, 4) 7.000000 0.000000 XL( 2, 5) 3.000000 0.000000 XL( 2, 6) 5.000000 0.000000 XL( 2, 7) 2.000000 0.000000 XL( 2, 8) 1.000000 0.000000 XL( 3, 4) 4.000000 0.000000 XL( 3, 5) 4.000000 0.000000 XL( 3, 6) 2.000000 0.000000 XL( 3, 7) 9.000000 0.000000 XL( 3, 8) 2.000000 0.000000 XL( 4, 5) 1.000000 0.000000 XL( 4, 6) 5.000000 0.000000 XL( 4, 7) 5.000000 0.000000 XL( 4, 8) 2.000000 0.000000 XL( 5, 6) 8.000000 0.000000 XL( 5, 7) 7.000000 0.000000 XL( 5, 8) 6.000000 0.000000 XL( 6, 7) 2.000000 0.000000 XL( 6, 8) 3.000000 0.000000 XL( 7, 8) 4.000000 0.000000 Y( 1, 2) 0.000000 -9.000000 Y( 1, 3) 0.000000 -3.000000 Y( 1, 4) 0.000000 -4.000000 Y( 1, 5) 0.000000 -2.000000 Y( 1, 6) 0.000000 -1.000000 Y( 1, 7) 0.000000 -5.000000 Y( 1, 8) 1.000000 -6.000000 Y( 2, 3) 0.000000 -1.000000 Y( 2, 4) 1.000000 -7.000000 Y( 2, 5) 0.000000 -3.000000 Y( 2, 6