例1.(选自《运筹学基础》P54.汽油混合问题,线性规划问题)
一种汽油的特性可用两个指标描述:其点火性用“辛烷数”描述,其挥发性用“蒸汽压力”描述。某炼油厂有四种标准汽油,设其标号分别为1,2,3,4,其特性及库存量列于下表1中,将上述标准汽油适量混合,可得两种飞机汽油,某标号为1,2,这两种飞机汽油的性能指标及产量需求列于表2中。
问应如何根据库存情况适量混合各种标准汽油,使既满足飞机汽油的性能指标,而产量又为最高。
表1
|
标准汽油 |
辛烷数 |
|
蒸汽压力(g/cm^2) |
库存量 |
|
1 |
107.5 |
|
7.11*10^(-2) |
380000 |
|
2 |
93.0 |
|
11.38*10^(-2) |
262200 |
|
3 |
87.0 |
|
5.69*10^(-2) |
408100 |
|
4 |
108.0 |
|
28.45*10^(-2) |
130100 |
(1 g/cm^2=98Pa)
表2
|
飞机汽油 |
辛烷数 |
|
蒸汽压力(g/cm^2) |
产量需求(L) |
|
1 |
>=91 |
|
<=9.96*10^(-2) |
越多越好 |
|
2 |
>=100 |
|
<=9.96*10^(-2) |
>=250000 |
建模过程略(详见《运筹学基础》P54—55)
目标函数:max z=x1+x2+x3+x4
约束条件:x5+x6+x7+x8>=250000
x1+x5<=380000
x2+x6<=265200
x3+x7<=408100
x4+x8<=130100
2.85x1-1.42x2+4.27x3-18.49x4>=0
2.85x5-1.42x6+4.27x7-18.49x8>=0
16.5x1+2.0x2-4.0x3+17x4>=0
7.5x5-7.0x6-13.0x7+8.0x8>=0
xj>=0(j=1,2...,8)
用LINDO6.1的运算结果:
LP OPTIMUM FOUND AT STEP 6
OBJECTIVE FUNCTION VALUE
1) 933400.0
VARIABLE VALUE REDUCED COST
X1 161351.734375 0.000000
X2 265200.000000 0.000000
X3 408100.000000 0.000000
X4 98748.265625 0.000000
X5 218648.265625 0.000000
X6 0.000000 0.000000
X7 0.000000 0.000000
X8 31351.734375 0.000000
ROW SLACK OR SURPLUS DUAL PRICES
2) 0.000000 -1.000000
3) 0.000000 1.000000
4) 0.000000 1.000000
5) 0.000000 1.000000
6) 0.000000 1.000000
7) 0.000000 0.000000
8) 43454.000000 0.000000
9) 3239024.250000 0.000000
10) 1890675.875000 0.000000
NO. ITERATIONS= 6
RANGES IN WHICH THE BASIS IS UNCHANGED:
OBJ COEFFICIENT RANGES
VARIABLE CURRENT ALLOWABLE ALLOWABLE
COEF INCREASE DECREASE
X1 1.000000 0.000000 1.154137
X2 1.000000 INFINITY 0.000000
X3 1.000000 INFINITY 0.000000
X4 1.000000 0.000000 0.000000
X5 0.000000 1.154137 0.000000
X6 0.000000 0.000000 INFINITY
X7 0.000000 0.000000 INFINITY
X8 0.000000 0.000000 0.000000
RIGHTHAND SIDE RANGES
ROW CURRENT ALLOWABLE ALLOWABLE
RHS INCREASE DECREASE
2 250000.000000 186222.062500 234752.984375
3 380000.000000 234752.984375 15247.017578
4 265200.000000 30601.410156 265200.000000
5 408100.000000 156685.250000 10176.581055
6 130100.000000 2350.135254 36184.207031
7 0.000000 43454.000000 669046.000000
8 0.000000 43454.000000 INFINITY
9 0.000000 3239024.250000 INFINITY
10 0.000000 1890675.875000 INFINITY
用LINGO8.0的运算结果:
Global optimal solution found at iteration: 8
Objective value: 933400.0
Variable Value Reduced Cost
X1 264937.9 0.000000
X2 135702.1 0.000000
X3 408100.0 0.000000
X4 124660.0 0.000000
X5 115062.1 0.000000
X6 129497.9 0.000000
X7 0.000000 0.000000
X8 5440.011 0.000000
Row Slack or Surplus Dual Price
1 933400.0 1.000000
2 0.000000 -1.000000
3 0.000000 1.000000
4 0.000000 1.000000
5 0.000000 1.000000
6 0.000000 1.000000
7 0.000000 0.000000
8 43454.00 0.000000
9 5129700. 0.000000
10 0.000000 0.000000
用excel的运算结果:
|
x1 |
264937.9423 |
|
|
约束条件1 |
250000 |
|
x2 |
135702.0686 |
|
|
约束条件2 |
380000 |
|
x3 |
408100 |
|
|
约束条件3 |
265200 |
|
x4 |
124659.9891 |
目标函数 |
933400 |
约束条件4 |
408100 |
|
x5 |
115062.0577 |
|
|
约束条件5 |
130100 |
|
x6 |
129497.9314 |
|
|
约束条件6 |
0 |
|
x7 |
0 |
|
|
约束条件7 |
43454 |
|
x8 |
5440.010913 |
|
|
约束条件8 |
5129700 |
|
|
|
|
|
约束条件9 |
-1.45519E-10 |
以上可以看出用三个软件计算的目标函数值相同,即z=933400。
可变函数值xj(j =1,2...,8),LINDO6.1与LINGO8.0和excel的结果不同,LINGO8.0和excel的结果相同。
由此可见,此题至少有二个最优解。
嗯. 应该是这样.
楼主您好!能否将您的lindo和lingo软件给我传一下?谢谢!
email:neu_zq@126.com
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