运筹学课程设计-模板1(2)

Minf(x)=0.68x1+0.72x2+0.23x3+0.22(1-0.1)x4+0.37x5+0.32x6+1.54(1+0.1)x7+0.38x8+23.0x9+0.56(1+0.1)x10+1.12(1+0.1)x11+0.42x12

该目标函数是用各种原料的单价乘以使用量，结果为总成本。要使总成本最小，只需该目标函数取得极小值即可，这与研究问题的目的是相一致的。

2.4限制条件的确定 2.4.1 原料总量约束

x1+x2+x3+x4+x5+x6+x7+x8+x9+x10+x11+x12=1000 2.4.2营养成分保证约束

3.35x1+3.08x2+1.78x3+2.10x4+2.40x5+1.62x6+2.80x7+1.61x8≥2700 78x1+114x2+142x3+117x4+402x5+360x6+450x7+170x8≥135000 78x1+114x2+142x3+117x4+402x5+360x6+450x7+170x8≤145000 16x1+22x2+95x3+72x4+49x5+113x6+108x8≤45000

2.3x1+3.4x2+6.0x3+6.5x4+24.1x5+8.1x6+29.1x7+10.6≥5600 1.2x1+1.7x2+2.3x3+2.7x4+5.1x5+7.1x6+11.8x7+2.2x8+980x9≥2600 0.7x1+0.6x2+0.3x3+1.0x4+3.2x5+5.3x6+63x7+4.0x8+140x10≥5000 1000x12=3700

2.4.3部分原料总量约束 x1≥400 x2≥100 x3≥100 x4≤150 x5≤100 x6≥30 x7≥50 x8≥30

2.4.4其他保证约束 x4≥0 x5≥0 x9≥0

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x10≥0 x11≥0 x12≥0

2.5模型的建立

综合以上各步工作，可以得出该问题的具体模型如下

Minf(x)=0.68x1+0.72x2+0.23x3+0.22x4+0.37x5+0.32x6+1.54x7+0.38x8+23.0x9+0.56x10+1.12x11+0.42x12

st

x1+x2+x3+x4+x5+x6+x7+x8+x9+x10+x11+x12=1000

3.35x1+3.08x2+1.78x3+2.10x4+2.40x5+1.62x6+2.80x7+1.61x8≥2700 78x1+114x2+142x3+117x4+402x5+360x6+450x7+170x8≥135000 78x1+114x2+142x3+117x4+402x5+360x6+450x7+170x8≤145000 2.3x1+3.4x2+6.0x3+6.5x4+24.1x5+8.1x6+29.1x7+10.6x8≥5600 16x1+22x2+95x3+72x4+49x5+113x6+108x8≤45000

1.2x1+1.7x2+2.3x3+2.7x4+5.1x5+7.1x6+11.8x7+2.2x8+980x9≥2600 0.7x1+0.6x2+0.3x3+1.0x4+3.2x5+5.3x6+63x7+4.0x8+140x10≥5000 1000x12=3700 x1≥400 x2≥100 x3≥100 x4≤150 x5≤100 x6≥30 x7≥50 x8≥30 x4≥0 x5≥0 x9≥0 x10≥0

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x11≥0 x12≥0

价格变动后的模型如下

Minf(x)=0.68x1+0.72x2+0.23x3+0.22(1-10%)x4+0.37x5+0.32x6+1.54(1+10%)x7+0.38x8+23.0x9+0.56(1+10%)x10+1.12(1+10%)x11+0.42x12

st

x1+x2+x3+x4+x5+x6+x7+x8+x9+x10+x11+x12=1000

3.35x1+3.08x2+1.78x3+2.10x4+2.40x5+1.62x6+2.80x7+1.61x8≥2700 78x1+114x2+142x3+117x4+402x5+360x6+450x7+170x8≥135000 78x1+114x2+142x3+117x4+402x5+360x6+450x7+170x8≤145000 2.3x1+3.4x2+6.0x3+6.5x4+24.1x5+8.1x6+29.1x7+10.6x8≥5600 16x1+22x2+95x3+72x4+49x5+113x6+108x8≤45000

1.2x1+1.7x2+2.3x3+2.7x4+5.1x5+7.1x6+11.8x7+2.2x8+980x9≥2600 0.7x1+0.6x2+0.3x3+1.0x4+3.2x5+5.3x6+63x7+4.0x8+140x10≥5000 1000x12=3700 x1≥400 x2≥100 x3≥100 x4≤150 x5≤100 x6≥30 x7≥50 x8≥30 x4≥0 x5≥0 x9≥0 x10≥0 x11≥0

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x12≥0

3 模型的求解及解的分析

3.1 模型的最优解

利用线性规划计算软件Lindo进行求解,结果如下： LP OPTIMUM FOUND AT STEP 15 OBJECTIVE FUNCTION VALUE 1) 556.5981

VARIABLE VALUE REDUCED COST X1 433.212891 0.000000 X2 100.000000 0.000000 X3 136.220139 0.000000 X4 150.000000 0.000000 X5 60.985413 0.000000 X6 30.000000 0.000000 X7 50.000000 0.000000 X8 30.000000 0.000000 X9 0.012054 0.000000 X10 5.869512 0.000000 X11 0.000000 1.385227 X12 3.700000 0.000000

ROW SLACK OR SURPLUS DUAL PRICES 2) 0.000000 0.265227 3) 0.000000 -0.283086 4) 10000.000000 0.000000 5) 0.000000 0.000458 6) 1014.458862 0.000000 7) 2509.395508 0.000000

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8) 0.000000 -0.023740 9) 0.000000 -0.005894 10) 0.000000 -0.000685 11) 33.212887 0.000000 12) 36.220139 0.000000 13) 0.000000 -0.121640 14) 0.000000 0.125659 15) 0.000000 -0.091712 16) 39.014587 0.000000 17) 0.000000 -0.567201 18) 0.000000 -0.191513 19) 150.000000 0.000000 20) 60.985413 0.000000 21) 3.700000 0.000000 22) 5.869512 0.000000 23) 0.000000 0.000000 24) 0.012054 0.000000

NO. ITERATIONS= 15

RANGES IN WHICH THE BASIS IS UNCHANGED:

OBJ COEFFICIENT RANGES

VARIABLE CURRENT ALLOWABLE ALLOWABLE COEF INCREASE DECREASE X1 0.680000 0.153615 0.244557 X2 0.720000 INFINITY 0.121640 X3 0.230000 0.154773 0.155868 X4 0.220000 0.125659 INFINITY

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