Asst2SolnsW10

Asst2SolnsW10 - CO372 Assignment 2 Winter 2010 Solutions 1...

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Unformatted text preview: CO372 Assignment 2, Winter 2010 Solutions 1 2 Chapter 1 Optimization Exercise 1.10 Let C be a ( n,n ) positive definite matrix and let A be a ( m,n ) matrix having full row rank. Define x = (1 , 1 ,..., 1) , u = (1 , 1 ,..., 1) and b = Ax . (a) Show how to choose c so that x is optimal for (1.4) with multiplier vector u . (b) Computationally verify Part (a) by modifying Example1p2.m (Figure 1.4) to solve (1.4) with C = diag (1 , 2 , 3 , 4 , 5) , A = • 1 2 3 4 5 6 7 8 9 10 ‚ . Solution: (a) C , A , x , b = Ax and u are all specified. The second optimality condition of Theorem 1.2 is satisfied provided- c- Cx = A u , or, c =- Cx- A u. 3 CHAPTER 1. OPTIMIZATION (b) The modification of Example1p2.m (Figure 1.4, text) to solve this prob- lem is named Exer1p10.m and is shown in Figure 1.1. 1 %Exer1p10.m Exercise 1.10(b) 2 n = 5; m = 2; 3 x0 = [ 1 1 1 1 1 ]' 4 u = [ 1 1 ]' 5 BigC = [1 0 0 0 0; 0 2 0 0 0 ; 0 0 3 0 0; 0 0 0 4 0; ... 6 0 0 0 0 5 ] 7 A = [1 2 3 4 5 ; 6 7 8 9 10 ] 8 b = A * x0 9 c =- BigC * x0- A' * u 10 11 checkdata(BigC,1.e- 6); 12 H = vertcat([BigC,A'] , [A,zeros(m,m)]) 13 rhs = vertcat(- c,b) 14 y = Hˆ- 1 * rhs % Solve Hy = rhs. 15 x = y(1:n) % x = optimal solution 16 u = y(n+1:n+m) % u = multipliers for Ax = b Figure 1.1 Exercise 1.10(b): Exer1p10.m. The output from running Exer1p10.m gives c = (- 8 ,- 11 ,- 14 ,- 17 ,- 20) and b = (15 , 40) . Solving the linear equations of Theorem 1.3 (lines 12 to 16 of Figure 1.1) with data A , C , c and b produces the output x = (1 , 1 , 1 , 1 , 1 , ) and u = (1 , 1) as expected. ♦ 4 Chapter 2 The Efficient Frontier Exercise 2.1 Show β 2 = 0 if and only if μ is a multiple of l . Solution: Recall that β 2 = h 1 Σ h 1 and h 1 = Σ- 1 μ- l Σ- 1 μ l Σ- 1 l ¶ Σ- 1 l. Suppose first ( ⇒ ) β 2 = h 1 Σ h 1 = 0. Since Σ is positive definite, we have h 1 = 0. Therefore, Σ- 1 μ = l Σ- 1 μ l Σ- 1 l ¶ Σ- 1 l, which implies that μ = l Σ- 1 μ l Σ- 1 l ¶ l, so that μ is indeed a multiple of l . Next suppose ( ⇐ ) μ = θl for some constant θ . Then h 1 = θ Σ- 1 l- θ l Σ- 1 l l Σ- 1 l ¶ Σ- 1 l 5 CHAPTER 2. THE EFFICIENT FRONTIER = θ Σ- 1 l- θ Σ- 1 l = 0 , which implies that β 2 = h 1 Σ h 1 = 0. / Exercise 2.2 Verify equation (2.15) ( β 1 = 0)....
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Asst2SolnsW10 - CO372 Assignment 2 Winter 2010 Solutions 1...

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