运筹学实验报告案例二
运筹学实验报告案例二
安阳师范学院
数学与统计学院
实
验
报
告
实验课程名称:运筹学
实验设计题目:配料问题
专业:数学与应用数学
班级:13级二班
学生:常俊建 130800003 学生:刘翠宇 130800004 学生:李燃 130800022
配料问题
一、问题的描述
某饲料公司生产鸡混合饲料,每千克饲料所需营养质量要求如表C-4所示。
表C-4每千克饲料所需营养质量要求
公司计划使用的原料有玉米、小麦、麦麸、米糠、豆饼、菜子饼、鱼粉、槐叶粉、DL-蛋氨酸、骨粉、碳酸钙和食盐等12种。各原料的营养成分含量及价格见表C-5。
表C-5原料的营养成分含量及价格
公司根据原料来源,还要求1吨混合饲料中原料含量为:玉米不低于400kg、小麦不低于100kg、麦麸不低于100kg、米糠不超过150kg、豆饼不超过100kg、菜子饼不低于30kg、鱼粉不低于50kg、槐叶粉不低于30kg,DL-蛋氨酸、骨粉、碳酸钙适量。
(1)按照肉用种鸡公司标准,求1kg混合饲料中每种原料各配多少成本最低,建立数学模型并求解。
(2)按照肉用种鸡国家标准,求1kg混合饲料中每种原料各配多少成本最低。
(3)公司采购了一批花生饼,单价是0.6元/kg,代谢能到有机磷的含量分别为(2.4,38,120,0,0.92,0.15,0.17),求肉用种鸡成本最低的配料方案。
(4)求产蛋鸡的最优饲料配方方案。
(5)公司考虑到未来鱼粉、骨粉和碳酸钙将要涨价,米糠将要降价,价格变化率r,试对两种产品配方方案进行灵敏度分析。
都是原价的%
说明:以上5个问题独立求解和分析,如在问题(3)中只加花生饼,其他方案则不加花生饼。
二、问题的分析与符号说明
2.1模型的分析
设公司计划使用的原料有玉米、小麦、麦麸、米糠、豆饼、菜子饼、鱼粉、槐叶粉、DL-蛋氨酸、骨粉、碳酸钙和食盐的用量分别为()1,2,,12i x i =L ,若将相应的原料单价分别用()1,2,,12i c i =L ,来表示,则其总成本可以用下面的线性函数来表示 即
12
1i i i w c x ==∑.
2.2模型的符号说明
三、模型建立与求解
4.1问题(1)的模型建立与求解
根据问题(1)所给数据及问题要求可列出约束条件,所以可建立混合饲料配料计划的线性规划模型如下:
12345678
9101112
123456781234567min 0.680.720.230.220.370.32 1.540.38230.56 1.120.423.35 3.08 1.78 2.10 2.40 1.62 2.80 1.61 2.73.35 3.08 1.78 2.10 2.40 1.62 2.80 1.6z x x x x x x x x x x x x x x x x x x x x x x x x x x x =++++++++++++++++++≥+++++++812345678123456812345678
12345678113578114142117402360450170145162295724911310845
2.3
3.4 6.0 6.52
4.18.129.110.6
5.6
1.2 1.7
2.3 2.7 5.17.111.8 2.29x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x ≥+++++++≤++++++≤+++++++≥++++++++91234567810111234567810121
23456789101112
1234580 2.60.70.60.3 1.0 3.2 5.3634300400300.30.34101358.427414051000 3.7
1
0.4,0.1,0.10.15x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x ≥+++++++++≥++++++++≥=+++++++++++=≥≥≥≤≤67
80.10.030.050.030,(1,2, (12)
j x x x x j ??
???
???
??
???
?????
???≥??≥?≥??
≥=? 问题(1)的lingo 程序如下:
Min=0.68*x1+0.72*x2+0.23*x3+0.22*x4+0.37*x5+0.32*x6+1.54*x7+0.38*x8+23*x9+ 0.56*x10+1.12*x11+0.42*x12;
3.35*x1+3.08*x2+1.78*x3+2.10*x4+2.40*x5+1.62*x6+2.80*x7+1.61*x8>=2.7; 78*x1+114*x2+142*x3+117*x4+402*x5+360*x6+450*x7+170*x8>=135; 78*x1+114*x2+142*x3+117*x4+402*x5+360*x6+450*x7+170*x8<=145; 16*x1+22*x2+95*x3+72*x4+49*x5+113*x6+108*x8<=45;
2.3*x1+
3.4*x2+6.0*x3+6.5*x4+2
4.1*x5+8.1*x6+29.1*x7+10.6*x8>=
5.6; 1.2*x1+1.7*x2+2.3*x3+2.7*x4+5.1*x5+7.1*x6+11.8*x7+2.2*x8+980*x9>=2.6; 0.7*x1+0.6*x2+0.3*x3+1.0*x4+3.2*x5+5.3*x6+63*x7+4.0*x8+300*x10+400*x11>=30; 0.3*x1+0.34*x2+10.0*x3+13.0*x4+5.0*x5+8.4*x6+27*x7+4.0*x8+140*x10>=5; 1000*x12=3.7;
X1+x2+x3+x4+x5+x6+x7+x8+x9+x10+x11+x12=1; X1>=0.4; x2>=0.1; x3>=0.1; x4<=0.15; x5<=0.1; x6>=0.03; x7>=0.05; x8>=0.03;
问题(1)Lingo 程序的结果:
Global optimal solution found.
Objective value: 0.6553693
Infeasibilities: 0.000000
Total solver iterations: 9
Variable Value Reduced Cost
X1 0.5385030 0.000000
X2 0.1000000 0.000000
X3 0.1000000 0.000000
X4 0.000000 0.1446276
X5 0.7213126E-01 0.000000
X6 0.3000000E-01 0.000000
X7 0.5000000E-01 0.000000
X8 0.3000000E-01 0.000000
X9 0.3233949E-03 0.000000
X10 0.4263719E-01 0.000000
X11 0.3270518E-01 0.000000
X12 0.3700000E-02 0.000000
Row Slack or Surplus Dual Price
1 0.6553693 -1.000000
2 0.000000 -0.5218799
3 0.000000 -0.2339449E-03
4 10.00000 0.000000
5 14.51952 0.000000
6 0.3329203 0.000000
7 0.000000 -0.2461224E-01
8 0.000000 -0.5600000E-02
9 4.247413 0.000000
10 0.000000 -0.1540000E-02
11 0.000000 1.120000
12 0.1385030 0.000000
13 0.000000 -0.1607394
14 0.000000 -0.3295455
15 0.1500000 0.000000
16 0.2786874E-01 0.000000
17 0.000000 -0.3059075
18 0.000000 -0.4502367
19 0.000000 -0.5434558
所以按照肉用种鸡公司标准,1kg混合饲料中玉米约需要0.54kg,小麦约需要0.1kg,麦麸约需要0.1kg,米糠约需要0kg,豆饼约需要0.072kg,菜子饼约需要0.03kg,鱼粉约需要0.05kg,槐叶粉约需要0.03kg,DL蛋氨酸约需要0.0003kg,骨粉约需要0.042kg,碳酸钙约需要0.032kg,食盐约需要0.0037kg,此时成本最低为约为0.655元。
4.2问题(2)的模型建立与求解
根据问题(2)所给数据及问题要求可列出约束条件,所以可建立混合饲料配料计划成本最低的线性规划模型如下:
12345678
9101112
123456781234567min 0.680.720.230.220.370.32 1.540.38230.56 1.120.423.35 3.08 1.78 2.10 2.40 1.62 2.80 1.61 2.73.35 3.08 1.78 2.10 2.40 1.62 2.80 1.6z x x x x x x x x x x x x x x x x x x x x x x x x x x x =++++++++++++++++++≥+++++++812345678123456781234568
123456781 2.8
7811414211740236045017013578114142117402360450170145162295724911310850
2.3
3.4 6.0 6.52
4.18.129.110.6
5.6
x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x ≤+++++++≥+++++++≤++++++≤+++++++≥1234567891234567810111234567810111231.2 1.7 2.3 2.7 5.17.111.8 2.2980 2.5
0.70.60.3 1.0 3.2 5.3634300400230.70.60.3 1.0 3.2 5.3634300400400.30.341013x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x ++++++++≥+++++++++≥+++++++++≤+++4567810
12345678101212345678910111212
34
567
858.4274140 4.6
0.30.34101358.427414051000 3.7
10.40.10.10.150.10.030.050.030,1,2,...,j x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x j +++++≥++++++++≤=+++++++++++=≥≥≥≤≤≥≥≥≥=()
12???
??
???
???
??
???
??
???
????????
?????
?问题(2)的lingo 程序如下:
Min =0.68*x1+0.72*x2+0.23*x3+0.22*x4+0.37*x5+0.32*x6+1.54*x7+0.38*x8+23*x9+ 0.56*x10+1.12*x11+0.42*x12;
3.35*x1+3.08*x2+1.78*x3+2.10*x4+2.40*x5+1.62*x6+2.80*x7+1.61*x8>=2.7; 3.35*x1+3.08*x2+1.78*x3+2.10*x4+2.40*x5+1.62*x6+2.80*x7+1.61*x8<=2.8; 78*x1+114*x2+142*x3+117*x4+402*x5+360*x6+450*x7+170*x8>=135; 78*x1+114*x2+142*x3+117*x4+402*x5+360*x6+450*x7+170*x8<=145; 16*x1+22*x2+95*x3+72*x4+49*x5+113*x6+108*x8<50;
2.3*x1+
3.4*x2+6.0*x3+6.5*x4+2
4.1*x5+8.1*x6+29.1*x7+10.6*x8>=
5.6; 1.2*x1+1.7*x2+2.3*x3+2.7*x4+5.1*x5+7.1*x6+11.8*x7+2.2*x8+980*x9>=2.5; 0.7*x1+0.6*x2+0.3*x3+1.0*x4+3.2*x5+5.3*x6+63*x7+4.0*x8+300*x10+400*x11>=23; 0.7*x1+0.6*x2+0.3*x3+1.0*x4+3.2*x5+5.3*x6+63*x7+4.0*x8+300*x10+400*x11<=40;
0.3*x1+0.34*x2+10.0*x3+13.0*x4+5.0*x5+8.4*x6+27*x7+4.0*x8+140*x10>=4.6;
0.3*x1+0.34*x2+10.0*x3+13.0*x4+5.0*x5+8.4*x6+27*x7+4.0*x8+140*x10<=5; 1000*x12=3.7;
X1+x2+x3+x4+x5+x6+x7+x8+x9+x10+x11+x12=1;
X1>=0.4;
x2>=0.1;
x3>=0.1;
x4<=0.15;
x5<=0.1;
x6>=0.03;
x7>=0.05;
x8>=0.03;
问题(2)Lingo程序的结果:
Global optimal solution found.
Objective value: 0.6246589
Infeasibilities: 0.000000
Total solver iterations: 9
Variable Value Reduced Cost
X1 0.4888733 0.000000
X2 0.1000000 0.000000
X3 0.1000000 0.000000
X4 0.5954138E-01 0.000000
X5 0.8930732E-01 0.000000
X6 0.3000000E-01 0.000000
X7 0.5000000E-01 0.000000
X8 0.3000000E-01 0.000000
X9 0.2869694E-04 0.000000
X10 0.6262596E-02 0.000000
X11 0.4228671E-01 0.000000
X12 0.3700000E-02 0.000000
Row Slack or Surplus Dual Price
1 0.6246589 -1.000000
2 0.000000 -0.3847177
3 0.1000000 0.000000
4 10.00000 0.000000
5 0.000000 0.1400426E-03
6 15.18499 0.000000
7 1.019734 0.000000
8 0.000000 -0.2411185E-01
9 0.000000 -0.4374035E-02
10 17.00000 0.000000
11 0.4000000 0.000000
12 0.000000 0.8756890E-03
13 0.000000 -0.1049614E-02
14 0.000000 0.6296142
15 0.8887329E-01 0.000000
16 0.000000 -0.1373318
17 0.000000 -0.1466902
18 0.9045862E-01 0.000000
19 0.1069268E-01 0.000000
20 0.000000 -0.1897662
21 0.000000 -0.6189834
22 0.000000 -0.3469865
所以按照肉用种鸡国家标准,1kg混合饲料中玉米约需要0.49kg,小麦约需要0.1kg,麦麸约需要0.1kg,米糠约需要0.06kg,豆饼约需要0.09kg,菜子饼约需要0.03kg,鱼粉约需要0.05kg,槐叶粉约需要0.03kg,DL蛋氨酸约需要0.000029kg,骨粉约需要0.0063kg,碳酸钙约需要0.042kg,食盐约需要0.0037kg,此时成本最低为约为0.625元。
4.3问题(3)的模型建立与求解
可建立肉用种鸡成本最低的配料方案模型如下
12345678
910111213
123456781312345min 0.680.720.230.220.370.32 1.540.38230.56 1.120.420.63.35 3.08 1.78 2.10 2.40 1.62 2.80 1.61 2.4 2.73.35 3.08 1.78 2.10 2.40 1.6z x x x x x x x x x x x x x x x x x x x x x x x x x x x =++++++++++++++++++++≥+++++67813123456781312345678131
23456813
122 2.80 1.61 2.4 2.8
78114142117402360450170381357811414211740236045017038145162295724911310812045
2.3
3.4 6.x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x +++≤++++++++≥++++++++≤+++++++≤++3456781234567891312345678101113123450 6.52
4.18.129.110.65
1.2 1.7
2.3 2.7 5.17.111.8 2.29800.92 2.6
0.70.60.3 1.0 3.2 5.36343004000.15300.30.3410135x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x +++++≥+++++++++≥++++++++++≥+++++()
67810131212345678910111213123
45
6788.42741400.1751000 3.7
10.40.10.10.150.10.030.050.030,1,2,...,13j x x x x x x x x x x x x x x x x x x x x x x x x x x x x j ???
??
??????
??
++++≥??=?
?
++++++++++++=??≥?
≥??≥?≤??≤??≥?
≥??≥?≥=??
问题(3)的lingo 程序如下:
Min =0.68*x1+0.72*x2+0.23*x3+0.22*x4+0.37*x5+0.32*x6+1.54*x7+0.38*x8+23*x9+ 0.56*x10+1.12*x11+0.42*x12+0.6*x13;
3.35*x1+3.08*x2+1.78*x3+2.10*x4+2.40*x5+1.62*x6+2.80*x7+1.61*x8+2.4*x13>=2.7;
78*x1+114*x2+142*x3+117*x4+402*x5+360*x6+450*x7+170*x8+38*x13>=135;
78*x1+114*x2+142*x3+117*x4+402*x5+360*x6+450*x7+170*x8+38*x13<=145;
16*x1+22*x2+95*x3+72*x4+49*x5+113*x6+108*x8+120*x13<=45;
2.3*x1+
3.4*x2+6.0*x3+6.5*x4+2
4.1*x5+8.1*x6+29.1*x7+10.6*x8>=
5.6;
1.2*x1+1.7*x2+
2.3*x3+2.7*x4+5.1*x5+7.1*x6+11.8*x7+2.2*x8+980*x9+0.92*x13>=2.6;
0.7*x1+0.6*x2+0.3*x3+1.0*x4+3.2*x5+5.3*x6+63*x7+4.0*x8+300*x10+400*x11+0.15*x13>=30;
0.3*x1+0.34*x2+10.0*x3+13.0*x4+5.0*x5+8.4*x6+27*x7+4.0*x8+140*x10+0.17*x13>=5;
1000*x12=3.7;
X1+x2+x3+x4+x5+x6+x7+x8+x9+x10+x11+x12+x13=1;
X1>=0.4;
x2>=0.1;
x3>=0.1;
x4<=0.15;
x5<=0.1;
x6>=0.03;
x7>=0.05;
x8>=0.03;
问题(3)Lingo程序的结果:
Global optimal solution found.
Objective value: 0.6553693
Infeasibilities: 0.000000
Total solver iterations: 11
Variable Value Reduced Cost
X1 0.5385030 0.000000
X2 0.1000000 0.000000
X3 0.1000000 0.000000
X4 0.000000 0.1446276
X5 0.7213126E-01 0.000000
X6 0.3000000E-01 0.000000
X7 0.5000000E-01 0.000000
X8 0.3000000E-01 0.000000
X9 0.3233949E-03 0.000000
X10 0.4263719E-01 0.000000
X11 0.3270518E-01 0.000000
X12 0.3700000E-02 0.000000
X13 0.000000 0.4351151
Row Slack or Surplus Dual Price
1 0.6553693 -1.000000
2 0.000000 -0.5218799
3 0.000000 -0.2339449E-03
4 10.00000 0.000000
5 14.51952 0.000000
6 0.3329203 0.000000
7 0.000000 -0.2461224E-01
8 0.000000 -0.5600000E-02
9 4.247413 0.000000 10 0.000000 -0.1540000E-02 11 0.000000 1.120000 12 0.1385030 0.000000 13 0.000000 -0.1607394 14 0.000000 -0.3295455 15 0.1500000 0.000000 16 0.2786874E-01 0.000000 17 0.000000 -0.3059075 18 0.000000 -0.4502367 19 0.000000 -0.5434558
所以公司采购花生饼后,肉用鸡成本最低为0.655元,其中玉米约需要0.539kg ,小麦约需要0.1kg ,麦麸约需要0.1kg ,米糠约需要0kg ,豆饼约需要0.072kg ,菜子饼约需要0.03kg ,鱼粉约需要0.05kg ,槐叶粉约需要0.03kg ,DL 蛋氨酸约需要0.00032kg ,骨粉约需要0.0426kg ,碳酸钙约需要0.0327kg ,食盐约需要0.0037kg ,花生饼需要0kg,
4.4问题(4)的模型建立与求解
根据问题(4)所给数据及问题要求可列出约束条件,所以可建立产蛋鸡的最优饲料配方方案的线性规划模型如下:
12345678
9101112
1234567812345678min 0.680.720.230.220.370.32 1.540.38230.56 1.120.423.35 3.08 1.78 2.10 2.40 1.62 2.80 1.61 2.6578114142117402360450170151
1z x x x x x x x x x x x x x x x x x x x x x x x x x x x x =++++++++++++++++++≥+++++++≥123456812345678123456789
1234567810622957249113108252.3 3.4 6.0 6.524.18.129.110.6 6.81.2 1.7 2.3 2.7 5.17.111.8 2.29806
0.70.60.3 1.0 3.2 5.3634300x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x ++++++≤+++++++≥++++++++≥+++++++++()111234567810121234567891011121234
5678400330.30.34101358.42741403
1000310.40.10.10.150.1
0.03
0.050.03
0,1,2, (12)
x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x j ???
??
???
≥??++++++++≥?
?=?
+++++++++++=??
≥??≥?
≥??≤?≤?≥≥≥≥=???
??
??
问题(4)的lingo程序如下:
Min=0.68*x1+0.72*x2+0.23*x3+0.22*x4+0.37*x5+0.32*x6+1.54*x7+0.38*x8+23*x9+0.56*x10+1.12*x11+0 .42*x12;
3.35*x1+3.08*x2+1.78*x3+2.10*x4+2.40*x5+1.62*x6+2.80*x7+1.61*x8>=2.65;
78*x1+114*x2+142*x3+117*x4+402*x5+360*x6+450*x7+170*x8>=151;
16*x1+22*x2+95*x3+72*x4+49*x5+113*x6+108*x8<=25;
2.3*x1+
3.4*x2+6.0*x3+6.5*x4+2
4.1*x5+8.1*x6+29.1*x7+10.6*x8>=6.8;
1.2*x1+1.7*x2+
2.3*x3+2.7*x4+5.1*x5+7.1*x6+11.8*x7+2.2*x8+980*x9>=6;
0.7*x1+0.6*x2+0.3*x3+1.0*x4+3.2*x5+5.3*x6+63*x7+4.0*x8+300*x10+400*x11>=33;
0.3*x1+0.34*x2+10.0*x3+13.0*x4+5.0*x5+8.4*x6+27*x7+4.0*x8+140*x10>=3;
1000*x12=3;
X1+x2+x3+x4+x5+x6+x7+x8+x9+x10+x11+x12=1;
X1>=0.4;
x2>=0.1;
x3>=0.1;
x4<=0.15;
x5<=0.1;
x6>=0.03;
x7>=0.05;
x8>=0.03;
问题(4)Lingo程序的结果:
Global optimal solution found.
Objective value: 0.8603805
Infeasibilities: 0.000000
Total solver iterations: 7
Variable Value Reduced Cost
X1 0.4168750 0.000000
X2 0.1000000 0.000000
X3 0.1000000 0.000000
X4 0.000000 2.471006
X5 0.000000 1.402233
X6 0.3000000E-01 0.000000
X7 0.2394888 0.000000
X8 0.3000000E-01 0.000000
X9 0.2035492E-02 0.000000
X10 0.7860067E-01 0.000000
X11 0.000000 0.5600000
X12 0.3000000E-02 0.000000
Row Slack or Surplus Dual Price
1 0.8603805 -1.000000
2 0.000000 -0.2535015
3 30.78623 0.000000
4 0.000000 0.4729421E-01
5 2.628938 0.000000
6 0.000000 -0.2289796E-01
7 6.328810 0.000000
8 16.00135 0.000000
9 0.000000 0.1400000E-03
10 0.000000 -0.5600000
11 0.1687500E-01 0.000000
12 0.000000 -0.3807617
13 0.000000 -3.659052
14 0.1500000 0.000000
15 0.1000000 0.000000
16 0.000000 -4.530998
所以产蛋鸡按照最优饲料配方,其中玉米约需要0.42kg,小麦约需要0.1kg,麦麸约需要0.1kg,米糠约需要0kg,豆饼约需要0kg,菜子饼约需要0.03kg,鱼粉约需要0.24kg,槐叶粉约需要0.03kg,DL蛋氨酸约需要0.002kg,骨粉约需要0.0786kg,碳酸钙约需要0kg,食盐约需要0.003kg,此时成本最低为约为0.86元。
4.5问题(5)的模型建立与求解
灵敏度分析问题一:
结果:
Ranges in which the basis is unchanged:
Objective Coefficient Ranges
Current Allowable Allowable Variable Coefficient Increase Decrease
变量当前目标函数系数允许增加量允许减少量
X1 0.6800000 0.1130246 1.505273
X2 0.7200000 INFINITY 0.1607394
X3 0.2300000 INFINITY 0.3295455
X4 0.2200000 INFINITY 0.1446276
X5 0.3700000 0.3285352 0.8097285E-01
X6 0.3200000 INFINITY 0.3059075
X7 1.540000 INFINITY 0.4502367
X8 0.3800000 INFINITY 0.5434558
X9 23.00000 18.71411 24.12000 X10 0.5600000 0.7428457E-01 0.3680286
X11 1.120000 0.4223341 0.9987023E-01
Righthand Side Ranges
Row Current Allowable Allowable RHS Increase Decrease
当前右边常数项允许增加量允许减少量
2 2.700000 0.2711689E-01 0.3903722E-01
3 135.0000 9.645910 5.132235
4 145.0000 INFINITY 10.00000
5 45.00000 INFINITY 14.51952
6 5.600000 0.3329203 INFINITY
7 2.600000 7.432973 0.3169270
8 30.00000 3.033867 3.270518
9 5.000000 4.247413 INFINITY
11 0.9963000 0.1090173E-01 0.7584667E-02
12 0.4000000 0.1385030 INFINITY
13 0.1000000 0.1588975 0.1000000
14 0.1000000 0.2034075E-01 0.2804781E-01
15 0.1500000 INFINITY 0.1500000
16 0.1000000 INFINITY 0.2786874E-01
17 0.3000000E-01 0.2391997E-01 0.2993016E-01
18 0.5000000E-01 0.3067834E-01 0.2204139E-01
19 0.3000000E-01 0.1893127E-01 0.2711439E-01 可分析得:
通过改变问题一的变量x7(鱼粉),x10(骨粉),x11(碳酸钙),可得出结论:
①仅当x7(鱼粉)涨价时,变化率满足r%>=0%可以没有限制的波动,均不会使得原有的原料配方方案不会发生变化;
②仅当x10(骨粉)涨价时,0%<=r%<=13.2%时原方案不会变化,超过这个变化率波动范围时,公司就要对配料方案进行调整;
③仅当x11(碳酸钙)涨价时,0%<=r%<=37.7%时,原配方不会变化,若在这个范围之外,公司就要对配料方案进行调整;
灵敏度分析问题二:
结果:
Ranges in which the basis is unchanged:
Objective Coefficient Ranges
Current Allowable Allowable Variable Coefficient Increase Decrease
X1 0.6800000 0.1823228 0.1721718
X2 0.7200000 INFINITY 0.1373318
X3 0.2300000 INFINITY 0.1466902
X4 0.2200000 0.1160232 0.5415697E-01
X5 0.3700000 0.4121340E-01 2.622442
X6 0.3200000 INFINITY 0.1897662
X7 1.540000 INFINITY 0.6189834
X8 0.3800000 INFINITY 0.3469865
X9 23.00000 82.17487 14.64407
X10 0.5600000 0.1143759 3.131764
X11 1.120000 4.230653 0.1525134 Righthand Side Ranges
Row Current Allowable Allowable
RHS Increase Decrease
2 2.700000 0.6278208E-01 0.2860993E-01
3 2.800000 INFINITY 0.1000000
4 135.0000 10.00000 INFINITY
5 145.0000 3.000801 10.00000
6 50.00000 INFINITY 15.18499
7 5.600000 1.019734 INFINITY
8 2.500000 15.70639 0.2801508E-01
9 23.00000 6.410773 2.979736
10 40.00000 INFINITY 17.00000
11 4.600000 0.4000000 INFINITY
12 5.000000 7.364819 0.4000000
14 0.9963000 0.7449340E-02 0.1602693E-01
15 0.4000000 0.8887329E-01 INFINITY
16 0.1000000 0.1179889 0.1000000
17 0.1000000 0.4709369E-01 0.2877853E-01
18 0.1500000 INFINITY 0.9045862E-01
19 0.1000000 INFINITY 0.1069268E-01
20 0.3000000E-01 0.1676645E-01 0.1383065E-01
21 0.5000000E-01 0.3809742E-02 0.7971571E-02
22 0.3000000E-01 0.4383041E-01 0.1859184E-01 通过改变问题二的变量x7(鱼粉),x10(骨粉),x11(碳酸钙),可得出结论:
①仅当x7(鱼粉)涨价时,变化率满足r%>=0%可以没有限制的波动,均不会使得原有的原料配方方案不会发生变化;
②仅当x10(骨粉)涨价时,0%<=r%<=19.6%时原方案不会变化,超过这个变化率波动范围时,公司就要对配料方案进行调整;
③仅当x11(碳酸钙)涨价时,0%<=r%<=377.68%时,原配方不会变化,若在这个范围之外,公司就要对配料方案进行调整;