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According to Bertsimas' text, the standard form of a LP problem is:


According to Vanderbei's text, the standard form of a LP problem is:


So, what is the standard form of a linear programming (LP) problem? Thanks.

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Bertsimas' text refers to Introduction to Linear Optimization. Vanderbei's text refers to Linear Programming: Foundations and Extensions. –  dwstu Oct 28 '12 at 4:11
Every author has his/her own convention. Some prefer the minimization formulation; some prefer the maximization form, and... you get the idea. Go with what you're comfortable with. –  J. M. Oct 28 '12 at 4:49
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1 Answer

up vote 2 down vote accepted

I have seen both the $\min$ and $\max$ forms of an LP frequently, it seems to be an author preference sort of thing. The only difference is a minus sign in the objective ($-c^Tx$ instead of $c^Tx$).

Regarding the constraints, I have more often seen the first form (your Bertsimas reference) referred to as standard or canonical. The two forms are equivalent in some sense.

Since $Ax=b$ can be written as the pair of inequality constraints $Ax \leq b$ and $(-A)x \leq (-b)$, it is clear that the first form can be expressed directly as a problem of the second form.

The inequality $Ax\leq b$ can be written as a combination of an equality $Ax+ \sigma = b$ and an inequality $\sigma \geq 0$. Hence by increasing the number of variables (ie, using the variables $x$ and $\sigma$), we can express the second form as a problem of the first form, ie, $\begin{bmatrix} A & I \end{bmatrix} \pmatrix{x \\ \sigma} = b$, $\pmatrix{x \\ \sigma} \geq 0$.

The problem $\min \{ c^T x | A x \leq b \}$ is sometimes referred to as an inequality form LP. Again, it is equivalent to the other two forms.

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Thanks @copper.hat. Perhaps it is indeed a matter of preference. In Bertsimas' own words "we will often use the general form $ \mathbf{Ax} \geq b $ to develop the theory of linear programming. However, when it comes to algorithms, and especially the simplex and interior point methods, we will be focusing on the standard form $ \mathbf{Ax} = b, \mathbf{x} \geq 0 $, which is computationally more convenient." –  dwstu Oct 28 '12 at 7:51
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