e62 Introduction to Optimization Fall 2016 Professor Benjamin Van Roy Homework 1 Solutions

Transcription

1 e62 Introduction to Optimization Fall 26 Professor Benjamin Van Roy 267 Homework Solutions A. Python Practice Problem The script below will generate the required result. fb_list = #this list will contain the series fb_list.append() fb_list.append() for i in range(2,): new_val = fb_listi- + fb_listi-2 fb_list.append(new_val) message = "25th element of the series : " + str(fb_list25) print (message) message = "5th element of the series : " + str(fb_list5) print (message) message = "th element of the series : " + str(fb_list) print (message) This is the result of running the code above: 25th element of the series : th element of the series : th element of the series: Problem 2 The script below will generate the required result. import numpy as np import e62

2 c = np.matrix(,2) A = np.matrix(2,4,8,) b = np.matrix(8,4) opt_sol, opt_val = e62.linprog(a,b,c) print ("optimal solution:") print (opt_sol) print ("optimal objective value:") print (opt_val) This is the result of running the code above: optimal solution:. 2. optimal objective value: 4. B. Linear Algebra Review Part I Problem Expanding out the first four expressions, we have: a T x = x 2 = 2 x () a T x = x 2 = 2 x + 2 (2) b T x = x 2 = 2x (3) b T x = x 2 = 2x + (4) Each of these corresponds to a line in the plane. Expanding out expressions 5 and 6, we have: a b x =, T a b x =, 2 T x x 2 x x 2 = = 2 (5) (6)

3 Each of these is a system with a unique solution. Their respective solutions are x = 4 3 x =. 2 3 We thus obtain the following graph: and x 2 2: a T x= : a T x= 5 6 x 4: b T x= 3: b T x= where the shaded region represents all x R 2 such that a T x. Problem 2 Part Finding a b such that Ax = b has no solution is equivalent to finding a vector, b R 3, such that b is not in C(A). One easy choice is: b = (7) The only way to get the first two b values in some linear combination of the columns of A is to set x =, x 2 =. But, this leads to = in the third row equation. So, Ax = b cannot have a solution in this case. Part 2 Finding a non-zero b such that Ax = b has a solution is equivalent to finding a vector, b R 3, such that b is in C(A). One easy choice is: b = 2 (8)

4 which leads to a solution of x =. Problem 5 By definition, S is a subspace of R N if and only if the following are true : a) For any x S and y S, x + y S b) For any x S and scalar α, αx S c) S Now, given that U and V are both subspaces in R N for some N, let s look at U V. For any x U V and y U V, x and y must each be elements of both U and V by the definition of the intersection operator. Thus, (x + y) U and (x + y) V since U and V are both subspaces and thus satisfy property above. Thus (x + y) U V for any such x and y. Also, for any x U V and any scalar α, αx must be in both U and V since these are subspaces and thus satisfy property 2 above. Thus, αx U V. Finally, we have U V since is an element of both U and V. Thus, by the definition above, U V is a subspace of R N. On the other hand, ( U ) V is not necessarily ( a subspace ) in R N. We can show this by counterexample. Let U = span and V = span. Now take x = and y =. Note that x U V and y U V by the definition of the union operator. But, we have x + y = which is not an element of U V. Thus, property above does not hold for U V and this cannot not a subspace. Problem 7 Yes, it is possible for {x, y, z} to be linearly dependent. For example, take x =, y =, and z =. We have that x T y = and y is not a multiple of z. However, z = x + y so the three vectors are linearly dependent. This definition can be found in an introductory Linear Algebra text, such as Linear Algebra with Applications by Otto Bretchler.

5 Part II Problem a An example is: A = Problem b An example is: A = Problem c An example is: A = For this matrix A, for all values of y, the vector specified below is a solution: y + x = y C. Contingent Claims Problem a Let M =. For the stock, the payoff vector a R M is given by: a m = m. (9) For the zero-coupon bond, the payoff vector a 2 R M is given by: a 2 m =. () For the European call option, the payoff vector a 3 R M is given by: a 3 m = max(m 5, ) ()

6 For the European put option, the payoff vector a 4 R M is given by: a 4 m = max(5 m, ) (2) For the straddle, the payoff vector a 5 R M is given by: a 5 m = + max(m 5, 5 m) (3) Problem b Let p j denote the price of the contingent claim with payoffs a j. Note that, a 5a 2 a 3 + a 4 =, and therefore, the initial investment must be zero for this portfolio to satisfy a no arbitrage condition. Since the initial investment is given by p 5p 2 p 3 + p 4 =, with p = $5, p 2 = $.9, p 3 = $, we conclude that p 4 = $5 Problem c Note that a 2 a 3 a 4 + a 5 =. Consequently, the payoff of the straddle can be replicate according to a 5 = a 2 + a 3 + a 4. In other words, the straddle is replicated by short-selling one zero-coupon bond, purchasing one European put option, and purchasing one European call option.