《运筹学(胡运权)第五版课后答案-运筹作业.docx》由会员分享，可在线阅读，更多相关《运筹学(胡运权)第五版课后答案-运筹作业.docx（36页珍藏版）》请在三一文库上搜索。
1、47 页 1.1bX22x1+x2=2用图解法找不到满足所有约束条件的公共范围，所以该问题无可行解47 页 1.1d无界解1.2 (b)约束方程的系数矩阵A= 1234(21P1 P21P3P4 )基基解是含RJ行解目标函数值X1X2X3X4P1P2-411/200否P1P32/5011/50是43/5P1P4-1/30011/6否P2P301/220是5P2P40-1/202否P3P40011是5最优解 A=(01/220)T 和(0011)T49页13题设Xij为第i月租j个月的面积minz=2800x11+2800x21+2800x31+2800x41+4500x12+4500x22+4
2、500x32+6000x13 +6000x23+7300x14s.t.x11+x12+x13+x14 当5x12+x13+x14+x21+x22+x23 m0( x13+x14+x22+x23+x31+x32 或0x14+x23+x32+x41当2L Xij10用excel求解为：用LINDO求解:LP OPTIMUM FOUND A T STEPOBJECTIVE FUNCTION VALUE1)118400.0VARIABLEVALUEREDUCED COST0.0000001.000000X113.0000000.000000X210.0000002800.000000X318.0000
3、000.000000X410.0000001100.000000X120.0000001700.000000X220.0000001700.000000X320.0000000.000000X130.000000400.000000X230.0000001500.000000X1412.0000000.000000ROW SLACK OR SURPLUS DUAL PRICES2)0.000000-2800.0000003)2.0000000.0000004)0.000000-2800.0000005)0.000000-1700.000000NO. ITERATIONS=3F 依 Edit S
4、olw Reports Wind。丙 Hdp口 ：；： o 门3 -tJR+陶阳住|国画 画XU 立国叫 FH1)HB4cX, 0VARtAELFVALLEREDITED COtT10. oocooo1. OODOtXJXII3 0000000. OOQOODQ.(X用MQ?SWJh ototqo仃工 OM0OOo. ooooooX410. 00000。ZLOO. 000000X120. 0000001700.000000220. oooooo-700.0000优X320 ooootoo.ooooooQ.OOOOOQ-400. OODOQO上巾0.0000001500. 000000X1412
5、. 0000003. 00000CIF OFTTMCI FQl?l5xl2+xl3-xl4+a2L-x22+x23;10xlW14+x22tK23+i31+x3220xl4-23+x32+i4112|end|答 若使所费租借费用最小，需第一个月租一个月租期300平方米，租四个月租期1200平方米，第三个月租一个月租期 800平方米，50页14题设al, a2, a3, a4, a5分别为在A1, A2, B1, B2, B3加工的I产品数量，bl, b2, b3分别为在A1, A2, B1加工的II产品数量，cl为在A2, B2上加工的田产品数 量。则目标函数为maxz= (1.25-0.25
6、)( a1+a2+a3)+( 2-0.35) b3+( 2.8-0.5)c1 -0.05 (a1+b1)-0.03 (a2+b2+c1)- 0.06 (a3+b3)-0.11(a4+c1)-0.05a5=0. 95a1+0. 97a2+0. 94a3+1.5b3+2.1c1-0.05b1-0.11a4-0.05a5s.t.5a1+10b10用lindo求解得:LP OPTIMUM FOUND AT STEP 6OBJECTIVE FUNCTION V ALUE1)16342.29VARIABLE A1 A2 A3 B3 C1 B1 A4 A5 B2VALUE1200.0000000.00000
7、0285.71429410000.0000000.0000000.000000342.857147571.42858910000.000000REDUCED COST 0.000000 9.640000 0.000000 0.00000015.900000 0.230000 0.000000 0.000000 0.000000ROW2)3)4)5)6)7)8)SLACK OR SURPLUS0.0000000.0000000.0000005628.5712890.0000000.0000000.000000DUAL PRICES 0.168000 1.500000 0.075000 0.000
8、000 0.008571 0.110000 -1.500000NO. ITERATIONS=计算lindo截屏Rf tdh MvtMvtaw faDjtf&HlAl I *刈日-I比图 a&B史al 业L? 0FTIW3! FOL? AT 昨 6CejECTJVE FIXCTIOS valIEU州3429尿 iLi c ded”班1 0rg由10 WE+Ol 94a341 ：bj-. c l-0. Ozil-O. llaJ-t 05c5J,-AftJAl产斜即clM利而认AVALUESEDUCED COSTL2OT. 1X)00000.0000000 twofflfl厚汨口阿工5, 71429
9、4认 OOOOOOLOOM OOftOOOa ooooooO.MOOOO13.000000O. OK)WOO必 23QOOO342. S5TU75 OOOOOO5；L.42Sd8a0. OOOWD100W. MONO0.000000加卜10bk啡OM 冷muE幽li-l-llcl -7OM7s5-JOOOal-62-43-41-a5iQbl-b2-b3-0F飙 SLAtK 顽 StKPULS2) 0.明喊i。如0.MOOOO410.woooa5) oflsa.a715B96) 0?)仇阊00(2W0.OQODOQM, ITERAT1然S=6DIAL FRICES 0.14BOOO L.5000
10、00 OTnftTfl 0. OQOTOO 。，皿打 0. noooo -L.30DWD2.1a:对偶问题为：maxz=2y1+3y2+5y3s.t.y1+2y2+y3 23y3+y2+4y30, y 222x1+x2+3x30,x3无约束因此本问题的对偶问题的对偶问81 页 2.12a)设x1,x2,x3分别为A, B, C产品数量 maxz=3x1+x2+4x3s.t.6x1+3x2+5x30453x1+4x2+5x3030x1,x2,x30 用lomdo求解为LP OPTIMUM FOUND AT STEP 2OBJECTIVE FUNCTION V ALUE1)27.00000VARI
11、ABLE X1 X2 X3X1,X2,X3VALUE 5.000000 0.000000 3.000000 0.000000REDUCED COST 0.000000 2.000000 0.000000 0.000000ROW SLACK OR SURPLUSDUAL PRICES2) 0.0000000.2000003) 0.0000000.6000004) 0.0000000.000000NO. ITERATIONS= 2max 3xHx2 + 4x36xl+3x2+5x3-45 3xl+4x2+5x3_0endLP OPTIMUM FOUND AT STEP 2OBJECTIVE FUN
12、CTION VALUE1)27.00000VARIABLE VALUEREDUCED COSTXI5.0000000.000000X20.0000002.000000X33.0000000,000000X1T X2, X30.0000000. 000000ROW SLACK OR SIRPLIJS DUAL PRICES 2)0,0000000.2000003) 0.0000000.6000004) 0.0000000.000000NO. ITERATIONS= 2最大生产计划为 A 生产 5 个单位， C 生产 3 个单位b)LP OPTIMUM FOUND AT STEP2OBJECTIV
13、E FUNCTION V ALUE1)27.00000VARIABLEX1X2X3X1,X2,X3VALUE5.0000000.0000003.0000000.000000REDUCED COST 0.000000 2.000000 0.000000 0.000000ROW SLACK OR SURPLUS DUAL PRICES2) 0.0000000.2000003) 0.0000000.6000004) 0.0000000.000000NO. ITERATIONS=2RANGES IN WHICH THE BASIS IS UNCHANGED:VARIABLEALLOWABLEDECRE
14、ASEX1X2X3X1,X2,X3COEF3.0000001.0000004.0000000.000000OBJ COEFFICIENT RANGESCURRENTALLOWABLEINCREASE1.8000002.0000001.0000000.0000000.600000 INFINITY 1.500000 INFINITYALLOWABLEINCREASE15.0000007.500000INFINITYRIGHTHAND SIDE RANGESROWCURRENTALLOWABLERHSDECREASE245.00000015.000000330.00000015.00000040.
15、0000000.000000Reports Windauw until Lied jLP OPTIMUM F01TFD AT STEP 21)27 00000V1PIABLEVALUEREM皿 COSTfllE CCOOOOO.ODOOCnK2C C OOOOOi.ooooao3 00000(1(J.QOUOCO1 . IZ, *3r coaoaoo. n-iDa dmax 3xl+x2-4x3 s. t.6x1-3x2+5x3045 3xLYx2 + 5x3二 30 xL x2, x3=0 end|1 二 F - -nx X一 z- X KICURRENTALLEABLECOEFINCREA
16、SEDECREASE.a.COCiOCOl.SOOOOO0 bODODQl.OOCOUO2 ooaoooIMF1M1TI4 ooccinnj QOOQOQ1 5QDODO0 . OOCIOOO匚 OOODOUIflFIHITVOBJ H近FFTGENT PiKGESVaiCF THE HAEIE ISUNCHiilTGECE 二口SLACK CR SCEF1USUUiil FR工匚E号2o.ooooao0.20000D,：)oconoao0.600000o.ooooaoO.OODDDDETGETEAND SIDE RiJIGEBCUBEEHTaLCPLaLLDVABLE广力INCREASEDE
17、CREASIib IDOC UL UIt iJUOOOU1 OOUU JO刈 GOCOCOIC OQOQOO7 BODODQC DOOOQO匚 nnoannTUN二 TV可知A产品的利润变化范围6. 8,2.4，上述计划不变c)设x4为产品D的数量maxz=3x1+x2+4x3+3x4s.t.6x1+3x2+5x3+8x40 453 3x1+4x2+5x3+2x40 30I x1,x2,x3 ,x40用lomdo求解为LP OPTIMUM FOUND AT STEP0OBJECTIVE FUNCTION V ALUE1)27.50000VARIABLEVALUEREDUCED COSTX10.
18、0000000.100000X20.0000001.966667X35.0000000.000000X42.5000000.000000ROW SLACK OR SURPLUS DUAL PRICES2) 0.0000000.2333333) 0.0000000.566667NO. ITERATIONS=0LP 0PI1MM FO：,NJ AT STEPmax 3xl+x2+4x3+3x4VARIABLE XI X2 * X4VALLE0. 0000000.0000005.0000002.500000REDICED COST 0. 100000 1. 9666B； 0. OOOODD 0. 0
19、00000ROW SLACK OR SURPLUSDIAL PRICES2)o. oooooo0. 2333333)0.0000000, 5666674)0. 0000000. 0000005)0 0000000. 0000006)5.0000000. OOOOQO7)2.5000000. 000000XO. ITER.ATIOXS=nOBJECTIVE Flyer I OX V.LUEi)27.500006xl-3x2+5x3+8x4=453x1+4x2+5k3+2x4=0x2=0x3-0 rndx1=x2=0,x3=5,x4=2.5,禾润为 27.5安排生产D有利，新最有生产计划为d)ma
20、xz=3x1+x2+4x3-0.4y s.t.厂 6x1+3x2+5x30453 3x1+4x2+5x3-y0用lomdo求解为LP OPTIMUM FOUND AT STEP0OBJECTIVE FUNCTION V ALUE1)30.00000REDUCED COST 0.600000 1.800000 0.000000 0.000000VARIABLE X1 X2 X3YVALUE0.0000000.0000009.00000015.000000ROW SLACK OR SURPLUS DUAL PRICES2) 0.0000000.4000003) 0.0000000.400000NO
21、. ITERATIONSE3ct&ports wirdflrt,LP OPTIMLM FOLXD AT STEP 0OBJECTIVE KATTrOX VAT 11)VARIABLEXI2330. 00000VALUE0.0000000.0000009,00000015,000000REDUCED COST 0.600000 1.800000 0. 000000 0.000000隆,:L”： ：| iHmax 3xl*x2+4x3-0+ 4ys. t.6xl3x2+5x3=453xl+4x2-5x3-y-0tc2=0y3=0y0 end|DUAL PRICES 0.400000 0.40000
22、0 0. 000000 0. 000000 0. 000000 0.000000R0K SLACK OK 5TRPLUF2) 0.0000003) 0.0000004) 0.0000005) 0.0000006) 9,0000007) 15.000000NO, ITERATIOK5= 0可知购进原材料15个单位为宜4.1a) 设yi= 1 第i组条件起作用M 一充分大正数J 0 第i组条件不起作用x1+x25+(1-y2)M y1+y2=1 y1,y2=0 或 1b)设yi= 1 第i组条件起作用1 0第i组条件不起作用x=0y1 - x=3y2 x=5y2x= x=7y4 y1+y2+y3+
23、y4=1、y1,y2,y3,y4=0 或 1c) 设yi= 1为假定取伯 500 0 为假定取值x=01 x=0y1I x50-(1-y2)M y1+y2=1L y1,y2=0 或 1d)设yi= 1第i组条件起作用i=1,2 则i=1,2 则0 0 第i组条件不起作用- x11-(1-y1)M x24+(1-y2)M y1+y2=1 j y1,y2=0 或 1e)设yi= 1第i组条件起作用/ 0第i组条件不起作用x x1+x2-(1-y1)Mx12+(1-y3)Mx3+x46+(1-y4)M y1+y2+y3+y4 2i y1,y2,y3,y4=1 或 04.2minz=j1 cjxj泮1
24、1 xj=5x1+x8=1x7+x8=1s.t.x3+x5 1A x4+x5 1 x5+x6+x7+x82xj= i 1选择钻探第sj井位T 0否4.5设xij为第i种泳姿用第名运动员4 x5minz=2_i=1 2_j=1 aijxij s.t.x x11+x12+x13+x14+x15=1 x21+x22+x23+x24+x25=1 x31+x32+x33+x34+x35=1 x41+x42+x43+x44+x45=1 x11+x21+x22+x23=1x12+x22+x32+x42=1 x13+x23+x33+x43=1 x14+x24+x34+x44=1I x15+x25+x35+x4
25、5=1 xij=1 或 0 (i=1,2,3,4 j=1,2,3,4,5)由excel计算得出；张游仰泳，王游蛙泳，赵游自由泳，预期总成绩为126.2s.iA k B L D E | F | t设E目标单元精1*S3 阉匚yjg；a 3*敬33.32S.53S.93, 433.6用于1 OBAAtO小值5值E7，；死用=由泳四L 226.429.fi2s.m31. 1可鹫里冗荏汨1:怙制印|_选击2_且践张王尚的毒电）】，泳xll112x13xl4xlG?EH5iiniB =一堆 H.4制 12tB|L9:p113、=3121:杆”；| 人一，“-、1121x23黄次08IGtlbilCfLB
26、 VJBMg弓甑员也比工lit i3l球i351 - -J _ .域L纥J第第,整 J 1n由泳 Ui142K必箕M145赵银张王周求和、泳I00001 = 1稣010001 = 1R泳001001 = 1目由泳o00101 = 1沫1111D= = = =11J11300000331000038,900D002S.50士蚌A7. 7黜1轴由为海0int一途且33. 3-缴338 &M.433.6自由泳sa 226.4叫6弟.531. 1理01市的端二料 xli 泪】钱 112k23张 xia x2a王 14 2i同 xl5 熔望以米M挖划-螃，可用足.Hi目的因*技藏比#U保春柞|下於铝i+
27、1 MBBM MB *( TW1短为用值8%值】决用msx31x32x33前x33 蹴 .力修可EiL增. 一用助位自由泳x42X*k44x43打钱张王宸京和gQ100=100010=010001=I由法100D01v131110111I1QQ33. S0QQ00M.700狙30D029. 20000总计29. 2明弭B34.70mini12(5.|26) ,E 点:d2-=4,d3-=0 min2d2-+d3-=8, F 点:d2-=0,d3-=4, min2d2-+d3-=4，故选 F八、程序法优先欲别约束工】约束建正偏差d-负偏差d-约束关系203010040=400424二240030
28、 二300050=5000cindl-Lind：尤先级别约束正约束建正楣差d-243014负偏差d-约束040关系=4000024=240030300Q403050灯并求解找到一鼾，可清正所育的狗束及最优状沅i.士屈值不吝振青0,fflg就网赖焉慝同怖立为回信W)即清保有方徐痔)一,优先凝别 dlB约束Ml24口EEIF蟒.差小 优愤差d“约束10。美乐d3金40243050LILSL9niirdl- mind4- min2d2!破圈法6.4a2避圈法2最小部分树166.4b10121012最小部分树32172 页 6.11红色曲线为使用一年卖出蓝色曲线为使用两年卖出绿色曲线为使用三年卖出紫色
29、曲线为使用四年卖出最短路程为 3.7万元，路径为 v0-v1-v4或v0-v2-v4或v0-v1-v2-v4三种方案分别为：第一年年初买新车，年末卖掉再买新车，一直用到第四年年末卖掉；第一年出买新车，用两年后于第二年末卖掉再买新车，用两年于第四年末卖掉；第一年出买新车，年末卖掉后再买新车，第二年末卖掉再买新车，再用两年于第四年年末卖掉。2. 80.8v12.6I 3d1.721.71.4v2v3v41.10.9I 0 I I TN1. 82.9由图可知，若摩托车最多使用三年，答案仍然不变6.14b根据题意，先给发点vs标号，由弧(vs,v2)得对v2进行标号(vs,1);弧(v2,v1)对v1
30、标号(v2,1);弧 (vs,v3)得对 v3 进行标号(vs,1);弧(v3,v4)对 v4 标号(v3,1);弧(v3,v4) 对 v4 标号(v3,1);弧(v4,v5)对 v5 标号(v4,1);弧(v5,vt)对 vt 标号(v5,1); 反向追踪找出增广链最大流为 14,最小割集为 (v1,v3) (v2,v3) (v2,vt) (v1,vt) 6.14c最大流13,无最小割集193 页-7.2 表 7-9194 页 7.3b假定最初事件在时刻零出现，则有t ES(1,2)= t ES(1,3)= t ES (1,4)=0t EF (1,2)= t ES (1,2)+t(1,2)=
31、2t EF (1,3)=4t EF (1,4)=8t ES (2,5)=2t EF (2,5)=5t ES (5,8)=8t EF (5,8)=15t es (8,9)=15t EF (8,9)=20t es (5,7)=5t ef (5,7)=7t ES (4,7)=8t EF (4,7)=11t ES (4,6)=8t EF (4,6)=10t ES (3,6)=4t EF (3,6)=12t ES (6,7)=12t EF (6,7)=13t ES (7,9)=13t EF (7,9)=15t ES(6,9)=12t EF (6,9)=19最短周期为 20t LF(6,9)= t LF(
32、8,9)= t LF (7,9)=20t LS(8,9)=15t LS(7,9)=16t LS(6,9)=13t LF(5,8)=15t LS(5,8)=8t LF(5,7)=16t LS(5,7)=14t LS(4,7)=10t LF(4,7)=13t ls(6,7)=15 t lf(6,7)=16 t ls(4,6)=10 t lf(4,6)=13 t ls(3,6)=5t lf(3,6)=13 t ls(1,3)=1t lf(1,3)=5t ls(1,4)=0 t lf(1,4)=8t lf(2,5)=8 t ls(2,5)=5t lf(1,2)=5t ls(1,2)=3作业(ij )t
33、 (i,j)t ES(i,j)t EF(i,j)t LS(i,j)t LF(i,j)Ri,j)F(i,j)12345678(1, 2)2023530(1, 3)4041510(1, 4)8080800(,5)3255830(3, 6)841251310(4, 5)0888800(4, 6)2810101332(4, 7)3811101322(5, 7)7571416968)781581500(6, 7)11213151630(6,9)71219132011(7,9)41315162055(8,9)51520152000关键路线为(1, 4), (4, 5), (5, 8), (8, 9)课堂作业1.柴油机生产计划设x1,x2,x3,x4分别为一二三四季度计划minz= 11.4x1+11.55x2+11.3x3+11.45x4 s.t./x1 3x235Jx330x410 _* F：E啜grlrt*ri.Bin 11. 4xl+lL 55x2+11. 3x3+11. 45x4k, txk1=23i2=35r3- -30xl=10xl+x2+i3+x470tn(lLP OPTIMLM FOUND AT STEP 3OBJECTIVE FiyCTTON VALLE1) 796,2500VARIABLE VALUEREDUCE