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## Re: Linear Programing

**From**: |
Etienne Grossmann |

**Subject**: |
Re: Linear Programing |

**Date**: |
Fri, 24 Nov 2000 08:05:19 +0000 |

**User-agent**: |
WEMI/1.13.7 (Shimada) FLIM/1.13.2 (Kasanui) Emacs/20.7 (i386-debian-linux-gnu) (with unibyte mode) |

Hello,
Jeff's program does
max f'*x subject to A*x == b and x >= 0
x
And you want :
LP Linear programming.
X=LP(f,A,b) solves the linear programming problem:
min f'x subject to: Ax <= b
x
I think you can transform your problem into a problem that Jeff's
program can solve. The reasoning (no warranty of correctness : I would
not at all be surprised if there were some pitfall I fell in) is :
Your problem is equivalent to (set y = A*x - b, y should be >= 0)
max f'*x subject to A*x - y == b and x >= 0 and y>=0
x
which is equiv. to (rewrite in vectorial form)
max [f',0]*[x;y] subject to [A,-eye]*[x;y] == b and [x;y]>=0
x
equiv. to
(set x = x1 - x2 with x1 >= 0 and x2 >= 0, idem for y = y1 - y2)
max [f',-f',0,0]*[x1;x2;y1;y2]
x
subject to [A,-A,-eye,-eye]*[x1;x2;y1;y2] == b
and [x1;x2;y1;y2]>=0
Which is the kind of problem that Jeff's function solves.
Etienne
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