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RE: ConstrainedMin with negative variables

  • To: mathgroup at smc.vnet.net
  • Subject: [mg35148] RE: [mg35137] ConstrainedMin with negative variables
  • From: "DrBob" <majort at cox-internet.com>
  • Date: Wed, 26 Jun 2002 01:09:23 -0400 (EDT)
  • Reply-to: <drbob at bigfoot.com>
  • Sender: owner-wri-mathgroup at wolfram.com

It's usual to constrain LP variables to be non-negative because it
simplifies the algorithm immensely and reduces the incidence of
unbounded problems.  It's not a real limitation, however, as the
following demonstrates.  First the result you obtained:

objective = -7/3 x1 - 13/3 x2;
constraints = {x1 + 5 x2 <= 0, -1 <= x1 <= 1, -1 <= x2 <= 1};
vars = Variables[objective];
ConstrainedMin[objective, constraints, vars]

{0, {x1 -> 0, x2 -> 0}}

Here's the substitution that generalizes it for you, and a solution in
terms of the new variables:

substitute = {x1 -> y1 - z1, x2 -> y2 - z2};
newVars = Variables[objective /. substitute];
ConstrainedMin[objective /. substitute, constraints /. substitute,
newVars];
solution = Last[%]

{y1 -> 1, y2 -> 0, z1 -> 0, z2 -> 1/5}

Finally, in terms of the original variables:

vars /. substitute /. solution

{1, -(1/5)}

That's the solution you wanted.

Bobby Treat

-----Original Message-----
From: Markus Kolöchter [mailto:koloechter at web.de] 
To: mathgroup at smc.vnet.net
Subject: [mg35148] [mg35137] ConstrainedMin with negative variables

Hello ng,

I'm trying to implement the method of Zoutendijk, which is to solve a
nonlinear programming problem.
Therefor I have to solve linear subproblems with the simplex method.
But the Mathematica function "ConstrainedMin" assumes that all variables
are
non-negative.

For example, I have the following input:

    ConstrainedMin[-7/3 x1 - 13/3 x2, {x1 + 5 x2 <= 0, -1 <= x1 <= 1, -1
<=
x2 <= 1}, {x1, x2}]

The Mathematica output is:

    {0, {x1 -> 0, x2 -> 0}}

I think the proper result should be {-22/15, {x1 -> 1, x2 -> -1/5}}.

If anyone has an idea, please let me know!

Regards,

Markus Kolöchter







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