数学实验--Lingo软件及其应用wergewywr5u7

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Lingo软件及其应用

班级 数学092 姓名李荣 学号 200912010206

(一)

1.求解整数规划

min

z 3x1 x2 3x3 3x4 x5 x6 3x7

3x1 3x2 2x3 x4 x5 50 x 2x x 3x 20 2456

x3 x5 2x7 15 xi 0

给出最优解及最优值。

程序:

model: , sets:

ba/1..7/:x,c,a1,a2,a3; bb/1,2,3/:b; endsets data:

c=3 1 3 3 1 1 3; b=50 20 15; a1=3 3 2 1 1 0 0; a2=0 1 0 2 1 3 0; a3=0 0 1 0 1 0 2; enddata

min=@sum(ba(I):c(I)*x(I)); @sum(ba(I):a1(I)*x(I))>=b(1); @sum(ba(I):a2(I)*x(I))>=b(2); @sum(ba(I):a3(I)*x(I))>=b(3); end

Global optimal solution found.

Objective value: 26.66667 Total solver iterations: 4

Row Slack or Surplus Dual Price 1 26.66667 -1.000000 2 0.000000 -0.3333333 3 6.666667 0.000000 4 0.000000 -0.6666667

2. 求解混合规划:

min

z 3x1 x2 3x3 3x4 x5

x2 x3 x4 x5 50 x x x x 20 1345 x1 x2 x4 x5 40

x1 x2 x3 x4 40 xi取整数i=1,2,3,4,5

2,3,4,5 0 xi 30i=1,

model: 1. sets:

ba/1..5/:x,c,a1,a2,a3,a4; bb/1,2,3,4/:b; endsets data:

c=3 1 3 3 1; b=50 20 40 40; enddata

min=@sum(ba(i):c(i)*x(i)); @for(bb(J):

@sum(ba(i)|i#ne#j:x(i))>=b(j); );

Global optimal solution found.

Objective value: 50.00000 Total solver iterations: 9 Variable Value Reduced Cost X( 1) 0.000000 3.000000 X( 2) 30.00000 0.000000 X( 3) 0.000000 2.000000 X( 4) 0.000000 2.000000 X( 5) 20.00000 0.000000 Row Slack or Surplus Dual Price

1 50.00000 -1.000000 2 0.000000 -1.000000 3 0.000000 0.000000 4 10.00000 0.000000 5 10.00000 0.000000

((二)

3. 求解混合规划

min

z cijxij

i 1j 1

34

x11 x12 x13 x14 30 2768

x x x x 40其中:c 7869 21222324

84210 x31 x32 x33 x34 50 x x x x 100j 1,2,3,4

2j3j4j

1j

2,3j 1,2,3,4 0 5 xiji=1,

(要求:必须利用集合语言描述,

a) 前三个约束用三个语句表达,

b) 第4-7的约束用sum和for用1个语句完成, c) 最后12个约束用1个语句完成 d) 目标函数用一个语句完成。)

e) 请将结果填入下面的表格中,不能用人工填写的方式完成(用ole函数完成)。

model: sets:

aa/1..3/:b1; bb/1..4/:b2; cc(aa,bb):x,c; endsets data:

c=2 7 6 8 7 8 6 9 8 4 2 10; b1=30 40 50 ;

b2=100 100 100 100 ;

@ole(E:\ddd.xls,x)=x;

enddata

min=@sum(cc(i,j):c(i,j)*x(i,j)); @for(aa(i):

@sum(bb(j):x(i,j))>=b1(i); );

@for(bb(j):

@sum(aa(i):x(i,j))>=b2(j); );

@for(cc(i,j): @gin(x(i,j)); ); End

100

0 0

4.

0 0 100

0 0 100

60 40 0

maxz 3x1 4x2 8x3 100y1 150y2 200y3 2x1 4x2 8x3 500

2x1 3x2 4x3 300

x1 2x2 3x3 1003x 5s.t. 1x2 7x3

700

x

1 200y1 x2 150y

x23 300y3 xj 0且为整数,j 1,2,3 yj 0或1,j 1,2,3

要求:1-4约束用一个语句完成,5-7用一个语句完成。 model: sets:

aa/1,2,3/:x,y,c1,c2,a1,a2,a3,a4,f; bb/1,2,3,4/:b; endsets data:

c1=3 4 8;

c2=-100 -150 -200; a1=2 4 8; a2=2 3 4; a3=1 2 3; a4=3 5 7;

f=200 150 700; enddata

max=@sum(aa(i):c1(i)*x(i)+c2(i)*y(i)); @sum(aa(i):a1(i)*x(i))<=b(1); @sum(aa(i):a2(i)*x(i))<=b(2); @sum(aa(i):a3(i)*x(i))<=b(3); @sum(aa(i):a4(i)*x(i))<=b(4); @for(aa(i):

x(i)<=f(i)*y(i); );

@for(aa(i): @gin(x(i)); @gin(x(i)); ); end

Variable Value

X( 1) 1.234568 X( 2) 1.234568 X( 3) 1.234568

Y( 1) 1.234568 Y( 2) 1.234568 Y( 3) 1.234568 Row Slack or Surplus

1 0.000000 2 1.234568 3 1.234568 4 1.234568 5 1.234568 6 1.234568 7 1.234568 8 1.234568

5.某小卖部6天卖出热茶的杯数(y)与当天气温(x)之间是线性相关的。数据如下表

model: sets:

c/1..6/:x,y; endsets data:

x=26 18 13 10 4 -1; y=20 24 34 38 50 64; enddata

min=@sum(c(i):(a*x(i)+b-y(i))^2); @free(a); @free(b); End

Local optimal solution found.

Objective value: 81.09091 Extended solver steps: 5 Total solver iterations: 20

Row Slack or Surplus Dual Price 1 81.09091 -1.000000

试用最小二乘法求出线性回归方程。(用集合语言描述)

6. 某公司有6个建筑工地要开工,每个工地的位置(用平面坐标系a,b表示,距离单位:km)及水泥日用量d(t)由下表给出.目前有两个临时料场位于A(5,1),B(2,7),日储量各有20t.假设从料场到工地之间均有直线道路相连.。

试制定每天的供应计划,即从A,B两料场分别向各工地运送多少水泥,可使总的吨千米数最小。

请建立模型并求解。(用集合语言描述); model:

sets:

a/1..6/:x,y,d,d1,d2; ab/1,2/:t; endsets data:

x=1.25 8.75 0.5 5.75 3 7.25; y=1.25 0.78 4.75 5 6.5 7.25; d=3 5 4 7 6 11; t=20 20; enddata

min=@sum(a(i):d1(i)*((x(i)-5)^2+(y(i)-1)^2)^(1/2)+d2(i)*((x(i)-2)^2+(y(i)-7)^2)^(1/2)); @sum(a(i):d1(i))<=t(1); @sum(a(i):d2(i))<=t(2); @for(a(i):d1(i)+d2(i)=d(i)); End

Global optimal solution found.

Objective value: 135.2722 Total solver iterations: 1

Row Slack or Surplus Dual Price 1 135.2722 -1.000000 2 4.000000 0.000000 3 0.000000 1.386716 4 0.000000 -3.758324 5 0.000000 -3.756448 6 0.000000 -4.090880 7 0.000000 -4.069705 8 0.000000 -2.504750 9 0.000000 -6.642665

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