1、enddatan = size (nodes);max = sum(arcs: c * x);sum (arcs (i,j)| i #eq# 1 : x(i,j) = 1;for (nodes (i)| i #ne# 1 #and# i #ne# n: sum(arcs(i,j): x(i,j) - sum(arcs(j,i): x(j,i)=0);sum (arcs (j,i)| i #eq# n : x(j,i) = 1;for (arcs: bin(x);End得到结果如下:Global optimal solution found. Objective value: 139.0000
2、Objective bound: Infeasibilities: 0.000000 Extended solver steps: 0 Total solver iterations: Variable Value Reduced Cost N 16.00000 0.000000 C (a2, b3) 17.30000 0.000000 C (a2, b1) -20.20000 0.000000 C (b3, c4) 15.70000 0.000000 C (b3, c1) -30.20000 0.000000 C (b1, c2) 18.40000 0.000000 C (b1, c1) -
3、0.2000000 0.000000 C (c4, d5) 13.80000 0.000000 C (c4, d1) -50.20000 0.000000 C (c2, d3) 17.30000 0.000000 C (c2, d1) -20.20000 0.000000 C (c1, d2) 18.40000 0.000000 C (c1, d1) -0.2000000 0.000000 C (d5, e1) 12.20000 0.000000 C (d5, e6) -70.20000 0.000000 C (d3, e4) 15.70000 0.000000 C (d3, e1) -30.
4、20000 0.000000 C (d2, e3) 17.30000 0.000000 C (d2, e1) -20.20000 0.000000 C (d1, e2) 18.40000 0.000000 C (d1, e1) -0.2000000 0.000000 C (e6, f) 5.000000 0.000000 C (e4, f) 30.00000 0.000000 C (e3, f) 50.00000 0.000000 C (e2, f) 60.00000 0.000000 C (e1, f) 80.00000 0.000000 X (a2, b3) 1.000000 -17.30
5、000 X (a2, b1) 0.000000 20.20000 X (b3, c4) 1.000000 -15.70000 X (b3, c1) 0.000000 30.20000 X (b1, c2) 0.000000 -18.40000 X (B1, c1) 0.000000 0.2000000 X (c4, d5) 1.000000 -13.80000 X (c4, d1) 0.000000 50.20000 X (c2, d3) 0.000000 -17.30000 X (c2, d1) 0.000000 20.20000 X (c1, d2) 0.000000 -18.40000
6、X (c1, d1) 0.000000 0.2000000 X (d5, e1) 1.000000 -12.20000 X (d5, e6) 0.000000 70.20000 X (d3, e4) 0.000000 -15.70000 X (d3, e1) 0.000000 30.20000 X (d2, e3) 0.000000 -17.30000 X (d2, e1) 0.000000 20.20000 X (d1, e2) 0.000000 -18.40000 X (d1, e1) 0.000000 0.2000000 X (e6, f) 0.000000 -5.000000 X (e
7、4, f) 0.000000 -30.00000 X (e3, f) 0.000000 -50.00000 X (e2, f) 0.000000 -60.00000 X (e1, f) 1.000000 -80.00000Row Slack or Surplus Dual Price 1 0.000000 0.000000 2 139.0000 1.000000 3 0.000000 0.000000 4 0.000000 0.000000 5 0.000000 0.000000 6 0.000000 0.000000 7 0.000000 0.000000 8 0.000000 0.0000
8、00 9 0.000000 0.000000 10 0.000000 0.000000 11 0.000000 0.000000 12 0.000000 0.000000 13 0.000000 0.000000 14 0.000000 0.000000 15 0.000000 0.000000 16 0.000000 0.000000 17 0.000000 0.000000 18 0.000000 0.000000分析结果容易得出最佳的路径为a2-b3-c4-d5-e1-f,因此得出结论,设备的最优更新策略应该是使用5年。2. 运输问题有甲、乙和丙三个城市,每年分别需要煤炭320万吨、25
9、0万吨和350万吨,由A, B两个煤矿负责供应.已知煤矿年产量A为400万吨,B为450万吨,从两煤矿至各城市煤炭运价如表4.2所示.由于需求大于供应,经协商平衡,甲城市在必要时可少供应0-30万吨,乙城市需求量须全部满足,丙城市需求量不少于270万吨。试求将甲、乙两矿煤炭全部分配出去,满足上述条件又使总运费最低的调运方案。根据题意认为该问题是一个运输线性规划的典型问题。设甲、乙、丙三个城市的煤炭获取量(Customer)1, 2, 3;A、B两个煤矿的煤炭贮存(Warehouse)表达为A, B;煤矿与各城市之间的煤炭运价为P,用x表示决策点。Model:!2 Warehouse, 3 Cu
10、stomer Transportation Problem; Warehouse / A, B/: Supply; Customer / 1, 2, 3/: Gain; Routes (Warehouse, Customer):Endsets The objective;OBJ min = sum (Routes: The supply constraints;for (Warehouse (i): SUPsum (Customer (j): x (i,j) 290; x (i, 1) 270;350; Here are the parameters; Supply = 400, 450; G
11、ain = 320, 250, 380; c = 15, 18, 22,21, 25, 16;Enddataend 得到如下结果: 14650.00 4Variable Value Reduced CostSUPPLY (A) 400.0000 0.000000SUPPLY (B) 450.0000 0.000000GAIN (1) 250.0000 0.000000GAIN (2) 290.0000 0.000000GAIN (3) 310.0000 0.000000C (A, 1) 15.00000 0.000000C (A, 2) 18.00000 0.000000C (A, 3) 22
12、.00000 0.000000C (B, 1) 21.00000 0.000000C (B, 2) 25.00000 0.000000C (B, 3) 16.00000 0.000000X (A, 1) 150.0000 0.000000X (A, 2) 250.0000 0.000000X (A, 3) 0.000000 12.00000X (B, 1) 140.0000 0.000000X (B, 2) 0.000000 1.000000X (B, 3) 310.0000 0.000000因此得到结论:由A矿向甲乙丙三座城市的送矿量(万吨)为150、250、0;由B矿向甲乙丙三座城市的送矿
13、量(万吨)为140、0、310,此时总运输费最小,为14650万元。3、生产计划与库存管理(1)某公司生产一种除臭剂,它在1至4季度的生产成本、生产量及订货量表4.3所示.如果除臭剂在生产当季没有交货,保管在仓库里除臭剂每盒每季度还需1元钱的储存费用。如果某个季度的货物供应量不足,则允许延期交货,延期交货的罚金是每盒每季度3元。请公司希望制定一个成本最低(包括储存费用和罚金)的除臭剂的生产计划,问各季度应生产多少?(2)如果产品不允许延期交货,则公司考虑工人加班,已知加班生产出产品的成本要比原成本高出20%,且每季度加班最多生产2万盒.问:在这种情况下,将如何安排生产,使总成本最少?(1) 根
14、据题意,将此问题转化为运输问题,将每季度的生产量看作是供方,将每季度的订货量看作是需求方,将不属于本季度供货的量视为其他季度运输过来的,那么具体思路作图如下:红色字样为第一季度运输每件费用,6+1红色字样为季度1生产向4个季度的运送量;绿色字样为季度2生产向4个季度的运送量;蓝色字样为季度2生产向4个季度的运送量;黑色字样为季度2生产向4个季度的运送量。列出配送供需表格:季度1需求季度2需求季度3需求季度4需求季度1配送单价(万元/万盒)56季度2配送单价87季度3配送单价129季度4配送单价15设Cij为从季度生产Ai到季度需求Bj的配送单价,Xij为从季度生产Ai到季度需求Bj的运输量,因
15、此总的费用为:第i个季度的运出量应该小于或等于该季度的生产量:第j个季度的运入量应该等于该季度的需求量:根据此模型写出Lingo程序并求出最优解:4 Warehouse, 4 Customer Transportation Problem; Warehouse /1.4/: a; Customer /1.4/: b; a= 13, 15, 15, 13; b= 10, 14, 20, 8; c = 5, 6, 7, 8, 8, 5, 6, 7, 12, 9, 6, 7, 15, 12, 9, 6; sum (Customer (j): X (i,j)= a(i);for (Customer (
16、j): DEM sum (Warehouse (i): X (i,j)= b(j); 294.0000 7Variable Value Reduced CostA (1) 13.00000 0.000000A (2) 15.00000 0.000000A (3) 15.00000 0.000000A (4) 13.00000 0.000000B (1) 10.00000 0.000000B (2) 14.00000 0.000000B (3) 20.00000 0.000000B (4) 8.000000 0.000000C (1, 1) 5.000000 0.000000C (1, 2) 6
17、.000000 0.000000C (1, 3) 7.000000 0.000000C (1, 4) 8.000000 0.000000C (2, 1) 8.000000 0.000000C (2, 2) 5.000000 0.000000C (2, 3) 6.000000 0.000000C (2, 4) 7.000000 0.000000C (3, 1) 12.00000 0.000000C (3, 2) 9.000000 0.000000C (3, 3) 6.000000 0.000000C (3, 4) 7.000000 0.000000C (4, 1) 15.00000 0.0000
18、00C (4, 2) 12.00000 0.000000C (4, 3) 9.000000 0.000000C (4, 4) 6.000000 0.000000X (1, 1) 10.00000 0.000000X (1, 2) 3.000000 0.000000X (1, 3) 0.000000 0.000000X (1, 4) 0.000000 4.000000X (2, 1) 0.000000 4.000000X (2, 2) 11.00000 0.000000X (2, 3) 4.000000 0.000000X (2, 4) 0.000000 4.000000X (3, 1) 0.0
19、00000 8.000000X (3, 2) 0.000000 4.000000X (3, 3) 15.00000 0.000000X (3, 4) 0.000000 4.000000X (4, 1) 0.000000 8.000000X (4, 2) 0.000000 4.000000X (4, 3) 1.000000 0.000000X (4, 4) 8.000000 0.000000Row Slack or Surplus Dual PriceOBJ 294.0000 -1.000000SUP (1) 0.000000 2.000000SUP (2) 0.000000 3.000000S
20、UP (3) 0.000000 3.000000SUP (4) 4.000000 0.000000DEM (1) 0.000000 -7.000000DEM (2) 0.000000 -8.000000DEM (3) 0.000000 -9.000000DEM (4) 0.000000 -6.000000分析结果易知,最佳的方案为:第一季度生产14万盒,给出10万盒满足第一季度要求,库存3万盒,不拖欠;第二季度生产15万盒,给出11万盒,连同第一季度库存3万盒加起来为14万盒满足第二季度要求,库存4万盒,不拖欠;第三季度生产15万盒,全部给出,连同第二季度库存4万盒加起来19万盒,拖欠1万盒;
21、第四季度生产9万盒,还清第三季度拖欠1万盒,剩下8万盒满足第四季度需求。此方案下总成本最小,为294万元。(2) 分析题目,可以将工人加班的费用视为除本季生产外为接下来其他季度供应货物的费用。参考如下配送供需表格:-7.2对于Lingo程序的改变,只需将每季度生产量扩充为+2万盒即可,对于表格中午单价的位置,使用极大数105来表示无穷大, a= 15, 17, 17, 15; c = 5, 6, 6, 6, 100000, 5, 6, 6, 1000000, 1000000, 6, 7.2, 1000000, 1000000, 1000000, 6; x (i,j)= b(j); Global
22、 optimal solution found. 288.0000 8A (1) 15.00000 0.000000A (2) 17.00000 0.000000A (3) 17.00000 0.000000A (4) 15.00000 0.000000C (1, 1) 5.000000 0.000000C (1, 2) 6.000000 0.000000C (1, 3) 6.000000 0.000000C (1, 4) 6.000000 0.000000C (2, 1) 100000.0 0.000000C (2, 2) 5.000000 0.000000C (2, 3) 6.000000
23、 0.000000C (2, 4) 6.000000 0.000000C (3, 1) 1000000 0.000000C (3, 2) 1000000 0.000000C (3, 3) 6.000000 0.000000C (3, 4) 7.200000 0.000000C (4, 1) 1000000 0.000000C (4, 2) 1000000 0.000000C (4, 3) 1000000 0.000000C (4, 4) 6.000000 0.000000X (1, 1) 10.00000 0.000000X (1, 2) 0.000000 1.000000X (1, 3) 0.0
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