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管理运筹学课后习题详解内蒙古工业大学国际商学院张剑二九年一月2第2章线性规划的图解法X2X15336A(12/7,15/7)00051X1X2071A(02,06)2(1)有唯一最优解A点,对应最优目标函数值Z36。(2)无可行解。045X1X25882023X1X2071(3)有无界解。3241(1)可行域为0,3,A,3围成的区域。(2)等值线为图中虚线所示。(3)如图,最优解为A点(12/7,15/7),对应最优目标函数值Z69/7。3(4)无可行解。012X1X221(5)无可行解。22X1X26804可行域4(6)最优解A点(20/3,8/3),最优函数值Z92/3。0812X1X261682可行域A(20/3,8/3)3(1)标准形式4(2)标准形式(3)标准形式4解(1)标准形式50X1X23225416求解0518259432121SXX5标准形式210426319463121SXX069X1X261024可行域A(36,24)6最优解为A点13266,84523112变为变化。斜率由)(如右图XXC0624X1X2101628可行域A(3,7)1047模型(1)X1150,X2150;最有目标函数值Z103000。6(2)第2、4车间有剩余。剩余分别为330、15,均为松弛变量。(3)四个车间对偶价格分别为50、0、200、0。如果四个车间加工能力都增加1各单位,总收益增加5002000250。(4)产品1的价格在0,500变化时,最优解不变;产品2的价格在4000,变化时,最优解不变。(5)根据(4)中结论,最优产品组合不变。8模型(1)XA4000,XB10000,回报金额60000。(2)模型变为XA18000,XB3000。即基金A投资额为180005090万,基金B投资额为300010030万。7第3章线性规划问题的计算机求解89第4章线性规划在工商管理中的应用10111213141516第5章单纯形法1可行解A、C、E、F;基本解A、B、F;基本可行解A、F。2(1)标准形式(2)有两个变量的值取0。由于有三个基变量、两个非基变量,非基变量最优解中取0。(3)解2,064,2106214820114201600216008312412241SXA(4)将X1S2代入约束方程组中可得。1,0,321SXS02021064321121SXA将对应的向量化作,即的排序是根据标准化后,321,SX321,SX对应向量中单位向量的位置而定的,两者为一一对应的关系。(5)此解不是基本可行解。由于基本可行解要求基变量的值全部为非负。173(1)解(2)该线性规划的标准型为(3)初始解的基为,初始解为,此时,321S20,54,0目标函数值为0。(4)第一次迭代,入基变量为X2,出基变量为S3。4(1)单纯形法0,92473MA431121XZX1X2X3X4次数XBCB4100BX301310770X40420197/4Z000041000X3005/211/419/41X1411/201/49/4Z420101019,4,4321X18(2)图解法0X1X2726745225可行域A(9/4,0)5(1)解0,48412235MAX653215313XXXZX1X2X3X4X5X6次数XBCB1285000BX403211002020/3X5011101011110X601241001484Z00000012850000X40013/4101/488X5002/311/12011/12721/21X11211/31/12001/12412Z124100104400148X28013/4101/4832/3X50005/122/311/125/342X112101/61/301/64/3Z128440000140080X2801011/59/51/105X350018/512/51/543X1121009/52/51/52Z12853/512/521/50003/512/521/5841984,52,321ZX(2)解0,5824MIN643213131XXXXFX1X2X3X4X5X6次数XBCB121000BX402211004X50122010840X6011100155Z0000001210000X405/21011/208X311/21101/2041X601/22001/211Z1/21111/203/21011/2044,32FX6解0,16245MA5431431132AXXMAXZX1X2X3X4X5X6次数XBCB51300MBA1M142101105/20X5012101016ZM4M2MM0M5M14M32MM0010MX211/411/21/401/45/2101X503/2021/211/22114Z1/411/21/401/419/405/21/40M1/45/22X151421011020X5006111166Z52010505019750M550400611116Z5105050011205M此问题有无界解。7(1)解0,8123MAX54321152XXMXZX1X2X3X4X5次数XBCB31200MBX30221001111/20X5M1101188ZMM0MM3M12M0M08MX212111/20011/21X5M201/2115/2Z122M126M/2MM92M06M/2M0665M/225,54321XX将本解代入所有约束中发现,不满足约束2,所以本题无可行解。(2)解0,281034MIN87654321857421687654321XXXXMXXXFX1X2X3X4X5X6X7X8次数XBCB43000MMMBX6M21/21001001050X7M1101001088COMMENT微微微微1有错误21X8M1000100122Z4M3M/2MMMMMM44M33M/2MMM00020MX6M01/210210263X7M0101101166143M/2MM3M4MM3M4033M/2MM43M004M412M8X5001/41/2011/201312X7M03/41/2101/210342X1411/41/2001/200520Z413M/42M/2M02M/2M0023M/42M/2M023M/20M203MX50002/31/312/31/312X23012/34/302/34/3043X14102/31/302/31/304Z432/316/302/38/300011/316/30M2/3M8/3M2828,4,8765421ZXX(4)解0,162844MA75431653121XXXZX1X2X3X4X5X6X7次数XBCB31200M00BX5M422110041X60240001020100X704820001164Z4M2M2MMM0024M12M12MM0004MX12121/20001/44X600010011/2121X70060110112Z2410001/24300M01/28,2,4,76531ZX22由于存在非基变量检验数为0,所以本题有无穷多解。23第6章单纯形法的灵敏度分析与对偶1(1)为非基变量,所以只要保证即可。1X011ZC。240241CC(2)为基变量,所以有2X6218,MAX0MIN022222CCAJJJJJJ(3)为非基变量,所以只要保证即可。2S0222SSSZC。808222SSSCC2解第五章习题5(2)最终表为X1X2X3X4X5X6次数XBCB121000BX405/21011/208X311/21101/2041X601/22001/211Z1/21101/203/21001/204(1)为非基变量,所以只要保证即可。1X11ZC。2211CC(2)为基变量,所以有3X24021,1,MINMAX0I03321333CCAJJJJJJ(3)为非基变量,所以只要保证即可。S0222SSSZC。102122SSSCC3(1)解250,0MINMAX11111BDXBDIIBIIB(2)解5050,310MINMAX22222BBDXDIIBIIB(3)解1500150,0MINMAX332333BBDXDIIBIIB254解X1X2X3X4X5X6次数XBCB121000BX405/21011/208X311/21101/2041X601/22001/211Z1/21101/203/21001/204(1)解4,80MIN0MAX11111BDXBDIIBIIB(2)解108,1MIN4,8MAX0I02222222BBBDXDXIIBIIB(3)解3333345,10MIN0MAXBDXBDIIBIIB5(1)解为基变量,所以有1X2663,1,MIN0MIN0AX111111CCACJJJJJJ当时,在上述范围内。所以,最优解不变。21C(2),。增加15个单位的原料不会5572B452B使原最优解变化。原材料的对偶价格为1。即增加一个单位的原材料可使总收益增加1。原料价格为067元。所以,有利。(3),。155B6031B(4)解1251332251321321ZCZPB由于检验数满足非正要求,最优解不变,所以不用修改生产计划。(5)解03523144452313141ZCZPB此时生产计划不需要调节,由于新产品的检验数为0。6答均为唯一最优解,根据计算机输出结果显示,如果松弛变量或剩余27变量为0且对应的对偶价格也为0,或存在取值为0的决策变量并且其相差值也为0时,可知此线性规划为无穷多组解。7(1)解0,1520MIN211YYTSF(2)解0,234610MAX1221YYTSZ8(1)解无约束321213321,0,050MINYYTSF(2)解无约束321321321,0,6MAXYYTSZ9解280,2843MAX65432121XXXTSZX1X2X3X4X5X6次数XBCB123000BX401111004X5011201080X600110012Z0000001230000X111111004X5002111041X600110012Z1111000321004X111001016X5000311202X200110012Z12210300510310,6,321FZ29第7章运输问题1(1)解最小元素法求初始调运方案销地甲乙丙丁12505030024004003350150500产地合计4002503502001200位势法求检验数销地1234U15250050300024001420400163073501505003产地合计4002503502001200V26172325闭回路法调整方案销地1234102505030024004003350150500产地合计4002503502001200求检验数销地1234U102502350300024006121440011319143501505003产地合计4002503502001200V21172325检验数都大于0,得到最优调运方案。运费为19800元。(2)解初始调运方案为30销地12345合计1505020030024002006003350150500产地合计4002503502002001400求检验数销地12345U11250235020002400200212322产地391435015033V121723250调整调运方案销地1234515050502400200产地3350150150求新的检验数销地12345U11250202005002400200182322产地312173501501500V121720250调整调运方案销地123451100200240015050产地3350150求新的检验数销地12345U112100222002024001502023502产地3101535025150231V121722252检验数都大于0,得到最优调运方案。运费为19650元。(3)解新的运价表为销地1234合计12117232530021015301940032321202250040000150产地合计5502503502001350最优调运方案(求解过程略)销地1234515025000300240000040030035015050041000050150产地合计5502503502001350运费为19600元。2解运价表11,22,3456合计104040505030404013002030307070905060350030606080804070504400407070404030704071005M0M000M0300合计1501501501003502002501501600求解可得11223456合计100500100001503002150150000200005003000015002500400400100000001005000100100000300合计150150150100350200250150150032此外,还有其他解如下11223456合计100500150001503002150150000150005003000015002500400400100000001005000100505000300合计150150150100350200250150150011223456合计100500001001503002150150000200005003000025001500400400100000001005000100100000300合计150150150100350200250150150011223456合计100500001501503002150150000150005003000030001000400400100000001005000100505000300合计1501501501003502002501501500运费为485元。3解运价表如下1234合计1600660720031,660720780032M700760042,M770830023MM650023,MM71503合计555217最优生产方案为331234合计1210031,300032040042,000223002023,00303合计5552174解运价表为甲乙ABCD合计甲01001502001802401600乙80080210601701700A15080060110801100B200210700140501100C180601101300901100D24017090508501100合计1100110014001300160012007700最优调运方案甲乙ABCD合计甲11000300200001600乙011000060001700A0011000001100B0001100001100C000010001001100D0000011001100合计1100110014001300160012007700调整后可得甲乙ABCD合计甲0030020000500乙00006000600A0000000B0000000C00001001000D0000000合计003002005001001100总运费表甲乙ABCD合计甲0045000400000085000乙00003600003600034A0000000B0000000C0000090009000D0000000合计004500040000360009000130000总运价为130000元。5解运价表为12345合计A5449526401100B5773696101000合计5003005506501002100最优调运方案12345合计A250300550001100B250006501001000合计5003005506501002100最低总成本为110700元。6(1)最小元素法确定的初始调运方案为123合计A87415B35925C00010合计20102050123合计A1515B1010525C1010合计20102050(2)表上作业法求最优调运方案调整运输方案并求检验数123合计123UA87415A107150B35925B101055C00010C10262合计20102050V20435123U123UA41150A63150B151061B20541C5254C2554V464V244123合计123合计A001515A006060B205025B6025085C05510C0000合计20102050合计602560145最优调运方案的总成本145元。(3)由于所有检验数大于0,所以存在唯一解。(4)解123合计123UA87415A107150B35925B101055C00020C20262合计30102060V204123U123UA41150A43150B151061B25361C15254C51054V464V444123合计123合计A001515A006060B250025B750075C510520C0000合计30102060合计75060135最优调运方案的总成本135元。36第8章整数规划1(1)(2)254173ZX43921ZX4182X39Z264021ZX无可行解39Z64037412ZX40521XZ22314672ZX751423ZX3142X0(3)61073421ZX69038421ZX620351X97601342ZX6153842ZX381052ZX1162Z3738394041424344454647第9章目标规划484950515253第10章动态规划1整个过程划分成4各阶段,设初始状态为KS(1)K4时阶段4本阶段各终点本阶段初始状态E到终点的最短距离本阶段最优终点D133ED244E(2)K3时阶段3本阶段各终点本阶段初始状态D1D2到终点的最短距离本阶段最优终点C13254595D1C237104488D2C33584488D1,D2(3)K2时阶段2本阶段各终点本阶段初始状态C1C2C3到终点的最短距离本阶段最优终点B156118311851311C1,C2B2538821084128C2B354981985139C1,C2(4)K1时本阶段各终点本阶段初始状态B1B2B3到终点的最短距离本阶段最优终点A113148513941313B1,B2则有最短路线长度为13。分别是(A,B1,C1,D1,E);(A,B2,C2,D2,E);(A,B1,C2,D2,E)。2按项目将整个过程划分为3个阶段分配给第K个项目到最后一个项目的资金。4KS1S分配给第K个项目的资金。X;KKXS1,MAX1KKKKSFSRSF54(1)K3时,3XSR3XS012343SF3X046460170701276762388883488884(2)K2时,32SFXSRX2S012342SF2X0464995950170491194652981190276491257052122466110712503884913776521287061131467111813704884913788521407661137707114146781241413(3)K1时,21SFXSR1XS012341SF1X41414718813751188125591841197119095761711903则有分配方案为(3,0,1)。3按月将整个过程划分为4个阶段为第K个月月初库存量。为第K个月的产量。KSKX;KKDXS1,MIN1KKKKSFXSRSF4,IINKKD(1)K4时55,4XSR4XS012344SFX0686831552232321306060(2)K3时,43SFXSR3XS012343SF3X01158158421414214331221241261222(3)K2时,32SFXSR2XS012342SF2X0224232243120821620820822191941981913164176181640(4)K1时,21SFXSR1XS012341SF1X02522562582522521,4最有生产策略(1,3,4,3);(4,0,4,3)。最低成本252。4按产品划分阶段,则共有4各阶段。为装载第K钟产品前,还可以装载的重量。为第K种产品的KSKX56装载数量。;KKKDXS1,MAX1KKKKSFSRSF;KKKQR,KDX(1)K3时,3XSR3XS0123SF3X0102031800418018015180180161801801718036018180360360291803603602101803603602(2)K2时,32SFXSR2XS01232SF2X000012301401401418014018005180140180061801402802802718032028032018360320280360093603202804204203103603204604204602(2)K1时57,21SFXSR1XS0123451SF1X104604604804804005005005最优策略(5,0,0)。5按年划分成5各阶段。为年初完好的机器数量。为第K年处于高负荷状态下工作的KSKX机器数量。状态转移方程8051KKKXSXS阶段指标函数6,R最有指标函数0,MAX610SFSFXSRSFKKKSKK(1)K5时,5X550101AXSSSFKS(2)K4时4440445440415MAX8061ASXSXXFSSFKKKSSSX(3)K3时333034330318518MAX064ASXSXSFXSXSFKKKSX5822202322024014MAX8586ASXSXFXSXSFKKKSSSX(5)K1时11101211013232MAX8054064ASXSXFXSXSFKKKSSSX由于125,代入。1S7905万元F6按工厂划分成4个阶段。第K期初剩余金额。为第K期投入的金额。KSKX状态转移方程KS1阶段指标函数DR(1)K4时,4XSR4XS01234564SF4X0000128281247472365653474744580805(2)K3时,43SFXSR3XS01234563SFX000010282818018281592047471828463903947030656518476539286761061672407474186583394786612889780788935080801874923965104614710878281069009010836085851880983974113616512678471259028118950951263(3)K2时,32SFXSR2X2S01234562SFX000010282825025280204747252853450455313067672547724528735705773240898925679245479257288565065921,2501081082589114456711257471146528947007011416012612625108133458913457671246547112702898730731342(4)K1时,21SFXSR1XS01234561SFX60134134201141344292134607313375531288528113900901340,1,2则有,最优策略为(0,2,3,1);(1,1,3,1);(2,1,2,1);(2,2,0,2),对应最优解134。7按照地区划分为3各阶段。第K期初可供分配的商店数。为第K期投建的商店数。KSKX状态转移方程KKS1阶段指标函数DR(1)K3时,地区1,3XSR3X0123453SFX603S000013312772312123414144515155(2)K2时,地区2,32SFXSR2XS0123452SFX00001033505512077538100101023012125712103131401414340141451217107171431716016171,2,350151551419101222147211631916016222(3)K1时,地区3,21SFXSR1XS0123451SFX5022224172171421910191051511011220则最优策略为(3,2,0)。8按年划分成5个阶段。为机器使用的年数。当年为更新费用,为使用年的机器维护TCOT费用。期初机器已使用的年数。本年机器是否更新。KSKXKRF,MIN10TGTORCTRIJIJI(1)K5时615R5XSRK5FX1513186066K2513188088K3513181101111K4513181801818K,R5513181801818R(2)K4时4R4XSRK4FX1512623681414K25126238111919K351262311182923R451262318213923R(3)K3时3R3XSRK3FX151214316192525K251214318233131K,R3512143111233431R(4)K2时2R2XSRK2FX151125416313737K251125418313939K(5)K1时1R1XSRK1FX151137536394545K62最优策略为(K,K,R,K,K)。9以周为单位划分成6个阶段。(1)K6时,66SF50F506F606F(2)K5时74325E60,57,5SSF当当当(3)K4时5430254E60,54,444SSF当当当(4)K3时45330253E60,453,33SSF当当当(5)K2时85240252E60,852,222SSF当当当(6)K1时15743051E6360,1575,1SSF当当当最优策略为在前4周如市价为500元时,立即购买。否则等待。在第五周市价为500或550时购买,否则等待。10按月划分为3个阶段。为期初有合格品;为期初无合格品。本期试制数量。0KS1KSKXKKXKXKSP321010,52KKXOC(1)K3时321523X3XS01234563SFX011513511179949469489819464(2)K2时2346952XX2XS01234562SFX001946988783837875933833(3)K1时1328521X1XS0123451SFX1839038197968148597963第一月、第二月试制3个,第三月试制4个。6411按月划分成4个阶段。期初库存。本期订货量。KSKX(1)K4时444101,SXSXF(2)K3时,34S33343090,SXSXSXF(3)K2时,2S22324708,SXSXF(4)K1时,12S13504045,21SXX

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