运筹学(胡运权)课后答案(清华大学出版社).pdf
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运筹学(胡运权)课后答案(清华大学出版社).pdf
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运筹学教程第二版运筹学教程第二版习题解答习题解答运筹学教程运筹学教程SchoolofManagementSchoolofManagementpagepage226January20116January20111.1用图解法求解下列线性规划问题。
并指出问题具有惟一最优解、无穷多最优解、无界解还是无可行解。
0,422664.32min)1(21212121xxxxxxstxxZ0,124322.23max)2(21212121xxxxxxstxxZ85105120106.max)3(212121xxxxstxxZ0,23222.65max)4(21212121xxxxxxstxxZ运筹学教程运筹学教程SchoolofManagementSchoolofManagementpagepage336January20116January2011是一个最优解无穷多最优解3,31,10,422664.32min)1(2121212121ZxxxxxxxxstxxZ该问题无解0,124322.23max)2(21212121xxxxxxstxxZ运筹学教程运筹学教程SchoolofManagementSchoolofManagementpagepage446January20116January201116,6,1085105120106.max)3(21212121ZxxxxxxstxxZ唯一最优解该问题有无界解0,23222.65max)4(21212121xxxxxxstxxZ运筹学教程运筹学教程SchoolofManagementSchoolofManagementpagepage556January20116January20111.2将下述线性规划问题化成标准形式。
.,0,2321422245243min)1(43214321432143214321无约束xxxxxxxxxxxxxxxxstxxxxZ无约束321321321321,0,0624322min)2(xxxxxxxxxstxxxZ运筹学教程运筹学教程SchoolofManagementSchoolofManagementpagepage666January20116January2011.,0,2321422245243min)1(43214321432143214321无约束xxxxxxxxxxxxxxxxstxxxxZ0,232142222455243max64241321642413215424132142413214241321xxxxxxxxxxxxxxxxxxxxxxxstxxxxxZ运筹学教程运筹学教程SchoolofManagementSchoolofManagementpagepage776January20116January2011无约束321321321321,0,0624322min)2(xxxxxxxxxstxxxZ0,6243322max43231214323121323121323121xxxxxxxxxxxxxxstxxxxZ运筹学教程运筹学教程SchoolofManagementSchoolofManagementpagepage886January20116January20111.3对下述线性规划问题找出所有基解指出哪些是基可行解并确定最优解。
6,1,0031024893631223max)1(6153214321321jxxxxxxxxxxxstxxxZj)4,1(,0322274322325min)2(432143214321jxxxxxxxxxstxxxxZj运筹学教程运筹学教程SchoolofManagementSchoolofManagementpagepage996January20116January20116,1,0031024893631223max)1(6153214321321jxxxxxxxxxxxstxxxZj基可行解x1x2x3x4x5x6Z03003.503001.5080300035000.7500022.252.25运筹学教程运筹学教程SchoolofManagementSchoolofManagementpagepage10106January20116January2011)4,1(,0322274322325min)2(432143214321jxxxxxxxxxstxxxxZj基可行解x1x2x3x4Z00.5205001152/5011/5043/5运筹学教程运筹学教程SchoolofManagementSchoolofManagementpagepage11116January20116January20111.4分别用图解法和单纯形法求解下述线性规划问题并对照指出单纯形表中的各基可行解对应图解法中可行域的哪一顶点。
0,825943.510max)1(21212121xxxxxxstxxZ运筹学教程运筹学教程SchoolofManagementSchoolofManagementpagepage12126January20116January20110,24261553.2max)2(21212121xxxxxxstxxZ运筹学教程运筹学教程SchoolofManagementSchoolofManagementpagepage13136January20116January2011l.5上题
(1)中若目标函数变为maxZ=cx1+dx2讨论c,d的值如何变化使该问题可行域的每个顶点依次使目标函数达到最优。
解得到最终单纯形表如下Cjcd00CB基bx1x2x3x4dx23/2015/14-3/4cx1110-2/1410/35j00-5/14d+2/14c3/14d-10/14c运筹学教程运筹学教程SchoolofManagementSchoolofManagementpagepage14146January20116January2011当c/d在3/10到5/2之间时最优解为图中的A点当c/d大于5/2且c大于等于0时最优解为图中的B点当c/d小于3/10且d大于0时最优解为图中的C点当c/d大于5/2且c小于等于0时或当c/d小于3/10且d小于0时最优解为图中的原点。
运筹学教程运筹学教程SchoolofManagementSchoolofManagementpagepage15156January20116January2011式中1c13,4c26,-1a113,2a125,8b112,2a215,4a226,10b214,试确定目标函数最优值的下界和上界。
0,.max21222212112121112211xxbxaxabxaxastxcxcZl.6考虑下述线性规划问题运筹学教程运筹学教程SchoolofManagementSchoolofManagementpagepage16166January20116January2011最优值上界为210,14421221.63max21212121xxxxxxstxxZ解上界对应的模型如下c,b取大a取小运筹学教程运筹学教程SchoolofManagementSchoolofManagementpagepage17176January20116January2011最优值下界为6.40,1064853.4max21212121xxxxxxstxxZ解下界对应的模型如下c,b取小a取大运筹学教程运筹学教程SchoolofManagementSchoolofManagementpagepage18186January20116January2011l.7分别用单纯形法中的大M法和两阶段法求解下列线性规划问题并指出属哪类解。
该题是无界解。
3,1,00222623max)1(3231321321jxxxxxxxxstxxxZj运筹学教程运筹学教程SchoolofManagementSchoolofManagementpagepage19196January20116January20116,0,54,590,623824.32min)2(3212121321321ZxxxxxxxxxxstxxxZ最优解之一该题是无穷多最优解。
运筹学教程运筹学教程SchoolofManagementSchoolofManagementpagepage20206January20116January2011517,0,1,59,524,1,042634334max)3(43214213212121ZxxxxjxxxxxxxxxstxxZj该题是唯一最优解运筹学教程运筹学教程SchoolofManagementSchoolofManagementpagepage21216January20116January2011该题无可行解。
3,1,052151565935121510max)4(321321321321jxxxxxxxxxxstxxxZj运筹学教程运筹学教程SchoolofManagementSchoolofManagementpagepage22226January20116January20111.8已知某线性规划问题的初始单纯形表和用单纯形法迭代后得到下面表格试求括弧中未知数al值。
项目X1X2X3X4X5X46(b)(c)(d)10X51-13(e)01CjZja-1200X1(f)(g)2-11/20X54(h)(i)11/21CjZj0-7jk(l)b=2,c=4,d=-2,g=1,h=0,f=3,i=5,e=2,l=0,a=3,j=5,k=-1.5运筹学教程运筹学教程SchoolofManagementSchoolofManagementpagepage23236January20116January20111.9若X
(1)、X
(2)均为某线性规划问题的最优解证明在这两点连线上的所有点也是该问题的最优解。
也是最优解。
所以也是可行解且满足两点连线上的点对于任何满足和设XXCXCXaCaXCXaCaXCXCXaaXXXaXbAXXCZXXTTTTTTTT,)1()1(,100max)2()2()2()1()2()1()2()1()2()1(运筹学教程运筹学教程SchoolofManagementSchoolofManagementpagepage24246January20116January20111.10线性规划问题maxZCX,AXbX0设X0为问题的最优解。
若目标函数中用C*代替C后问题的最优解变为X*求证(C*-C)(X*-X0)00)()()(;0max;0max0*00*0*00XXCXXCXXCCXCXCXCZXCXCXCXZX的最优解故是的最优解故是运筹学教程运筹学教程SchoolofManagementSchoolofManagementpagepage25256January20116January20111.11考虑线性规划问题0,)(75232)(24.42min432143214214321xxxxiixxxxixxxstxxxxZ模型中为参数要求
(1)组成两个新的约束(i)(i)+(ii)(ii)(ii)一2(i)根据(i)(ii)以x1,x2为基变量列出初始单纯形表运筹学教程运筹学教程SchoolofManagementSchoolofManagementpagepage26266January20116January20111)(23)(32431xxiixxxiCja21-4CB基bx1x2x3x4ax13+2011-12x21-10-10j003-aa-4运筹学教程运筹学教程SchoolofManagementSchoolofManagementpagepage27276January20116January2011
(2)在表中假定0则为何值时x1,x2为问题的最优基变量解如果=0则当3a4时x1,x2为问题的最优基变量(3)在表中假定3则为何值时x1,x2为问题的最优基。
解如果a=3则当-11时x1,x2为问题的最优基变量。
运筹学教程运筹学教程SchoolofManagementSchoolofManagementpagepage28286January20116January20111.12线性规划问题maxZCXAXbX0如X*是该问题的最优解又0为某一常数分别讨论下列情况时最优解的变化。
(1)目标函数变为maxZCX
(2)目标函数变为maxZ(C+)X(3)目标函数变为maxZC/*X约束条件变为AXb。
解:
(1)最优解不变;
(2)C为常数时最优解不变否则可能发生变化。
(3)最优解变为:
X/。
运筹学教程运筹学教程SchoolofManagementSchoolofManagementpagepage29296January20116January20111.13某饲养场饲养动物出售设每头动物每天至少需700g蛋白质、30g矿物质、100mg维生素。
现有五种饲料可供选用各种饲料每kg营养成分含量及单价如下表所示。
饲料蛋白质(g)矿物质(g)维生素(mg)价格元/kg1310.50.2220.51.00.7310.20.20.446220.35180.50.80.8运筹学教程运筹学教程SchoolofManagementSchoolofManagementpagepage30306January20116January2011要求确定既满足动物生长的营养需要又使费用最省的选用饲料的方案。
(建立这个问题的线性规划模型不求解)5,4,3,2,1,01008.022.05.0305.022.05.0700186238.03.04.07.02.0min5,4,3,2,1,54321543215432154321ixxxxxxxxxxxxxxxxxxxxxZiixii种饲料数量表示第设运筹学教程运筹学教程SchoolofManagementSchoolofManagementpagepage31316January20116January20111.14某医院护士值班班次、每班工作时间及各班所需护士数如下页表格所示。
班次工作时间所需护士数人16:
0010:
0060210:
0014:
0070314:
0018:
0060418:
0022:
0050522:
002:
002062:
006:
0030运筹学教程运筹学教程SchoolofManagementSchoolofManagementpagepage32326January20116January2011
(1)若护士上班后连续工作8h该医院最少需多少名护士以满足轮班需要且为整数班开始上班的护士人数表示第设,6,5,4,3,2,1,0302050607060min65,4,3,2,1,655443322161654321ixxxxxxxxxxxxxxxxxxxZiixii运筹学教程运筹学教程SchoolofManagementSchoolofManagementpagepage33336January20116January2011
(2)若除2200上班的护士连续工作8h外(取消第6班)其他班次护士由医院排定上1-4班的其中两个班则该医院又需多少名护士满足轮班需要。
解:
第5班一定要30个人运筹学教程运筹学教程SchoolofManagementSchoolofManagementpagepage34346January20116January20114,3,2,1,10,02,1,502,1,602,1,702,1,6030min4,3,2,1,44434241444443342241143433323133443333223113242322212244233222211214131211114413312211114321jiyxyyyyyxyxyxyxyyyyyyxyxyxyxyyyyyyxyxyxyxyyyyyyxyxyxyxyxxxxZiixijii变量是第四班约束第三班约束第二班约束第一班约束班开始上班的护士人数表示第设运筹学教程运筹学教程SchoolofManagementSchoolofManagementpagepage35356January20116January20111.15艘货轮分前、中、后三个舱位它们的容积与最大允许载重量见后面的表格。
现有3种货物待运已知有关数据列于后面的表格。
又为了航运安全前、中、后舱的实际载重量大体保持各舱最大允许载重量的比例关系。
具体要求前、后舱分别与中舱之间载重量比例的偏差不超过15前、后舱之间不超过10。
问该货轮应装载ABC各多少件运费收入才最大?
试建立这个问题的线性规划模型。
运筹学教程运筹学教程SchoolofManagementSchoolofManagementpagepage36366January20116January2011商品数量件每件体积(m3/件)每件重量(t/件)运价元/件A6001081000B100056700C80075600项目前舱中舱后舱最大允许载重量t200030001500容积m3400054001500运筹学教程运筹学教程SchoolofManagementSchoolofManagementpagepage37376January20116January2011MAX=1000X(1,1)+X(1,2)+X(1,3)+700X(2,1)+X(2,2)+X(2,3)+600X(3,1)+X(3,2)+X(3,3)SUBJECTTOX(i,j)表示第商品i在舱j的装载量i,j=1,2,3商品数量约束1X(1,1)+X(1,2)+X(1,3)=6002X(2,1)+X(2,2)+X(2,3)=10003X(3,1)+X(3,2)+X(3,3)=800运筹学教程运筹学教程SchoolofManagementSchoolofManagementpagepage38386January20116January2011商品容积约束410X(1,1)+5X(2,1)+7X(3,1)=4000510X(1,2)+5X(2,2)+7X(3,2)=5400610X(1,3)+5X(2,3)+7X(3,3)=1500最大载重量约束78X(1,1)+6X(2,1)+5X(3,1)=200088X(1,2)+6X(2,2)+5X(3,2)=300098X(1,3)+6X(2,3)+5X(3,3)=1500运筹学教程运筹学教程SchoolofManagementSchoolofManagementpagepage39396January20116January2011重量比例偏差约束108X(1,1)+6X(2,1)+5X(3,1)=2/31-0.158X(1,2)+6X(2,2)+5X(3,2)128X(1,3)+6X(2,3)+5X(3,3)=1/21-0.158X(1,2)+6X(2,2)+5X(3,2)148X(1,3)+6X(2,3)+5X(3,3)=3/41-0.18X(1,1)+6X(2,1)+5X(3,1)运筹学教程运筹学教程SchoolofManagementSchoolofManagementpagepage40406January20116January20111.16某厂生产I两种食品现有50名熟练工人每名熟练工人每h可生产食品110kg或食品6kg。
由于需求量将不断增长(见下页表格)该厂计划到第8周末前培训出50名新工人组织两班生产。
已知一名工人每周工作40h一名熟练工人用2周时间可培训出不多于3名新工人(培训期间熟练工人和被培训人员均不参加生产)。
熟练工人每周工资360元新工人培训期间工资每周120元新工人培训结束后工作每周工资240元且生产效率同熟练工人。
培训过渡期工厂将安排部分熟练工人加班加班1h另加付12元。
又生产食品不能满足订货需求推迟交货的赔偿费分别为食品I为0.50元(kg周)食品为0.60元(kg周)。
工厂应如何全面安排使各项费用总和最小试建立线性规划模型。
运筹学教程运筹学教程SchoolofManagementSchoolofManagementpagepage41416January20116January2011周次食品12345667810101212161616202067.28.410.81212121212设x(i)y(i)表示从事两个产品生产的人数xx(i)yy(i)表示从事生产两个产品的加班小时数f1(i),f2(i)表示两个产品推迟交货的数量r1(i),r2(i)表示两个产品的需求数量w(i),n(i)分别表示开始从事培训工作的人数和新接受培训的工人人数。
运筹学教程运筹学教程SchoolofManagementSchoolofManagementpagepage42426January20116January2011MIN=360X(i)+360Y(i)+360W(i)+12XX(i)+12yy(i)+0.5f1(i)+0.6f2(i)+(120+120)n(i)+240(7-i)n(i)n(i)=nx(i)+ny(i)N(8)=0-3W(i)+N(i)=0XX(i)=1000YY(i)=1000运筹学教程运筹学教程SchoolofManagementSchoolofManagementpagepage43436January20116January2011400X(i)+10XX(i)=116000240y(i)+6yy(i)=79200400*x
(1)+10*xx
(1)+f1
(1)=10000;400*(x
(1)+x
(2)+10*(xx
(1)+xx
(2)+f1
(2)=20000;for(a(i)|i#ge#3#and#i#le#s:
400*x
(1)+400*x
(2)+10*xx
(1)+10*xx
(2)+sum(a(j)|j#le#i#and#j#gt#2:
400*(x(j)+nx(j-2)+10*xx(j)+f1(i)=sum(a(j)|j#le#i:
r1(j);f1(s)=0;运筹学教程运筹学教程SchoolofManagementSchoolofManagementpagepage44446January20116January2011240*y
(1)+6*yy
(1)+f2
(1)=6000;240*(y
(1)+y
(2)+6*(yy
(1)+yy
(2)+f2
(2)=13200;for(a(i)|i#ge#3#and#i#le#s:
240*y
(1)+240*y
(2)+6*yy
(1)+6*yy
(2)+sum(a(j)|j#le#i#and#j#gt#2:
240*(y(j)+ny(j-2)+6*yy(j)+f2(i)=sum(a
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