C let p be the optimal value of the lp show that n

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Unformatted text preview: ) Then the matrix Z defined above will not be dual feasible. In this problem we will show how to construct ˆ 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 . Z= t ˆ (a) Verify that when x is on the central path, we have Z = Z . ˆ (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 . ˆ (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...
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