# IE 5531 Midterm #2 Solutions

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1 IE 5531 Midterm #2 s Prof. John Gunnar Carlsson November 9, 2011 Before you begin: This exam has 9 pages and a total of 5 problems. Make sure that all pages are present. To obtain credit for a problem, you must show all your work. if you use a formula to answer a problem, write the formula down. Do not open this exam until instructed to do so. 1

2 Problem 1: True or False (10 points) State whether each of the following statements is true or false. Assume all auxiliary information given is correct. For each part, only your answer, which should be one of True or False, will be graded. Any explanation will not be read. 1. Although its worst-case running time is faster than that of the simplex method, the ellipsoid method is rarely used because it performs poorly in practical situations. True. 2. One disadvantage of the Wolfe line-search conditions is that, in any given iteration, we may erroneously remove local minimizers from the search space. False. (local minimizers have φ (α) = 0, which is never discarded) 3. At a point satisfying the KKT conditions, if a Lagrange multiplier λ of a constraint is 0, then that constraint must be inactive. False. Minimizing x 2 subject to x 0 has a zero Lagrange multiplier but the constraint is active at optimality.. Consider the problem minimize f (x) a T x b where f (x) is a convex function. If x satises a T x < b and f (x ) = 0, then x must be a globally optimal solution to our problem. True. s.t. 5. Newton's method, applied to the function below, should converge to the marked root of the function if our initial guess lies in the shaded interval. True. 2

3 Problem 2: Line search conditions (20 points) Suppose that you are performing a line search to nd an unconstrained minimum of the function f (x) in R n. At the current iteration, you have chosen your descent direction d k and you are currently trying to determine the optimal step length α k. Sketch the ranges of feasible step lengths in the gure below for the Goldstein conditions. You may assume that φ (0) = f (x k ) T d k = 1, as suggested by the gure. Assume that c = 1/ and consequently that 1 c = 3/. 3

4 Problem 3: Cobb-Douglas Utility Function (20 points) The Cobb-Douglas production function is widely used in economics to represent the relationship between inputs and outputs of a rm. It takes the form Y = AL α K β where Y represents output, L labor, and K capital. The parameters α and β are constants that determine how production is scaled. The Cobb-Douglas function can also be applied to utility maximization and takes the general form N i=1 xαi i. Consider the following utility maximization problem: maximize u (x) = x α 1 x 1 α 2 s.t. p 1 x 1 + p 2 x 2 w Here p is a price vector for x 1 and x 2 and w represents a user's budget. Assume that p > 0 and w > Perform an appropriate transformation to turn this into a minimization problem with a convex objective function. Taking logarithms and multiplying by 1 we obtain the equivalent convex problem minimize [α log x 1 + (1 α) log x 2 ] s.t. p 1 x 1 + p 2 x 2 w 2. Write the KKT conditions and nd an explicit solution for x as a function of α, p, and w. (Hint: can you assume that certain constraints must be tight? Can you assume that certain constraints must be slack?) Are these conditions sucient for optimality? The KKT conditions for the transformed problem are ( ) ( ) ( ) ( α/x 1 p λ (1 α)/x 0 + λ 2 p 1 + λ ) = 0 λ 0 (p 1 x 1 + p 2 x 2 w) = 0 λ 1 x 1 = 0 λ 2 x 2 = 0 p 1 x 1 + p 2 x 2 w The constraint p 1 x 1 + p 2 x 2 w must be tight because the objective function of the transformed problem is monotonically decreasing in x 1 and x 2. The sign constraints on x 1 and x 2 are slack because otherwise the objective function of the transformed problem is, and thus λ 1 = λ 2 = 0. The conditions above therefore tell us that α/x 1 + λ 0 p 1 = 0 (1 α)/x 2 + λ 0 p 2 = 0 p 1 x 1 + p 2 x 2 = w which admits the unique solution λ 0 = 1/w so that x 1 = αw/p 1 and x 2 = (1 α)w/p Find the corresponding Lagrange multiplier for the budget constraint and give an interpretation of it. The Lagrange multiplier is λ 0 = 1/w; the economic interpretation is that it represents the marginal benet of one extra unit of budget.

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6 Problem : A linearly constrained problem (25 points) optimization problem: Consider the following linearly constrained convex minimize x 2 1 6x 1 + x 3 2 3x 2 s.t. x 1 + x Derive the KKT conditions for this problem. The KKT conditions are 2x λ 0 λ 1 = 0 3x λ 0 λ 2 = 0 λ 0 (x 1 + x 2 1) = 0 λ 1 x 1 = 0 λ 2 x 2 = 0 x 1 + x Determine whether or not x = (1/2, 1/2) is an optimal solution. If x = (1/2, 1/2), then by complementary slackness we must have λ 1 = λ 2 = 0. However, the rst two equations tell us that 2x λ 0 λ 1 = 0 3x λ 0 λ 2 = 0 so that 2(1/2) 6 + λ 0 = 3(1/2) λ 0 = 0, which cannot happen, and thus (1/2, 1/2) is not optimal. 3. Given that the optimal solution lies on a corner point of the feasible region, nd the optimal solution to the above problem. The feasible region has three corner points: (0, 0), (1, 0), and (0, 1). By inspection we nd that the optimal point is (1, 0). 6

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8 Problem 5: Interior point method for LP (25 points) Suppose we want to solve the following linear program: minimize x 1 + x 2 s.t. x 1 + x 2 + x 3 = 1 x 1, x 2, x 3 0 We will use a barrier method on the inequality constraints (not the primal-dual potential function!) to solve this problem. 1. Write the KKT conditions for the barrier-function relaxation to this linear program. The problem is and the KKT conditions are minimize x 1 + x 2 µ (log x 1 + log x 2 + log x 3 ) 1 µ/x 1 + λ 0 = 0 1 µ/x 2 + λ 0 = 0 µ/x 3 + λ 0 = 0 x 1 + x 2 + x 3 = 1 x 1 + x 2 + x 3 = 1 2. Show that for every µ > 0, the optimal solution to the barrier-function relaxation is: x 1 (µ) = 1 + 3µ 1 + 9µ 2 2µ x 2 (µ) = x 1 (µ) x 3 (µ) = 1 2x 1 (µ) s.t. The KKT conditions give us four equations in four unknowns. Isolating λ 0 we can write 1 µ/x 1 = 1 µ/x 2 = µ/x 3 x 1 + x 2 + x 3 = 1 which tells us that x 2 = x 1 at optimality. Since x 1 + x 2 + x 3 = 1 this tells us that x 3 = 1 2x 1. To solve for x 1 we write 1 µ/x 1 = µ/x 3 = µ/(1 2x 1 ) which gives a quadratic equation in x 1. Solving this equation gives us x 1 = 1 + 3µ ± 1 + 9µ 2 2µ We have two options for the ± sign, since both give feasible solutions. However, the objective function at x 1 = x 2 = 1+3µ+ 1+9µ 2 2µ is strictly greater than the objective function at x 1 = x 2 = 1+3µ 1+9µ 2 2µ, and the desired result follows. 3. Show that as µ 0, the optimal solution to the barrier problem approaches the unique optimal solution.. It is straightforward to verify that lim µ 0 (x 1 (µ), x 2 (µ), x 3 (µ)) = (0, 0, 1).. (Bonus; 5 points) Suppose that our objective function is just to minimize x 1. Where does the solution converge to now? Using the same approach the solution converges to (0, 1/2, 1/2). 8

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### Part 1. The Review of Linear Programming In the name of God Part 1. The Review of Linear Programming 1.5. Spring 2010 Instructor: Dr. Masoud Yaghini Outline Introduction Formulation of the Dual Problem Primal-Dual Relationship Economic Interpretation March 8, 200 MATH 408 FINAL EXAM SAMPLE EXAM OUTLINE The final exam for this course takes place in the regular course classroom (MEB 238) on Monday, March 2, 8:30-0:20 am. You may bring two-sided 8 page Solutions to review problems for Midterm #1 First: Midterm #1 covers Chapter 1 and 2. In particular, this means that it does not explicitly cover linear algebra. Also, I promise there will not be any proofs.