Linear Programming Lecture Notes - Penn State Personal Web Server
Linear Programming Lecture Notes - Penn State Personal Web Server
Linear Programming Lecture Notes - Penn State Personal Web Server
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Likewise beginning with the KKT conditions for the dual problem (Expression 9.15 -<br />
9.17) we can write:<br />
∗ ∗<br />
w A − v ≥ c<br />
Primal Feasibility<br />
w ∗ ≥ 0<br />
⎧<br />
⎪⎨<br />
Ax<br />
Dual Feasibility<br />
⎪⎩<br />
∗ + s ∗ = b<br />
x ∗ ≥ 0<br />
s ∗ ≥ 0<br />
∗ ∗<br />
(v ) x = 0<br />
Complementary Slackness<br />
w ∗ s ∗ = 0<br />
Here we substitute v ∗ for w ∗ A − c in the complementary slackness terms. Thus we can see<br />
that the KKT conditions for the primal and dual problems are equivalent. This completes<br />
the proof. <br />
Remark 9.8. Notice that the KKT conditions for the primal and dual problems are<br />
equivalent, but the dual feasibility conditions for the primal problem are identical to the<br />
primal feasibility conditions for the dual problem and vice-versa. Thus, two linear programming<br />
problems are dual to each other if they share KKT conditions with the primal and<br />
dual feasibility conditions swapped.<br />
Exercise 67. Compute the dual problem for a canonical form minimization problem:<br />
⎧<br />
⎪⎨<br />
min cx<br />
P s.t. Ax ≥ b<br />
⎪⎩<br />
x ≥ 0<br />
Find the KKT conditions for the dual problem you just identified. Use the result from<br />
Exercise 63 to show that KKT conditions for Problem P are identical to the KKT conditions<br />
for the dual problem you just found.<br />
Lemma 9.9 (Strong Duality). There is a bounded optimal solution x ∗ for Problem P if<br />
and only if there is a bounded optimal solution w ∗ for Problem D. Furthermore, cx ∗ = w ∗ b.<br />
Proof. Suppose that there is a solution x ∗ for Problem P . Let s ∗ = b − Ax ∗ . Clearly<br />
s ∗ ≥ 0.<br />
By Theorem 8.7 there exists dual variables w ∗ and v ∗ satisfying dual feasibility and<br />
complementary slackness. Dual feasibility in the KKT conditions implies that:<br />
(9.26) v ∗ = w ∗ A − c<br />
We also know that w ∗ , v ∗ ≥ 0. Complementary Slackness (from Theorem 8.7) states that<br />
v ∗ x ∗ = 0. But v ∗ is defined above and we see that:<br />
(9.27) v ∗ x ∗ = 0 =⇒ (w ∗ A − c) x = 0<br />
Likewise, since we have s ∗ = b − Ax ∗ . Complementary slackness assures us that<br />
w ∗ (b − Ax ∗ ) = 0. Thus we see that:<br />
(9.28) w ∗ (b − Ax ∗ ) = 0 =⇒ w ∗ s ∗ = 0<br />
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