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! 
Here's my solution using a binary variable delta  maximize y s.t. C0.7x >= 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! answered 13 Mar '15, 07:33 Rajeev N :) 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

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