《管理运筹学》第三版习题答案(韩伯棠教授)
x 1 =
20
92 3
f 有唯一解
8 函数值为3 x2 =
3
3、解:
a 标准形式:
max f = 3x 1 + 2 x 2 + 0s 1 + 0 s2 + 0s 3
9 x 1 + 2 x 2 + s 1 = 30 3 x 1 + 2 x 2 + s 2 = 13 2 x 1 + 2 x 2 + s 3 = 9 x 1 , x 2 , s 1 , s 2 , s 3 ≥ 0
b 标准形式:
max f = 4 x 6 x 0s 0s
c 标准形式:
max f = x + 2x 2x 0s 0s
1
2
2
1
2
'
'
''
1
3
1
2
3 x 1 x 2 s 1 = 6 x 1 + 2 x 2 + s 2 = 10 7 x 1 6 x2 = 4 x 1 , x 2 , s 1, s 2 ≥ 0
3 x 1 5 x' 2 5 x 2 + s 1 = 70 + ' 2x1 5x2 + 5x2 = 50 3x1 + 2 ' ' x 2 2 x 2 s 2 = 30 x 1 , x 2 , x 2, s 1, s 2 ≥ 0
4 、解:
标准形式: max z = 10 x + 5 x + 0s + 0 s
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2
1
2
'
'
'
3x + 4x + s = 9 5x + 2x + s = 8 x , x , s , s ≥ 0
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2
1
2
1
2
2
1
2
1
s = 2, s = 0
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2