运筹学第三版（刁在钧）光盘中的内容

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第二章 线性规划

本章, 我们介绍三种解决线性规划问题的软件:

第一种: MATLAB软件中的optimization toolbox中的若干程序;

第二种: LINDO软件;

第三种: LINGO软件.

1. MATLAB程序说明程序名: lprogram执行实例：

file:///C:/DOCUME~1/ADMINI~1/LOCALS~1/Temp/msohtml1/01/clip_image002.gif

在命令窗口的程序执行过程和结果如下：

the program is with the linear programming

Please input the constraints number of the linear programming m=7

m =7

Please input the variant number of the linear programming n=4

n =4

Please input cost array of the objective function c(n)_T=[-2,-1,3,-5]'

c =
-2

-1

8 b# a1 m1 N9 g$ F! S' [2 ~3

" v  B, ?+ s6 c! F+ [, o% i" A" t-5

Please input the coefficient matrix of the constraints A(m,n)=[1,2,4,-1;2,3,-1,1;

1,0,1,1;-1,0,0,0;0,-1,0,0;0,0,-1,0;0,0,0,-1]

A =
1
; S6 |7 D- m- v2 _* s5 H& O) e2
. y' h# j; r& w" n1 g4 P% g4
-1

; Z' g1 D; w2 z5 R: k" ~: P3 g2
3
-1
1

1
% k( r' _0 z( J0
1
1

-1
) U8 e( ^) h$ X' e0
2 Q- g8 L+ l3 h/ r! f( ^0
0

0
-1
0
0

0
/ K0 L; w6 w4 l) Z3 N1 P& R0
* _" ~6 b- d! z4 J( x! G  i-1
- q$ C  h0 W0 _3 D4 J0

0
0
0 Z; P+ ~5 x. N; a7 S/ L1 ?0
* N7 s1 f( @2 U5 w; T. E0 P0 p-1

Please input the resource array of the program b(m)_T=[6,12,4,0,0,0,0]'

b =
6

12

4

0 F/ j( O! M! X. ^1 E8 ?0

0

$ R$ N: R2 T, o/ B0

0

Optimization terminated successfully.

The optimization solution of the programming is:

x =
0.0000

2.6667

-0.0000

) ]' Q; c( n9 W7 z' d6 c4.0000

The optimization value of the programming is:

opt_value = -22.6667

注: 红色字表示计算机的输出结果.

程序的相关知识:

Solve a linear programming problem

file:///C:/DOCUME~1/ADMINI~1/LOCALS~1/Temp/msohtml1/01/clip_image003.gif

where f, x, b, beq, lb, and ub are vectors and A and Aeq are matrices.

相关的语法:

x = linprog(f,A,b,Aeq,beq)

x = linprog(f,A,b,Aeq,beq,lb,ub)

x = linprog(f,A,b,Aeq,beq,lb,ub,x0)

x = linprog(f,A,b,Aeq,beq,lb,ub,x0,options)

[x,fval] = linprog(...)

[x,fval,exitflag] = linprog(...)

[x,fval,exitflag,output] = linprog(...)

[x,fval,exitflag,output,lambda] = linprog(...)

解释:

linprog solves linear programming problems.

x = linprog(f,A,b) solves min f'*x such that A*x <= b.

x = linprog(f,A,b,Aeq,beq) solves the problem above while additionally satisfying the equality constraints Aeq*x = beq. Set A=[] and b=[] if no inequalities exist.

x = linprog(f,A,b,Aeq,beq,lb,ub) defines a set of lower and upper bounds on the design variables, x, so that the solution is always in the range lb <= x <= ub. Set Aeq=[] and beq=[] if no equalities exist.

x = linprog(f,A,b,Aeq,beq,lb,ub,x0) sets the starting point to x0. This option is only available with the medium-scale algorithm (the LargeScale option is set to 'off' using optimset). The default large-scale algorithm and the **x algorithm ignore any starting point.

x = linprog(f,A,b,Aeq,beq,lb,ub,x0,options) minimizes with the optimization options specified in the structure options. Use optimset to set these options.

[x,fval] = linprog(...) returns the value of the objective function fun at the solution x: fval = f'*x.

[x,lambda,exitflag] = linprog(...) returns a value exitflag that describes the exit condition.

[x,lambda,exitflag,output] = linprog(...) returns a structure output that contains information about the optimization.

[x,fval,exitflag,output,lambda] = linprog(...) returns a structure lambda whose fields contain the Lagrange multipliers at the solution x.

2．LINDO 程序说明程序名：linear执行实例：

file:///C:/DOCUME~1/ADMINI~1/LOCALS~1/Temp/msohtml1/01/clip_image005.gif

在命令窗口键入以下内容：

max 10x+15y ！也可以直接解决min问题

subject to

x<10

y<12

x+2y<16

end
！注释符号; 系统默认为自变量>0, 若不要求用free命令.

% ~; c: @2 Z2 Y/ `  |) o！在出来report windows之前可选择显示对此规划进行灵敏度分析等

按solve键, 在reports window中出现以下内容:

LP OPTIMUM FOUND AT STEP
2

OBJECTIVE FUNCTION VALUE

1)
7 o" H/ b. Z6 l5 U. s2 j% ]145.0000

* P4 {6 L3 w5 V- G( h/ ~VARIABLE
+ Q& g% C5 e, O: rVALUE
, f- {# f% x# W; Q' j8 AREDUCED COST

7 N. A, a& V' i4 F& b* |X
10.000000
; s3 }% ^% l: p% A/ c0.000000

Y
3.000000
0.000000

# H7 X4 N5 \" `( ]ROW
9 G! u* T5 p* F% p5 t. a; |SLACK OR SURPLUS
# ?8 l, ]: H8 g) m, h  KDUAL PRICES

1 c9 @* q" K" {, f9 U2)
0.000000
2.500000

( v: |3 E  \) W3)
7 b/ D6 B3 A. r7 `# p  R; {9.000000
. f8 y% u8 a  X0.000000

4)
0.000000
7.500000

NO. ITERATIONS=
2

RANGES IN WHICH THE BASIS IS UNCHANGED:

OBJ COEFFICIENT RANGES

& L# V/ M9 }+ c, aVARIABLE
CURRENT
ALLOWABLE
ALLOWABLE

# g  ]" K- W/ D6 DCOEF
+ }( B; m9 a( D9 xINCREASE
4 V/ X& F9 v7 A. D5 D5 vDECREASE

X
10.000000
INFINITY
) U' f' F: O4 S* i" v& U4 U2.500000

Y
9 f' [9 i0 A7 h; B" ~! G% ^15.000000
6 p. a! A8 O( e2 o* F" q$ t5.000000
9 a) F; W- L+ A* {" a15.000000

  l0 ~: W% Z9 pRIGHTHAND SIDE RANGES

ROW
CURRENT
+ @" K4 @# T% T; k; Y! XALLOWABLE
ALLOWABLE

" {, D% t8 Z) N5 }/ h5 cRHS
INCREASE
DECREASE

2
9 Y) u" S5 h5 M$ m10.000000
6.000000
2 n1 b' v: g; q9 Q, r- a' F5 W3 R* P10.000000

4 q# t, H( N& o4 @+ R8 M3
12.000000
INFINITY
6 U/ @% e+ T. `2 K  B8 a+ Z9.000000

8 k) N: x. C5 T4
" l8 W) e% N4 J5 r& }7 Y16.000000
18.000000
6.000000

3．LINGO 程序说明3.1 程序名: linearp1（求极小问题）linearp1运行实例:

file:///C:/DOCUME~1/ADMINI~1/LOCALS~1/Temp/msohtml1/01/clip_image007.gif

在model window中输入以下语句：

min=5*x1+21*x3;

x1-x2+6*x3-x4=2;

x1+x2+2*x3-x5=1;

按运行按钮在solution
: G1 m* J* m* Z  kreport 窗口得到以下结果：

8 y  H! L; l% k0 z# bGlobal optimal solution found at iteration:
2

Objective value:
7 r, F( B* S7 D4 l+ E3 E7.750000

- o4 N- Y( P. h3 I8 F. Z$ [Variable
Value
9 O- k! H+ i8 GReduced Cost

* ]2 E% f0 e: b4 F( Y( g; ]X1
1 ?! F3 d! r; R1 x0.5000000
0.000000

; a0 z" O1 E' U1 IX3
7 n# T7 V8 X) P7 |$ u! B* G0.2500000
! }( n7 C6 }# z% P0.000000

5 B( _" i7 N9 \, i
X2
0.000000
0.5000000

X4
% ~: p' m% Y5 q# ~% n4 J% Y0.000000
6 d1 ^% y+ i" `; X+ e! w; v: K+ m8 Q2.750000

( \5 K- i$ w2 f8 ]+ h2 A" v/ I  RX5
! B2 R  V/ G( W( n1 C8 A1 b0.000000
; ~! `) ?/ Y/ Z2.250000

2 a* q' y, M, E1 A; l/ tRow
Slack or Surplus
2 j. Q7 R& {! A  JDual Price

1
( d# x% d$ c2 y! S( D5 F" e7.750000
! t8 F, j5 Z' P- V9 k0 Z: F2 a-1.000000

2
0 O* s6 O" ]/ j( S, @/ |0 _0.000000
-2.750000

2 J1 a8 m; ?& w9 E( l2 \8 r( v1 }3
0.000000
-2.250000

3.2 程序名: linearp2（求极大问题）linearp2运行实例：

file:///C:/DOCUME~1/ADMINI~1/LOCALS~1/Temp/msohtml1/01/clip_image009.gif

在model window中输入以下语句：

max=100*x+150*y;
! this is a commnent;

x<=100;

y<=120;

x+2*y<=160;

按运行按钮在solution report 窗口得到以下结果：

  Global optimal solution found at iteration:
+ `5 s1 j8 a9 u- Z2

Objective value:

# e# }& V  `; |' X14500.00

Variable
& p- g. \+ c9 _, jValue
Reduced Cost

+ a0 j! s1 \7 x0 j9 {8 Y% k2 TX
100.0000
* O. q2 S8 l. _( F0.000000

Y
; C& U+ \7 K  T! R& k+ M30.00000
  H) }# Y  ]+ Q' c0.000000

Row
Slack or Surplus
Dual Price

5 e- o" v1 x/ Q6 P5 e+ c1
14500.00
1.000000

+ F% X" K; ~, t, Q2
6 B2 U  U. V5 H+ v6 W0.000000
- [% T8 a" l/ s& _( }2 y25.00000

3
' i! j/ G( l  C  Y2 e( G+ z- b90.00000
! _% `" J" g2 {: j9 b: }; ]! q( y0.000000

4
0.000000

75.00000

loooog12

先下下来看看