华南理工大学工商管理学院运筹学课后习题及答案3

目目第性规划础1习题1第二深入性规划8习题8第三性规划的理19习题19第四整规划23习题23第运FL题40习题40第56习题56第八态规划63习题63I第性规划础第性规划础习题、应用FL题的模1、某养鸡场饲养肉鸡设每鸡每天少需100蛋白、12、60生。现有饲料用饲料每千营养'及价如表110表110饲料'本表饲料蛋白生价元/千1316052203100832048064527151608315FL如在足肉鸡营养需的前提下济地配饲料”本FL题的性规划模型。案解第I饲料的购XII15则本FL题的性规划模型MINZ05X108X206X3X415X5ST3X12X22X35X416X5必须下。100X103X204X32X408X5126X110X28X37X43X560XI0I15必须有变的负约2、某利用条生产L1L2生产产P1P2。产'别由心组装而价价的110元。表111了生产生产需及生产的每月用的总小。表111产生产的价表产产P1产P2用心AB心CD生产100300200500140生产200400200500245价250150400100配产的后政每生产1产P1P2需额外生产2备。FL该应如安生产现月额”本FL题的性规划模型。案1第性规划础解下表由生产生产的案表11变产产P1产P2心AB心CD生产1X1AX1BX1CX1D生产2X2AX2BX2CX2D本FL题完整模型MAXZ740X1AX2A750X1CX2CST3X1AX2AX1BX2B03X1CX2CX1DX2D0003X1A002X1B005X1C001X1D640004X2A002X2B005X2C002X2D645XIJ0I12JABCD必须有变的负约3、某司在生产产足顾需。现FI知下月三地的需FL如安从而得司的总运本低”表112了的生产能及到三地送的流本元/。表112运表产地运地地1地2地3生产能17007506504002850550450600需350250400本FL题的性规划模型。案解XIJI地J送的。则本FL题的模型MINZ700X11750X12650X13850X21550X22450X23STX11X12X136400X21X22X236“等于需的FL题则必须6600X11X21下“350X11X22250X13X23400XIJ0I12J123必须有变的负约4、某司提4型的料产M1、M2、M3M4型产的场价如表113。些料由3原料原料C1、C2C3的配理而2第性规划础原料原料在的配比例如表113购价如表114。表113型的配价型配比例价元/斤M1C1少于40120C2I于20C3I于5M2C1I于1090C2I于30C3少于50M3C1I于1070C2少于60M4C1I于3050C2I于40C3I于40表1143原料的场价原料C1C2C3价503040假设4产应本月的购10000元FL该司本月应如购如生产得I利润”本FL题的性规划模型。案解XIJI1234J123MI原料CJ的。则本FL题的性规划模型MAXZ120X11X12X1390X21X22X2370X31X32X3350X41X42X4350X11X21X31X4130X12X22X32X4240X13X23X33X43STX1104X11X12X13X12602X11X12X13X136005X11X12X13X21601X21X22X23X22603X21X22X23X2305X21X22X23X31601X31X32X33X3206X31X32X33X41603X41X42X43X42604X41X42X43X43604X41X42X434PI13PJ1PJPJ表第J原料的场价XIJ610000XIJ0I1234J123必须有变的负约5、某司产从三A1、A2A3运往四A6、A7、A8A911了行的运本千元/斤A4A5'心的字3第性规划础'别表的应的需斤。A1A3A4A5A6A7A8A9234325365634200150350300250450A2356430011流网FL如安运总运少”本FL题的性规划模型。案解从AI到AJ的运XIJAI到AJ有运I125J459。则本FL题的本FL题的性规划模型MINZ35X162X143X156X244X254X343X352X465X473X486X494X563X576X585X59STX16X14X15250X24X25300X34X35450X14X24X34X46X47X48X49X15X25X35X56X57X58X59约条容掉X46X56200X47X57150X48X58350X49X59300XIJ0I15J196、SH地产集有闲资金20元拟在来5年进行外资。了保证资金安全提了下4的资资1借资每年年资当年年末本利107资2国内金资每年年资次年年末本利118资3地每年年资3年本利130资4股权资能在第3年年资资能过10元第5年年末本利155。假存在资”险忽利FL该集应如安资计划得第5年年末拥有的本利总额”本FL题的性规划模型。案XIJI15J14第I年年用于第J资的资额。FL题得到每年年的资额及年底的如下表4第性规划础案表12年年资总额年末本利总额1X11X12X13107X112X21X22X23107X21118X123X31X32X33X34107X31118X22130X134X41X42X43107X41118X32130X235X51X52X53107X51118X42130X33155X34本FL题的性规划模型MAXZ107X51118X42130X33155X34STX11X12X136等下20X21X22X236107X11X31X32X33X346107X21118X12X41X42X436107X31118X22130X13X51X52X536107X41118X32130X23X34610XIJ0I15J14必须有变的负约7、某生产的需用到3规每子需长60CM、40CM30CM的2、62。在场购些规的现长110CM的割需的规但每割1次发生1CM的长。FL如果100该的少有I少条长110CM的用去场购”本FL题的性规划模型。案解110CM长的割60CM、40CM30CM共有5“下表案表135割“割“得到规60CM40CM30CM11102101301240205003XI取第I“割的110CM的则本FL题的性规划模型MINZX1X2X3X4X5STX1X2必须200X1X32X4600X22X33X5200XI0I15必须有变的负约5第性规划础8、JM司家于互网的化妆司该司每月需租用存。FI知来4月的储需如表115租金按的租用间计租金价表116。表115储需月140,000230,000320,000450,000表116租的租金租用长月每月租金元160210031354170现JM司需租签来4月的租用该细化到月租租用储例如在2月租10,000租1月20,000的3月。FLJM司应如租用计划租金支少”本FL题的性规划模型提设XIJ第I月租用租J月的储I14J14。案XIJ第I月租用租J月的储I14J14每月际用储如下表案表14月X11X12X13X14X21X22X23X24X31X32X33X34X41X42X43X441234表储当月用本FL题的模型MINZ60X11100X12135X13170X1460X21100X22135X23170X2460X31100X32135X33170X3460X41100X42135X43170X44STX11X12X13X14必须下40000X12X13X14X21X22X23X2430000X13X14X22X23X24X31X32X33X3420000X14X23X24X32X33X34X41X42X43X4450000XIJ0I14J14二、性规划FL题的解法计9、应用解法解下列性规划FL题6第性规划础1MAXZ2X1X2STX1X264X1X25X1X20案行FL题行解。2MAXZX13X2STX1X21X12X264X1X20案优解40优4。3MAXZ2X1X2STX26102X15X2630X1X26203X1X2636X1X20案优解150131813优31813。4MINZ2X14X2STX12X2615X1X26125X13X2645X1X20案优解39优42。7第二深入性规划第二深入性规划习题、标准纯形法的计1、下列性规划FL题变标准形“。1MINZ3X14X22X35X4ST4X1X22X3X42X1X23X3X46142X13X2X32X42X1X20X360X4案MAXW3X14X22X035X045X004ST4X1X22X03X04X0042X1X23X03X04X004X5142X13X2X032X042X004X62X1X2X03X04X004X5X602MAXZ2X13X2STX1X2632X1X22X10X2案MAXZ2X13X023X002STX1X02X002X332X1X02X002X42X1X02X002X3X402、下列性规划FL题的有本解的本行解优解。1MAXZX1X2ST2X13X2662X1X264X1X20案入松变X3X4模型变标准形“MAXZX1X2ST2X13X2X362X1X2X44X1X2X3X40约条2本解变2。案表21序变组本解行解目标X1X2T优解1X1X2T32100T52321T2X1X3T2020T220T3X1X4T3002T4X2X3T0460T5X2X4T0202T202T6X3X4T0064T000T8第二深入性规划2MINZ3X1X22X34X4ST2X13X2X32X412X1X2X32X48X1X2X3X40案模型标准化MAXW3X1X22X34X4ST2X13X2X32X412X1X2X32X48X1X2X3X40约条2本解变2。案表22序变组本解行解目标W里么写Z么写W能写ZX1X2X3X4T优解1X1X2T12400T2X1X3T2030430T3X1X4T4002T44002T4X2X3T0530T5X2X4T0203T140203T6X3X4T0025T160025T注由于涉及纯形法解本题的目标用化MAXW而接计。3、应用纯形表法解下列性规划FL题。1MAXZX15X22X3STX1X2X3616X12X2X36322X13X22X3660X1X2X30案案表23CJ152000BCBXBX1X2X3X4X5X65X201013130162X30017121121431X1100112512143CJ000512231214Z839第二深入性规划本例在开'解即现化案需确本FL题有优解XX1X2X3X4X5X6T3163000TZ832MINZ3X13X2X3STX1X2612X1X23X36143X1X2X3616X1X2X30案标准化后用纯形表解结果如下下表目标转化MAXW3X13X2X3的纯形表如接MIN解检验。案表24CJ331000BCBXBX1X2X3X4X5X63X20105418384341X3001121414323X110014183854CJ000521414W3712本FL题有优解XX1X2X3X4X5X6T5443432000TZ注如果转了目标在后应转Z。37123MAXZ4X15X24X3STX1X2X368X13X2X36213X12X2X3615X1X2X30案标准化后用纯形表解结果如下案表25CJ454000BCBXBX1X2X3X4X5X64X3001741412545X2010121201324X110014141214CJ00072120有0检验Z381210第二深入性规划本FL题有I必须确优解优解XX1X2X3X4X5X6T1413254000TZ38124MINZX12X2X3STX1X212X1X364X1X2X30案在第1、2约条'别入余变X4松变X5MINZX12X2X3STX1X2X412X1X3X54X1X2X3X4X50接X2X5变'纯形表接解注下表小接解的纯形表如目标检验取。案表26CJ12100BCBXBX1X2X3X4X52X211010120X5101014CJ10120Z242X20111181X1101014CJ000有0检验21Z20本FL题有I必须确优解优解XX1X2X3X4X5T48000TZ204、'别应用M法阶法解下列性规划FL题。1MAXZX12X23X3ST2X13X25X3102X15X27X315X1X2X30案M法标准化后在第1、2约条'别入人变X5X6FL题的目标改写MAXZX12X23X3MX5MX6注人FL题写特别M的纯形表解结果如下11第二深入性规划案表27CJ1230MMBCBXBX1X2X3X4X5X61X11527200121520X40221115CJ012120M12MZ152本FL题有优解XX1X2X3X4X5X6T15200500TZ152阶法第阶构造辅FL题MINWX5X6注辅FL题写特别目标应MIN而MAX纯形表解结果如下案表28CJ000011BCBXBX1X2X3X4X5X60X32757100171570X447470115757CJ000011W0优表W0即人变全0变进入第二阶。第二阶去掉第阶优表的人变原'FL题的目标系纯形表解结果如下案表29CJ1230BCBXBX1X2X3X41X1152720152注2解过右常得现负0X4注1得调整变的次序下02215CJ012120Z152本FL题有优解XX1X2X3X4T152005TZ1522MAXZ12X115X210X3STX1X22X35X13X25X36915X16X25X3615X1X2X30案M法标准化后在第1约条入人变X5目标改写MAXZ12X115X210X3MX5注人FL题写特别M的纯形表解结果如下12第二深入性规划案表210CJ1215100M00BCBXBX1X2X3X4X5X6X7MX5043800117163801210X3039801003161803212X1191600011611632CJ02784380M0M0218716M58380MZ3312M优表留有0的人变X5本FL题行解必须确而写“解”、“界解”的。阶法第阶构造辅FL题MINWX5注辅FL题写特别目标应MIN而MAX纯形表解结果如下案表211CJ0000100BCBXBX1X2X3X4X5X6X71X504380011716380120X303980100316180320X1191600011611632CJ04380010716380W12辅FL题优表留有0的人变X5原'FL题行解必须确而写“解”、“界解”的。3MINZX1X2X3STX12X23X3122X1X24X38X1X2X30案M法在第1、2约条'别入人变X4X5目标改写MINZX1X2X3MX4MX5注人FL题写特别M的纯形表解结果如下案表212CJ111MMBCBXBX1X2X3X4X51X30310115110851X111110025310365CJ015015M25MZ285本FL题有优解XX1X2X3X4X5T36508500TZ285阶法13第二深入性规划第阶构造辅FL题MINWX4X5注辅FL题写特别目标应MIN而MAX纯形表解结果如下案表213CJ00011BCBXBX1X2X3X4X50X30310115110850X111110025310365CJ00011W0优表W0即人变全0变进入第二阶。第二阶去掉第阶优表的人变原'FL题的目标系得到的纯形表接优表案表214CJ111BCBXBX1X2X31X303101851X1111100365CJ0150Z285本FL题有优解XX1X2X3T365085TZ2854MINZ2X14X2STX14X21X1X21X1X20案M法标准化入人变X4X6则目标改写MINZ2X14X2MX4MX6注人FL题写特别M的纯形表解结果如下案表215CJ240M0MBCBXBX1X2X3X4X5X64X20115151515252X1101515454535CJ006525M45M45MZ25后表小比准则效入变X5在约阵列的'全负人FL题有界解。又表的变人变本FL题有界解必须确而写解、行解的。阶法14第二深入性规划第阶构造辅FL题MINWX4X6注辅FL题写特别目标应MIN而MAX纯形表解结果如下案表216CJ000101BCBXBX1X2X3X4X5X60X20115151515250X1101515454535CJ0002511W0优表W0即人变全0变进入第二阶。第二阶去掉第阶优表的人变原'FL题的目标系得到下纯形表案表217CJ2400BCBXBX1X2X3X54X2011515252X110154535CJ006545Z25X5入变小比准则效本FL题有界解必须确而写解、行解的。5、某的性规划FL题解过得到下纯形表表229CJ311011BCBXBX1X2X3X4X5X6X3201X5332X6613CJ、知常。表229白的'完整'别当、足么条有下结1当前表优表但有I优解2下本行解化解3本FL题有界解4当前本解进优化的行解X2入变X5变5当前的本行解化解。案解纯形表缺'完整有案表218CJ311011BCBXBX1X2X3X4X5X61X32100011X53302101X6601013CJ620100Z215第二深入性规划1足86620此“掉下6/2本FL题有I优解。2足8006060060则该性规划FL题有界解。4需足当前解行解X2的检验正于X1的检验后正X2入变小比现在X5在行需'X6在行的系正当前表能判有界解。当0足8002026308000当60足86002026660806064660综即足8000806064660注组“子'解I形“。5前提1、由当前表能判FL题有界解2、有变取0即右有常16第二深入性规划0第2题的结果知案6600二、改进纯形法的计6、应用改进纯形法解下列性规划FL题本题得用表法解应体现变组迭点B1心计变检验择入变择变计入变列及右新常应用小比法然后得到新的变组。1MAXZ6X13X23X3ST2X1X2684X12X23X3614X12X2X3618X1X2X30案得用表法解优解X08003034T优ZCBB1B242MINZ10X19X2STX168X26105X13X245X1X20案得用表法解FL题标准化MINZ10X19X2MX6本题能入人变后再用改进纯形法计里人FL题的目标能写STX1X38X2X4105X13X2X5X645X1X2X3X4X5X60优解X853025300T优ZCBB1B9517第二深入性规划3MAXZX1X22X3STX12X2X36102X14X22X36402X13X2X3630X1X2X30案得用表法解优解X10307033000T优ZCBB1B13034MAXZ5X14X23X3STX1X2X36323X1X2X36842X13X2X3660X1X2X30案得用表法解优解X2615000T优ZCBB1B1497、应用改进纯形法证性规划FL题MAXZ4X15X22X33X4STX13X2X32X46102X1X23X3X4610X1X2X3X40存在X1X2变的优解。案得解的“来证证当X1X2变组变检验全负足优化条X1X2变组得到的本解行解。掉过则证完整。综结得证。18第三性规划的理第三性规划的理习题、理的应用1、FI知性规划FL题MAXZX1X2STX1X2X3622X1X2X361XI0I123应用理证该FL题有界解。案证写原FL题的FL题观察知FL题行解又原FL题行如X1X2X30行解原FL题行能少了“原FL题行”条由弱理及知原FL题应有界解即优解。2、写下列FL题的FL题1MINZ2X1X23X3X4STX1X2X3X4652X1X23X34X1X3X41X1X20X3X4案MAXW5Y14Y2Y3STY12Y2Y362Y1Y261Y13Y2Y33Y1Y31Y160Y2Y30注检查约条变的。2MINZ3X14X22X35X4ST4X1X22X3X42X1X23X3X46142X13X2X32X42X1X20X360X4案MAXW2Y114Y22Y3ST4Y1Y22Y363Y1Y23Y3642Y13Y2Y32Y1Y22Y35Y1Y260Y30注检查约条变的。3MAXZ2X13X2STX1X2632X1X22X10X2案MINW3Y12Y2STY12Y22Y1Y23Y10Y260注检查约条变的。3、FI知性规划FL题MAXZ2X1X219第三性规划的理STX26102X15X2660X1X26183X1X2644XI0I12的优解X113X25用理FL题的优解。案Y10Y20Y3Y4124、FI知性规划FL题MINW2X13X25X32X43X5STX1X22X3X43X542X1X23X3X4X53XJ0J12345的FL题的优解YY1Y24535Z5用互松理原FL题的优解。案X1X51X2X3X40二、纯形法5、用纯形法解下列性规划FL题1MINZX12X23X34X4STX13X22X33X4302X1X23X3X420XJ0J1234案优解X16X28Z222MINZ4X112X218X3STX13X332X22X35XJ0J123案优解X10X232X31Z363MINZ6X13X22X3ST6X13X25X3103X1X22X34XJ0J123案优解X32Z420第三性规划的理4MINZ6X13X25X34X4ST4X12X23X32X442X14X27X3X453X18X2X34X44XJ0J1234案有I优解纯形表优解X112X21X7112Z6必须确有I优解三、灵FL'规划6、有性规划FL题MAXZ2X13X2X3STX1X2X363X14X27X369XJ0J123先的优解然后在条变的前提下'别讨下列A目标变X1的系C1的变化优解保变B当右常B39T变B069T讨优解的变化C第1约条右常B1的变化优保变D约条X1的系列由11T变12T讨优解的变化E增加约条X12X23X3610讨优解的变化。案解A解FL题过得优表案表31CJ23100BCBXBX1X2X3X4X52X1101431313X201213132CJ0035313Z8C1在343间优解变。B优变但优解变X15X21X3X4X50B小题如果从原'模型解而用“BB1B判即结果正确“效案”。CB1在949间优变。DFL题应新解过。新的优解X1X232X3X4X50D小题有些法新解而仍X1X2变计有变的检验判优结果正确E新的约条够约此优解变。此处判新约加入原优表计的“I此”7、某司生产A、B、C三产需资源劳原料优生产计划表总利润。下的性规划模型有于FL题。MAXZ3X1X25X3ST6X13X25X3645劳3X14X25X3630原料X1X2X3021第三性规划的理X1X2X3产A、B、C的产利润元劳原料的'别小千。入松变X4X5模型的优表如表329。表329优纯形表CJ31500BCBXBX1X2X3X4X53X11130131355X301115253CJ03001Z30下列FL题A当前优解变条下产A的利润间。C12的优解。B当用原料增80千优解。C当前优解下劳原料的影子价'别I少”D场应缺额外购原料用于扩生产须支付高于原价08元/千的价FL司应购入扩生产”I购入I少优生产组保变”如果价高于原价15元/千结果又应该么”E由于技进产B的原料减15千影优解”么”F增加“检查”约条2X1X23X3615优解变化”如有变化新的优解。案解A36C166产A的利润间36当C12优解X36X1X2X50X415优Z30。B当原料增加到80有BB1B13131525458035323此原优解行B填入原优表用纯形法解新的优解过得X39X1X2X40X535优Z45B小题如果从原'模型解而在B后再从优纯形表开'解即结果正确“效案”。C劳原料的影子价'别01即JC4J0JC5J1D由B小题类似的法得当原料B2的应在45245优组变。当原料高于原价08元/千应购入小于影子价080整后“变负及01整约能Y1AY1BY2AY2B012、某型社临街的“FL店每天的营间800到2400社早、晚的需天员的需如427。024681012141618202422246810时间人数427FL店的员需该店的员'类。第类正“员'别在38小800到1600、1200到2000及1600到2400薪14元/小规正“员23第四整规划能少于3人第二类CURRENCY1点在800到2400的任间薪12元/小。FL应如雇用正“员CURRENCY1点在人资源本小的础足需”本FL题的整规划模型。案解X1X2X3表3的正“员的间CURRENCY1点的人设间800120012001600160020002000220022002400注CURRENCY1点的正“员CURRENCY1点人Y1Y2Y3Y4Y5则得到间注正“员的员的总需总人如下表案表41间800120012001600160020002000220022002400人需481064员总X1Y1X1X2Y2X2X3Y3X3Y4X3Y5本FL题的模型MINZ148X1148X2148X3124Y1124Y2124Y3122Y4122Y5STX1Y14X1X2Y28X2X3Y310X3Y46X3Y54X13X1X23X2X33X33X1X2X3Y1Y2Y3Y4Y50整变负、整约能3、某司计划在、南三地网点总共有7备地点AII17择。现设的网点必须足下条在地A1A2A3三备地点I择地点设网点在地A4A5备地点少择地点设网点在南地A6A7备地点能设网点于场的考如果案择了A2地点必须择在A5设网点。若在备地点AI设网点需资BI万元每年得利润CI万元。FL如果总资B万元在哪些备地点设网点得I的利润”本FL题的模型。案24第四整规划解XII17表XI1在第I地设网点0在第I地设网点本FL题的模型如下MAXZ7PI1CIXISTX1X2X362X4X51X6X71X2X5607PI1BIXI6BXI01I12701整约能4、某司有10条9表422了10条行的行总小小。表422行表备行12345678910ABCDEFGHI行间4656746557注“”表该从10条择3条既能够足行总间少的又能够9少1次。本FL题的01整规划模型。案解01变XII110XI1择I0择I则本FL题的模型MINZ4X16X25X36X47X54X66X75X85X97X10ST10XI1XI325第四整规划X1X3X91X2X3X4X5X101X3X5X6X71X1X8X101X5X6X101X2X8X91X1X4X7X81X4X6X91X1X2X5X71XI01I121001整约能5、某小提顾提的需生产小提价固本需而。由于的技有12人该能挑'甚能'完的。目前到来自3家交乐的小提表423了此关的本价元。表423123把345价元/把300040001000固本元400030000技需人/把232FL应接I少台得I的利润”本FL题的整规划模型。案解X1X2X3该完的考固本得到本FL题的整规划模型MAXZ3000X14000Y14000X23000Y21000X3STX163X264X3652X13X22X3612X16M1Y1X26M2Y2X1X2X30整01整约能Y1Y2016、FL题5如果小提的价固而由的确的总价。表424了体元。26第四整规划表424的总价123000011000100010002300020002000350004000300046000500057000FL此应接I少台得I的利润”本FL题的01整规划模型。案解01变XIJ表XIJ1IJ把0IJ把例如X101表10把。知有3有X10X11X12X131X20X21X22X23X241X30X31X32X33X34X351本FL题的模型MAXZ1000X113000X125000X1340001X10里的40001X10表该0产生固本下1000X212000X224000X236000X2430001X201000X312000X323000X335000X347000X35STX10X11X12X131X20X21X22X23X241X30X31X32X33X34X3512X114X126X133X216X229X2312X242X314X326X338X3410X35612X10X11X12X13X20X21X22X23X24X30X31X32X33X34X350101整约能7、某地级下辖的8邻的镇资过4生水处理用于集处理8镇的生水水处理的设计处理能20万/小。目前镇郊有用于设水处理的规划用地征用本如表425。表4258镇的用的地万元镇12345678本10012020016018090140150水处理后须镇水流处理需在镇间挖设沟管道。了地条水管道设本有差表426。27第四整规划表4268镇设水管道的设本元/千小镇123456781200040001000224003000450038000240018004240024001900540002000400062200360014007400030002700827002800表缺表行的案。在镇I郊的水镇I间需管道即本0。镇的水总按常住人粗估计的比例每千人每小产生500的生水镇常住人表427。表4278镇的常住人千人镇12345678人1002009018015012060130FL应如如设管道济地足8镇水处理需”本FL题的整规划模型。案解本题涉及固本。先计镇每小造水的千/小下表。案表42镇12345678人1002009018015012060130水501004590756030651变01变YII18YI1在镇I水处理0在镇I水处理XIJIJ18镇IXIJ的水到镇J的水。2目标总设本括管道本固本2镇I郊的水处理水需考管道本则目标MINZ02X1204X1401X16024X2403X26045X2808X31024X35018X37024X43024X45019X4804X5202X5604X57022X63036X64014X6704X7103X74027X78027X82028X85100Y1120Y2200Y3160Y4180Y590Y6140Y7150Y828第四整规划3约条镇的水总X11X12X14X1650X22X24X26X28100X31X33X35X3745X43X44X45X4890X52X55X56X5775X63X64X66X6760X71X74X77X7830X82X85X8865水处理20万/小水处理能约X11X31X716200X12X22X52X826200X33X43X636200X14X24X44X64X746200X35X45X55X856200X16X26X56X666200X37X57X67X776200X28X48X78X886200水处理的固本约X11X31X716MY1X12X22X52X826MY2X33X43X636MY3X14X24X44X64X746MY4X35X45X55X856MY5X16X26X56X666MY6X37X57X67X776MY7X28X48X78X886MY8I4间水处理即Y1Y2Y3Y4Y5Y6Y7Y864变的约YI01XIJ0IJ18完整模型。8、某生产AB型的产生产过须过甲、FL、丙三流水间加FL间有条加效的流水FL1FL2关表428。29第四整规划表428产加甲FL丙产利润FL1FL2万元产A332225产B754140间额250150120100FI知FL间的条流水能任条用FL如安生产得的利润。本FL题的整规划模型。案解本题有互的约条。设生产A、B的X1X2YI表用流水FLII12YI1用流水FLI0用流水FLI本FL题的完整模型MAXZ25X140X2ST3X17X262502X1X261003X15X26150M11Y12X14X26120M21Y2Y1Y21X1X20整变负、01整约能Y1Y2019、下FL题化01整规划FL题MAXZX12X25X3STJX110X23X3J252X1X23X3615XI0I123案本FL题模01整规划FL题如下MINZ19S1112S2134S3120S41STX2X1P1M1Y12X1X2P2MY12X3X1P1M1Y13X1X3P3MY1331第四整规划X4X1P1M1Y14X1X4P4MY14X3X2P2M1Y23X2X3P3MY23X4X2P2M1Y24X2X4P4MY24X4X3P3M1Y34X3X4P4MY34S11S12X1P1D152520S21S22X2P2D220222S31S32X3P3D3153520S41S42X4P4D4165034X1X2X3X4S11S12S21S22S31S32S41S420Y12Y13Y14Y23Y24Y340101整约能11、E流司每天凌GP日报从郊的印刷运往的5'点由5'点全有报、购报的'发报。在'避了交拥后有如表430的6条运择表运从印刷发次FL'点后再到印刷。表430备运运FL顺序113424353125423551426135FI知印刷'点间的行CURRENCY1离如表431。表431离表里印刷'点1'点2'点3'点4'点5印刷10131598'点11016495'点2131671110'点3154789'点4981076'点5851095FL择哪些运在足报送'点的前提下的总行CURRENCY1里”本FL题的01整规划模型提本题模思第4题类似过需自行计运的行CURRENCY1离。案解运的行CURRENCY1离运FL的'点01变XII16XI1择I0择I32第四整规划本FL题的模型如下MINZ31X133X244X337X440X531X6STX1X3X5X61X3X4X51X1X2X4X61X1X2X51X2X3X4X61XI01I1601整约能12、某少年体校游泳教领3生加少年游泳锦标赛比赛规则每能I能加2比赛而比赛允许校的I赛。报绩教估计3加5比赛的的能性如表432。表4323生加比赛的能性目50蛙泳50蝶泳50自由泳50仰泳100自由泳139656966572648424902235945553150FL教应该安哪些生加哪些比赛目能该校的总比赛绩”本FL题的整规划模型。案解01整变XIJ表第I加第J比赛PIJ表第I加第J比赛能够的能性即表。此有MAXZ3PI15PJ1PIJXIJST5XJ1XIJ62I1233XI1XIJ61J12345XIJ01I123J1234501整约能13、428了某地高小型用元。V1V2V3V4V5V6V7V835247492821655242833第四整规划FL某司计划从V1入高从V8下高应该如择道现行本低”本FL题的01整规划模型提结第的转运FL题性规划模型此FL题运1的转运FL题。案解如果VIVJ邻存在由VIVJ的边IJ12801变XIJXIJ0择从VI到VJ的1择从VI到VJ的提本FL题的性规划模型MINZ3X125X132X144X237X255X264X369X372X438X472X582X65X676X685X78STX12X13X141X12X23X25X26X13X23X43X36X37X14X43X47X25X65X58X26X36X65X67X68X37X47X67X78X58X68X781XIJ01IJ18存在由VIVJ的边01整约能14、某电信司需在如429的“”字形住的街道安装币电话每影'街道。了便用户电信司希望任人I需走到邻的街到台币电话。例如429点的行人到星在到币电话。从本的角考电信司I在每条道安装台电话。另外政的设计规减少交叉的碍币电话能够安装在交叉括转角、三岔字例如三角形的能币电话的安装地点。建筑物道路429FI知每街道I能安装台币电话FL少需安装I少台币电话能足”此FL题的01整规划模型。案解住的每条道进行编01变应。34第四整规划12345678910111213141516171819202122案41FL题知道如果在街道1安装币电话则邻的街道2、5、6用安装电话即X1X2X5X61整理得到本FL题的01整规划模型MINZ22PI1XISTX1X2X5X61X1X2X3X6X71X2X3X4X7X81X3X4X8X91X1X5X10X141X1X2X6X10X11X151X2X3X7X11X12X161X3X4X8X12X13X171X4X9X13X181X5X6X10X11X14X151X6X7X10X11X12X15X161X7X8X11X12X13X16X171X8X9X12X13X17X181X5X10X14X191X6X10X11X15X19X201X7X11X12X16X20X211X8X12X13X17X21X221X9X13X18X221X14X15X19X201X15X16X19X20X211X16X17X20X21X221X17X18X21X221XI01I12215、下表了2组英'词每组词皆6。每词由4字母组。表4332组12词123456第1组HEATPASTPROFSTOPFOOTAREA第2组FORTHOPESPARTHATTREESTAR35第四整规划2组英'词际仅由9字母A、E、F、H、O、P、R、S、T组。现在用字19表9字母得2组英'词的字接近。词HEAT例如果字母H、E、A、T'别用1、2、3、4替那么词HEAT的字101234。本FL题模派FL题写派本表提字19“派”9字母。案解FL题先计2组每字母的现“差得到案表43AEFHOPRST第1组422143224第2