《运筹学》实验报告Word文档格式.docx
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30.0000000.5000000
410.000000.000000
故最优解为x1=0,x2=8,x3=0,x4=-6,最优值为2。
2.该问题Lingo中的代码如下,
min=150*(x1+x2+x3)+80*(y1+y2+y3);
500*x1<
=5000;
1000*x1+500*x2<
=9000;
1500*x1+1000*x2+500*x3<
=12000;
2000*x1+1500*x2+1000*x3+500*y1<
=16000;
2500*x1+2000*x2+1500*x3+1000*y1+500*y2<
=18500;
3000*x1+2500*x2+2000*x3+1500*y1+1000*y2+500*y3<
=21500;
3500*x1+3000*x2+2500*x3+2000*y1+1500*y2+1000*y3<
=25500;
4000*x1+3500*x2+3000*x3+2500*y1+2000*y2+1500*y3<
=30000;
4000*x1+4000*x2+3500*x3+2500*y1+2500*y2+2000*y3<
=33500;
4000*x1+4000*x2+4000*x3+2500*y1+2500*y2+2500*y3>
=36000;
2000*x1+1500*x2+1000*x3+500*y1>
3500*x1+3000*x2+2500*x3+2000*y1+1500*y2+1000*y3>
x1+x2+x3+y1+y2+y3<
=11;
求解可得解报告,
Globaloptimalsolutionfound.
Objectivevalue:
1350.000
Infeasibilities:
0.000000
Totalsolveriterations:
5
VariableValueReducedCost
X13.0000000.000000
X20.0000000.000000
X36.0000000.000000
Y10.00000027.50000
Y20.00000027.50000
Y30.0000000.000000
RowSlackorSurplusDualPrice
11350.000-1.000000
23500.0000.000000
36000.0000.000000
44500.0000.000000
54000.0000.000000
62000.0000.000000
7500.00000.000000
80.0000000.000000
90.0000000.5500000E-01
10500.00000.000000
110.000000-0.6500000E-01
120.000000-0.5500000E-01
134000.0000.000000
142.0000000.000000
VariableValue
X13.000000
X20.000000
X36.000000
Y10.000000
Y20.000000
Y30.000000
故最优解为x1=3,x2=0,x3=6,y1=0,y2=0,y3=0,最优值为1350。
3.
设从班次i开始上班的人数为xi,i=1,2……,6.
Minz=x1+x2+x3+x4+x5+x6
x6+x1>
=60
x1+x2>
=70
x2+x3>
x3+x4>
=50
x4+x5>
=20
x5+x6>
=30
xi>
=0,xi为整数,i=1,2,……,7.
Min=x1+x2+x3+x4+x5+x6;
x6+x1>
=60;
x1+x2>
=70;
x2+x3>
x3+x4>
=50;
x4+x5>
=20;
x5+x6>
=30;
150.0000
4
X160.000000.000000
X210.000000.000000
X350.000000.000000
X40.0000000.000000
X530.000000.000000
X60.0000000.000000
1150.0000-1.000000
20.0000000.000000
30.000000-1.000000
40.0000000.000000
50.000000-1.000000
610.000000.000000
70.000000-1.000000
X160.00000
X210.00000
X350.00000
X40.000000
X530.00000
X60.000000
故最优解为x1=60,x2=10,x3=50,x4=0,x5=30.x6=0.最优值为150.
实验二.简单运输问题数学模型的Lingo软件求解
一.实验目的:
熟悉运输问题的数学模型,掌握简单运输问题数学模型的Lingo软件求解的方法,掌握解报告的内容。
二.实验内容:
1.用Lingo求解教材P94例1;
2.如果将销地B2的需要量改为10吨,其余不变,重新求解;
3.如果在例1中产地3的产量改为8,其余不变,重新求解。
三.实验要求:
1.写出数学模型;
2.在Lingo中输入求解的程序;
3.求解得到解报告;
4.写出最优解和最优值;
四.写出实验报告:
其数学模型为:
min=3x11+11x12+3x13+10x14+x21+9x22+2x23+8x24+7x31+4x32+10x33+5x34;
x11+x12+x13+x14=7;
x21+x22+x23+x24=4;
x31+x32+x33+x34=9;
x11+x21+x31=3;
x12+x22+x32=6;
x13+x23+x33=5;
x14+x24+x34=6;
其编程为:
model:
sets:
As/A1..A3/:
a;
Bs/B1..B4/:
b;
links(As,Bs):
c,x;
endsets
min=@sum(links(I,J):
C(I,J)*x(I,J));
@for(As(I):
@sum(Bs(J):
x(I,J))=a(I));
@for(Bs(J):
@sum(As(I):
x(I,J))=B(J));
data:
a=749;
b=3656;
c=311310
1928
74105;
enddata
end
其可行解报告为,
85.00000
7
X(A1,B1)0.0000000.000000
X(A1,B2)0.0000002.000000
X(A1,B3)5.0000000.000000
X(A1,B4)2.0000000.000000
X(A2,B1)3.0000000.000000
X(A2,B2)0.0000002.000000
X(A2,B3)0.0000001.000000
X(A2,B4)1.0000000.000000
X(A3,B1)0.0000009.000000
X(A3,B2)6.0000000.000000
X(A3,B3)0.00000012.00000
X(A3,B4)3.0000000.000000
185.00000-1.000000
20.000000-3.000000
40.0000002.000000
50.0000000.000000
60.000000-6.000000
70.0000000.000000
80.000000-7.000000
故总运费最小为85元。
2.
将销地B2的需要量改为10吨,其余不变,其编程为,
x(I,J))<
b(J));
b=31056;
82.00000
6
X(A1,B2)0.0000001.000000
X(A2,B2)0.0000001.000000
X(A3,B1)0.00000010.00000
X(A3,B2)9.0000000.000000
X(A3,B3)0.00000013.00000
X(A3,B4)0.0000001.000000
182.00000-1.000000
20.000000-10.00000
30.000000-8.000000
40.000000-4.000000
50.0000007.000000
61.0000000.000000
70.0000007.000000
83.0000000.000000
故最小运费是82.
产地3的产量改为8,其余不变,其编程为,
x(I,J))>
a(I));
x(I,J))=b(J));
a=748;
8
20.000000-5.000000
30.000000-3.000000
41.0000000.000000
50.0000002.000000
60.000000-4.000000
70.0000002.000000
80.000000-5.000000
实验三.大型线性规划模型的求解与编程
掌握求解大型线性规划模型Lingo软件的编程的基本方法。
1.在Lingo中编程求解教材P55习题2.2
(1)的线性规划数学模型;
2.用Lingo编程求解教材P52例12的数学模型。
3.建立教材P57习题2.9的数学模型并用Lingo编程求解。
1.给出所求解问题的数学模型;
2.给出Lingo中的编程程序;
3.能给出最优解和最优值;
4.指出哪些约束是取等式和哪些约束取不等式。
该线性规划的数学模型为:
minz=-3x1+4x2-2x3+5x4;
is/1..3/:
js/1..4/:
links(is,js):
min=@sum(js(J):
c(J)*x(J));
@sum(js(J):
a(1,J)*x(J))=b
(1);
a(2,J)*x(J))<
=b
(2);
a(3,J)*x(J))>
=b(3);
@free(x(4));
c=-34-25;
b=-2142;
a=4-12-1
113-1
-23-12;
enddata
其可行解报告为:
故最优解为x1=0,x2=8,x3=0,x4=-6,最优值为2.
其中第一个约束条件为等式,第二个约束条件和第三个约束条件为不等式。
2.
Minz=0x1+0.1x2+0.2x3+0.3x4+0.8x5;
x1+2x2+x4=100;
2x3+2x4+x5=100;
3x1+x2+2x3+3x5=100;
x1,x2,x3,x4,x5>
=0;
js/1..5/:
@for(is(I):
a(I,J)*x(J))=b(I));
c=00.10.20.30.8;
b=100100100;
a=12010
00221
31203;
16.00000
VariableValueReducedCost
X
(1)30.000000.000000
X
(2)10.000000.000000
X(3)0.0000000.000000
X(4)50.000000.000000
X(5)0.0000000.7400000
116.00000-1.000000
20.000000-0.6000000E-01
30.000000-0.1200000
40.0000000.2000000E-01
故最优解为x1=30,x2=10,x3=0,x4=50,x5=0,最优值为16.
其编程为,
!
Min=x1+x2+x3+x4+x5+x6;
=60;
=70;
=50;
=20;
is/1..6/:
js/1..6/:
a(I,J)*x(J))>
=b(I));
c=111111;
b=607060502030;
a=100001
110000
011000
001100
000110
000011;
X
(1)60.000000.000000
X(3)50.000000.000000
X(4)0.0000000.000000
X(5)30.000000.000000
X(6)0.0000000.000000