University:
Massachusetts Institute of Technology
Course:
15.501/516 | Accounting
Academic year:

2023
Views:

125

Pages:

3
Author:

Beckett Jones

Introduction to Optimization 1. Col. Max 2 1 2 Row Min 2 1 2 3 3 Since max (row min) = min (column max) = 2, row choosing row 1 and column choosing column 1 is a saddle point. Value of the game is 2 units to the row player. 2. 4 5 5 8 Row Min 4 6 7 6 9 6 5 7 5 4 4 6 6 5 5 5 Col. 6 Max 7 6 9 Since max (row min) = min (col max) = 6, row choosing row 2 and column choosing column 1 or column 3 are saddle points. Value of game is 6 units to the row player. 3. Since the Witch wants to maximize the length of Max's route, we let the Witch be the row player. We obtain the following reward matrix: Witch Leaves Unblocked Atl-St.L and Nash.-St. L. Atl.-NO and Nash.-St. L Atl.-St. L and Nash.-NO Atl.-NO and Nash.-NO Column Maximum Max Goes To Atlanta Nashville 1600 1800 1700 1800 1600 1400 1700 1400 1700 1800 Row Min 1600 1700 1400 1400 Since max(row min) = min(column max) = 1700 we find that Max going to Atlanta and the Witch leaving the Atlanta-New Orleans and Nashville-St. Louis roads unblocked (or blocking Atlanta-St. Louis and Nashville-New Orleans) is a saddle point. Total length of Max's trip will be 1700 miles. 4. To find optimal strategies we add one to each element of the reward matrix. This ensures that we may assume that all variables are non-negative in the LP's of the row and column player. Thus we deal with the following reward matrix: Row Player's LP max v st (1) v≤3x1/2 (2) v≤3x2/2 (3) v≤2(1 - x1 - x2) All variables ≥0 3/2 0 0 0 3/2 0 0 0 2 Column Player's LP min w st (4) w≥3y1/2 (5) w≥3y2/2 (6) w≥2 - 2y1 - 2y2 All variables ≥0 Since we are given that x1, x2, and x3 are all positive, (4)-(6) all hold with equality. This yields w = 6/11, y1 = y2 = 4/11, y3 = 1 - y1 - y2 = 3/11. This implies that (1)-(3) hold with equality, yielding x1 = x2 = 4/11, x3 = 3/11, v = 6/11. Thus value of original game is 6/11 - 1 = -5/11. Optimal strategy for each player is (4/11, 4/11, 3/11). Suppose the column player plays (1/2, 1/2, 0). If row plays (1, 0, 0), row will obtain an expected reward of 1/2(1/2) + 1/2(-1) = -1/4. This exceeds the value of the game (-5/11). 5. a) If Bill were to invest $1 in C, he would get $0.7 if N is elected; if Bill were to invest $1 in E, he would get $0.9 if B is elected. Therefore, Bill would invest in D, and get at least $1.1 for each $1 invested. b) The nonlinear program is max min{1.5 p C + 1.1 p D + 0.9 p E ,1.1 p C + 1.3 p D + 1.2 p E ,0.7 pC + 1.2 p D + 1.4 p E } s.t. pC + p D + p E = 1 p C , p D , p E ≥ 0. After converting to an LP, we obtain the answer: max s.t. z pC + p D + p E = 1 pC , p D , p E ≥ 0 z ≤ 1.5 pC + 1.1 p D + 0.9 p E z ≤ 1.1 pC + 1.3 p D + 1.2 p E z ≤ 0.7 pC + 1.2 p D + 1.4 p E . c) Let p be the proportion of money invested in stock C. If B is elected, the payoff is 1.5p+1.1(1-p). If K is elected, the payoff is 1.1p+1.3(1-p). The worst case payoff is the lower of these two. The best worst case payoff is when these two payoffs are equal, that is when 1 1.5 p + 1.1(1 − p) = 1.1 p + 1.3(1 − p) ⇔ 0.2 p = 0.6 ⇔ p = . 3 Hence, the answer is to invest 1/3 of $500,000 in stock C, and 2/3 in stock D. The following plot illustrates the reasoning above. return on investment 1.5 1.5 1.4 1.4 1.3 1.3 1.2 1.2 1.1 0 1.1 .1 .2 .3 .4 .5 p .6 .7 .8 .9 1