运筹学-菜篮子工程

1、运筹学-菜篮子工程某市的菜篮子工程某市是一个人口不到15万的小城市，根据该市的蔬菜种植情况, B、C三地设三个收购点，再由收购点分别送到全市 况，A、B、C三个收购点每天收购量分别为 200、 各个菜市场每天需求量及发生供应短缺时的损失见表 距离见表2，设从收购点到各菜市场蔬菜调运费用为分别在A、8个蔬菜市场。按照常年情170 和 160 (单位:100kg)，1。从收购点到个菜市场的1 元/ (100kg 100m)。菜市场1234_5_678每天需求量(100kg)75608070100559080短缺损失(元/100kg)10851010858表1各菜市场每天需求量及短缺损失表2收购点到

2、各菜市场的距离距离(单位：100m)菜市场12345678收A488191162216购B14771612162317占八、C20191114615510(1) 为该市设计一个从收购点到菜市场的定点供应方案，使蔬菜调运费用与预期 的短缺损失之和最小。(2) 若规定各菜市场短缺量一律不超过需求量的20%，重新设计定点供应方案(3) 为了满足城市居民的蔬菜供应，该市的领导规划增加蔬菜种植面积，试问增 产的蔬菜每天应分别向A、B、C三个收购点各供应多少最为经济合理。参考答案:(1) 调运总费用3890,短缺总损失720,总费用为4610 (方案略).(2) 调运总费用4208,短缺总费用598,总费

3、用为4806 (方案略)。(3) 增产蔬菜80,每天都向C这个收购点供应最经济。(1) 基本假设设刈为第i个集散点向第j个菜市场供应蔬菜 的数量，lij为第i个集散点到第j个菜市场供应 蔬菜的距离，bj为第j个市场每天的需求量，dj 为第j个市场每天的短缺损失，ai为第i个集散 点每天的收购量，cij为第i个集散点向第j个菜 市场的单位运费(i=123,j=1.8)(2) 约束条件、Xj = a (i=1,2,3)iXij <=b 0=仁8)iXj>=0(i=1,2,3;j=1.8)(3) 对各问求解分析第一问目标函数minz=.c*Xj*|j+ j dj(bi- Xj)(i=1,

4、2,3;j=i j1. 8)程序：Model:Sets:Jsd/1.3/:a;Csc/1.8/:b;dqss/1.8/:d;J_c(jsd,csc):x,c,l;En dsetsData:a=200,170,160;b=75,60,80,70,100,55,90,80;d=10,8,5,10,10,8,5,8;1=4,8,8,19,11,6,22,16,14,7,7,16,12,16,23,17,20,19,11,14,6,15,5,10;c=1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1;En ddatafor(jsd(i):st1sum(cs

5、c(j):x(i,j)=a(i);for(csc(j):st2 sum(jsd(i):x(i,j)<=b(j);objmi n=sum(jsd(i):sum(csc(j):c(i,j)*x(i,j)* l(i,j)+sum(csc(j):d(j)*(b(j)-sum(jsd(i):x(i,j) )；End运行结果：Global optimal soluti on found.Objective value:4610.000In feasibilities:0.000000Total solver iterati ons:9Model Class:LPTotal variables:24N

6、on li near variables:0In teger variables:0Total con stra in ts:12Non li near con stra in ts:0Total non zeros:70Non li near non zeros:0VariableValueReduced CostA( 1)200.00000.000000A( 2)170.00000.000000A( 3)160.00000.000000B( 1)75.000000.000000B( 2)60.000000.000000B( 3)80.000000.000000B( 4)70.000000.

7、000000B( 5)100.00000.000000B( 6)55.000000.000000B( 7)90.000000.000000B( 8)80.000000.000000D( 1)10.000000.000000D( 2)8.0000000.000000D( 3)D( 4)D( 5)D( 6)D( 7)D( 8)X( 1,1)X( 1,2)X( 1,3)X( 1,4)X( 1,5)X( 1,6)X( 1,7)X( 1,8)X( 2, 1)X( 2, 2)X( 2, 3)X( 2, 4)X( 2, 5)X( 2, 6)X( 2, 7)X( 2, 8)X( 3, 1)X( 3, 2)X(

8、 3, 3)X( 3, 4)X( 3, 5)X( 3, 6)X( 3, 7)X( 3, 8)C( 1, 1)C( 1, 2)C( 1, 3)C( 1,4)C( 1, 5)C( 1, 6)C( 1, 7)C( 1, 8)C( 2, 1)C( 2, 2)C( 2, 3)C( 2, 4)C( 2, 5)C( 2, 6)5.00000010.0000010.000008.0000005.0000008.00000075.000000.00000040.000000.00000030.0000055.000000.0000000.0000000.00000060.0000040.0000070.0000

9、00.0000000.0000000.0000000.0000000.0000000.0000000.0000000.00000070.000000.00000090.000000.0000001.0000001.0000001.0000001.0000001.0000001.0000001.0000001.0000001.0000001.0000001.0000001.0000001.0000001.0000000.0000000.0000000.0000000.0000000.0000000.0000000.0000000.0000000.0000002.0000000.0000000.0

10、0000012.000001.00000011.000000.0000000.0000000.0000002.00000011.0000014.000003.00000021.0000016.000008.0000002.0000000.00000014.000000.0000000.0000000.0000000.0000000.0000000.0000000.0000000.0000000.0000000.0000000.0000000.0000000.0000000.0000000.0000000.000000C( 2, 7)1.0000000.000000C( 2, 8)1.00000

11、00.000000C( 3, 1)1.0000000.000000C( 3, 2)1.0000000.000000C( 3, 3)1.0000000.000000C( 3, 4)1.0000000.000000C( 3, 5)1.0000000.000000C( 3, 6)1.0000000.000000C( 3, 7)1.0000000.000000C( 3, 8)1.0000000.000000L(1, 1)4.0000000.000000L( 1, 2)8.0000000.000000L( 1, 3)8.0000000.000000L( 1,4)19.000000.000000L( 1,

12、 5)11.000000.000000L( 1, 6)6.0000000.000000L( 1, 7)22.000000.000000L( 1, 8)16.000000.000000L( 2, 1)14.000000.000000L( 2, 2)7.0000000.000000L( 2, 3)7.0000000.000000L( 2, 4)16.000000.000000L( 2, 5)12.000000.000000L( 2, 6)16.000000.000000L( 2, 7)23.000000.000000L( 2, 8)17.000000.000000L( 3, 1)20.000000

13、.000000L( 3, 2)19.000000.000000L( 3, 3)11.000000.000000L( 3, 4)14.000000.000000L( 3, 5)6.0000000.000000L( 3, 6)15.000000.000000L( 3, 7)5.0000000.000000L( 3, 8)10.000000.000000RowSlack or SurplusDual PriceST1( 1)0.000000-7.000000ST1( 2)0.000000-6.000000ST1( 3)0.000000-2.000000ST2( 1)0.00000013.00000S

14、T2( 2)0.0000007.000000ST2( 3)0.0000004.000000ST2( 4)0.0000000.000000ST2( 5)0.0000006.000000ST2( 6)0.0000009.000000ST2( 7)0.0000002.000000ST2( 8)80.000000.000000OBJ 4610.000-1.000000结果分析最优运输方案为A运往菜市场1蔬菜数量为75 公斤，运往菜市场3蔬菜数量为40公斤，运往 菜市场5蔬菜数量为30公斤，运往菜市场6蔬 菜数量为55公斤；B运往菜市场2蔬菜数量为 60公斤，运往菜市场3蔬菜数量为40公斤，运 往菜市场

15、4蔬菜数量为70公斤；C运往菜市场 5蔬菜数量为70公斤，运往菜市场7蔬菜数量 为90公斤，最小损失为4610元第二问目标函数minz=o*Xij*|ij+ j dj(b- Xij)(i=1,2,3；j=i j1. 8)新增一个约束条件：X8 = bj( j=1.8)程序Model:Sets:Jsd/1.3/:a;Csc/1.8/:b;dqss/1.8/:d;J_c(jsd,csc):x,c,l;En dsetsData:a=200,170,160;b=75,60,80,70,100,55,90,80;d=10,8,5,10,10,8,5,8;1=4,8,8,19,11,6,22,16,14,

16、7,7,16,12,16,23,17,20,19,11,14,6,15,5,10;c=1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1;En ddatafor(jsd(i):st1sum(csc(j):x(i,j)=a(i);for(csc(j): st2sum(jsd(i):x(i,j)<=b(j);for(csc(j):st3sum(jsd(i):x(i,j)>=0.8*b(j);objmi n=sum(jsd(i):sum(csc(j):c(i,j)*x(i,j)* l(i,j)+sum(csc(j):d(j)*(b(j)-sum

17、(jsd(i):x(i,j) );End运行结果：Global optimal soluti on found.Objective value:4806.000In feasibilities:0.000000Total solver iterati ons:11Model Class:LPTotal variables:24Non li near variables:0In teger variables:0Total con stra in ts:20Non li near con stra in ts:0Total non zeros:94Non li near non zeros:0V

18、ariableValueReduced CostA( 1)200.00000.000000A( 2)170.00000.000000A( 3)160.00000.000000B( 1)75.000000.000000B( 2)60.000000.000000B( 3)80.000000.000000B( 4)70.000000.000000B( 5)100.00000.000000B( 6)55.000000.000000B( 7)90.000000.000000B( 8)80.000000.000000D( 1)10.000000.000000D( 2)8.0000000.000000D(

19、3)D( 4)D( 5)D( 6)D( 7)D( 8)X( 1,1)X( 1,2)X( 1,3)X( 1,4)X( 1,5)X( 1,6)X( 1,7)X( 1,8)X( 2, 1)X( 2, 2)X( 2, 3)X( 2, 4)X( 2, 5)X( 2, 6)X( 2, 7)X( 2, 8)X( 3, 1)X( 3, 2)X( 3, 3)X( 3, 4)X( 3, 5)X( 3, 6)X( 3, 7)X( 3, 8)C( 1, 1)C( 1, 2)C( 1, 3)C( 1,4)C( 1, 5)C( 1, 6)C( 1, 7)C( 1, 8)C( 2, 1)C( 2, 2)C( 2, 3)C(

20、 2, 4)C( 2, 5)C( 2, 6)5.00000010.0000010.000008.0000005.0000008.00000075.000000.00000010.000000.00000060.0000055.000000.0000000.0000000.00000060.0000054.0000056.000000.0000000.0000000.0000000.0000000.0000000.0000000.0000000.00000024.000000.00000072.0000064.000001.0000001.0000001.0000001.0000001.0000

21、001.0000001.0000001.0000001.0000001.0000001.0000001.0000001.0000001.0000000.0000000.0000000.0000000.0000000.0000000.0000000.0000000.0000000.0000002.0000000.0000000.00000012.000001.00000011.000000.0000000.0000000.0000002.00000011.0000014.000003.00000021.0000016.000008.0000002.0000000.00000014.000000.

22、0000000.0000000.0000000.0000000.0000000.0000000.0000000.0000000.0000000.0000000.0000000.0000000.0000000.0000000.0000000.000000C( 2, 7)1.0000000.000000C( 2, 8)1.0000000.000000C( 3, 1)1.0000000.000000C( 3, 2)1.0000000.000000C( 3, 3)1.0000000.000000C( 3, 4)1.0000000.000000C( 3, 5)1.0000000.000000C( 3,

23、6)1.0000000.000000C( 3, 7)1.0000000.000000C( 3, 8)1.0000000.000000L(1, 1)4.0000000.000000L( 1, 2)8.0000000.000000L( 1, 3)8.0000000.000000L( 1,4)19.000000.000000L( 1, 5)11.000000.000000L( 1, 6)6.0000000.000000L( 1, 7)22.000000.000000L( 1, 8)16.000000.000000L( 2, 1)14.000000.000000L( 2, 2)7.0000000.00

24、0000L( 2, 3)7.0000000.000000L( 2, 4)16.000000.000000L( 2, 5)12.000000.000000L( 2, 6)16.000000.000000L( 2, 7)23.000000.000000L( 2, 8)17.000000.000000L( 3, 1)20.000000.000000L( 3, 2)19.000000.000000L( 3, 3)11.000000.000000L( 3, 4)14.000000.000000L( 3, 5)6.0000000.000000L( 3, 6)15.000000.000000L( 3, 7)

25、5.0000000.000000L( 3, 8)10.000000.000000RowSlack or SurplusDual PriceST1( 1)0.000000-1.000000ST1( 2)0.0000000.000000ST1( 3)0.0000004.000000ST2( 1)0.0000007.000000ST2( 2)0.0000001.000000ST2( 3)16.000000.000000ST2( 4)14.000000.000000ST2( 5)16.000000.000000ST2( 6)0.0000003.000000ST2( 7)18.000000.000000

26、ST2( 8)16.000000.000000ST3( 1)15.000000.000000ST3( 2)12.000000.000000ST3( 3)0.000000-2.000000ST3( 4)0.000000-6.000000ST3( 5)4.0000000.000000ST3( 6)11.000000.000000ST3( 7)0.000000-4.000000ST3( 8)0.000000-6.000000OBJ 4806.000-1.000000最优运输方案为A运往菜市场1蔬菜数量为75 公斤，运往菜市场3蔬菜数量为10公斤，运往 菜市场5蔬菜数量为60公斤，运往菜市场6蔬 菜数

27、量为55公斤；B运往菜市场2蔬菜数量为 60公斤，运往菜市场3蔬菜数量为54公斤，运 往菜市场4蔬菜数量为56公斤；C运往菜市场 5蔬菜数量为24公斤，运往菜市场7蔬菜数量 为72公斤，运往菜市场8蔬菜数量为64公斤， 最小损失为4806元第三问因为根据题意，各菜市场的需求量大于各收购点 的收购量，所以要增大收购量来满足短缺的需 求。与第一问比较分析，应使收购量大于目前的量，使需求量得到满足，不再产生短缺损失，因 此重新建立模型为:Min z=cj*xj*|j(i=123;j=1.8)Xj>=ai(i=1,2,3)xj = bj(i=1,2,3)Xj>=0(i=123;j=1.8)

28、程序:Model :Sets :Jsd/1.3/:a;Csc/1.8/:b;J_c(jsd,csc):x,c,l;En dsetsData :a=200,170,160;b=75,60,80,70,100,55,90,80;l=4,8,8,19,11,6,22,16,14,7,7,16,12,16,23,17,20,19,11,14,6,15,5,10;c=1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1;En ddatafor (jsd(i):st1sum(csc(j):x(i,j)>=a(i);for (csc(j):st2sum(jsd

29、(i):x(i,j)=b(j);obj min =sum(jsd(i):sum(csc(j):c(i,j)*x(i,j)*l(i,j);EndGlobal optimal soluti on found.Objective value:In feasibilities:4770.0000.000000Total solver iterati ons:12LPModel Class:Total variables:24Non li near variables:0In teger variables:0Total con stra ints:12Non li near con stra in t

30、s:0Total non zeros:72Non li near non zeros:0VariableValueReduced CostA( 1)200.00000.000000A( 2)170.00000.000000A( 3)160.00000.000000B( 1)75.000000.000000B( 2)60.000000.000000B( 3)80.000000.000000B( 4)70.000000.000000B( 5)100.00000.000000B( 6)55.000000.000000B( 7)90.000000.000000B( 8)80.000000.000000

31、X( 1,1)75.000000.000000X( 1,2)40.000000.000000X( 1,3)0.0000000.000000X( 1,4)0.0000002.000000X( 1,5)30.000000.000000X( 1,6)55.000000.000000X( 1,7)0.00000012.00000X( 1,8)0.0000001.000000X( 2, 1)0.00000011.00000X( 2, 2)20.000000.000000X( 2, 3)80.000000.000000X( 2, 4)70.000000.000000X( 2, 5)0.0000002.00

32、0000X( 2, 6)0.00000011.00000X( 2, 7)0.00000014.00000X( 2, 8)0.0000003.000000X( 3, 1)0.00000021.00000X( 3, 2)0.00000016.00000X( 3, 3)0.0000008.000000X( 3, 4)X( 3, 5)X( 3, 6)X( 3, 7)X( 3, 8)C( 1,1)C( 1, 2)C( 1, 3)C( 1,4)C( 1, 5)C( 1, 6)C( 1, 7)C( 1, 8)C( 2, 1)C( 2, 2)C( 2, 3)C( 2, 4)C( 2, 5)C( 2, 6)C(

33、 2, 7)C( 2, 8)C( 3, 1)C( 3, 2)C( 3, 3)C( 3, 4)C( 3, 5)C( 3, 6)C( 3, 7)C( 3, 8)L(1, 1)L( 1, 2)L( 1, 3)L( 1,4)L( 1, 5)L( 1, 6)L( 1, 7)L( 1, 8)L( 2, 1)L( 2, 2)L( 2, 3)L( 2, 4)L( 2, 5)L( 2, 6)L( 2, 7)0.00000070.000000.00000090.0000080.000001.0000001.0000001.0000001.0000001.0000001.0000001.0000001.0000001.0000001.0000001.0000001.0000001.0000001.0000001.0000001.0000001.0000001.0000001.0000001.0000001.0000001.0000001.0000001.0000004.0000008.0000008.00000019.0000011.000006.00000022.0000016.0000014.000007.0000007.00000016.0000012.0000016.0000023.000002.0000000.000