胡运权《运筹学教程》习题答案(第一章)[1]

2011/03/21

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= 2 x1+ 3 x 2 4 x1+ 6 x 2≥ 6 st . 2 x1+ 2 x 2≥ 4 x,x≥ 0 1 2

第一章习题解答 (1) min Z= 2 x1+ 3 x 2 4 x1+ 6 x 2≥ 6 st . 2 x1+ 2 x 2≥ 4 x,x≥ 0 1 2

(1)

( 2)

max Z= 3 x1+ 2 x 2 2 x1+ x 2≤ 2 st . 3 x1+ 4 x 2≥ 12 x, x≥ 0 1 2

无穷多最优解， 1 x1= 1, x 2=, Z= 3是一个最优解 3

( 3)

max Z= x1+ x 2 6 x1+ 10 x 2≤ 120 st . 5≤ x1≤ 10 5≤ x≤8 2

( 4)

max Z= 5 x1+ 6 x 2 2 x1 x 2≥ 2 st . 2 x1+ 3 x 2≤ 2 x,x≥ 0 1 2

max Z= 3 x 1+ 2 x 2 2 x1+ x 2≤ 2 (2) st . 3 x 1+ 4 x 2≥ 12 x, x≥ 0 1 2该问题无解

第一章习题解答 max Z= x1+ x 2 6 x1+ 10 x 2≤ 120 ( 3) st . 5≤ x1≤ 10 5≤ x≤8 2 唯一最优解，唯最优解 x1= 10, x 2= 6, Z= 16 max Z= 5 x1+ 6 x 2 2 x1 x 2≥ 2 ( 4) st . 2 x1+ 3 x 2≤ 2 x,x≥0 1 2 该问题有无界解

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= 3 x1+ 4 x 2 2 x3+ 5 x 4 4 x1 x 2+ 2 x3 x 4= 2 x+ x x+ 2 x≤ 14 2 3 4 st 1 . 2 x1+ 3 x 2+ x3 x 4≥ 2 x1, x 2, x3≥ 0, x 4无约束 min st x 1 Z= 2 x1 2 x 2+ 3 x 3 x1+ x 2+ x 3= 4 2 x1+ x 2 x 3≤ 6≤ 0, x 2≥ 0, x 3无约束

(1)

(2)

第一章习题解答 minZ= 3x1+ 4x2 2x3+ 5x4 4x1 x2+ 2x3 x4= 2 x+ x x+ 2x≤ 14 (1) 4 . st 1 2 3 2x1+ 3x2+ x3 x4≥ 2 x1, x2, x3≥ 0, x4无约束 max Z= 3 x1 4 x 2+ 2 x3 5 x 41+ 5 x 42 4 x1+ x 2 2 x3+ x 41 x 42= 2 x+ x x+ 2 x 2 x+ x= 14 1 2 3 41 42 5 st 2 x1+ 3 x 2+ x3 x 41+ x 42 x6= 2 x1, x 2, x3, x 41, x 42, x 6≥ 0 (2)

第一章习题解答 min st x 1 Z= 2 x1 2 x 2+ 3 x 3 x1+ x 2+ x 3= 4 2 x1+ x 2 x 3≤ 6≤ 0, x 2≥ 0, x 3无约束

max Z= 2 x1+ 2 x 2 3 x 31+ 3 x 32 x1+ x 2+ x 31 x 32= 4 st 2 x1+ x 2 x 31+ x 32+ x 4= 6 x1, x 2, x 31, x 32, x 4≥ 0