《供应链管理课程实验报告(啤酒游戏)》山东交通学院Word下载.doc
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2012年4月7日
三、实验目的和要求
通过该试验,掌握供应链网络规划的数学建模方法,能够运用Lingo优化软件求解。
要求:
遵守实验纪律,实验前完成预习和数学建模工作。
四、实验内容
建立供应链网络规划问题的优化模型,编写相应的Lingo优化程序并上机运行调试,对实验结果进行分析。
五、实验仪器、设备及材料:
计算机、WindowsXP系统和Lingo10.0软件。
六、实验步骤
1、针对供应链网络规划问题进行数学建模;
2、编写相应的Lingo优化程序;
3、在计算机上运行Lingo软件,得出供应链网络优化方案;
4、对实验结果进行分析。
七、实验结果及分析
1、数学模型
ModelData:
Vendors:
Warehouses:
V1
V2
V3
V4
V5
V6
V7
V8
Capacity:
WH1
6
2
7
4
5
9
60
WH2
3
8
55
WH3
1
51
WH4
43
WH5
41
WH6
52
Demand:
35
37
22
32
38
2、LINGO程序
MODEL:
!
A6Warehouse8VendorTransportationProblem;
SETS:
WAREHOUSES:
CAPACITY;
VENDORS:
DEMAND;
LINKS(WAREHOUSES,VENDORS):
COST,VOLUME;
ENDSETS
Hereisthedata;
DATA:
!
setmembers;
WAREHOUSES=WH1WH2WH3WH4WH5WH6;
VENDORS=V1V2V3V4V5V6V7V8;
attributevalues;
CAPACITY=605551434152;
DEMAND=3537223241324338;
COST=62674259
49538582
52197433
76739271
23957265
55228143;
ENDDATA
Theobjective;
MIN=@SUM(LINKS(I,J):
COST(I,J)*VOLUME(I,J));
Thedemandconstraints;
@FOR(VENDORS(J):
@SUM(WAREHOUSES(I):
VOLUME(I,J))=
DEMAND(J));
Thecapacityconstraints;
@FOR(WAREHOUSES(I):
@SUM(VENDORS(J):
VOLUME(I,J))<
=
CAPACITY(I));
END
3、运行结果
Globaloptimalsolutionfound.
Objectivevalue:
664.0000
Totalsolveriterations:
15
VariableValueReducedCost
CAPACITY(WH1)60.000000.000000
CAPACITY(WH2)55.000000.000000
CAPACITY(WH3)51.000000.000000
CAPACITY(WH4)43.000000.000000
CAPACITY(WH5)41.000000.000000
CAPACITY(WH6)52.000000.000000
DEMAND(V1)35.000000.000000
DEMAND(V2)37.000000.000000
DEMAND(V3)22.000000.000000
DEMAND(V4)32.000000.000000
DEMAND(V5)41.000000.000000
DEMAND(V6)32.000000.000000
DEMAND(V7)43.000000.000000
DEMAND(V8)38.000000.000000
COST(WH1,V1)6.0000000.000000
COST(WH1,V2)2.0000000.000000
COST(WH1,V3)6.0000000.000000
COST(WH1,V4)7.0000000.000000
COST(WH1,V5)4.0000000.000000
COST(WH1,V6)2.0000000.000000
COST(WH1,V7)5.0000000.000000
COST(WH1,V8)9.0000000.000000
COST(WH2,V1)4.0000000.000000
COST(WH2,V2)9.0000000.000000
COST(WH2,V3)5.0000000.000000
COST(WH2,V4)3.0000000.000000
COST(WH2,V5)8.0000000.000000
COST(WH2,V6)5.0000000.000000
COST(WH2,V7)8.0000000.000000
COST(WH2,V8)2.0000000.000000
COST(WH3,V1)5.0000000.000000
COST(WH3,V2)2.0000000.000000
COST(WH3,V3)1.0000000.000000
COST(WH3,V4)9.0000000.000000
COST(WH3,V5)7.0000000.000000
COST(WH3,V6)4.0000000.000000
COST(WH3,V7)3.0000000.000000
COST(WH3,V8)3.0000000.000000
COST(WH4,V1)7.0000000.000000
COST(WH4,V2)6.0000000.000000
COST(WH4,V3)7.0000000.000000
COST(WH4,V4)3.0000000.000000
COST(WH4,V5)9.0000000.000000
COST(WH4,