运筹学上机实验报告

运筹学上机实验报告-18-运筹学上机实验报告一、投资计划问题某地区在今后3年内有4种投资机会，第一种是在3年内每年年初投资，年底可获利润20%，并可将本金收回。第二种是在第一年年初投资，第二年年底可获利50%，并可将本金收回，但该项投资金额不超过2百万元。第三种是在第二年年初投资，第三年年底收回本金，并获利60%，但该项投资金额不超过1.5百万元。第四种是在第三年年初投资，第三年年底收回本金，并可获利40%，但该项投资金额不超过1百万元。现在该地区准备了3百万元资金，如何制定投资方案，使到第三年年末本利的和最大？解：设x1表示第一种投资方案的投资额,x2第二种投资方案的投资额,x3第三种投资方案的投资额,x4第四种投资方案的投资额。输入程序：max=x1*1.2+x2*1.5+(x1+x3)*1.2+x4*1.6+(x1+x3+x5)*1.2+x6*1.4;x1+x2+x3+x4+x5+x5+x6=3;x2<2;x4<1.5;x6<1;end计算机计算结果：Globaloptimalsolutionfound.Objectivevalue:10.80000Totalsolveriterations:0VariableValueReducedCostX13.0000000.000000X20.0000002.100000X30.0000001.200000X40.0000002.000000X50.0000006.000000X60.0000002.200000RowSlackorSurplusDualPrice110.800001.00000020.0000003.60000032.0000000.00000041.5000000.00000051.0000000.000000二、配料问题某冶炼厂计划炼制含甲、乙、丙、丁4种金属成分的合金1吨，4种金属的含量比例为：甲不少于23％，乙不多于15％，丙不多于4％，丁介于35％～65％之间，此外不允许有其他成分。该厂准备用6种不同等级的矿石熔炼这种合金，各种矿石中的杂质在熔炼中废弃。现将每种矿石中的4种金属含量和价格列表如下，试计算如何选配各种矿石才能使合金的原料成本达到最低。金属含量和价格矿石品种金属甲金属乙金属丙金属丁矿石含金属（％）矿石含杂质（％）矿石价格（元/吨）矿石10.250.100.100.250.700.3023矿石20.400.000.000.300.700.3020矿石30.200.100.000.300.600.4018矿石40.000.150.050.200.400.6010矿石50.200.200.000.400.800.2027矿石60.080.050.100.170.400.6012解：设x1为矿石1的需求量,x2为矿石2的需求量,x3为矿石3的需求量,x4为矿石4的需求量,x5为矿石5的需求量,x6为矿石6的需求量。输入程序：min=23*x1+20*x2+18*x3+10*x4+27*x5+12*x6;0.25*x1+0.4*x2+0.2*x3+0.2*x5+0.08*x6>0.23;0.1*x1+0.1*x3+0.15*x4+0.2*x5+0.05*x6<0.15;0.1*x1+0.05*x4+0.1*x6<0.04;0.25*x1+0.3*x2+0.3*x3+0.2*x4+0.4*x5+0.17*x6>0.35;0.25*x1+0.3*x2+0.3*x3+0.2*x4+0.4*x5+0.17*x6<0.65;0.25*x1+0.4*x2+0.2*x3+0.2*x5+0.08*x6+0.1*x1+0.1*x3+0.15*x4+0.2*x5+0.05*x6+0.1*x1+0.05*x4+0.1*x6+0.25*x1+0.3*x2+0.3*x3+0.2*x4+0.4*x5+0.17*x6=1;end计算机计算结果：Globaloptimalsolutionfound.Objectivevalue:27.42857Totalsolveriterations:4VariableValueReducedCostX10.0000005.857143X20.97142860.000000X30.0000000.8571429X40.80000000.000000X50.0000004.142857X60.0000003.428571RowSlackorSurplusDualPrice127.42857-1.00000020.15857140.00000030.3000000E-010.00000040.00000028.5714350.10142860.00000060.19857140.00000070.000000-28.57143三、下料问题有一批500cm长的条材，要截成98cm长的毛坯1000根、78cm长的毛坯2000根。现有6种下料方法，每种方法截出两种毛坯的根数和残料的长度列表如下，要求计算如何下料可使所用条材根数最少。6种下料方法比较下料方法98cm毛坯根数78cm毛坯根数残料（cm）方法15010方法24130方法33260方法42370方法51512方法60632解：设x1为方法1所用的条材根数,x2为方法2所用的条材根数,x3为方法3所用的条材根数,x4为方法4所用的条材根数,x5为方法5所用的条材根数,x6为方法6所用的条材根数输入程序：min=x1+x2+x3+x4+x5+x6;5*x1+4*x2+3*x3+2*x4+x5=1000;x2+2*x3+3*x4+5*x5+6*x6=2000;end计算机计算结果：Globaloptimalsolutionfound.Objectivevalue:520.0000Totalsolveriterations:0VariableValueReducedCostX1120.00000.000000X20.0000000.4000000E-01X30.0000000.8000000E-01X40.0000000.1200000X5400.00000.000000X60.0000000.4000000E-01RowSlackorSurplusDualPrice1520.0000-1.00000020.000000-0.200000030.000000-0.1600000四、资源分配问题某个中型的百货商场对售货人员的需求经过统计分析如下表所示。时间所需售货员人数星期日28人星期一15人星期二24人星期三25人星期四19人星期五31人星期六28人为了保证销售人员充分休息，售货人员每周工作五天，工作的五天连续。休息两天，并要求休息的两天是连续的。问应该如何安排售货人员的作息，既能满足工作需要，又使配备的售货人员的人数最少？解：mon,tue,wen,thu,fri,sat,sun依次表示每天安排的售货员人数输入程序：sets:days/mon..sun/:required,start;endsetsdata:required=15242519312828;enddatamin=@sum(days:start);@for(days(j):@sum(days(i)|i#le#5:start(@wrap(j+i+2,7)))>=required(j));end计算机计算结果：Globaloptimalsolutionfound.Objectivevalue:36.00000Totalsolveriterations:5VariableValueReducedCostREQUIRED(MON)15.000000.000000REQUIRED(TUE)24.000000.000000REQUIRED(WED)25.000000.000000REQUIRED(THU)19.000000.000000REQUIRED(FRI)31.000000.000000REQUIRED(SAT)28.000000.000000REQUIRED(SUN)28.000000.000000START(MON)8.0000000.000000START(TUE)0.0000000.000000START(WED)12.000000.000000START(THU)0.0000000.3333333START(FRI)11.000000.000000START(SAT)5.0000000.000000START(SUN)0.0000000.000000RowSlackorSurplusDualPrice136.00000-1.00000029.0000000.00000030.000000-0.333333340.000000-0.333333351.0000000.00000060.000000-0.333333370.0000000.00000080.000000-0.3333333五、计算如下运输问题：销销地单位运费产地B1B2B3B4B5B6B7B8产量A16267425960A24953858255A35219743351A47673927143A52395726541A65522814352需求量3537223241324338解：a1,a2,a3,a4,a5,a6是6个产地，b1,b2,b3,b4,b5,b6,b7,b8是8个销地输入程序：sets:need/a1,a2,a3,a4,a5,a6/:re;coul/b1,b2,b3,b4,b5,b6,b7,b8/:requ;link(need,coul):cost,num;endsetsdata:cost=626742594953858252197433767392712395726555228143;re=605551434152;requ=3537223241324338;enddatamin=@sum(link:cost*num);@for(need(i):@sum(coul(j):num(i,j))<=re(i));@for(coul(j):@sum(need(i):num(i,j))>=requ(j));end计算机计算结果：Rows=15Vars=48No.integervars=0(allarelinear)Nonzeros=158Constraintnonz=96(96are+-1)Density=0.215Smallestandlargestelementsinabsvalue=1.0000060.0000No.<:6No.=:0No.>:8,Obj=MIN,GUBs<=8Singlecols=0Optimalsolutionfoundatstep:24Objectivevalue:664.0000VariableValueReducedCostRE(A1)60.000000.0000000RE(A2)55.000000.0000000RE(A3)51.000000.0000000RE(A4)43.000000.0000000RE(A5)41.000000.0000000RE(A6)52.000000.0000000REQU(B1)35.000000.0000000REQU(B2)37.000000.0000000REQU(B3)22.000000.0000000REQU(B4)32.000000.0000000REQU(B5)41.000000.0000000REQU(B6)32.000000.0000000REQU(B7)43.000000.0000000REQU(B8)38.000000.0000000COST(A1,B1)6.0000000.0000000COST(A1,B2)2.0000000.0000000COST(A1,B3)6.0000000.0000000COST(A1,B4)7.0000000.0000000COST(A1,B5)4.0000000.0000000COST(A1,B6)2.0000000.0000000COST(A1,B7)5.0000000.0000000COST(A1,B8)9.0000000.0000000COST(A2,B1)4.0000000.0000000COST(A2,B2)9.0000000.0000000COST(A2,B3)5.0000000.0000000COST(A2,B4)3.0000000.0000000COST(A2,B5)8.0000000.0000000COST(A2,B6)5.0000000.0000000COST(A2,B7)8.0000000.0000000COST(A2,B8)2.0000000.0000000COST(A3,B1)5.0000000.0000000COST(A3,B2)2.0000000.0000000COST(A3,B3)1.0000000.0000000COST(A3,B4)9.0000000.0000000COST(A3,B5)7.0000000.0000000COST(A3,B6)4.0000000.0000000COST(A3,B7)3.0000000.0000000COST(A3,B8)3.0000000.0000000COST(A4,B1)7.0000000.0000000COST(A4,B2)6.0000000.0000000COST(A4,B3)7.0000000.0000000COST(A4,B4)3.0000000.0000000COST(A4,B5)9.0000000.0000000COST(A4,B6)2.0000000.0000000COST(A4,B7)7.0000000.0000000COST(A4,B8)1.0000000.0000000COST(A5,B1)2.0000000.0000000COST(A5,B2)3.0000000.0000000COST(A5,B3)9.0000000.0000000COST(A5,B4)5.0000000.0000000COST(A5,B5)7.0000000.0000000COST(A5,B6)2.0000000.0000000COST(A5,B7)6.0000000.0000000COST(A5,B8)5.0000000.0000000COST(A6,B1)5.0000000.0000000COST(A6,B2)5.0000000.0000000COST(A6,B3)2.0000000.0000000COST(A6,B4)2.0000000.0000000COST(A6,B5)8.0000000.0000000COST(A6,B6)1.0000000.0000000COST(A6,B7)4.0000000.0000000COST(A6,B8)3.0000000.0000000NUM(A1,B1)0.00000005.000000NUM(A1,B2)19.000000.0000000NUM(A1,B3)0.00000005.000000NUM(A1,B4)0.00000007.000000NUM(A1,B5)41.000000.0000000NUM(A1,B6)0.00000002.000000NUM(A1,B7)0.00000002.000000NUM(A1,B8)0.000000010.00000NUM(A2,B1)1.0000000.0000000NUM(A2,B2)0.00000004.000000NUM(A2,B3)0.00000001.000000NUM(A2,B4)32.000000.0000000NUM(A2,B5)0.00000001.000000NUM(A2,B6)0.00000002.000000NUM(A2,B7)0.00000002.000000NUM(A2,B8)0.00000000.0000000NUM(A3,B1)0.00000004.000000NUM(A3,B2)11.000000.0000000NUM(A3,B3)0.00000000.0000000NUM(A3,B4)0.00000009.000000NUM(A3,B5)0.00000003.000000NUM(A3,B6)0.00000004.000000NUM(A3,B7)40.000000.0000000NUM(A3,B8)0.00000004.000000NUM(A4,B1)0.00000004.000000NUM(A4,B2)0.00000002.000000NUM(A4,B3)0.00000004.000000NUM(A4,B4)0.00000001.000000NUM(A4,B5)0.00000003.000000NUM(A4,B6)5.0000000.0000000NUM(A4,B7)0.00000002.000000NUM(A4,B8)38.000000.0000000NUM(A5,B1)34.000000.0000000NUM(A5,B2)7.0000000.0000000NUM(A5,B3)0.00000007.000000NUM(A5,B4)0.00000004.000000NUM(A5,B5)0.00000002.000000NUM(A5,B6)0.00000001.000000NUM(A5,B7)0.00000002.000000NUM(A5,B8)0.00000005.000000NUM(A6,B1)0.00000003.000000NUM(A6,B2)0.00000002.000000NUM(A6,B3)22.000000.0000000NUM(A6,B4)0.00000001.000000NUM(A6,B5)0.00000003.000000NUM(A6,B6)27.000000.0000000NUM(A6,B7)3.0000000.0000000NUM(A6,B8)0.00000003.000000RowSlackorSurplusDualPrice1664.00001.00000020.00000003.000000322.000000.000000040.00000003.00000050.00000001.00000060.00000002.00000070.00000002.00000080.0000000-4.00000090.0000000-5.000000100.0000000-4.000000110.0000000-3.000000120.0000000-7.000000130.0000000-3.000000140.0000000-6.000000150.0000000-2.000000六、目标规划某单位领导在考虑单位职工的升级调资方案时，依次遵守以下规定：（1）不超过年工资总额60000元；（2）每级的人数不超过定编规定的人数；（3）Ⅱ、Ⅲ级的升级面尽可能达到现有人数的20％；（4）Ⅲ级不足编制的人数可录用新职工，又Ⅰ级的职工中有10％要退休。有关资料汇总于下表中，问该领导应如何拟定一个满意的方案。等级工资额（元/年）现有人数编制人数Ⅰ20001012Ⅱ15001215Ⅲ10001515合计3742解：设x1,x2分别表示提升到Ⅰ级，Ⅱ级的人数，x3表示录用到Ⅲ级的新职工人数输入程序：min=10000*d1p+100*(d2p+d3p+d4p)+d5m+d6m;2000*(9+x1)+1500*(12-x1+x2)+1000*(15-x2+x3)+d1p-d1m=60000;9+x1+d2p-d2m=12;12-x1+x2+d3p+d3m=15;15-x2+x3+d4p-d4m=15;x1+d5p-d5m=2.4;x2+d6p-d6m=3;end计算机计算结果：Rows=7Vars=15No.integervars=0(allarelinear)Nonzeros=33Constraintnonz=22(19are+-1)Density=0.295Smallestandlargestelementsinabsvalue=1.0000010000.0No.<:0No.=:6No.>:0,Obj=MIN,GUBs<=2Singlecols=6Optimalsolutionfoundatstep:1Objectivevalue:0.5999999VariableValueReducedCostD1P0.000000010000.00D2P0.000000099.00000D3P0.0000000100.0000D4P0.0000000100.0000D5M0.59999990.0000000D6M0.00000001.000000X13.0000000.0000000X20.00000000.0000000X37.5000000.0000000D1M0.00000000.0000000D2M0.00000001.000000D3M6.0000000.0000000D4M7.5000000.0000000D5P0.00000001.000000D6P3.0000000.0000000RowSlackorSurplusDualPrice10.59999991.00000020.00000000.000000030.0000000-1.00000040.00000000.000000050.00000000.000000060.00000001.00000070.00000000.0000000七、整数规划有一部货车每天沿着公路给4个零售店卸下6箱货物，如果各零售店出售该货物所得到的利润如下表所示，试求在各零售店卸下几箱货物，能够获得利润最大？其值是多少？零零售店利润箱数1234000001423426455376764788657986671086解：输入程序：Max=0x1+4x2+6x3+7x4+7x5+7x6+7x7+0x8+2x9+4x10+6x11+8x12+9x13+10x14+0x15+3x16+5x17+7x18+8x19+8x20+8x21+0x22+4x23+5x24+6x25+6x26+6x27+6x28x1+x2+x3+x4+x5+x6+x7=1x8+x9+x10+x11+x12+x13+x14=1x15+x16+x17+x18+x19+x20+x21=1x22+x23+x24+x25+x26+x27+x28=10x1+1x2+2x3+3x4+4x5+5x6+6x7+0x8+1x9+2x10+3x11+4x12+5x13+6x14+0x15+1x16+2x17+3x18+4x19+5x20+6x21+0x22+1x23+2x24+3x25+4x26+5x27+6x28=6End计算机计算结果：INT28LPOPTIMUMFOUNDATSTEP9OBJECTIVEVALUE=17.0000000NEWINTEGERSOLUTIONOF17.0000000ATBRANCH0PIVOT9RE-INSTALLINGBESTSOLUTION...OBJECTIVEFUNCTIONVALUE1)17.00000VARIABLEVALUEREDUCEDCOSTX10.0000000.000000X20.000000-4.000000X31.000000-6.000000X40.000000-7.000000X50.000000-7.000000X60.000000-7.000000X70.000000-7.000000X80.0000000.000000X90.000000-2.000000X101.000000-4.000000X110.000000-6.000000X120.000000-8.000000X130.000000-9.000000X140.000000-10.000000X150.0000000.000000X161.000000-3.000000X170.000000-5.000000X180.000000-7.000000X190.000000-8.000000X200.000000-8.000000X210.000000-8.000000X220.0000000.000000X231.000000-4.000000X240.000000-5.000000X250.000000-6.000000X260.000000-6.000000X270.000000-6.000000X280.000000-6.000000ROWSLACKORSURPLUSDUALPRICES2)0.0000000.0000003)0.0000000.0000004)0.0000000.0000005)0.0000000.0000006)0.0000000.000000NO.ITERATIONS=9BRANCHES=0DETERM.=1.000E0八、图论设备更新问题某企业使用一台设备，在每年年初，企业领导部门就要决定是购置新的，还是继续使用旧的。若购置新设备，就要支付一定的购置费用；若继续使用旧设备，则需支付一定的维修费用。现在的问题是如何制定一个几年之内的设备更新计划，使得总的支付费用最少。我们用一个五年之内要更新某种设备的计划为例。若已知改种设备在各年年初的价格为：第1年第2年第3年第4年第5年1111121213还已知使用不同时间（年）的设备所需要的维修费用为：使用年数0-11-22-33-44-5维修费用5681118解：输入程序：data:n=6;enddatasets:ci/1..n/:f;ro(ci,ci)/1,21,31,41,51,62,32,42,52,63,43,53,64,54,65,6/:d,p;endsetsdata:d=162230415916223041172331172318;enddataf(n)=0;@for(ci(i)|i#lt#n:f(i)=@min(ro(i,j):d(i,j)+f(j)););@for(ro(i,j):p(i,j)=@if(f(i)#eq#d(i,j)+f(j),1,0));end计算机计算结果：Feasiblesolutionfound.Totalsolveriterations:0VariableValueN6.000000F(1)53.00000F(2)41.00000F(3)31.00000F(4)23.00000F(5)18.00000F(6)0.000000D(1,2)16.00000D(1,3)22.00000D(1,4)30.00000D(1,5)41.00000D(1,6)59.00000D(2,3)16.00000D(2,4)22.00000D(2,5)30.00000D(2,6)41.00000D(3,4)17.00000D(3,5)23.00000D(3,6)31.00000D(4,5)17.00000D(4,6)23.00000D(5,6)18.00000P(1,2)0.000000P(1,3)1.000000P(1,4)1.000000P(1,5)0.000000P(1,6)0.000000P(2,3)0.000000P(2,4)0.000000P(2,5)0.000000P(2,6)1.000000P(3,4)0.000000P(3,5)0.000000P(3,6)1.000000P(4,5)0.000000P(4,6)1.000000P(5,6)1.000000RowSlackorSurplus10.00000020.00000030.00000040.00000050.00000060.00000070.00000080.00000090.000000100.000000110.000000120.000000130.000000140.000000150.000000160.000000170.000000180.000000190.000000200.000000210.000000九、下图是一个电力输送网（单位：兆瓦）。现在要把电力从发电站Vs输送到地区Vt处，每条边上的数字表示每条输送线的最大输送能力。要求制定一个最优方案，将电力从Vs输送到Vt，使得输送电力达到最大。99128571085589710V6V5V2VtV4V3VsV1解：设节点1到节点8表示电力输送网的8个节点，CAP,FLOW表示弧容量和弧流量输入程序：!maximumflowproblem;model:SETS:NODES/1..8/;ARCS(NODES,NODES)/1,21,31,42,32,53,53,64,34,65,86,76,87,88,1/:CAP,FLOW;ENDSETSMAX=FLOW(8,1);@FOR(ARCS(I,J):FLOW(I,J)<CAP(I,J));@FOR(NODES(I):@SUM(ARCS(J,I):FLOW(J,I))=@SUM(ARCS(I,J):