運(yùn)籌學(xué)上機(jī)實(shí)驗(yàn)報(bào)告

運(yùn)籌學(xué)上機(jī)實(shí)驗(yàn)報(bào)告-18-運(yùn)籌學(xué)上機(jī)實(shí)驗(yàn)報(bào)告一、投資計(jì)劃問題某地區(qū)在今后3年內(nèi)有4種投資機(jī)會(huì)，第一種是在3年內(nèi)每年年初投資，年底可獲利潤20%，并可將本金收回。第二種是在第一年年初投資，第二年年底可獲利50%，并可將本金收回，但該項(xiàng)投資金額不超過2百萬元。第三種是在第二年年初投資，第三年年底收回本金，并獲利60%，但該項(xiàng)投資金額不超過1.5百萬元。第四種是在第三年年初投資，第三年年底收回本金，并可獲利40%，但該項(xiàng)投資金額不超過1百萬元?，F(xiàn)在該地區(qū)準(zhǔn)備了3百萬元資金，如何制定投資方案，使到第三年年末本利的和最大？解：設(shè)x1表示第一種投資方案的投資額,x2第二種投資方案的投資額,x3第三種投資方案的投資額,x4第四種投資方案的投資額。輸入程序：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計(jì)算機(jī)計(jì)算結(jié)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、配料問題某冶煉廠計(jì)劃煉制含甲、乙、丙、丁4種金屬成分的合金1噸，4種金屬的含量比例為：甲不少于23％，乙不多于15％，丙不多于4％，丁介于35％～65％之間，此外不允許有其他成分。該廠準(zhǔn)備用6種不同等級(jí)的礦石熔煉這種合金，各種礦石中的雜質(zhì)在熔煉中廢棄?，F(xiàn)將每種礦石中的4種金屬含量和價(jià)格列表如下，試計(jì)算如何選配各種礦石才能使合金的原料成本達(dá)到最低。金屬含量和價(jià)格礦石品種金屬甲金屬乙金屬丙金屬丁礦石含金屬（％）礦石含雜質(zhì)（％）礦石價(jià)格（元/噸）礦石10.250.100.100.250.700.3023礦石20.400.000.000.300.700.3020礦石30.200.100.000.300.600.4018礦石40.000.150.050.200.400.6010礦石50.200.200.000.400.800.2027礦石60.080.050.100.170.400.6012解：設(shè)x1為礦石1的需求量,x2為礦石2的需求量,x3為礦石3的需求量,x4為礦石4的需求量,x5為礦石5的需求量,x6為礦石6的需求量。輸入程序：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計(jì)算機(jī)計(jì)算結(jié)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、下料問題有一批500cm長的條材，要截成98cm長的毛坯1000根、78cm長的毛坯2000根。現(xiàn)有6種下料方法，每種方法截出兩種毛坯的根數(shù)和殘料的長度列表如下，要求計(jì)算如何下料可使所用條材根數(shù)最少。6種下料方法比較下料方法98cm毛坯根數(shù)78cm毛坯根數(shù)殘料（cm）方法15010方法24130方法33260方法42370方法51512方法60632解：設(shè)x1為方法1所用的條材根數(shù),x2為方法2所用的條材根數(shù),x3為方法3所用的條材根數(shù),x4為方法4所用的條材根數(shù),x5為方法5所用的條材根數(shù),x6為方法6所用的條材根數(shù)輸入程序：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計(jì)算機(jī)計(jì)算結(jié)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、資源分配問題某個(gè)中型的百貨商場對售貨人員的需求經(jīng)過統(tǒng)計(jì)分析如下表所示。時(shí)間所需售貨員人數(shù)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人為了保證銷售人員充分休息，售貨人員每周工作五天，工作的五天連續(xù)。休息兩天，并要求休息的兩天是連續(xù)的。問應(yīng)該如何安排售貨人員的作息，既能滿足工作需要，又使配備的售貨人員的人數(shù)最少？解：mon,tue,wen,thu,fri,sat,sun依次表示每天安排的售貨員人數(shù)輸入程序：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計(jì)算機(jī)計(jì)算結(jié)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、計(jì)算如下運(yùn)輸問題：銷銷地單位運(yùn)費(fèi)產(chǎn)地B1B2B3B4B5B6B7B8產(chǎn)量A16267425960A24953858255A35219743351A47673927143A52395726541A65522814352需求量3537223241324338解：a1,a2,a3,a4,a5,a6是6個(gè)產(chǎn)地，b1,b2,b3,b4,b5,b6,b7,b8是8個(gè)銷地輸入程序：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計(jì)算機(jī)計(jì)算結(jié)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、目標(biāo)規(guī)劃某單位領(lǐng)導(dǎo)在考慮單位職工的升級(jí)調(diào)資方案時(shí)，依次遵守以下規(guī)定：（1）不超過年工資總額60000元；（2）每級(jí)的人數(shù)不超過定編規(guī)定的人數(shù)；（3）Ⅱ、Ⅲ級(jí)的升級(jí)面盡可能達(dá)到現(xiàn)有人數(shù)的20％；（4）Ⅲ級(jí)不足編制的人數(shù)可錄用新職工，又Ⅰ級(jí)的職工中有10％要退休。有關(guān)資料匯總于下表中，問該領(lǐng)導(dǎo)應(yīng)如何擬定一個(gè)滿意的方案。等級(jí)工資額（元/年）現(xiàn)有人數(shù)編制人數(shù)Ⅰ20001012Ⅱ15001215Ⅲ10001515合計(jì)3742解：設(shè)x1,x2分別表示提升到Ⅰ級(jí)，Ⅱ級(jí)的人數(shù)，x3表示錄用到Ⅲ級(jí)的新職工人數(shù)輸入程序：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計(jì)算機(jī)計(jì)算結(jié)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、整數(shù)規(guī)劃有一部貨車每天沿著公路給4個(gè)零售店卸下6箱貨物，如果各零售店出售該貨物所得到的利潤如下表所示，試求在各零售店卸下幾箱貨物，能夠獲得利潤最大？其值是多少？零零售店利潤箱數(shù)1234000001423426455376764788657986671086解：輸入程序：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計(jì)算機(jī)計(jì)算結(jié)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、圖論設(shè)備更新問題某企業(yè)使用一臺(tái)設(shè)備，在每年年初，企業(yè)領(lǐng)導(dǎo)部門就要決定是購置新的，還是繼續(xù)使用舊的。若購置新設(shè)備，就要支付一定的購置費(fèi)用；若繼續(xù)使用舊設(shè)備，則需支付一定的維修費(fèi)用?，F(xiàn)在的問題是如何制定一個(gè)幾年之內(nèi)的設(shè)備更新計(jì)劃，使得總的支付費(fèi)用最少。我們用一個(gè)五年之內(nèi)要更新某種設(shè)備的計(jì)劃為例。若已知改種設(shè)備在各年年初的價(jià)格為：第1年第2年第3年第4年第5年1111121213還已知使用不同時(shí)間（年）的設(shè)備所需要的維修費(fèi)用為：使用年數(shù)0-11-22-33-44-5維修費(fèi)用5681118解：輸入程序：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計(jì)算機(jī)計(jì)算結(jié)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、下圖是一個(gè)電力輸送網(wǎng)（單位：兆瓦）?，F(xiàn)在要把電力從發(fā)電站Vs輸送到地區(qū)Vt處，每條邊上的數(shù)字表示每條輸送線的最大輸送能力。要求制定一個(gè)最優(yōu)方案，將電力從Vs輸送到Vt，使得輸送電力達(dá)到最大。99128571085589710V6V5V2VtV4V3VsV1解：設(shè)節(jié)點(diǎn)1到節(jié)點(diǎn)8表示電力輸送網(wǎng)的8個(gè)節(jié)點(diǎn)，CAP,FLOW表示弧容量和弧流量輸入程序：!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):