hw5_2011_FinalSolutions

hw5_2011_FinalSolutions - MS&E 211 Fall 2011...

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

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

Unformatted text preview: MS&E 211 Fall 2011 Linear and Nonlinear Optimization Prof. Yinyu Ye Assignment 5 Solutions Problem 1: Portfolio Management (12+8=20pts) a) We have the following model: min 2 x 2 1- 2 x 1 x 2 + 3 x 2 2 subject to x 1 + 2 x 2 1 . 2 x 1 + x 2 = 1 x 1 1 x 2 2 For a local minimizer, the KKT conditions are: ( x 1 + 2 x 2- 1 . 2) = 0 4 x 1- 2 x 2- - = 1 6 x 2- 2 x 1- 2 - = 2 1 x 1 = 0 2 x 2 = 0 x 1 + 2 x 2 1 . 2 x 1 + x 2 = 1 x 1 x 2 1 2 From the model constraints, we know x 2 . 2 , so 2 = 0 . Then we have two cases for 1 : Case 1: 1 > . Then x 1 = 0 , so x 2 = 1 , and the objective value is 3. Case 2: 1 = 0 . Then (i) if > , then we get x 1 + 2 x 2 = 1 . 2 , so x 1 = 0 . 8 , x 2 = 0 . 2 , so the objective value is 1.68. Or (ii) = 0 , then 4 x 1- 2 x 2 = = 6 x 2- 2 x 1 , and we get x 1 = 4 / 7 , x 2 = 3 / 7 , and the objective value is 5 / 7 . This gives us the global minimum (5pts). b) The new model is: 1 min 2 x 2 1- 2 x 1 x 2 + 3 x 2 2 subject to x 1 + 2 x 2 2 . 2 1 x 2 . 5 2 x 1 + x 2 = 1 The KKT conditions are: 1 ( x 1 + 2 x 2- 2 . 2) = 0 4 x 1- 2 x 2- 1- = 0 6 x 2- 2 x 1- 2 1- + 2 = 0 2 (0 . 5- x 2 ) = 0 x 1 + 2 x 2 2 . 2 x 2 . 5 x 1 + x 2 = 1 1 2 Problem 2 (20 pts) This aggregate social problem is a linearly constrained problem as (5pts): min- 50log(3 x 11 + 2 x 12 )- 80log(2 x 21 + 3 x 22 ) s.t. x 11 + x 21 = 4 x 12 + x 22 = 3 x Let y 1 and y 2 be the Lagrange multipliers for the first and second equality constraints. The first order (necessary and sufficient) optimality conditions are (5pts):- 150 3 x 11 + 2 x 12- y 1- s 11 = 0- 100 3 x 11 + 2 x 12- y 2- s 12 = 0- 160 2 x 21 + 3 x 22- y 1- s 21 = 0- 240 2 x 21 + 3 x 22- y 2- s 22 = 0 x ij s ij = 0 , i,j s ij , i,j x 11 + x 21 = 4 x 12 + x 22 = 3 x 2 Guess x 12 = s 11 = s 21 = s 22 = 0 . Then from feasibility x 22 = 3 , so that:- 150 3 x 11- y 1 = 0- 160 2 x 21 + 9- y 1 = 0- 240 2 x 21 + 9- y 2 = 0 From the first and second equations we see 5 x 1 1 = 16 2 x 21 +9 so that we have x 11 = 85 26 and x 21 = 19 26 from the fact that they sum up to 4. Then, we have y 1 =- 260 17 and y 2 = 3 2 y 1 =- 390 17 . Thus, the equilibrium price is 260 17 for good 1 and 390 17 for good 2. (10pts) Please note, when you change a max f ( x ) problem to min- f ( x ) and solve, you will get the correct x values that solve your original problem. However, the dual variables will be opposite in sign since your primal objective function is now...
View Full Document

Page1 / 7

hw5_2011_FinalSolutions - MS&E 211 Fall 2011...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online