Practice Problems 7 Solutions

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Information about Practice Problems 7 Solutions

Published on March 6, 2014

Author: anhtuantran509

Source: slideshare.net

Practice Problems 7: Solutions 1. Consider using Lagrange’s Method to solve the following optimization problem: minimize: 3x1 − 4x2 + x3 − 2x4 subject to: −x2 + x2 + x2 = 1 2 3 4 3x2 + x2 + 2x2 = 6 1 3 4 F (x, λ) = 3x1 − 4x2 + x3 − 2x4 + λ1 −x2 + x2 + x2 − 1 + λ2 3x2 + x2 + 2x2 − 6 2 3 4 1 3 4 T 3 + 6λ2 x1   −4 − 2λ1 x2    1 + 2λ1 x3 + 2λ2 x3    DF (x, λ) =  x 4λ  −2 + 2λ12 4 + 2 2 x4   −x2 + x3 + x4 − 1  2 3x2 + x2 + 2x2 − 6 1 3 4  Finding a critical point, solving DF (x, λ) = 0, is difficult because DF (x, λ) is not a linear system. 2. Compute the gradient of the Lagrangian for the maximum expected return portfolio subject to risk = (20%)2 maximize: µT w subject to: eT w = 1 wT Σw = (0.20)2 F (w, λ) = µT w + λ1 eT w − 1 + λ2 wT Σw − 0.04  T T µ + λ1 eT + 2λ2 wT Σ  eT w − 1 DF (w, λ) =  T w Σw − 0.04 Note: Unlike the lecture example of finding the minimum variance portfolio, DF (w, λ) = 0 is not a linear system in this case, so the solution must be computed numerically. 3. Second order Taylor polynomial around S0 for C(S) in terms of ∆ and Γ: 2 C (S0 ) + (S − S0 ) 2 (S − S0 ) ∂ 2 C (S − S0 ) ∂C (S0 ) = C (S0 ) + (S − S0 ) ∆ (S0 ) + (S0 ) + Γ (S0 ) ∂S 2 ∂S 2 2

1 1+x −1 f (x) = (1 + x)2 4. f (x) = f (x) = 2 (1 + x)3 f (x) = −6 (1 + x)4 ... f (n) (x) = (−1)n n! (1 + x)n+1 Taylor series expansion: T (x) = (x − 0)2 (x − 0)3 −6 (x − 0)n (−1)n n! 1 −1 2 + + + ··· + + (x − 0) 1+0 (1 + 0)2 2 (1 + 0)3 6 (1 + 0)4 n! (1 + x)n+1 = 1 − x + x2 − x3 + · · · + (−1)n xn Radius of convergence: 1 1 R= = =1 1/k 1 lim |ak | 1 1 1 where lim |ak | /k = lim |(−1)k | /k = lim 1 /k = 1 k→∞ k→∞ k→∞ k→∞ Where does T (x) = 1 ? 1+x rn rn rn (−1)n n! = lim max |f (n) (z)| = lim max = 0 for r < n→∞ (1 − r)n+1 n→∞ n! z∈[−r,r] n→∞ n! z∈[−r,r] (1 + z)n+1 1 T (x) = for |x| < 1 2 1+x lim 5. Rt = Pt − Pt−1 Pt−1 rt = log Pt Pt−1 ⇒ = log Rt Pt−1 = Pt − Pt−1 (Rt + 1) Pt−1 Pt−1 ⇒ Rt Pt−1 + Pt−1 = Pt ⇒ 1 2 Pt = (Rt + 1) Pt−1 = log (Rt + 1) Note: Rt is actually a first order Taylor polynomial of rt around the point a = 0. Let Rt = 0.01, then rt = log(1.01) ≈ 0.00995 6. Find the order of the Taylor polynomial needed to compute e0.25 to six digits of accuracy. x f (x) − Pn (x) = a (x − t)n (n+1) f (t)dt n! Let a = 0 and x = 0.25 0.25 Find n such that f (0.25) − Pn (0.25) = 0 n 1 2 3 4 5 (0.25 − t)n t e dt < 10−6 n! Error 3.40 × 10−2 2.78 × 10−3 1.73 × 10−4 8.49 × 10−6 3.52 × 10−7 We need a 5th order Taylor polynomial to compute e0.25 to six digits of accuracy.

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