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1、 运筹学教程（第二版）习题解答 运筹学教程第一章习题解答1.1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。min Z = 2x + 3x2max Z = 3x + 2x2114x + 6x 62x + x 21st. 2x + 2x 4121st. 3x + 4x 121x , x 012(1)(2)22x1, x 022max Z = x + x2max Z = 5x + 6x2116x +10x 120 2x - x 21st. 5 x 1021st.- 2x + 3x 212(3) (4) 125 x 82x1, x 02page 26 Ja
2、nuary 2011School of Management 运筹学教程第一章习题解答min Z = 2x + 3x214x + 6x 612(1)st.2x + 2x 41x1, x 022无穷多最优解，1x1 = 1, x = , Z = 3是一个最优解23max Z = 3x + 2 x 212 x + x 212st . 3x + 4 x 12(2)12x , x 012该问题无解page 36 January 2011School of Management 运筹学教程第一章习题解答max Z = x + x216x +10x 1201st. 5 x 102(3) 15 x 82唯一
3、最优解， x =10, x = 6,Z =1612max Z = 5x + 6x21 2x - x 21st. - 2x + 3x 21x1, x 02(4) 22该问题有无界解page 46 January 2011School of Management 运筹学教程第一章习题解答1.2 将下述线性规划问题化成标准形式。min Z = -3x + 4x - 2x + 5x41234x - x + 2x - x = -21234x + x - x + x 141 2 3- 2x + 3x + x - x 2.1 2 3 4x , x , x 0, x 无约束24(1)st1234min Z =
4、 2 x - 2 x + 3x312 - x + x + x = 4123(2)st- 2 x + x - x 61 2 3x 0, x 0, x 无约束123page 56 January 2011School of Management 运筹学教程第一章习题解答minZ =-3x +4x -2x +5x41234x -x +2x -x =-212342 14 .x +x -x + x (1)1234st-2x +3x +x -x 21234x ,x ,x 0,x无约束1234max Z = 3x - 4x + 2x - 5x + 5x4212341 - 4x + x - 2x + x -
5、x = 21x1 + x - x + 2x - 2x + x = 14- 2x + 3x + x - x + x - x = 22341422341425st12341426x1, x , x , x , x , x 02341426page 66 January 2011School of Management 运筹学教程第一章习题解答min Z = 2x - 2x + 3x312 - x + x + x = 4123(2)st - 2x + x - x 61x 0, x 0, x无约束23123max Z = 2x + 2x - 3x + 3x321231 - x + x + x - x
6、= 41st 2x + x - x + x + x = 6231321231324x1, x , x , x , x 0231324page 76 January 2011School of Management 运筹学教程第一章习题解答1.3 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。max Z = 3x + x + 2x31212 x + 3x + 6x + 3x = 918x + x - 4x + 2x = 10234(1)1235st 3x - x = 016x 0（, j = 1,L ,6）jmin Z = 5x - 2x + 3x + 2x41x 2x 3x
7、 4x 72st 2x + 2x + x + 2x = 32+ +33= +14(2)1234x 0,( j = 1,L 4)jpage 86 January 2011School of Management 运筹学教程第一章习题解答max Z = 3x + x + 2x31212 x + 3x + 6x + 3x = 918x + x - 4x + 2x = 10234(1)1235st 3x - x = 016x 0（, j = 1,L ,6）j基可行解x10x23x30x40 3.5 0x5x6Z33000 1.5 085000000300.75 02 2.25 2.25page 96
8、January 2011School of Management 运筹学教程第一章习题解答min Z = 5x - 2x + 3x + 2x41x 2x 3x 4x 72st 2x + 2x + x + 2x = 32+ +33= +14(2)1234x 0,( j = 1,L 4)j基可行解x10x20.50x32x40Z501152/5011/5043/5page 106 January 2011School of Management 运筹学教程第一章习题解答1.4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。max Z = 1
9、0 x + 5x213x + 4x 912(1)st. 5x + 2x 81x1, x 022page 116 January 2011School of Management 运筹学教程第一章习题解答max Z = 2x + x213x + 5x 1512st. 6x + 2x 242x1, x 0(2)12page 126 January 2011School of Management 运筹学教程第一章习题解答l.5上题(1)中，若目标函数变为 max Z = cx +1dx，讨论c,d的值如何变化，使该问题可行域的每个顶2点依次使目标函数达到最优。解：得到最终单纯形表如下：Cjc d00CBb x x21x3x4基d x 3/2 0 125/14-3/4c x 1 1 01-2/1410/35sj0 0 -5/14d+2/14c 3/14d-10/14cpage 136 January 2011School of Management 运筹学教程第一章习题解答当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
