发布时间 : 星期五 文章运筹学1至5章习题参考答案更新完毕开始阅读
maxZ?10x1?15x2?5x1?3x2?9?(3)??5x1?6x2?15??2x1?x2?5??x1、x2、x3?0
【解】大M法。数学模型为
maxZ?10x1?15x2?Mx6?5x1?3x2?x3?9??5x?6x?x?15?124??2x1?x2?x5?x6?5?xj?0,j?1,2,,6?15 0 0 0 X2 X3 X4 X5 3 1 0 0 6 0 1 0 1 0 0 -1 15 0 0 0 1 0 0 -1 3/5 0 0 1/5 9 1 1 0 -1/5 -2/5 0 -1 9 -2 0 0 -1/5 -2/5 0 -1
C(j) Basis C(i) X3 X4 X6 0 0 -M 10 X1 [5] -5 2 10 2 1 0 0 0 -M X6 0 0 1 0 0 0 0 1 0 0 R. H. S. Ratio 9 15 5 0 0 9/5 24 7/5 18 0 1.8 M 2.5 C(j)-Z(j) * Big M X1 X4 X6 10 0 -M C(j)-Z(j) * Big M 0 因为X6>0,原问题无可行解。 两阶段法
第一阶段：数学模型为
minZ?x6?5x1?3x2?x3?9??5x?6x?x?15 ?124??2x1?x2?x5?x6?5?xj?0,j?1,2,,6?0 0 0 0 X2 X3 X4 X5 3 1 0 0 6 0 1 0 1 0 0 -1 -1 0 0 1 3/5 0 0 1/5 9 1 1 0 C(j) Basis C(i) X3 X4 X6 X1 X4 0 0 1 0 0 0 X1 [5] -5 2 -2 1 0 1 X6 0 0 1 0 0 0 R. H. S. Ratio 9 15 5 5 9/5 24 1.8 M 2.5 14 C(j)-Z(j) X6 1 0 -1/5 -2/5 0 0 -1 1 1 0 7/5 C(j)-Z(j) 0 1/5 2/5 因为X6>0,原问题无可行解。图解法如下：
maxZ?4x1?2x2?5x3?6x1?x2?4x3?10? (4) ?3x1?3x2?5x3?8??x1?2x2?x3?20?xj?0,j?1,2,3?
【解】大M法。X7是人工变量，数学模型为
maxZ?4x1?2x2?5x3?Mx7?6x1?x2?4x3?x4?10?3x?3x?5x?x?8 ?1235??x1?2x2?2x3?x6?x7?20?xj?0,j?1,2,,7?Cj CB 0 0 －M XB X4 X5 X7 4 X1 2 X2 5 X3 0 X4 0 X5 0 X6 －M R.H.S. X7 Ratio 10 6 3 1 4 M -1 -3 [2] 2 2M 4 -5 1 5 M 1 C(j)-Z(j) * Big M 1 －1 1 －1 10 8 20 0 0 2 X4 13/2 X5 X2 9/2 1/2 3 C(j)-Z(j) * Big M 5 0 2 X3 13/9 X5 86/9 X2 -2/9 -25/9 C(j)-Z(j) * Big M 1 1 [9/2] -7/2 1/2 4 1 1 2/9 7/9 -1/9 -8/9 1 1 -1/2 -3/2 -1/2 1 1/2 3/2 1/2 -1 -1 1/9 4/9 -1 20 38 10 -1/9 -4/9 40/9 70/9 -17/9 17/9 482/9 13/9 -13/9 无界解。 两阶段法。第一阶段：
minZ?x7?6x1?x2?4x3?x4?10?3x?3x?5x?x?8 ?1235??x1?2x2?x3?x6?x7?20?xj?0,j?1,2,,7?Cj CB 0 0 1 XB X4 X5 X7 X1 X2 X3 0 X4 0 X5 0 X6 1 X7 R.H.S. Ratio 10 6 3 1 －1 13/2 9/2 1/2 -1 -3 [2] 4 -5 1 [9/2] -7/2 1/2 1 C(j)-Z(j) 0 0 2 X4 X5 X2 －2 －1 C(j)-Z(j) 1 1 1 1 －1 1 -1/2 -3/2 -1/2 1 1/2 3/2 1/2 1 10 8 20 20 38 10 第二阶段： Cj CB 0 0 1 XB X4 X5 X2 4 X1 2 X2 5 X3 0 X4 0 X5 0 X6 R.H.S. Ratio 13/2 9/2 1/2 C(j)-Z(j) 0 0 2 X3 X5 X2 C(j)-Z(j) 7/2 13/9 86/9 -2/9 -3 1 1 [9/2] -7/2 1/2 1 9/2 1 2/9 7/9 -1/9 -1 1 1 -1/2 -3/2 -1/2 20 38 10 1/2 -1/9 40/9 -17/9 482/9 -4/9 70/9 1 原问题无界解。
?21?1.13 在第1.9题中，对于基B???,求所有变量的检验数?j(j?1,?,4),并判断B是不
40??是最优基．
1??0?4??1【解】B??4,B???,
1?1????2????C?CBB?1A1??0?4??2310?
?(5,2,0,0)?(5,0)????14?201???1????2??5595?(5,2,0,0)?(5,?,0,)?(0,,0,?)2424??(0,,0,?), B不是最优基，可以证明B是可行基。
1.14已知线性规划
9254maxz?5x1?8x2?7x3?4x4?2x1?3x2?3x3?2x4?20 ?3x?5x?4x?2x?30?1234?x?0,j?1,,4?j?23?的最优基为B???，试用矩阵公式求（1）最优解；(2)单纯形乘子；
25??(3)N1及N3；(4)?1和?3。
【解】
?5?4?1B????1??23???4,C?(c,c)?(4,8,),则 ?B421?2??T?1(1)XB?(x4,x2)?Bb?(,5),最优解X?(0,5,0,),Z?50 (2)??CBB(3)
?152T52T?(1,1)