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Solve very simple linear program

asked 2011-04-27 17:12:51 +0200

kcrisman gravatar image

I have tried to figure out the docs at the optimization docs and cannot for the life of me figure out how to program this very simple LP. The problem is that everything has to be written as a minimization, and figuring out what should be minus is a real headache - especially when it doesn't work as expected!

This is in an interact.

def _(M = matrix(RR,[[1,-1],[-1,1]]),auto_update=False):
    f1(p) = (M[1,0] - M[0,0])*p+M[0,0]
    f2(p) = (M[1,1] - M[0,1])*p+M[0,1]
#    sol1 = find_maximum_on_interval(lambda y: min_symbolic(f1(y),f2(y)),0,1)[0]
    sol1 = linear_program(vector([0,1]),matrix([[-1,0],[1,0],[M[1,0] - M[0,0],-1],[M[1,1] - M[0,1],-1]]),vector([0,1,-M[0,0],-M[0,1]]))['x'][0]
    print sol1

E.g., M=[[1,2],[0,-1]] is just trying to maximize $y$ for below the lines $1-p$ and $2-3p$. (This is a basic zero-sum game I'm trying to demonstrate the solution of, for context.)

For the matrix [[1,2],[0,-1]], it should give p=0, but instead gives the minimum of the opposite.

It would be nice to not have to do quite as much work. Also, the find_maximum_on_interval function doesn't seem to give what I want here either.

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answered 2011-04-27 23:05:56 +0200

chaesloc gravatar image

updated 2011-04-27 23:10:01 +0200

You program is minimizing y above those lines. The problem is not unbounded only because of the additional constraints 0 < p < 1 given by the top 2 rows of G (=see below).

Using the notation in the article you linked, linear_program(c,G,h) minimizes c * x given G * x < h. You have c=[0,1] and x=[p,y], so you are minimizing y. You want c=[0,-1].

You have G * x = [-p,p,-p-y,-3p-y] < h, which gives y>-h[2]-p, y>-h[3]-3p. You want G*x=[-p,p,p+y,3p+y] and h=[0,1,1,2].

So you want c=[0,-1]; G=[[-1,0],[1,0],[1,1],[3,1]]; h=[0,1,1,2]. As for M, I have no clue what it is.

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Thanks much; I had already realized I needed to switch c, but for some reason had my inequalities goofed and was not switching the [1,1],[3,1] from my original [-1,-1],[-3,-1]. Don't worry about M - I know what it is :)

kcrisman gravatar imagekcrisman ( 2011-04-28 00:40:12 +0200 )edit

answered 2011-04-28 14:00:44 +0200

Nathann gravatar image

You may find the use of the MixedIntegerLinearProgram class easier :


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Well, but I wasn't looking for integer answers! Looks like that (floating-point) is supported as well, but it's sometimes hard to tell from this doc which is which in solving. Thanks!

kcrisman gravatar imagekcrisman ( 2011-04-28 17:03:06 +0200 )edit

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Asked: 2011-04-27 17:12:51 +0200

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Last updated: Apr 28 '11