Is it possible to write the robust form of equality constraints in linear programming models? Or it is possible only for inequality constraints?
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What do you call robust form? – gt6989b Nov 06 '15 at 03:47
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for example Ben-tal and Nemirovski or Soyster Method. – user223276 Nov 06 '15 at 07:00
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It only makes sense to talk about uncertain equalities if you have products between the uncertainty and the decision variables and thus have the possibility to zero out the uncertain terms suitably.
As an example, let $x$ be decision variable and $w$ the uncertainty, and study the equality $x_1 + x_2 = w ~\forall ~w\in\mathcal{W}$. This makes makes no sense since there is no way to assign a fixed value to $x$ which holds for any $w$. The equation $x_1 +x_2 = x_3 w ~\forall ~w\in\mathcal{W}$ is reasonable though, since a solution is given by $x_3=0$ and any pair satisfying $x_1+x_2=0$
A typical mistake in robust modeling is to read assignment as equality. The first constraint was perhaps meant to be $x_2$ is assigned the value $-x_1+w$. In other words, the expression $x_2$ is not a decision variable, but an affine function of the decision variable $x_1$ and uncertainty $w$. This mistake is very common in modelling of uncertain discrete-time systems, and similar problems with some kind of causality or direction.
Another point-of-view is perhaps that you want $x_1$ and $x_2$ to be functions of $w$ (policies) and not fixed decision variables. In the first case, you might thus have a linear policy $x_1 = c_0 + c_1 w$ and $x_2 = d_0+d_1w$, where $c$ and $d$ are decision variables satisfying $c_0+d_0=0$ and $c_1+d_1 = 1$.
Johan Löfberg
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Dear Johan, many thanks for your answer. I'm trying to write the robust form of following constraint: ax+by=c, in two situations: 1) when "a" and "b" are uncertain and "c" is certain and 2) when "a","b" and "c" are uncertain.All methods that I know are suitable when we have >= or <= equations and I don't know which method is appropriate for my problem. – user223276 Nov 07 '15 at 15:28
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The reason you don't find any methods is because your model doesn't make sense. $ax+by$ is not a fixed value, it is in the most easiest example an interval. You want that interval to be equal to the constant value $c$. You obviously cannot, unless $c$ is zero which you can solve by $x=y=0$. – Johan Löfberg Nov 07 '15 at 21:54
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@JohanLöfberg: but you could replace $ax+by=c$ by $f(x)=c$ where $f$ is an unknown perturbation of the function $ax+by$ in the max-norm, for example. – Peter Franek Dec 16 '15 at 16:51
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