运筹学软件lingo9.0下载（带附件）

mathtake

共享，有需要的下。求解线性规划最好用的软件。

mathtake

这次终于有了。

szr13132008

有没有license key？

mathtake

有，是破解版的。

371568491

就在这个网站上面有Lingo11的破解版，无限变量！！

majiancumt

lingo 与mathematica

majiancumt

lingo  教程

majiancumt

function y=mminterp(tab,col,val)
%MMINTERP 1-D Table Search by Linear Interpolation.
% Y=MMINTERP(TAB,COL,VAL) linearly interpolates the table
+ H: F% I4 b- I% {3 s6 a- G% TAB searching for the scalar value VAL in the column COL.
6 N- I; q* U( `# E9 T5 H% All crossings are found and TAB(:,COL) need not be monotonic.
% Each crossing is returned as a separate row in Y and Y has as
- A9 D5 u# j; l% many columns as TAB. Naturally, the column COL of Y contains
. f! j4 h5 V. T8 A/ X% the value VAL. If VAL is not found in the table, Y=[].

0 h. X. w& i. j+ {( J1 d% D.C. Hanselman, University of Maine, Orono ME,  04469
% 1/26/94
: f" p) M" ?( ?4 G  }% Copyright (c) 1996 by Prentice-Hall, Inc.

0 I0 O, \! ]* @. v* x/ d" K: Z[rt,ct]=size(tab);
if length(val)>1, error('VAL must be a scalar.'), end
9 Z: Z) r* W* O5 S9 {if col>ct|col<1,  error('Chosen column outside table width.'), end
if rt<2, error('Table too small or not oriented in columns.'), end
above=tab(:,col)>val;
below=tab(:,col)<val;
equal=tab(:,col)==val;
if all(above==0)|all(below==0), % handle simplest case
! ~# X; q7 ~" T. h        y=tab(find(equal),;
        return
end
pslope=find(below(1:rt-1)&above(2:rt)); %indices where slope is pos
nslope=find(below(2:rt)&above(1:rt-1)); %indices where slope is neg

) ^) v& f; M9 ^0 U, A0 S. P* J# |+ lib=sort([pslope;nslope+1]);        % put indices below in order
ia=sort([nslope;pslope+1]);        % put indices above in order
ie=find(equal);                                % indices where equal to val

0 {9 a( _& V) {  L4 w  i$ M& O[tmp,ix]=sort([ib;ie]);                % find where equals fit in result
2 L- W% X* {' ]ieq=ix>length(ib);                        % True where equals values fit
  y4 H  Y: i8 F/ c' K2 I  r2 Yry=length(tmp);                                % # of rows in result y
- n% T& H! Z# i/ V
; B" a3 @% M) L: n( s9 l: [y=zeros(ry,ct);                                % poke data into a zero matrix

4 J0 w4 \1 `& `) L  qalpha=(val-tab(ib,col))./(tab(ia,col)-tab(ib,col));
alpha=alpha(:,ones(1,ct));
+ l: s* t6 @. ~& w! O& R4 My(~ieq,=alpha.*tab(ia,+(1-alpha).*tab(ib,;        % interpolated values
. [: [% O# j( y" \: ~
y(ieq,=tab(ie,;                        % equal values
y(:,col)=val*ones(ry,1);        % remove roundoff error

majiancumt

function y=mminterp(tab,col,val)
& P6 @6 I* a7 M%MMINTERP 1-D Table Search by Linear Interpolation.
( h, D- b& f9 i( K% Y=MMINTERP(TAB,COL,VAL) linearly interpolates the table
1 ?$ U1 w+ s$ N9 H1 x, ?% TAB searching for the scalar value VAL in the column COL.
" x2 F* e0 I2 i# v1 i- Z9 n9 b- i% All crossings are found and TAB(:,COL) need not be monotonic.
6 h7 C/ b. A8 c% Each crossing is returned as a separate row in Y and Y has as
% many columns as TAB. Naturally, the column COL of Y contains
% the value VAL. If VAL is not found in the table, Y=[].
+ F/ r$ W% Q- n/ y$ ^0 \/ J
% D.C. Hanselman, University of Maine, Orono ME,  04469
5 u; F! _2 d' K5 W- Y/ W) Z6 q% 1/26/94
0 F/ w3 h( \5 N% i0 }- f& H% Copyright (c) 1996 by Prentice-Hall, Inc.
: N. \0 I6 U% [& d) t
1 W; o( }; j  @) ]. `" l' b[rt,ct]=size(tab);
% X, l8 Y$ U9 G! H4 bif length(val)>1, error('VAL must be a scalar.'), end
* b& d% }( O( Z3 M6 X3 wif col>ct|col<1,  error('Chosen column outside table width.'), end
if rt<2, error('Table too small or not oriented in columns.'), end
above=tab(:,col)>val;
below=tab(:,col)<val;
equal=tab(:,col)==val;
if all(above==0)|all(below==0), % handle simplest case
0 G: b4 v- f' F: ~: T. \        y=tab(find(equal),;
        return
end
pslope=find(below(1:rt-1)&above(2:rt)); %indices where slope is pos
nslope=find(below(2:rt)&above(1:rt-1)); %indices where slope is neg

ib=sort([pslope;nslope+1]);        % put indices below in order
ia=sort([nslope;pslope+1]);        % put indices above in order
ie=find(equal);                                % indices where equal to val

0 e/ z* o5 A6 Y# w[tmp,ix]=sort([ib;ie]);                % find where equals fit in result
ieq=ix>length(ib);                        % True where equals values fit
' \" u# E2 O: A4 d! V' Lry=length(tmp);                                % # of rows in result y
$ F% C1 S/ F. j3 ]# Z
y=zeros(ry,ct);                                % poke data into a zero matrix
  G. R7 a7 p3 {, j# {, F9 D5 D  T) h
+ M1 v  }  z2 s9 Nalpha=(val-tab(ib,col))./(tab(ia,col)-tab(ib,col));
alpha=alpha(:,ones(1,ct));
y(~ieq,=alpha.*tab(ia,+(1-alpha).*tab(ib,;        % interpolated values
! Q6 R! K( p5 N  b& w
, U+ T3 @7 i8 \$ }( m8 Iy(ieq,=tab(ie,;                        % equal values
y(:,col)=val*ones(ry,1);        % remove roundoff error

majiancumt

lingo 教程