This preview shows page 1. Sign up to view the full content.
Unformatted text preview: ) Then
the matrix Z deﬁned above will not be dual feasible. In this problem we will show how to construct
a dual feasible Z (which agrees with Z as given above when x is on the central path), from any
point x that is near the central path. Deﬁne X = W + diag(x), and let v = −∇2 φ(x)−1 ∇φ(x) be
the Newton step for the function φ deﬁned above. Deﬁne
ˆ 1 X −1 − X −1 diag(v )X −1 .
(a) Verify that when x is on the central path, we have Z = Z .
(b) Show that Zii = 1, for i = 1, . . . , n.
(c) Let λ(x) = ∇φ(x)T ∇2 φ(x)−1 ∇φ(x) be the Newton decrement at x. Show that
λ(x) = tr(X −1 diag(v )X −1 diag(v )) = tr(X −1/2 diag(v )X −1/2 )2 .
(d) Show that λ(x) < 1 implies that Z ≻ 0. Thus, when x is near the central path (meaning,
λ(x) < 1), Z is dual feasible.
9.5 Standard form LP barrier method. In the following three parts of this exercise, you will implement
a barrier method for solving the standard form LP
minimize cT x
subject to Ax...
View Full Document
- Fall '13
- The Aeneid