This presentation is the property of its rightful owner.
Sponsored Links
1 / 30

优化模型 PowerPoint PPT Presentation


  • 120 Views
  • Uploaded on
  • Presentation posted in: General

优化模型. 教学目的: 初步认识优化模型的基本形式及掌握线性规划模型的建模及求解。 通过实例建模并求解 , 熟练掌握一些数学软件的使用。. 教学内容 : 简单介绍优化模型的基本概念和基本类型。 重点介绍优化模型中的线性规划模型。 线性规划模型建模实例及求解的实现。 布置本次课的练习与上机实验内容。.

Download Presentation

优化模型

An Image/Link below is provided (as is) to download presentation

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.


- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

Presentation Transcript


5588241

  • ,

  • :


5588241

1. ,


5588241

  • (1)

  • 2

  • 3


5588241

opt (opt(optimize)) s.t. () ()

Linear Programming,LP(NLP)


5588241

2.


5588241

3.1

3. ()


5588241

3.2


5588241

3.3 (1)2 (2)EXCELSolver,Matlab,LINDO/LINGO


5588241

3.41 4504,150.280100.0545

  • 1. x1 x22. Max f=80 x1+45 x23. 0.2 x1 +0.05 x2 4 15 x1 +10 x2 450


5588241

. x1, x2,()max f=80x1+45x20.2x1+0.05x24 15x1+10x2 450,x1 0, x2 0,


5588241

x2

0.2x1+0.05x2=4

x1

15x1+10x2=450

12


5588241

(),(()).


2 excel solver excel sumproduct excel

2EXCELSolverEXCELSUMPRODUCTEXCEL


5588241


5588241

3Matlab------- lp MinSub.to: Ax=b


5588241

max f=80 X1+45 X2

sub.to

0.2 X1+0.05 X24 15 X1+10 X2 450

X1 0, X2 0

min f=- 80 X1- 45 X2sub.to

0.2 X1+0.05 X24 15 X1+10 X2 450 X1 0, X2 0


5588241

c=[-80,-45];a=[0.2,0.05;15,10];b=[4,450];vlb=[0,0];vub=[];[x,lam]=lp(c,a,b,vlb,vub) vlb,vubx = 14.0000 24.0000lam = 100.0000 4.0000 0 0xlam


4 lindo lingo lp 4 1

4LINDO/LINGOLP41

41

LINDO SolveSolve(Ctrsl+S)LINDO42LINDOObjective2200


5588241

42


5588241

43

LP42

43N42LINDOWindow,Reports WindowWindowReports Window44


5588241

44

LP OPTIMUM FOUND AT STEP 2

OBJECTIVE FUNCTION VALUE 1) 2200.0002200.000LINDO11


5588241

VALUEX1=14.000000X2=24.000000 SLACK OR SURPLUS230 DUAL PRICESNO. ITERATIONS=22 13 5080905


3 lindo 1 gin general integer end 3 1

3LINDO1GINgeneral integerEND31)

31


5588241

4.1122(),.20000,8,15/12000,.,?


5588241

[1] http://www.mathrs.net/data/upload/shuxuejianmo/kejian/14.ppt#3

[2] .[M].

.20055.

[3] .LINDO/LINGO[M].

.2005,7.

[4] . MATLAB 5.X[M]..2000,5.

[5] .[M]..2005,12.

[6] http://www.mathrs.net/data/upload/yunchouxuejiangzuo.ppt


  • Login