Re: Problem using linprog

Hi Stephen,

The problem appears to be singular around the solution. A very quick
exploration shows me that if you replace your upper bound b by a very small
epsilon > 0, you get a stable result.

For example:
b = np.zeros(8) + 0.001

 fun: -0.11764011575264395
 message: 'Optimization terminated successfully.'
 nit: 6
 slack: array([0.        , 0.        , 0.40742577, 0.        ,
0.40742577, 0.        , 0.        , 0.        , 0.88235988])
 status: 0
 success: True
 x: array([0.        , 0.        , 0.        , 0.0834722 , 0.41811509,
0.11764012])

And for print(np.dot(A, result.x)) I get [ 0.001 0.001 -0.00307426 0.001
-0.00307426 0.001 0.001 0.001 ]

In the objective function, y_2 = y_4 = -3.0742577 * epsilon, and the other
6 values also converge towards zero when epsilon -> 0 .

If I read your problem correctly, your objective function is simply (-1)
times x_5, the last element of x. The approach above would converge towards
the trivial solution, x = 0, but your solution above minimizes f(x) by
maximizing x_5 at 1. If we pick out an x_5, then the problem collapses to a
new problems to find [x_0 ... x_4] so that A[:, 0:7] * [x_0 ... x_4]' < b,
where b is (-1) * the last column of your A. But the objective function is
now indeterminate, so there is nothing to optimize.

HTH,

Chris

On Mon, Nov 26, 2018 at 11:43 AM Montgomery-Smith, Stephen <
[email protected]> wrote:

> I am trying to solve a linear programming problem.  The constraint is of
> the form A.x <= 0.  But linprog gives an answer that doesn't satisfy the
> constraint.
>
> The attached program gives A.x as
>
> [-2.32109228  2.32017594  4.71436317  3.6433767  -4.26629574  2.32384597
>  -1.96166184 -4.96206197]
>
> which definitely doesn't satisfy the constraint.  Is this a bug, or some
> subtle floating point error?
>
> Program follows (also as attachment):
>
> from scipy.optimize import linprog
> import numpy as np
>
> A = [[0.5919650431077654, -0.5271408402306996, 0.6096719792636803,
> 1.2379670854947114, 0.2656040423387233, -0.972363043155988],
> [-0.5914974900295467, -0.5266568950860249, 0.6105433925177587,
> 1.258297461476007, -0.285688537323182, 0.9726089241528251],
> [-0.593015674004932, 0.5280764198909397, 0.6078385518701857,
> -1.1964319796886902, -0.2223431679788034, -0.9740888117098865],
> [0.5935986604093653, 0.5285277328950352, 0.6068764832493029,
> -1.1752312553140132, 0.19916734259906424, 0.976063912714949],
> [0.593015674004932, -0.5280764198909397, -0.6078385518701857,
> -1.1964319796886902, -0.2223431679788034, -0.9740888117098865],
> [-0.5935986604093653, -0.5285277328950352, -0.6068764832493029,
> -1.1752312553140132, 0.19916734259906424, 0.976063912714949],
> [-0.5919650431077654, 0.5271408402306996, -0.6096719792636803,
> 1.2379670854947114, 0.2656040423387233, -0.972363043155988],
> [0.5914974900295467, 0.5266568950860249, -0.6105433925177587,
> 1.258297461476007, -0.285688537323182, 0.9726089241528251]]
> e = [0, 0, 0, 0, 0, -1]
> bounds = [(None, None), (None, None), (None, None), (None, None), (None,
> None), (0, 1)]
> b = [0]*len(A)
> result = linprog(e, A_ub = A, b_ub = b, bounds = bounds)
> print np.matmul(A, result.x)
>
> _______________________________________________
> SciPy-User mailing list
> [email protected]
> https://mail.python.org/mailman/listinfo/scipy-user
>


-- 
Chris Waigl . [email protected] . [email protected]
http://eggcorns.lascribe.net . http://chryss.eu

_______________________________________________
SciPy-User mailing list
[email protected]
https://mail.python.org/mailman/listinfo/scipy-user