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I do not know much about this subject, but I am trying to learn a little.

In a book I have it says that a primal problem is:

max $c'X$

subject to

$AX \ge b$

$X \ge 0$

It says that the dual of this is:

maximize $b'Y$

subject to:

$A'y \le c$

$Y \ge 0$

However they did not prove the linear programming duality theorem.

However I found another text where they do this, but there they use that:

max $c'X$

$AX \le b$

Has a dual

min $b'Y$

$A'Y=c$

$y \ge 0$

Notice that in the third last onee I do not have a restriction on X, and in the last one we have an equality not an inequality, are these two representations equal? Is it a way to prove they are?

user119615
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1 Answers1

0

You can check out the following table. enter image description here

If the restrictions have the relation sign $\geq $ in the max-problem, then the dual-variables (min-Problem) has to be $y\leq 0$ (row 8 of the table)

And it should be $A'y\geq c$, because $x \geq 0$ (row 9 of the table)

callculus42
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