hw5_2011_FinalSolutions

# hw5_2011_FinalSolutions - MS&amp;amp;amp;E 211 Fall 2011...

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Unformatted text preview: MS&amp;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 &gt; . Then x 1 = 0 , so x 2 = 1 , and the objective value is 3. Case 2: 1 = 0 . Then (i) if &gt; , 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...
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## hw5_2011_FinalSolutions - MS&amp;amp;amp;E 211 Fall 2011...

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