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学年第一学期2018-2019《运筹学》实验报告
(五)班级交通运输171学号1000000000姓名*****日期
2018.
12.6Y1,
10.
0000005.
00000020.
0000005.000000Y lzY1/
30.
0000005.000000Y1,
40.
0000005.000000Y1/
50.
0000005.000000Y2,
11.
0000003.
00000020.
0000003.000000Y2,Y2,
30.
0000003.000000Y2,
40.
0000003.000000Y2,
50.
0000003.000000Y3,
11.
0000004.000000Y3,
20.
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30.
0000004.000000Y3,
40.
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50.
0000004.000000Y4,
10.
0000002.000000Y4,
20.
0000002.
00000030.
0000002.000000Y4,Y4,
40.
0000002.000000Y4,
50.
0000002.000000z1/
10.
0000005.000000Z lz
20.
0000005.000000Z1,
30.
0000005.000000Z1,
40.
0000005.000000Z lz
50.
0000005.000000Z2,
10.
0000004.000000Z2,
20.
0000004.000000Z2,
31.
0000004.000000Z2,
40.
0000004.
00000050.
0000004.000000z2,Z3,
10.
0000001.000000Z3,
20.
0000001.000000Z3,
30.
0000001.000000Z3,
40.
0000001.000000Z3,
50.
0000001.000000Z4,
10.
0000009.000000Z4,
20.
0000009.000000Z4,
30.
0000009.000000Z4,
40.
0000009.000000Z4,
50.
0000009.000000Row SlackorSurplus DualPrice
175.50000-
1.
0000002150.
00000.
000000350.
000000.
0000004200.
00000.
0000005100.
00000.
00000060.
0000000.
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0000000.
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0000000.
00000090.
0000000.
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000000380.
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000000410.
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000000420.
0000000.000000实验
三一、问题重述某班准备从5名游泳队员中选择4个人组成接力队,参加学校的4*100m混合泳接力比赛5名队员4种泳姿的百米平均成绩如下表所示单位秒,问应如何选拔队员组成接力队?甲乙丙T戊蝶泳
66.
857.
2787067.4仰泳
75.
66667.
874.271蛙泳
8766.
484.
669.
683.8自由泳
58.
65359.
457.
262.4进一步考虑如果最近队员丁的蛙泳成绩有较大退步,只有
75.2秒;而队员戊经过艰苦训练自由泳成绩有所进步,达到
57.5秒,组成接力队的方案是否应该调整?
二、模型假设及符号说明以i=l,2,3,4,5分别表示5名队员,j=l,2,3,4分别表示4种泳姿,以c”表示第i名队员第j种泳姿的最好成绩;设第i名队员是否使用第i种泳姿参与比赛为Xij,若第i名队员使用第j种泳姿参与比赛,则xu=l,对应地,若xu=O,则表示第i名队员不使用第j种泳姿参与比赛其中,每位成员最多可上场一次,且四种游泳方式都必须有队员使用
三、数学模型54min z=工=内i=l;=l,a=1/T,2,3,4s.t.,£x屋li=1,2,3,4,5勺二0或1
四、模型求解及结果分析模型一结果分析当为2=83=134=匕|=1时,即甲选手以自由泳,乙选手以蝶泳,丙选手以仰泳,丁选手以蛙泳参加比赛,所得成绩时间少,即z=
253.2秒;模型二结果分析当芭2=/3=马4=匕5=1时,即乙选手以蝶泳,丙选手以仰泳,丁选手以蛙泳,戊选手以自由泳参加比赛,所得成绩时间最短,即z=
257.7秒
五、附录程序模型一sets:yong/
1..4/;xuan/
1..5/;linkyong,xuan:c,x;endsetsdata:c=
66.
857.
2787067.
475.
66667.
874.
2718766.
484.
669.
683.
858.
65359.
457.
262.4;enddatamin=@sumlink,i,j:ci j*xi,j;z@forxuanj:@sumyongi:xi,j=1;©foryongi:@sumxuanj:xi,j=1;@forlinki,j:@binxi,j;模型一运算结果
66.
800000.000000Global optimalsolutionfound.
57.
200000.000000Objective value:
78.
000000.000000Objective bound:
70.
000000.000000Infeasibilities:
67.
400000.000000Extended solversteps:
75.
600000.000000Totalsolver iterations:
66.
000000.000000Variable
67.
800000.000000c lz
174.
200000.000000C1/
271.
000000.000000C1,
387.
000000.000000C1/
466.
400000.000000C5lz
84.
600000.000000c2,
169.
600000.000000c2,
283.
800000.000000c2,
358.
600000.000000C2,
453.
000000.000000C2,
559.
400000.000000C3,1C3,2c3,3c3,4c3,5c4,1c4,2C4,
3253.
2000253.
20000.00000000Value ReducedCostC4,
457.
200000.000000C
4562.
400000.000000ZX1,
10.
00000066.80000X1,
21.
00000057.20000X1,
30.
00000078.00000X1,
40.
00000070.00000X1,
50.
00000067.40000X
210.
00000075.60000ZX
220.
00000066.00000ZX2,
31.
00000067.80000X
240.
00000074.20000ZX
250.
00000071.00000ZX3,
10.
00000087.00000X3,
20.
00000066.40000X3,
30.
00000084.60000X3,
41.
00000069.60000X3,
50.
00000083.80000X
411.
00000058.60000ZX4,
20.
00000053.00000X4,
30.
00000059.40000X
440.
00000057.20000ZX4,
50.
00000062.40000Row SlackorSurplus DualPrice
1253.2000-
1.
00000020.
0000000.
00000030.
0000000.
00000040.
0000000.
00000050.
0000000.
00000061.
0000000.
00000070.
0000000.
00000080.
0000000.
00000090.
0000000.
000000100.
0000000.000000模型二sets:yong/
1..4/;xuan/
1..5/;linkyong,xuan:c,x;endsetsdata:c=
66.
857.
2787067.
475.
66667.
874.
2718766.
484.
675.
283.
858.
65359.
457.
257.5;enddatamin=@sumlinki,j:ci,j*xi,j;@forxuan j:@sumyongi:xi,j=1;@foryongi:@sumxuanj:xi,j=1;@forlinki j:@bin xi,j;z模型二运算结果Globaloptimalsolutionfound.Objectivevalue:
257.7000Objectivebound:Infeasibilities:
257.
70000.000000Extendedsolver steps:0Totalsolver iterations:0Variable ValueReducedCostC lz
166.
800000.000000C1,
257.
200000.000000C1,
378.
000000.000000C
470.
000000.000000lzC1/
567.
400000.000000C2,
175.
600000.000000c2,
266.
000000.000000C2,
367.
800000.000000C2,
474.
200000.000000C2,
571.
000000.000000C3,
187.
000000.000000C3,
266.
400000.000000C
384.
600000.0000003,C3,
475.
200000.000000C3,
583.
800000.000000c4,
158.
600000.000000C4,
253.
000000.000000C
359.
400000.0000004,C4,
457.
200000.000000C4,
557.
500000.000000X1/
10.
00000066.80000X1,
21.
00000057.20000X1/
30.
00000078.00000x lz
40.
00000070.00000x1/
50.
00000067.40000X2,
10.
00000075.60000X2,
20.
00000066.00000X2,
31.
00000067.80000X2,
40.
00000074.20000X2,
50.
00000071.00000X3,
10.
00000087.00000x3,
20.
00000066.40000X3,
30.
00000084.60000X3,
41.
00000075.20000X3,
50.
00000083.80000x4,
10.
00000058.60000X4,
20.
00000053.00000X4,
30.
00000059.40000x4,
40.
00000057.20000X4,
51.
00000057.50000Row Slackor Surplus DualPrice
1257.7000-
1.
00000021.
0000000.
00000030.
0000000.
00000040.
0000000.
00000050.
0000000.
00000060.
0000000.
00000070.
0000000.
00000080.
0000000.
00000090.
0000000.
000000100.
0000000.000000实验
一一、问题重述某昼夜服务的公共交通系统每天各时间段每4个小时为一个时段所需的值班人数如卜.表所示这些值班人员在某一时段开始上班后要连续工作8个小时包括轮流用膳时间问该公交系统至少需要多少名工作人员才能满足值班的需要?班次时间段所需人数16:00-10:0060210:00-14:0070314:00-18:0060418:00-22:0050522:00-2:002062:00-6:0030
二、模型假设及符号说明设该第i班次开始上班的工作人员的人数为Xi人,则第i班次上班的工作人员将在第i+l班次下班i=1,2,3,4,5,6
三、数学模型min z=x+x+x+x+x+xl23456r x4-^1606x+x7i2x+x6023sJ.+x504x+x204sx+x3056y之0,且一均为整数,i=1,2,...6
四、模型求解及结果分析Global optimalsolutionfound.Objective value:
150.0000Objective bound:
150.0000Infeasibilities:
0.000000Extended solversteps:0Totalsolver iterations:4Variable ValueReducedCostXI
60.
000001.000000X
210.
000001.000000X
350.
000001.000000X
40.
0000001.000000X
530.
000001.000000X
60.
0000001.000000Row Slack,orSurplusDualPrice
1150.0000-
1.
00000020.
0000000.
00000030.
0000000.
00000040.
0000000.
00000050.
0000000.
000000610.
000000.
00000070.
0000000.000000根据Ling程序运行结果分析可知当第i班次开始上班的工作人员排布如下时,所需人力最少,为150人班次123456人数6010500300
五、附录程序min=xl+x2+x3+x4+x5+x6;xl+x6=60;xl+x2=70;x2+x3=60;x3+x4=50;x4+x5=20;x5+x6=30;@ginxl;@ginx2;@ginx3;@ginx4;@ginx5;@ginx6;end实验
二一、问题重述某公司正在考虑在某城市开发一些销仕:代理业务经过测试,该公司已经确定了该城市未来5年的业务量,分别为400,500,600,700和800该公司已经初步物色了4家销售公司作为其代理候选企业,下表给出了该公司与每个候选企业建立代理关系的一次性费用,以及每个候选企业每年所能承揽的最大业务量和年运行费用问改公司应该与哪些候选企业建立代理关系?候选代理1候选代理2候选代理3候选代理4年最大业务量350250300200一次性费用(万元)100809070年运行费用(万元)
7.
54.
06.
53.0在此基础上继续讨论如果该公司目前已经与上述4个代理建立了代理关系并处于运行状态,但每年年初可以决定临时中断或重新恢复代理关系,每次临时中断或重新恢复代理关系的费用如下表所示问该公司应如何对这些代理进行业务调整?代理1代理2代理3代理4临时中断费用(万元)5342重新恢复费用(万元)5419
二、模型假设及符号说明模型一设第j年与候选代理i是否开始建立代理关系为Xij,若在第j年与该候选代理i开始建立代理关系,则xij=l,表示关系不中断,对应需要付清一次性费用,以及共(6-j)年的年运行费用;对应地,若Xij=O,则表示该代理公司在第j年尚未建立代理关系或在此之前已经建立了代理关系即七=J1,第/年与第,家代理开始建立代理关系.0,else5且有=1模型二模型假设公司已经与四家都有代理关系符号说明%=1一一第j年与第i个公司有代理关系马=0——笫j年与第i个公司没有代理关系为=1一一第j年与第i个公司中断代理关系z=I---第j年与第i个公司恢复代理关系/7z=0一一第j年与第i个公司不恢复代理关系1
三、数学模型模型一5555min z=102x\j+8°Z-v-+90Zx+70^x,2/3J4川7-1J-I J-I+
7.5g6-JA-ly+4^6-jx2;+
6.5g6一/刍/+3g6—力匕,J=I,=i;=i,=i350x+250%+300X+200x400n3I412222350》,「j+2502占4+300213,+200Xx4;2500六y=i j=i J=i i3503+250火%+300力为+200^x.600j4y=i;=i i J=I44443502%+252%+30Z%+20WX2700J=I;=i J=I J=I5555350Z XM+250Z工2,+302x3j+200Z X4J-800J=I,=iJ=I六i、%=0或1,i=123,4,J=1,2,345模型二5555min Z=
7.5^X1,+
4.0^x+
6.5^x+
3.0^2x2j3j4y7-i六j-i7-i+523+32y+4之y+y.2j3j47=1;=I7=1j=l5555+4,+Z2j+Z z3;+9Zz管八i j=i八i350x+250X+300X+200x400112I3I41350M2+250X+300X4-200x500351%+
250.+3000+223242200%600s工350%+250%+300处+200%700350不+250X+3000+200x800为一毛加工力+中,=123,4,2545j=0,1,2,3,4XGFJW Z«M,i=1,2,3,4,j=1,2,3,4
四、模型求解及结果分析模型一结果分析当%“=/1=匕4=1时,即公司在第一年初开始与代理
1、代理2建立代理关系,在第四年开始与代理4建立关系,此时为最少费用,即z=
313.5万元模型二结果分析当%=%i=l,Z23=l时,即公司在与所有代理保持代理关系的前提下,在第一年与代理
2、代理3中断代理,在第3年与代理2恢复代理关系,此时为最少费用,即z=
75.5万元
五、附录程序模型一运算代码sets:buss/
1..4/;year/
1..5/;linkbuss,year:c,x;endsetsdata:c=
3501007.
555250804.
034300906.
541200703.029;enddatamin=@sumlinki,j:ci,2*xi,j+@sumbussi:ci,3*5*xi,l+4*xi,2+3*xi,3+2*xi,4+xi,5/@forbussi:@sumyearj:xi,j=1;@sumbussi:ci l*xi,l=400;z@sumbussi:ci,l*xi,l+xi,2=500;@sum bussi:ci l*xi l+xi2+xi,3=600;z z z@sum bussi:ci,l*xi,l+xi,2+xi,3+xi,4=700;@sumbussi:ci,l*xi,l+xi,2+xi,3+xi,4+xi,5=800;@forlinki,j:@binxi,j;模型一运算结果Global optimalsolutionfound.Objective value:
313.5000Objective bound:
313.5000Infeasibilities:
0.000000Extended solversteps:0Totalsolver iterations:35Variable ValueReducedCostC1,
1350.
00000.000000clz
2100.
00000.000000c1/
37.
5000000.000000c1,
45.
0000000.000000C1/
55.
0000000.000000C2,
1250.
00000.000000C2,
280.
000000.000000C2,
34.
0000000.000000c2,
43.
0000000.000000C2,
54.
0000000.000000c3,
1300.
00000.000000C3,
290.
000000.000000C3,
36.
5000000.000000C3,
44.
0000000.000000C3,
51.
0000000.000000c4,
1200.
00000.000000C4,
270.
000000.000000c4,
33.
0000000.000000C4,
42.
0000000.000000C4,
59.
0000000.000000X1/
11.
000000137.5000x lz
20.
000000130.0000x1,
30.
000000122.5000X1,
40.
000000115.0000xlz
50.
000000107.5000X2,
11.
000000100.0000X2,
20.
00000096.00000X2,
30.
00000092.00000x2,
40.
00000088.00000X2,
50.
00000084.00000X3,
10.
000000122.5000x3,
20.
000000116.0000X3,
30.
000000109.5000X3,
40.
000000103.0000X3,
50.
00000096.50000X4,
10.
00000085.00000X4,
20.
00000082.00000x4,
30.
00000079.00000x4,
41.
00000076.00000X4,
50.
00000073.00000Row SlackorSurplus DualPrice
1313.5000-
1.
00000020.
0000000.
00000030.
0000000.
00000041.
0000000.
00000050.
0000000.
0000006200.
00000.
0000007100.
00000.
00000080.
0000000.
0000009100.
00000.
000000100.
0000000.000000模型二sets:buss/
1..4/;year/
1..5/:b;linkbuss,year:c,x,y,z;endsetsdata:c=
3501007.
555250804.
034300906.
541200703.029;b=400500600700800;enddatamin=@sumlinki,j:ci,3*x i,j+@sumlinki,j:ci,4*yi,j+@sumlinki,j:ci,5*zi,j;@foryearj:@sumbussi:ci,l*xi,j=bj;@forbussi:1-xi,1=y i,1;@forbussi:@foryearj|j#LE#4:xi j-xi,j+1=yi j+1;zz@forbussi:@foryearj|j#LE#4:xi,j+1-xi,j=zi,j+1;@forlinki,j:@bin xi,j;@forlink i,j:@binyi,j;@forlinki,j:@binzi,j;模型二运算结果Global optimalsolutionfound.Objective value:
75.50000Objective bound:
75.50000Infeasibilities:
0.000000Extended solversteps:0Totalsolver iterations:141Variable ValueReducedCostB
1400.
00000.000000B
2500.
00000.000000B
3600.
00000.000000B
4700.
00000.000000B
5800.
00000.000000C1/
1350.
00000.000000C lz
2100.
00000.000000C1,
37.
5000000.000000C lz
45.
0000000.000000c1/
55.
0000000.000000C2,
1250.
00000.000000c2,
280.
000000.000000C2,
34.
0000000.000000C2,
43.
0000000.000000C2,
54.
0000000.000000C3,
1300.
00000.000000c3,
290.
000000.000000C3,
36.
5000000.000000c3,
44.
0000000.000000C3,
51.
0000000.000000C4,
1200.
00000.000000C4,
270.
000000.000000C4,
33.
0000000.000000c4,
42.
0000000.000000C4,
59.
0000000.000000x lz
11.
0000007.500000X1/
21.
0000007.500000X1,
31.
0000007.500000X1/
41.
0000007.500000x lz
51.
0000007.500000X2,
10.
0000004.000000X2,
20.
0000004.000000x2,
31.
0000004.000000X2,
41.
0000004.000000X2,
51.
0000004.000000X3,
10.
0000006.500000X3,
20.
0000006.500000X3,
30.
0000006.500000x3,
40.
0000006.500000x3,
50.
0000006.500000X4,
11.
0000003.000000x4,
21.
0000003.000000X4,
31.
0000003.000000X4,
41.
0000003.000000X4,
51.
0000003.000000。