Hi,

I have the following condition:

\(x\) and \(y\) are continuous variables and \(\text{C}\) is a constant.

If \(\text{C} - 0.7x > 0\) then \(y = \min(\text{C},1.3x)\) else \(y = 0\).

In other words \(y = \min(\text{C}, 1.3x, \max(0, \text{M} \cdot \text{C} - 0.7x)) \)

And the objective is to maximize \(y\).

Is it possible to model this as a set of linear constraints without using a binary variable?

I want to stay in the LP regime and not create an MIP.

Thanks!

asked 13 Mar '15, 07:14

Rajeev%20N's gravatar image

Rajeev N
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accept rate: 0%

edited 13 Mar '15, 11:43

fbahr's gravatar image

fbahr ♦
4.6k716

It is important to say that y and in particular x are not restricted in sign, right? Otherwise always y < 0.7x. I don't think that this "either or" resp. "if then" can be formulated without a binary variable.

(13 Mar '15, 07:31) Marco Luebbecke ♦

Yes, I forgot to add that y>0. Otherwise it is possible

(13 Mar '15, 07:37) Rajeev N

Here's my solution using a binary variable delta -

maximize y

s.t.

C-0.7x >= M(delta -1)

y <= delta*C

y <= 1.3x

But I want to get rid of delta as it is slowing down my solver.

Thanks!

link

answered 13 Mar '15, 07:33

Rajeev%20N's gravatar image

Rajeev N
13129
accept rate: 0%

edited 13 Mar '15, 07:34

:-) then use a different solver; linear constraints alone will not help you here. Is this part of a larger linear program?

(13 Mar '15, 07:37) Marco Luebbecke ♦

:-) No, this is the complete formulation. I want this to converge in less than a minute with about 1000 x variables and y variables each, but of course, this is not getting there :-)

(13 Mar '15, 07:44) Rajeev N

Are you trying to solve 1000 independent problems as one big problem?

(13 Mar '15, 08:03) Rob Pratt

Forgot to add one more constraint -

x = a1z1 + a2z2 + a3*z3 where a1, a2, a3 are constants and z1, z2, z3 are decision variables. So these are not really 1000 independent problems

(16 Mar '15, 07:10) Rajeev N
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