n is optimal obviously x 0 and l x n x 1 so primal

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: . . , µn ) ≥ 0, we are to consider the minimization problem min −µ x s.t. x ≥ 0 , l x = 1 By Theorem 5.2, the necessary and sufficient conditions for optimality are x≥0, lx=1 µ = −v + ul for some u ∈ R , v ∈ Rn with v ≥ 0 vx=0 Now we check the point x∗ defined by x∗ = i 1 if i = k 0 if i = k , where k ∈ {1, . . . , n} is such that µk ≥ µi for all i = 1, . . . , n, is optimal. Obviously, x∗ ≥ 0 and l x = n x∗ = 1, so primal feasibility conditions i=1 i are satisfied. Let v ∈ Rn be defined by vi := µk − µi for i = 1, . . . , n. By the choice of index k, vi ≥ 0 for all i. It is also clear from the definition that v = µk l − µ, that is, µ = −v + µk l, so dual feasibility conditions are satisfied. Finally, v x∗ = n vi x∗ = vk x∗ = (µk − µk )x∗ = 0, so complementarity i k k i=1 slackness is satisfied. Therefore, x∗ is optimal for the given optimization problem. (5.7) Given the problem 1 min{−tµ x + x Σx | l x = 1, x ≥ 0} , 2 the optimality conditions for this problem are given by: l x = 1, x ≥ 0, tµ − Σx = cl − u, u ≥ 0, u x = 0. 8 We need to make an additional assumption that only one asset has the greatest expected return, say µ1 > µi for all i = 1. (Otherwise, the result is not necessarily true.) We claim that x∗ = (1, 0, . . . , 0) for all t large enough. We just need to show that x∗ satisfies the optimality conditions for some u. Notice that primal feasibility holds for x∗ and that complementary slackness implies that u1 = 0. Dual feasibility implies that c = ui + tµi − σi,1 , ∀i = 1, . . . , n, where σ1,1 . . Σ= . σn,1 ··· ... σ1,n . . . . · · · σn,n Since u1 = 0, we have c = tµ1 − σ1,1 , which implies tµ1 − σ1,1 = ui + tµi − σi,1 , ∀i = 1, . . . , n. Therefore, ui = t(µ1 − µi ) − (σ1,1 − σi,1 ), ∀i = 1, . . . , n. Since we must have ui ≥ 0 for all i, we get the following condition on t: t≥ σ1,1 − σi,1 , ∀i = 2, . . . , n. µ1 − µi Therefore, x∗ = (1, 0, . . . , 0) is optimal when t ≥ max σ1,1 − σi,1 | i = 2, . . . , n . µ1 − µi 9...
View Full Document

This document was uploaded on 04/10/2014.

Ask a homework question - tutors are online