lpsolve - unfeasible solution, but I have example of 1

问题

I'm trying to solve this in LPSolve IDE:

/* Objective function */
min: x + y;

/* Variable bounds */
r_1: 2x = 2y;
r_2: x + y = 1.11 x y;
r_3: x >= 1;
r_4: y >= 1;

but the response I get is:

Model name:  'LPSolver' - run #1
Objective:   Minimize(R0)

SUBMITTED
Model size:        4 constraints,       2 variables,            5 non-zeros.
Sets:                                   0 GUB,                  0 SOS.

Using DUAL simplex for phase 1 and PRIMAL simplex for phase 2.
The primal and dual simplex pricing strategy set to 'Devex'.

The model is INFEASIBLE
lp_solve unsuccessful after 2 iter and a last best value of 1e+030

How come this can happen when x=1.801801802 and y=1.801801802 are possible solutions here?

回答1:

How To Find The Solution

Let's do some math.

Your problem is:

min x+y
s.t. 2x = 2y
     x + y = 1.11 x y
     x >= 1
     y >= 1

The first constraint 2x = 2y can be simplified to x=y. We now substitute throughout the problem:

min 2*x
s.t. 2*x = 1.11 x^2
     x >= 1

And rearrange:

min 2*x
s.t. 1.11 x^2-2*x=0
     x >= 1

From geometry we know that 1.11 x^2-2*x makes an upward-opening parabola with a minimum less than zero. Therefore, there are exactly two points. These are given by the quadratic equation: 200/111 and 0.

Only one of these satisfies the second constraint: 200/111.

Why Can't I Find This Constraint With My Solver

The easy way out is to say it's because the x^2 term (x*y before the substitution is nonlinear). But it goes a little deeper than that. Nonlinear problems can be easy to solve as long as they are convex. A convex problem is one whose constraints form a single, contiguous space such that any line drawn between two points in the space stays within the boundaries of the space.

Your problem is not convex. The constraint 1.11 x^2-2*x=0 defines an infinite number of points. No two of these points can be connected by a straight line which stays in the space defined by the constraint because that space is curved. If the constraint were instead 1.11 x^2-2*x<=0 then the space would be convex because all points could be connected with straight lines that stay in its interior.

Nonconvex problems are part of a broader class of problems called NP-Hard. This means that there is not (and perhaps cannot) be any easy way of solving the problem. We have to be smart.

Solvers that can handle mixed-integer programming (MIP/MILP) can solve many non-convex problems efficiently, as can other techniques such as genetic algorithms. But, beneath the hood, these techniques all rely on glorified guess-and-check.

So your solver fails because the problem is nonconvex and your solver is neither smart enough to use MIP to guess-and-check its way to a solution nor smart enough to use the quadratic equation.

How Then Can I Solve The Problem?

In this particular instance, we are able to use mathematics to quickly find a solution because, although the problem is nonconvex, it is part of a class of special cases. Deep thinking by mathematicians has given us a simple way of handling this class.

But consider a few generalizations of the problem:

(a) a x^3+b x^2+c x+d=0
(b) a x^4+b x^3+c x^2+d x+e =0
(c) a x^5+b x^4+c x^3+d x^2+e x+f=0

(a) has three potential solutions which must be checked (exact solutions are tricky), (b) has four (trickier), and (c) has five. The formulas for (a) and (b) are much more complex than the quadratic formula and mathematicians have shown that there is no formula for (c) that can be expressed using "elementary operations". Instead, we have to resort to glorified guess-and-check.

So the techniques we used to solve your problem don't generalize very well. This is what it means to live in the realm of the nonconvex and NP-hard, and it's a good reason to fund research in mathematics, computer science, and related fields.

来源：https://stackoverflow.com/questions/57091654/lpsolve-unfeasible-solution-but-i-have-example-of-1

标签

linear-programming

lpsolve