1、数学建模全部作业一、图论(组合优化)和排列论实验解:设cij表示i年开始到j-1年结束购车的总消费,则有:C12=2.5+0.3-2.0=0.8,C13=2.5+0.3+0.5-1.6=1.7,C14=2.5+0.3+0.5+0.8-1.3=2.8,C15=2.5+0.3+0.5+0.8+1.2-1.1=4.2,C23=2.6+0.3-2.0=0.9,C24=2.6+0.3+0.5-1.6=1.8,C25=2.6+0.3+0.5+0.8-1.3=2.9,C34=2.8+0.3-2.0=1.1,C35=2.8+0.3+0.5-1.6=2,C45=3.1+0.3-2.0=1.4;建模如下:sets
2、: nodes/1.5/; arcs(nodes, nodes)|&1 #lt# &2: c, x;endsetsdata: c = 0.8 1.7 2.8 4.2 0.9 1.8 2.9 1.1 2.0 1.4;enddatan = size(nodes);min = sum(arcs: c * x);for(nodes(i)| i #ne# 1 #and# i #ne# n: sum(arcs(i,j): x(i,j) = sum(arcs(j,i): x(j,i);sum(arcs(i,j)| i #eq# 1 : x(i,j) = 1;LINGO运行如下: Global optimal
3、 solution found. Objective value: 3.700000 Total solver iterations: 0 Variable Value Reduced Cost X( 1, 2) 1.000000 0.000000 X( 2, 5) 1.000000 0.000000由计算结果分析可得,其最短路径为1-2-5,最小花费为3.7万元。即:该单位应该在第一年购买新设备,年末卖掉设备;第二年初更换新设备,一直用到第四年年末,再卖出。(1)假设每个季度分别生产万盒除臭剂,为第一季度后剩余量,为第二季度后的剩余量,为第三季度后的剩余量,为第四季度后的剩余量。其数学模型为
4、: Lingo语句:model:min =5*x1+5*x2+6*x3+6*x4+y1+y2+y3+y4; x1=10; x1=14; x2=20; x3=8; x4=13; y4=x4+y3-8; endLingo软件的计算结果: Global optimal solution found. Objective value: 292.0000 Total solver iterations: 0 Variable Value Reduced Cost X1 14.00000 0.000000 X2 15.00000 0.000000 X3 15.00000 0.000000 X4 8.000
5、000 0.000000 Y1 4.000000 0.000000 Y2 5.000000 0.000000 Y3 0.000000 0.000000 Y4 0.000000 7.000000 Row Slack or Surplus Dual Price 1 292.0000 -1.000000 2 4.000000 0.000000 3 0.000000 0.000000 4 0.000000 5.000000 5 5.000000 0.000000 6 0.000000 1.000000 7 0.000000 6.000000 8 0.000000 -2.000000 9 0.00000
6、0 1.000000 10 0.000000 5.000000 11 0.000000 0.000000 12 5.000000 0.000000 13 0.000000 6.000000由计算结果可得第一个季度应生产14万盒,第二季度应该生产15万盒,第三季度应该生产15万盒,第四季度应该生产8万盒除臭剂。最低费用为292万元。(2)在其他条件不改变的情况下将第一季度的生产量由14万盒变为13万盒时没有可行方案。数学模型: Lingo语句:model: sets: buyer/1.4/:need; producer/1.4/:provide; matrix(producer, buyer):
7、c,x; endsets min=sum(matrix:c*x); for(buyer(j): sum(producer(i): x(i,j)=need(j); for(producer(i): sum(buyer(j): x(i,j)=provide(i); z=sum(matrix:c*x); data: need=10 14 20 8; provide=13 15 15 13; c= 5 6 7 8 8 5 6 7 12 9 6 7 15 12 9 6;enddata endLingo软件的计算结果: Global optimal solution found. Objective va
8、lue: 294.0000 Total solver iterations: 6 Variable Value Reduced Cost Z 294.0000 0.000000 NEED( 1) 10.00000 0.000000 NEED( 2) 14.00000 0.000000 NEED( 3) 20.00000 0.000000 NEED( 4) 8.000000 0.000000 PROVIDE( 1) 13.00000 0.000000 PROVIDE( 2) 15.00000 0.000000 PROVIDE( 3) 15.00000 0.000000 PROVIDE( 4) 1
9、3.00000 0.000000 C( 1, 1) 5.000000 0.000000 C( 1, 2) 6.000000 0.000000 C( 1, 3) 7.000000 0.000000 C( 1, 4) 8.000000 0.000000 C( 2, 1) 8.000000 0.000000 C( 2, 2) 5.000000 0.000000 C( 2, 3) 6.000000 0.000000 C( 2, 4) 7.000000 0.000000 C( 3, 1) 12.00000 0.000000 C( 3, 2) 9.000000 0.000000 C( 3, 3) 6.00
10、0000 0.000000 C( 3, 4) 7.000000 0.000000 C( 4, 1) 15.00000 0.000000 C( 4, 2) 12.00000 0.000000 C( 4, 3) 9.000000 0.000000 C( 4, 4) 6.000000 0.000000 X( 1, 1) 10.00000 0.000000 X( 1, 2) 3.000000 0.000000 X( 1, 3) 0.000000 0.000000 X( 1, 4) 0.000000 4.000000 X( 2, 1) 0.000000 4.000000 X( 2, 2) 11.0000
11、0 0.000000 X( 2, 3) 4.000000 0.000000 X( 2, 4) 0.000000 4.000000 X( 3, 1) 0.000000 8.000000 X( 3, 2) 0.000000 4.000000 X( 3, 3) 15.00000 0.000000 X( 3, 4) 0.000000 4.000000 X( 4, 1) 0.000000 8.000000 X( 4, 2) 0.000000 4.000000 X( 4, 3) 1.000000 0.000000 X( 4, 4) 8.000000 0.000000 Row Slack or Surplu
12、s Dual Price 1 294.0000 -1.000000 2 0.000000 -7.000000 3 0.000000 -8.000000 4 0.000000 -9.000000 5 0.000000 -6.000000 6 0.000000 2.000000 7 0.000000 3.000000 8 0.000000 3.000000 9 4.000000 0.000000 10 0.000000 0.000000 由计算结果分析可得:第一季度生产万盒;第二季度生产万盒;第三季度生产万盒;第四季度生产万盒。最低费用为294万元。(3)如果不允许延期交货,需要工人加班。假设第季
13、度有万盒除臭剂是加班生产的,则有。数学模型如下: Lingo语句:model: sets: buyer/1.4/:need; producer/1.4/:provide; matrix(producer, buyer):c,x,y; endsets min=sum(matrix:c*(x+1.2*y); for(buyer(j): sum(producer(i): x(i,j)+y(i,j)=need(j); for(producer(i): sum(buyer(j): x(i,j)=provide(i); for(producer(i): sum(buyer(j): y(i,j)=2); z
14、=sum(matrix:c*(x+1.2*y); data: need=10 14 20 8; provide=13 15 15 13; c= 5 6 7 8 99 5 6 7 99 99 6 7 99 99 99 6;enddata endLingo软件的计算结果: Global optimal solution found. Objective value: 292.0000 Total solver iterations: 13 Variable Value Reduced Cost Z 292.0000 0.000000 NEED( 1) 10.00000 0.000000 NEED(
15、 2) 14.00000 0.000000 NEED( 3) 20.00000 0.000000 NEED( 4) 8.000000 0.000000 PROVIDE( 1) 13.00000 0.000000 PROVIDE( 2) 15.00000 0.000000 PROVIDE( 3) 15.00000 0.000000 PROVIDE( 4) 13.00000 0.000000 C( 1, 1) 5.000000 0.000000 C( 1, 2) 6.000000 0.000000 C( 1, 3) 7.000000 0.000000 C( 1, 4) 8.000000 0.000
16、000 C( 2, 1) 99.00000 0.000000 C( 2, 2) 5.000000 0.000000 C( 2, 3) 6.000000 0.000000 C( 2, 4) 7.000000 0.000000 C( 3, 1) 99.00000 0.000000 C( 3, 2) 99.00000 0.000000 C( 3, 3) 6.000000 0.000000 C( 3, 4) 7.000000 0.000000 C( 4, 1) 99.00000 0.000000 C( 4, 2) 99.00000 0.000000 C( 4, 3) 99.00000 0.000000
17、 C( 4, 4) 6.000000 0.000000 X( 1, 1) 10.00000 0.000000 X( 1, 2) 2.000000 0.000000 X( 1, 3) 0.000000 0.000000 X( 1, 4) 0.000000 2.000000 X( 2, 1) 0.000000 95.00000 X( 2, 2) 10.00000 0.000000 X( 2, 3) 5.000000 0.000000 X( 2, 4) 0.000000 2.000000 X( 3, 1) 0.000000 95.00000 X( 3, 2) 0.000000 94.00000 X(
18、 3, 3) 15.00000 0.000000 X( 3, 4) 0.000000 2.000000 X( 4, 1) 0.000000 94.00000 X( 4, 2) 0.000000 93.00000 X( 4, 3) 0.000000 92.00000 X( 4, 4) 8.000000 0.000000 Y( 1, 1) 0.000000 1.000000 Y( 1, 2) 0.000000 1.200000 Y( 1, 3) 0.000000 1.400000 Y( 1, 4) 0.000000 3.600000 Y( 2, 1) 0.000000 113.8000 Y( 2,
19、 2) 2.000000 0.000000 Y( 2, 3) 0.000000 0.2000000 Y( 2, 4) 0.000000 2.400000 Y( 3, 1) 0.000000 113.8000 Y( 3, 2) 0.000000 112.8000 Y( 3, 3) 0.000000 0.2000000 Y( 3, 4) 0.000000 2.400000 Y( 4, 1) 0.000000 113.8000 Y( 4, 2) 0.000000 112.8000 Y( 4, 3) 0.000000 111.8000 Y( 4, 4) 0.000000 1.200000 Row Sl
20、ack or Surplus Dual Price 1 292.0000 -1.000000 2 0.000000 -5.000000 3 0.000000 -6.000000 4 0.000000 -7.000000 5 0.000000 -6.000000 6 1.000000 0.000000 7 0.000000 1.000000 8 0.000000 1.000000 9 5.000000 0.000000 10 2.000000 0.000000 11 0.000000 0.000000 12 2.000000 0.000000 13 2.000000 0.000000 14 0.
21、000000 0.000000 由计算结果分析可得:第一季度生产万盒,说明不加班生产;第二季度生产万盒,说明需要加班生产2万盒;第三季度生产万盒,说明不加班生产;第四季度生产万盒,说明不加班生产。最低费用为292万元。(4)既考虑可以延期交货,也考虑可以加班。数学模型如下: Lingo语句:model: sets: buyer/1.4/:need; producer/1.4/:provide; matrix(producer, buyer):c,x,y; endsets min=sum(matrix:c*(x+1.2*y); for(buyer(j): sum(producer(i): x(i
22、,j)+y(i,j)=need(j); for(producer(i): sum(buyer(j): x(i,j)=provide(i); for(producer(i): sum(buyer(j): y(i,j)=2); z=sum(matrix:c*(x+1.2*y); data: need=10 14 20 8; provide=13 15 15 13; c= 5 6 7 8 8 5 6 7 12 9 6 7 15 12 9 6;enddata endLingo软件的计算结果: Global optimal solution found. Objective value: 292.000
23、0 Total solver iterations: 14 Variable Value Reduced Cost Z 292.0000 0.000000 NEED( 1) 10.00000 0.000000 NEED( 2) 14.00000 0.000000 NEED( 3) 20.00000 0.000000 NEED( 4) 8.000000 0.000000 PROVIDE( 1) 13.00000 0.000000 PROVIDE( 2) 15.00000 0.000000 PROVIDE( 3) 15.00000 0.000000 PROVIDE( 4) 13.00000 0.0
24、00000 C( 1, 1) 5.000000 0.000000 C( 1, 2) 6.000000 0.000000 C( 1, 3) 7.000000 0.000000 C( 1, 4) 8.000000 0.000000 C( 2, 1) 8.000000 0.000000 C( 2, 2) 5.000000 0.000000 C( 2, 3) 6.000000 0.000000 C( 2, 4) 7.000000 0.000000 C( 3, 1) 12.00000 0.000000 C( 3, 2) 9.000000 0.000000 C( 3, 3) 6.000000 0.0000
25、00 C( 3, 4) 7.000000 0.000000 C( 4, 1) 15.00000 0.000000 C( 4, 2) 12.00000 0.000000 C( 4, 3) 9.000000 0.000000 C( 4, 4) 6.000000 0.000000 X( 1, 1) 10.00000 0.000000 X( 1, 2) 3.000000 0.000000 X( 1, 3) 0.000000 0.000000 X( 1, 4) 0.000000 2.000000 X( 2, 1) 0.000000 4.000000 X( 2, 2) 10.00000 0.000000
26、X( 2, 3) 5.000000 0.000000 X( 2, 4) 0.000000 2.000000 X( 3, 1) 0.000000 8.000000 X( 3, 2) 0.000000 4.000000 X( 3, 3) 15.00000 0.000000 X( 3, 4) 0.000000 2.000000 X( 4, 1) 0.000000 10.00000 X( 4, 2) 0.000000 6.000000 X( 4, 3) 0.000000 2.000000 X( 4, 4) 8.000000 0.000000 Y( 1, 1) 0.000000 1.000000 Y(
27、1, 2) 0.000000 1.200000 Y( 1, 3) 0.000000 1.400000 Y( 1, 4) 0.000000 3.600000 Y( 2, 1) 0.000000 4.600000 Y( 2, 2) 1.000000 0.000000 Y( 2, 3) 0.000000 0.2000000 Y( 2, 4) 0.000000 2.400000 Y( 3, 1) 0.000000 9.400000 Y( 3, 2) 0.000000 4.800000 Y( 3, 3) 0.000000 0.2000000 Y( 3, 4) 0.000000 2.400000 Y( 4
28、, 1) 0.000000 13.00000 Y( 4, 2) 0.000000 8.400000 Y( 4, 3) 0.000000 3.800000 Y( 4, 4) 0.000000 1.200000 Row Slack or Surplus Dual Price 1 292.0000 -1.000000 2 0.000000 -5.000000 3 0.000000 -6.000000 4 0.000000 -7.000000 5 0.000000 -6.000000 6 0.000000 0.000000 7 0.000000 1.000000 8 0.000000 1.000000
29、 9 5.000000 0.000000 10 2.000000 0.000000 11 1.000000 0.000000 12 2.000000 0.000000 13 2.000000 0.000000 14 0.000000 0.000000 由计算结果分析可得:第一季度生产万盒,说明不加班生产;第二季度生产万盒,说明需要加班生产1万盒;第三季度生产万盒,说明不加班生产;第四季度生产万盒,说明不加班生产。最低费用为292万元。解答:分析题意可知,此为工作少于人数的情况。假设决策变量为,当第个人采用第种泳姿时,决策变量,否则,;用表示第个人采用第种泳姿时所花费的时间,因此,数学模型如下: Lingo语句:model:sets: person /Zhao, Qian, Li, Wang, Zhang, Sun/; swim /Die, Yang, Wa, ZiYou/; assign( person, swim) : t, x;endsetsdata: t = 54.7, 58.2, 52.1, 53.6, 56.4, 59.8, 62.2, 63.4, 58.2, 56.5, 59.7, 61.5, 69.1, 70.5, 65.3, 67.8, 68.4, 71.3, 52.2, 53.8
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