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I know how to formulate the dual problem to a standard minimisation Linear Program. However I came across this example which confused me slightly.

Write down the dual problem of:

Minimise $3x_1+5x_2+6x_3$

Subject to:

$2x_1+4x_2+7x_3-x_4=5$

$-3x_1+2x_2-6x_3+x_5=-4$

$x_1,x_2,x_3,x_4,x_5\geqslant0$

The answer to this is:

Maximise $5y_1-4y_2$

Subject to:

$2y_1-3y_2\le3$

$4y_1+2y_2\le5$

$7y_1-6y_2\le6$

$-y_1\le0, y_2\le0$

It is just the last line of the solution that confuses me, I assumed it was always $y_1\ge0, y_2\ge0$ so this would become $-y_1\le0, -y_2\le0$

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    $\begingroup$ Since the primal constraints are with "$=$" i.e. equations, the dual variables should be unrestricted. So the solution seems wrong to me. $\endgroup$ – dave Aug 16 '18 at 10:22

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