数学模型实验5.docx
- 文档编号:18318504
- 上传时间:2023-08-15
- 格式:DOCX
- 页数:21
- 大小:390.65KB
数学模型实验5.docx
《数学模型实验5.docx》由会员分享,可在线阅读,更多相关《数学模型实验5.docx(21页珍藏版)》请在冰点文库上搜索。
数学模型实验5
数学建模实验
五.最优化与存储模型试验
解:
(1)采用最小二乘法,编写程序,
Model:
sets:
trial/1..15/:
xi,y;
endsets
min=@sum(trial:
(a+b*@exp(c*xi)-y)^2);
@free(a);
@free(b);
@free(c);
data:
xi=2,5,7,10,14,19,26,31,34,38,45,52,53,60,65;
y=54,50,45,37,35,25,20,16,18,13,8,11,8,4,6;
enddata
End
解得,
Localoptimalsolutionfound.
Objectivevalue:
44.78049
Infeasibilities:
0.000000
Extendedsolversteps:
5
Totalsolveriterations:
64
ModelClass:
NLP
Totalvariables:
4
Nonlinearvariables:
3
Integervariables:
0
Totalconstraints:
2
Nonlinearconstraints:
1
Totalnonzeros:
4
Nonlinearnonzeros:
3
VariableValueReducedCost
A2.4301770.000000
B57.332090.000000
C-0.4460383E-01-0.2214543E-07
XI
(1)2.0000000.000000
XI
(2)5.0000000.000000
XI(3)7.0000000.000000
XI(4)10.000000.000000
XI(5)14.000000.000000
XI(6)19.000000.000000
XI(7)26.000000.000000
XI(8)31.000000.000000
XI(9)34.000000.000000
XI(10)38.000000.000000
XI(11)45.000000.000000
XI(12)52.000000.000000
XI(13)53.000000.000000
XI(14)60.000000.000000
XI(15)65.000000.000000
Y
(1)54.000000.000000
Y
(2)50.000000.000000
Y(3)45.000000.000000
Y(4)37.000000.000000
Y(5)35.000000.000000
Y(6)25.000000.000000
Y(7)20.000000.000000
Y(8)16.000000.000000
Y(9)18.000000.000000
Y(10)13.000000.000000
Y(11)8.0000000.000000
Y(12)11.000000.000000
Y(13)8.0000000.000000
Y(14)4.0000000.000000
Y(15)6.0000000.000000
RowSlackorSurplusDualPrice
144.78049-1.000000
可得,a=2.430177,b=57.33209,c=-0.4460383e-01.
线性回归方程为,y=2.430177+57.33209*exp(-0.4460383e-01*x).
(2)采用最小一乘法,编写程序,
Model:
sets:
trial/1..15/:
xi,y;
endsets
min=@sum(trial:
@abs(a+b*@exp(c*xi)-y));
@free(a);
@free(b);
@free(c);
data:
xi=2,5,7,10,14,19,26,31,34,38,45,52,53,60,65;
y=54,50,45,37,35,25,20,16,18,13,8,11,8,4,6;
enddata
End
解得,
Linearizationcomponentsadded:
Constraints:
60
Variables:
60
Integers:
15
Localoptimalsolutionfound.
Objectivevalue:
20.80640
Objectivebound:
20.80640
Infeasibilities:
0.000000
Extendedsolversteps:
11
Totalsolveriterations:
1565
ModelClass:
MINLP
Totalvariables:
64
Nonlinearvariables:
2
Integervariables:
15
Totalconstraints:
62
Nonlinearconstraints:
15
Totalnonzeros:
196
Nonlinearnonzeros:
30
VariableValueReducedCost
A3.3982670.000000
B57.114610.000000
C-0.4752126E-010.000000
XI
(1)2.0000000.000000
XI
(2)5.0000000.000000
XI(3)7.0000000.000000
XI(4)10.000000.000000
XI(5)14.000000.000000
XI(6)19.000000.000000
XI(7)26.000000.000000
XI(8)31.000000.000000
XI(9)34.000000.000000
XI(10)38.000000.000000
XI(11)45.000000.000000
XI(12)52.000000.000000
XI(13)53.000000.000000
XI(14)60.000000.000000
XI(15)65.000000.000000
Y
(1)54.000000.000000
Y
(2)50.000000.000000
Y(3)45.000000.000000
Y(4)37.000000.000000
Y(5)35.000000.000000
Y(6)25.000000.000000
Y(7)20.000000.000000
Y(8)16.000000.000000
Y(9)18.000000.000000
Y(10)13.000000.000000
Y(11)8.0000000.000000
Y(12)11.000000.000000
Y(13)8.0000000.000000
Y(14)4.0000000.000000
Y(15)6.0000000.000000
RowSlackorSurplusDualPrice
120.80640-1.000000
可得,a=3.398267,b=57.11461,c=-0.4752126e-01.
线性回归方程为,y=3.398267+57.11461*exp(-0.4752126e-01*x).
(3)最大偏差最小方法,编写程序,
Model:
sets:
trial/1..15/:
xi,yi;
endsets
min=@max(trial:
@abs(a+b*@exp(c*xi)-yi));
@free(a);
@free(b);
@free(c);
data:
xi=2,5,7,10,14,19,26,31,34,38,45,52,53,60,65;
yi=54,50,45,37,35,25,20,16,18,13,8,11,8,4,6;
enddata
End
解得,
Linearizationcomponentsadded:
Constraints:
91
Variables:
76
Integers:
30
Localoptimalsolutionfound.
Objectivevalue:
2.717410
Objectivebound:
2.717410
Infeasibilities:
0.1397950E-04
Extendedsolversteps:
65
Totalsolveriterations:
24712
ModelClass:
MINLP
Totalvariables:
80
Nonlinearvariables:
2
Integervariables:
30
Totalconstraints:
93
Nonlinearconstraints:
15
Totalnonzeros:
272
Nonlinearnonzeros:
30
VariableValue
A0.9242570
B56.38518
C-0.3916105E-01
XI
(1)2.000000
XI
(2)5.000000
XI(3)7.000000
XI(4)10.00000
XI(5)14.00000
XI(6)19.00000
XI(7)26.00000
XI(8)31.00000
XI(9)34.00000
XI(10)38.00000
XI(11)45.00000
XI(12)52.00000
XI(13)53.00000
XI(14)60.00000
XI(15)65.00000
YI
(1)54.00000
YI
(2)50.00000
YI(3)45.00000
YI(4)37.00000
YI(5)35.00000
YI(6)25.00000
YI(7)20.00000
YI(8)16.00000
YI(9)18.00000
YI(10)13.00000
YI(11)8.000000
YI(12)11.00000
YI(13)8.000000
YI(14)4.000000
YI(15)6.000000
RowSlackorSurplusDualPrice
12.717410-1.000000
可得,a=3.701164,b=54.54622,c=-0.4645980e-01.
线性回归方程为,y=3.701164+54.54622*exp(-0.4645980e-01*x).
解:
(1)设X11,X12为A类原油用于生产汽油和民用燃料油的桶数,X21,X22为B类原油用于生产汽油和民用燃料油的桶数。
建立模型并求解,
@gin(x11);@gin(x12);@gin(y11);@gin(y12);
max=248*(x11+x21)+199*(x12+x22);
x11+x12<=5000;
x21+x22<=10000;
(10*x11+5*x21)/(x11+x21)>=8;
(10*x12+5*x22)/(x12+x22)>=6;
解得,
Localoptimalsolutionfound.
Objectivevalue:
3230000.
Extendedsolversteps:
0
Totalsolveriterations:
82
VariableValueReducedCost
X113000.000-280.6667
X122000.000-199.0000
Y110.0000000.000000
Y120.0000000.000000
X212000.0000.000000
X228000.0000.000000
RowSlackorSurplusDualPrice
13230000.1.000000
20.0000000.000000
30.000000199.0000
40.000000-81666.67
50.0000000.000000
因此,最佳销售方式为,
汽油
民用燃料油
A
3000
2000
B
2000
8000
利润
3230000
(2)设X11,X12为A类原油用于生产汽油和民用燃料油的桶数,X21,X22为B类原油用于生产汽油和民用燃料油的桶数。
建立模型并求解,
@gin(x11);@gin(x12);@gin(y11);@gin(y12);@gin(z1);@gin(z2);
max=248*(x11+y11)+199*(x12+y12)+48*z1-z2;
x11+x12<=5000;
y11+y12<=10000;
(10*x11+5*y11)*(1+(z1/(x11+y11))^(0.5))/(x11+y11)>=8;
(10*x12+5*y12)*(1+0.6*(z2/(x12+y12))^(0.6))/(x12+y12)>=6;
(x11+y11)*0.05>=z1;
(x12+y12)*0.05>=z2;
解得,
Localoptimalsolutionfound.
Objectivevalue:
3756000.
Extendedsolversteps:
0
Totalsolveriterations:
64
VariableValue
X115000.000
X120.000000
Y1110000.00
Y120.000000
Z1750.0000
Z20.000000
RowSlackorSurplus
13756000.
20.000000
30.000000
40.1573787
5219.6022
60.000000
70.000000
因此,最优销售方式为
汽油
民用燃料油
A
5000
B
10000
SQ
750
利润
3756000
(3)由
(2)的优化结果可知,SQ需要购买750桶,大于400桶,所以销售方式不变,最大利润为379100元。
解:
(1)设订货费为CD,存储费为CP,存储率为D。
建立模型并求解,
C_D=81;D=600;C_P=0.03;
Q=(2*C_D*D/C_P)^0.5;
T=Q/D;
n=1/T;
TC=1/2*C_P*Q+C_D*D/Q;
解得,
Feasiblesolutionfound.
Totalsolveriterations:
0
VariableValue
C_D81.00000
D600.0000
C_P0.3000000E-01
Q1800.000
T3.000000
N0.3333333
TC54.00000
RowSlackorSurplus
10.000000
20.000000
30.000000
40.000000
50.000000
60.000000
70.000000
因此,最优方式为每隔三天洗1800条,每天54元。
(2)
因此,旅馆应该利用这项打折服务。
解:
设订货费CD,存储费CP,需求率为D。
建立模型并求解,
D=26000;C_P=7.3;C_D=15;
min=1/2*C_P*Q+C_D*D/Q;
解得,
Localoptimalsolutionfound.
Objectivevalue:
2386.210
Infeasibilities:
0.000000
Extendedsolversteps:
5
Totalsolveriterations:
192
ModelClass:
NLP
Totalvariables:
2
Nonlinearvariables:
1
Integervariables:
0
Totalconstraints:
2
Nonlinearconstraints:
1
Totalnonzeros:
2
Nonlinearnonzeros:
1
VariableValueReducedCost
D26000.000.000000
C_P7.3000000.000000
C_D15.000000.000000
Q326.87810.000000
RowSlackorSurplusDualPrice
10.000000-0.4588866E-01
20.000000-163.4391
30.000000-79.54035
42386.210-1.000000
可得,在购买模式下,全年最小库存费为2386.21元,订货次数为79.54此,每次订购326.8781件,取证为327件。
经济生产批量存储模型计算,同理,设订货费CD,存储费CP,需求率为D。
建立模型并求解,
D=26000;P=36500;C_P=7.3;C_D=20;
min=@SQRT(2*C_P*C_D*(1-D/P)*D);
解得,
Globaloptimalsolutionfound.
Objectivevalue:
1477.836
Infeasibilities:
0.000000
Totalsolveriterations:
0
ModelClass:
LP
Totalvariables:
1
Nonlinearvariables:
0
Integervariables:
0
Totalconstraints:
2
Nonlinearconstraints:
0
Totalnonzeros:
1
Nonlinearnonzeros:
0
VariableValueReducedCost
D26000.000.000000
P36500.000.000000
C_P7.3000000.000000
C_D20.000000.000000
RowSlackorSurplusDualPrice
10.0000000.4195323E-01
20.000000-0.5012882E-01
30.000000-101.2208
40.000000-36.94579
51477.836-1.000000
可得,在购买模式下,全年的最小库存费为1477.836元,每年组织生产36.94579件,每次生产7.3.733件,取证为704件。
综合可得,应采用自己生产的模式。
解:
(1)由题可得,航空公司预期损失包括两个部分:
多售票造成超员带来的损失和售票不足带来空座的损失。
建立模型并求解,
sets:
probability/1..9/:
p;
extra/1..8/:
loss;
endsets
data:
p=0.180.250.250.160.060.040.030.020.01;
enddata
k=1500;
h=1500*0.8+1500*1.05*0.2;
@for(extra(i):
loss(i)=h*@sum(probability(j)|j#LE#i:
(i-j+1)*p(j))+k*@sum(probability(l)|l#LE#(8-i):
(9-l-i)*p(10-l)));
解得,
Feasiblesolutionfound.
Totalsolveriterations:
0
ModelClass:
Totalvariables:
0
Nonlinearvariables:
0
Integervariables:
0
Totalconstraints:
0
Nonlinearconstraints:
0
Totalnonzeros:
0
Nonlinearnonzeros:
0
VariableValue
K1500.000
H1515.000
P
(1)0.1800000
P
(2)0.2500000
P(3)0.2500000
P(4)0.1600000
P(5)0.6000000E-01
P(6)0.4000000E-01
P(7)0.3000000E-01
P(8)0.2000000E-01
P(9)0.1000000E-01
LOSS
(1)2147.700
LOSS
(2)1944.150
LOSS(3)2494.350
LOSS(4)3526.950
LOSS(5)4740.450
LOSS(6)6074.550
LOSS(7)7499.100
LOSS(8)8983.950
RowSlackorSurplus
10.000000
20.000000
30.000000
40.000000
50.000000
60.000000
70.000000
80.000000
90.000000
100.000000
因此,航空公司多售出2张票时,可以使该公司的预期损失达到最小为1944.150元。
(2)建立模型并求解,
sets:
probability/1..9/:
p;
extra/1..8/:
profit;
endsets
data:
p=0.180.250.250.160.060.040.030.020.01;
enddata
N=150;k=1500;h=15
- 配套讲稿:
如PPT文件的首页显示word图标,表示该PPT已包含配套word讲稿。双击word图标可打开word文档。
- 特殊限制:
部分文档作品中含有的国旗、国徽等图片,仅作为作品整体效果示例展示,禁止商用。设计者仅对作品中独创性部分享有著作权。
- 关 键 词:
- 数学模型 实验