线性规划建模,求助!
2020-02-14
设6 个项目分包的工程量为x1,x2,x3,x4,x5,x6,收益为z
问题a解答如下
max z=32.4x1+35.8x2+17.75x3+14.8x4+18.2x5+12.35x6
s.t. :
10.5x1+8.3x2+10.2x3+7.2x4+12.3x5+9.2x6≤60
14.4x1+12.6x2+14.2x3+10.5x4+10.1x5+7.8x6≤70
2.2x1+9.5x2+5.6x3+7.8x4+8.3x5+6.9x6≤35
2.24x1+3.1x2+4.2x3+5.0x4+6.3x5+5.1x6≤20
1≥xi≥0(i=1,2,3,4,5,6)
问题b
max z=32.4x1+35.8x2+17.75x3+14.8x4+18.2x5+12.35x6
s.t. :
10.5x1+8.3x2+10.2x3+7.2x4+12.3x5+9.2x6≤60
14.4x1+12.6x2+14.2x3+10.5x4+10.1x5+7.8x6≤70
2.2x1+9.5x2+5.6x3+7.8x4+8.3x5+6.9x6≤35
2.24x1+3.1x2+4.2x3+5.0x4+6.3x5+5.1x6≤20
x2≥x6
1≥xi≥0(i=1,2,3,4,5,6)
问题c,需要引入剩余变量si
max z=32.4x1+35.8x2+17.75x3+14.8x4+18.2x5+12.35x6
s.t. :
10.5x1+8.3x2+10.2x3+7.2x4+12.3x5+9.2x6=60-s1
14.4x1+12.6x2+14.2x3+10.5x4+10.1x5+7.8x6=s1+70-s2
2.2x1+9.5x2+5.6x3+7.8x4+8.3x5+6.9x6=s2+35-s3
2.24x1+3.1x2+4.2x3+5.0x4+6.3x5+5.1x6=s3+20-s4
si≥0(i=1,2,3,4)
1≥xi≥0(i=1,2,3,4,5,6)
问题d
max z=32.4x1+35.8x2+17.75x3+14.8x4+18.2x5+12.35x6
s.t. :
10.5x1+8.3x2+10.2x3+7.2x4+12.3x5+9.2x6≤60+t1
14.4x1+12.6x2+14.2x3+10.5x4+10.1x5+7.8x6≤70+t2
2.2x1+9.5x2+5.6x3+7.8x4+8.3x5+6.9x6≤35+t3
2.24x1+3.1x2+4.2x3+5.0x4+6.3x5+5.1x6≤20+t4
1≥xi≥0(i=1,2,3,4,5,6)
ti≥0(i=1,2,3,4,)
经分析
当xi=1时收益最大
第一年第二年可用资金均可6个项目全包,
第三年第四年需要借用内部资金。
第三年借用的资金为2.2+9.5+5.6+7.8+8.3+6.9-35=5.3
第三年借用的资金为2.24+3.1+4.2+5.0+6.3+5.1-20=5.94
借用资金的收益即为第d问的最优解减去第a问的最优解
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