运筹学(胡运权)第五版课后答案-运筹作业

47页1.1b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123454321-1-2-6-5-4-3-2-1X2X12x1-x2=2-2x1+3x2=212344321X12x1+x2=23x1+4x2=120X21.2（b）约束方程的系数矩阵A=12342112P1P2P3P4基基解是否可行解目标函数值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+4500x32+6000x13+6000x23+7300x14s.t.x11+x12+x13+x14≥15x12+x13+x14+x21+x22+x23≥10x13+x14+x22+x23+x31+x32≥20x14+x23+x32+x41≥12Xij≥0用excel求解为：()用LINDO求解：LPOPTIMUMFOUNDATSTEP3OBJECTIVEFUNCTIONVALUE1)118400.0VARIABLEVALUEREDUCEDCOSTZ0.0000001.000000X113.0000000.000000X210.0000002800.000000X318.0000000.000000X410.0000001100.000000X120.0000001700.000000X220.0000001700.000000X320.0000000.000000X130.000000400.000000X230.0000001500.000000X1412.0000000.000000ROWSLACKORSURPLUSDUALPRICES2)0.000000-2800.0000003)2.0000000.0000004)0.000000-2800.0000005)0.000000-1700.000000NO.ITERATIONS=3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a1，a2，a3,a4,a5分别为在A1,A2,B1,B2,B3加工的Ⅰ产品数量，b1，b2，b3分别为在A1,A2,B1加工的Ⅱ产品数量，c1为在A2，B2上加工的Ⅲ产品数量。则目标函数为‘maxz=(1.25-0.25)(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+10b1≤60007a2+b2+12c1≤100006a3+8a3≤40004a4+11c1≤70007a5≤4000a1+a2-a3-a4-a5=0b1+b2-b3=0a1，a2，a3,a4,a5,b1，b2，b3,c1≥0用lindo求解得：LPOPTIMUMFOUNDATSTEP6OBJECTIVEFUNCTIONVALUE1)16342.29VARIABLEVALUEREDUCEDCOSTA11200.0000000.000000A20.0000009.640000A3285.7142940.000000B310000.0000000.000000C10.00000015.900000B10.0000000.230000A4342.8571470.000000A5571.4285890.000000B210000.0000000.000000ROWSLACKORSURPLUSDUALPRICES2)0.0000000.1680003)0.0000001.5000004)0.0000000.0750005)5628.5712890.0000006)0.0000000.0085717)0.0000000.1100008)0.000000-1.500000NO.ITERATIONS=6计算lindo截屏2.1a:对偶问题为：maxz=2y1+3y2+5y3s.t.y1+2y2+y3≤23y3+y2+4y3≤24y1+3y2+3y3=4y1≥0,y2≤0,y3无约束因为原问题的对偶问题的对偶问题仍是原问题，因此本问题的对偶问题的对偶问题为：minz=2x1+2x2+4x3s.t.x1+3x2+4x3≥22x1+x2+3x3≤3x1+4x2+3x3=5x1,x2≥0,x3无约束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+5x3≤453x1+4x2+5x3≤30x1,x2,x3≥0用lomdo求解为LPOPTIMUMFOUNDATSTEP2OBJECTIVEFUNCTIONVALUE1)27.00000VARIABLEVALUEREDUCEDCOSTX15.0000000.000000X20.0000002.000000X33.0000000.000000X1,X2,X30.0000000.000000ROWSLACKORSURPLUSDUALPRICES2)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)LPOPTIMUMFOUNDATSTEP2OBJECTIVEFUNCTIONVALUE1)27.00000VARIABLEVALUEREDUCEDCOSTX15.0000000.000000X20.0000002.000000X33.0000000.000000X1,X2,X30.0000000.000000ROWSLACKORSURPLUSDUALPRICES2)0.0000000.2000003)0.0000000.6000004)0.0000000.000000NO.ITERATIONS=2RANGESINWHICHTHEBASISISUNCHANGED:OBJCOEFFICIENTRANGESVARIABLECURRENTALLOWABLEALLOWABLECOEFINCREASEDECREASEX13.0000001.8000000.600000X21.0000002.000000INFINITYX34.0000001.0000001.500000X1,X2,X30.0000000.000000INFINITYRIGHTHANDSIDERANGESROWCURRENTALLOWABLEALLOWABLERHSINCREASEDECREASE245.00000015.00000015.000000330.00000015.0000007.50000040.0000000.000000INFINITY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+8x4≤453x1+4x2+5x3+2x4≤30x1,x2,x3,x4≥0用lomdo求解为LPOPTIMUMFOUNDATSTEP0OBJECTIVEFUNCTIONVALUE1)27.50000VARIABLEVALUEREDUCEDCOSTX10.0000000.100000X20.0000001.966667X35.0000000.000000X42.5000000.000000ROWSLACKORSURPLUSDUALPRICES2)0.0000000.2333333)0.0000000.566667NO.ITERATIONS=0安排生产D有利，新最有生产计划为x1=x2=0,x3=5,x4=2.5，利润为27.5d)maxz=3x1+x2+4x3-0.4ys.t.6x1+3x2+5x3≤453x1+4x2+5x3-y≤30x1,x2,x3,y≥0用lomdo求解为LPOPTIMUMFOUNDATSTEP0OBJECTIVEFUNCTIONVALUE1)30.00000VARIABLEVALUEREDUCEDCOSTX10.0000000.600000X20.0000001.800000X39.0000000.000000Y15.0000000.000000ROWSLACKORSURPLUSDUALPRICES2)0.0000000.4000003)0.0000000.400000NO.ITERATIONS=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0第i组条件不起作用x1+x2≤2-(1-y1)MM—充分大正数2x1+3x2≥5+(1-y2)My1+y2=1y1,y2=0或1b)设yi=1第i组条件起作用0第i组条件不起作用x=0y1x=3y2x=5y2x=7y4y1+y2+y3+y4=1y1,y2,y3,y4=0或1c)设yi=1为假定取值≥500为假定取值x=0x=0y1x≥50--(1-y2)My1+y2=1y1,y2=0或1d)设yi=1第i组条件起作用0第i组条件不起作用i=1,2则x1≤2+(1-y1)Mx2≥1-(1-y1)Mx2≤4+(1-y2)My1+y2=1y1,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i=1,2则x1+x2≤5-(1-y1)Mx1≤2-(1-y2)Mx3≥2+(1-y3)Mx3+x4≥6+(1-y4)My1+y2+y3+y4≥2y1,y2,y3,y4=1或04.2minz=∑cjxj10j=1∑xj10j=1=5x1+x8=1x7+x8=1s.t.x3+x5≤1x4+x5≤1x5+x6+x7+x8≤2xj=1选择钻探第sj井位0否4.5设xij为第i种泳姿用第名运动员minz=∑∑ai5j=1jxij4i=1s.t.x11+x12+x13+x14+x15=1x21+x22+x23+x24+x25=1x31+x32+x33+x34+x35=1x41+x42+x43+x44+x45=1x11+x21+x22+x23=1x12+x22+x32+x42=1x13+x23+x33+x43=1x14+x24+x34+x44=1x15+x25+x35+x45=1xij=1或0（i=1,2,3,4j=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.5.3c因为使mind1-,故在x1+x2=40的右侧，若使mind4+,则在x1+x2=50的左侧，即阴影区域，因为在阴影部分无法使2d2-+d3-最小，故比较E（20，30）,F（24，26）,E点:d2-=4,d3-=0min2d2-+d3-=8,F点:d2-=0,d3-=4,min2d2-+d3-=4,故选F点10203040505040302010d2+d3+d3-d2-d4+d4-d1-d1+EF程序法6.4a破圈法避圈法2166277143453234216627714345323488最小部分树166.4b最小部分树321526101012426673282181045815261010124266732821810458172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