IE 5531 Practice Midterm #2

Size: px

Start display at page:

Download "IE 5531 Practice Midterm #2"

Alyson May
5 years ago
Views:

1 IE 5531 Practice Midterm #2 Prof. John Gunnar Carlsson November 23, 2010 Problem 1: Nonlinear programming You are a songwriter who writes Top 40 style songs for the radio. Each song you write can be described by a feature vector x, which encodes information about the song (for example, number of times the word 'love' is used, number of guitar ris, average duration of a song, and so forth). Each song you write will be a hit with probability p (x) = exp ( c T x + d ) 1 + exp (c T x + d) where c and d are known parameters (this function is commonly used in logistic regression, which you may have seen previously; note that p lies between 0 and 1 by construction). In writing songs, you must also obey restrictions imposed by the local radio station, which we can express as the system Ax b. 1. Consider the problem of choosing x so as to maximize the probability of making a hit song while obeying the radio station's restriction. Then, write an equivalent minimization problem with a convex objective function f ( ), using a transformation of the objective function (hint: the function log (e t + 1) is convex). 2. Suppose that c = (1, 2) and d = 1. Sketch some level sets of f ( ). 3. Suppose that a hit song will generate a prot w T x + q, where q is a positive constant so that there exists a feasible x satisfying w T x+q > 0 (a non-hit song will generate a prot of 0). Write the problem of maximizing the expected prot due to this song. Is there an equivalent convex optimization problem for this? Take the negative logarithm of p (x) and you get f (x) := log (p (x)) = log ( exp ( c T x + d )) + log ( 1 + exp ( c T x + d )) = c T x + d + log ( 1 + exp ( c T x + d )) where we set t = c T x + d. By the hint this is a sum of convex functions and is therefore convex. Since f (x) depends only on c T x + d, the level sets of f (x) are just the level sets of c T x + d, which are just straight lines of the form x 1 + 2x = κ for various constants κ. The problem of maximizing the expected prot is maximize ( w T x + q ) p (x) Again, if we take the negative logarithm of this objective function, the problem is minimize log (( w T x + q ) p (x) ) = log ( w T x + q ) log p (x) = log ( w T x + q ) which is a sum of convex functions and therefore convex. c T x + d + log ( 1 + exp ( c T x + d )) 1

2 Problem 2: Multi-rm alliance revisited Suppose you are the mayor of Minneapolis and there are three rms in your city: rm 1 (capital), rm 2 (labor), and rm 3 (technology). The three rms are considering possible cooperations. Let x 1, x 2, x 3 denote the input of rms 1, 2, and 3, respectively. The payo function is f(x 1, x 2, x 3 ) = x 1 + x 3 + (x 1 + x 3 )(3x 2 2x 2 2). 1. Assume each input variable x 1, x 2, x 3 takes values in [0, 1]. That is, 0 x i 1 for i = 1, 2, 3. Derive the KKT condition for maximizing the social prot function f(x 1, x 2, x 3 ). Also show that x 1 = 1, x 2 = 0.75 and x 3 = 1 is an optimal solution. (Hint: KKT conditions are NOT sucient for this non-convex program. you should look for additional arguments to identify global optimality) 2. Suppose the social prot will be assigned to each rm proportionally to their input. So each rm i (i = 1 corresponds to rm 1, so on) will get prot x i x 1 + x 2 + x 3 f(x 1, x 2, x 3 ), i = 1, 2, 3. Suppose we do not consider the cost of the input. Show that if x 1 = 1, x 3 = 1 are xed, then x 2 = 0.75 is NOT an optimal strategy for rm 2 if rm 2 just aims to maximizes its own prot, which is x 2 x 2+2 f(1, x 2, 1). 3. From part 2, you may have realized that under this mechanism, the rms will never cooperate towards the social optimum. So, you changed the input rules that each rm either inputs x i = 1 or inputs x i = 0. For example, if rm 1 and rm 2 form a sub-alliance, then their total payo is f(1, 1, 0) = 2. We list the payos of all possible sub-alliances S {1, 2, 3} in the table below. S f(s) 0 {1} 1 {2} 0 {3} 1 {1, 2} 2 {2, 3} 2 {1, 3} 2 {1, 2, 3} 4 Table 1: Obviously, the grand alliance maximizes the social payo (= 4). We know a core is the set of payo allocation vectors (z 1, z 2, z 3 ) under the grand alliance, such that no subgroups can do better by deserting the grand alliance. Write out the expression of the core using the given data. Also show that this core is nonempty. 4. Due to the recent economic recession, your city has run out of budget, so you have to ask the grand alliance to pay a tax of T. However, you still want to maintain the grand alliance. In other words, you want to make sure that the core is nonempty. Under this condition, nd the maximal T you can charge the grand alliance. (The grand alliance has payo f({1, 2, 3}) T after tax, but we suppose any sub-alliance S {1, 2, 3} is exempt of tax and still has payo f(s).) First, note that 3x 2 2x for x 2 [0, 1]. Therefore, the social objective function is monotonically increasing as a function of x 1 and x 3 and therefore x 1 = 1 and x 3 = 1. Finally, it is easy to see that 3x 2 2x 2 2 is maximized at the point x 2 = 0.75 as desired. Next, suppose that x 1 = x 3 = 1 and consider the prot of rm 2. The derivative of rm 2's prot at the point x 2 = 0.75 is and therefore 0.75 is not a local minimizer of rm 2's prot. Indeed, if we set x 2 = 1 then rm 2 receives a prot of 4/3 as compared with a prot of 1.16 at x 2 = The core must satisfy z 1 + z 3 2 2

3 It is non-empty because setting z 1 = 1.5, z 2 = 1, and z 3 = 1.5 is in the core (for example). Finally, to determine the maximum tax, we can formulate the linear problem maximize T z 1 + z 2 + z 3 + T = 4 z 1 + z 3 2 Omitting redundant constraints, substituting T = 4 (z 1 + z 2 + z 3 ), and removing the constant 4 from the objective function, this simplies to s.t. minimize z 1 + z 2 + z 3 s.t. Note that and implies z 1 + z 2 + z 3 3. Therefore a lower bound on the new objective function is 3, which implies a tax of T = 1 unit. This is indeed feasible, by setting z 1 = z 2 = z 3 = 1, and therefore the optimal tax is T = 1 unit. 3

4 Problem 3: Interior point methods In addition to solving linear programming problems, the barrier function method also works for solving quadratic programming problems. Consider the following quadratic programming problem: minimize (x 1 3) 2 x 2 s.t. 2 x 1 x 2 0 x 1 0 (1) 1. Solve this quadratic programming problem using the KKT conditions. Next, consider the following optimization problem: minimizeφ µ (x 1, x 2 ) = (x 1 3) 2 x 2 µ(log(2 x 1 x 2 ) + log x 1 ) 2. Find the unconstrained minimal solution of φ µ (x 1, x 2 ) for any given µ 0. Note: Since x 1, x 2 are functions of µ, the optimal solution can be written as (x 1 (µ), x 2 (µ)), where µ is a parameter. 3. For the above problem, would the minimum solution be a local or global optimum? Why? 4. Verify that (x 1 (µ), x 2 (µ)) converges to the solutions as µ 0. The KKT conditions are 2 (x 1 3) + λ 1 λ 2 = λ 1 = 0 λ 1 (2 x 1 x 2 ) = 0 λ 2 x 1 = 0 Since there are only two constraints, by trial and error we nd that the optimal solution has x 1 = 5/2 and x 2 = 0.5. The optimality conditions for the barrier problem are ( 1 2x µ 1 ) 2 x 1 x 2 x 1 = 0 µ x 1 x 2 = 0 which tells us that x 1 (µ) = µ 4 and x 2 (µ) = 2 x 1 (µ) µ. The minimizer is a global minimizer because the objective function is convex. Finally, we see that as µ 0, we nd that x 1 (µ) 5/2 and x 2 (µ) 0.5 as desired. 4

5 Problem 4: KKT system Consider the nonlinear program minimize xy (x 3) 2 + (y 2) 2 = 1 1. Write the KKT conditions for optimality of this problem. 2. We know Newton's method can be used to nd roots of an equation. How can it be applied to nd the KKT points here? To that end, apply one iteration of the Newton's method starting with the point (x 0, y 0, λ) = (1, 1, 1). s.t. The KKT conditions, plus the feasibility condition, are y + 2λ (x 3) = 0 x + 2λ (y 2) = 0 (x 3) 2 + (y 2) 2 1 = 0 The Jacobian matrix of this system is J = 2λ 1 x 1 2λ y 2 (x 3) 2 (y 2) 0 the iteration scheme is where x = (x, y, λ) and f (x, y, λ) = x k+1 = x k J 1 f (x k ) y + 2λ (x 3) x + 2λ (y 2) (x 3) 2 + (y 2) 2 1 Computing one iteration with x 0 = (1, 1, 1) we nd that x 1 = (2.33, 0.33, 1.5). 5

IE 5531 Midterm #2 Solutions

IE 5531 Midterm #2 Solutions 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,