# How to linearise a non-linear constraint?

 0 Hi, I am working on an optimisation model where some of the constraints (the way I see it) have to be non-linear. The constraints are all in this form: Ax - yz <= 0 where A is a set of constant parameters, z is a continuous and non-negative variable, and x and y are binary variables. Help would be much appreciated, -E asked 28 Apr '12, 07:48 Schmeivind 1●1●1 accept rate: 0% yeesian 846●2●10 I just noticed that I had mixed up the variables. Some more info on the bounds and their interdependencies is perhaps also useful. Here is the correct problem: z and x are both continuous, non-negative variables, where x's lower bound is 0, upper bound is reliant on A and z. z has a lower and an upper bound that follows from the corresponding value of A and x. A is a set of parameters. Thanks to the person below. It helped quite a bit in terms of how to the problem, but the bounds are still quite unclear to me. (28 Apr '12, 08:26) Schmeivind First, please note that depending on problem definition, having lower and upper bounds of z would perhaps just result in less solution time and less numerical instability. Therefore, it's not always necessary to compute the lower and upper bounds of variables. Second, if lower and upper bounds of z depends on value of x, then I think you should write a separate constraint for this dependency. If this dependency has a special structure, you might be able to compute the lower and upper bounds of z. Otherwise, maybe you should just use the big-M as your upper bound. (28 Apr '12, 11:08) Ehsan ♦

 2 You might see here for a discussion of both cases where (z) is bounded or not. answered 28 Apr '12, 08:03 Ehsan ♦ 4.8k●3●12●24 accept rate: 16%
 2 You can take a look at this Integer Programming Tricks pdf provided by @fbahr in this thread. answered 28 Apr '12, 10:33 kunigami 93●1●7 accept rate: 0%
 1 You can find an alternative description how to transform a product of variables into a system of linear inequations here beginning at page 43. Cheers, Mike answered 12 Oct '12, 06:15 stegger 41●2 accept rate: 0% Ehsan ♦ 4.8k●3●12●24
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