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midterm2006solutions - MS&E 246: Game Theory with...

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Unformatted text preview: MS&E 246: Game Theory with Engineering Applications Feryal Erhun & Ramesh Johari Midterm Exam Winter, 2006 Solutions 1. (10 points; 2.5 points each) Select true or false; you do not need to justify your answer. (a) In an extensive form game, every SPNE is an NE and every NE is an SPNE. False – every SPNE is an NE, but not every NE is necessarily an SPNE. (b) Every dynamic game of perfect information has an SPNE in pure strategies. True – it can be found using backward induction. (c) If M is a weakly dominant strategy for player 1 and L is a weakly dominant strategy for player 2, then (M,L) is a NE. True – each is a best response to the other. (d) Every game has at least one efficient NE. False – see the Prisoner’s Dilemma. 2. (30 points) Consider the strategic form games and answer each question separately. (a) Consider the following game. Player 2 L T Player 1 M B (1,1) (1,1) (0,0) R (0,0) (2,1) (2,1) i. Does player 1 have a strictly/weakly dominant strategy? Answer: (3 points) Player 1 does not have a strictly dominant strategy. However, he has a weakly dominant strategy, M. ii. Does player 2 have a strictly/weakly dominant strategy? Answer: (3 points) Player 2 does not have a strictly/weakly dominant strategy. iii. Is this game dominance solvable? Answer: (4 points) This game is not dominance solvable. (b) Consider the following game. 1 Player 2 L T Player 1 M B (3,2) (2,0) (1,1) M (4,0) (3,3) (0,2) R (1,1) (0,0) (2,3) i. Iteratively eliminate all the strictly dominated strategies. Answer: (5 points) For player 1, T strictly dominates M, hence we eliminate M of player 1. Now, R strictly dominates M, hence we eliminate M of player 2. We obtain the following. Note that, for part (iii), we are going to use this game as a rational player will not play a dominated strategy as a part of a NE. L T B (3,2) (1,1) R (1,1) (2,3) ii. State the rationality/knowledge assumptions corresponding to each elimination. Answer: (5 points) For the first elimination, we assume that player 1 is rational so he will not play a strictly dominated strategy. For the second elimination, we assume that player 2 is rational and she knows that player 1 is rational. iii. Find all the Nash equilibria. (Don’t forget the mixed-strategy equilibrium!) Answer: (5 points) Pure strategy N.E.: (T,L) and (B,R). Mixed strategy NE is (2/3T+1/3B, 1/3L+2/3R). (c) Consider the following game. Player S S1 F1 Player F F2 F3 F4 (3,1) (2,2) (2,2) (1,3) S2 (2,2) (3,1) (1,3) (2,2) S3 (2,2) (1,3) (3,1) (2,2) S4 (1,3) (2,2) (2,2) (3,1) Check if the following mixed strategy is a Nash equilibrium of this game: Player F: Mix between F3 and F4 with equal probabilities. Player S: Mix between S1 and S2 with equal probabilities. Answer: (5 points) The strategy given above is not a NE, since player F can do better by playing F1 and F2 when player S mixes between S1 and S2. 3. (30 points) Feryal and Ramesh are trying to split $1. They have K days to reach a settlement (assume K > 2). If they cannot, then they will both get nothing. 2 They negotiate as follows: On day 1, Feryal puts an offer on the table for Ramesh: she will keep the whole $1, and give Ramesh nothing. Afterwards, on even days, (k = 2, 4, 6, ...), Ramesh wakes up, looks at the offer that Feryal put on the table the previous day, and decides whether to accept the offer (in which case the game ends) or make a counter offer. Similarly, on subsequent odd days (k = 3, 5, 7, ...), Feryal wakes up, looks at the offer that Ramesh put on the table the previous day, and decides whether to accept the offer (in which case the game ends) or make a counter offer. At day k , if the player who moves decides to reject the previous offer and make a counter offer, the new counter offer is as follows: the offering player keeps gives the remaining 1 − 1 k 1 k to herself/himself, and to the other player. (a) Assume that K is even and plot the extensive form game. Answer: (10 points) (1,0) accept (1/2,1/2) accept (1/3,2/3) accept 1 1 , 1 − K −2 K −2 1 1 ,1 − K −1 K −1 accept accept (0,0) reject F1 R2 F3 reject reject R4 reject F(K-1) RK reject (b) Assume that K is even and find a subgame perfect Nash equilibrium using backward induction. Will your answer change if K is odd? Answer: (10 points) Backwards induction shows that on any day k > 2, whoever plays would prefer to accept the offer currently on the table. However, when k = 2, Ramesh rejects the offer Feryal made on the first day, and proposes the counter offer (1/2, 1/2). Feryal accepts this on day 3, and both get a payoff of 1/2. Thus the backward induction solution is the following pair of strategies: Feryal accepts the offer currently on the table on any day where she has a move (k = 3, 5, 7, ...). Ramesh rejects the offer on the table on day 2, and accepts the offer on the table on any subsequent day (k = 4, 6, ...). Note: To give the backward induction solution you 3 must give this pair of strategies – it is not enough to just give their payoffs, or only the actions they actually end up playing! (c) When K is even, how many strategies does Feryal have? How about Ramesh? Answer: (10 points) Feryal has 1 action on day 1, and 2 actions on the remaining K/2 − 1 odd days. Ramesh has 2 actions on the K/2 even days. Thus Feryal has 2K/2−1 strategies, and Ramesh has 2K/2 strategies. 4. (30 points) Two children, Alice and Bob, are fighting over who will get to play a video game first. Their mother is threatening to take the game away if the children cannot come to an agreement. Their are three possible outcomes: (1) Neither child gets the game. (2) Alice gets to play the game first. (3) Bob gets to play the game first. Both kids get payoff zero if outcome 1 occurs. Alice prefers outcome 2 twice as much as outcome 3. Bob prefers outcome 3 three times as much as outcome 2. In addition to deterministically picking an outcome, the family has access to a random number generator, so they can randomize over the outcomes in any way that they like. (a) Suppose that the payoff to Alice at outcome 2 is equal to twice the payoff to Bob at outcome 3. Define complete payoffs for Alice and Bob (as a function of the outcome) that are consistent with the relationships given above. Answer: (5 points) Any payoffs such that ΠA (1) = 0, ΠA (2) = 2a, ΠA (3) = a, ΠB (1) = 0, ΠB (2) = a/3, and ΠB (3) = a. (b) Suppose that the children evaluate a randomized choice over the outcomes by computing their expected payoff. Give the set of all payoff pairs that are achievable for the children, and describe the Pareto efficient payoff pairs. Answer: (5 points) The achievable payoff pairs are displayed in Figure 1. Pareto efficient points are: {(ΠA , ΠB ) : ΠA = a + ap, ΠB = a − 2ap/3, 0 ≤ p ≤ 1} (c) What randomization over the three choices gives the max-min fair solution? Answer: (5 points) Max-min fair (in this case) means we equalize the payoffs, so p = 0. 4 A 2a a a/3 a B Figure 1: The achievable payoff pairs. (d) What randomization over the three choices gives the utilitarian solution? Answer: (5 points) The utilitarian solutions requires that we maximize the sum of payoffs, so p = 1. (e) What randomization over the three choices gives the Nash bargaining solution? (You should assume the status quo point is where both children receive payoff zero.) Answer: (5 points) To find the Nash bargaining solution, we maximize log ΠA + log ΠB over all achievable payoffs, so p = 1/4. (f) Would your answers in (c), (d), and (e) change if Alice’s payoff at outcome 2 is not necessarily equal to twice the payoff to Bob at outcome 3? Answer: (5 points) The answers to (c) and (d) might change, but the answer to (e) would not. 5 ...
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This note was uploaded on 02/02/2009 for the course MS&E 246 taught by Professor Johari during the Winter '07 term at Stanford.

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