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I know if a dual is unbounded then the primal is unfeasible and vice versa, but I don't know why they can't both be unbounded. Is it because it's impossible to have linear constraints that are unbounded in the direction of the gradient and the opposite direction?

Gooby
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2 Answers2

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Well, it is because of the weak duality theorem

Tab1e
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  • I thought that's why if the primal is unbounded then the dual is infeasible – Gooby Oct 17 '19 at 06:10
  • Given primal max and dual min, any feasible solution for the primal is a lower bound for the solution to the dual. Conversely, any feasible solution for the dual is an upper bound for the primal. We clearly cannot have the first quantity strictly greater than the second, which is what would be necessary for both LPs to be unbounded. – Math1000 Oct 17 '19 at 06:17
  • @DylanY Well, if a linear program is infeasible, it can't be unbounded since it has to be feasible to be unbounded. – Tab1e Oct 17 '19 at 08:26
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Note: although unbounded primal is sufficient for an infeasible dual, it is not a necessary condition. Because there can be a case when both are infeasible. When one is infeasible we can say that the other is either unbounded or infeasible.

Given primal (P) normal max-LP: \begin{equation*} \begin{aligned} & \underset{\vec{x}}{\text{max}} & & z=\vec{c}^T\vec{x} \\ & \text{subject to} & & A\vec{x} \leq \vec{b}, \\ & & &\,\,\,\,\vec{x} \geq \vec{0} \end{aligned} \end{equation*}

and dual (D) normal min-LP: \begin{equation*} \begin{aligned} & \underset{\vec{y}}{\text{min}} & & w=\vec{b}^T\vec{y} \\ & \text{subject to} & & A\vec{y} \geq \vec{c}, \\ & & & \,\,\,\,\vec{y} \geq \vec{0} \end{aligned} \end{equation*}

Lemma 1 (weak duality): If $\vec{x}$ is feasible for (P) and $\vec{y}$ is feasible for (D), we have $z=\vec{c}^T\vec{x}\leq\vec{b}^T\vec{y}=w$.

Proof:

$\vec{x}$ is feasible for (P) $\Rightarrow$ $A\vec{x} \leq \vec{b}$, $\vec{x}\geq\vec{0}$

$\vec{y}$ is feasible for (D) $\Rightarrow$ $A\vec{y} \geq \vec{c}$, $\vec{y}\geq\vec{0}$

$\vec{y}^T(A\vec{x}\leq\vec{b}) \Rightarrow \vec{y}^T A\vec{x}\leq\vec{y}^T\vec{b}={w}$

$(A^T\vec{y}\geq\vec{c})^T \Rightarrow (\vec{y}^TA\geq\vec{c}^T)\vec{x} \Rightarrow \vec{y}^TA\vec{x} \geq\vec{c}^T\vec{x}=z$

combining, we get $z=\vec{c}^T\vec{x}\leq\vec{b}^T\vec{y}=w\,\,\square$

Lemma 2 (Strong Duality): if $z=\vec{c}^T\vec{x}=\vec{b}^T\vec{y}=w$ for feasible $\vec{x}$,$\vec{y}$ in (P) and (D), respectively, then $\vec{x}$,$\vec{y}$ are optimal for (P) and (D), respectively.

Proof: All z values lie below all w values (Lemma 1). Hence when z=w, we get optimality for both.

Lemmas 3&4: If (P) is unbounded, then (D) is infeasible. Similarly if (D) is unbounded, (P) is infeasible.

Explanation: If (P) is unbounded, we can push z up without limits. Hence there are no finite w values, i.e, there are no feasible $\vec{y}$ for (D).

John Karasev
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