ch12_solutions_solved edit

ch12_solutions_solved edit - Solutions to Chapter 12...

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Solutions to Chapter 12 Exercises SOLVED EXERCISES S1. (a) The number choosing X should decrease to move the population division between X and Y away from the unstable center equilibrium. (b) Because the line for action X is above the line for action Y when 100 people choose X, and because we expect the number choosing X to decrease from this point [see part (a) ], it must be the case that the lines in the diagram represent costs associated with each action. Thus, when 100 people are choosing X, the cost of X exceeds the cost of Y, so people will want to switch from X to Y. S2. (a) This game is a prisoners’ dilemma since s ( n ) is greater than p ( n + 1) for all n . Any single player can always raise his payoff by switching from participating to shirking. (b) T ( n ) = n * p ( n ) + (100 – n )* s ( n ) = n 2 + (100 – n )(4 + 3 n ). (c) Plugging the appropriate values into the text’s Equation (12.1) yields T ( n + 1) – T ( n ) = n + 1 – 4 – 3 n + n ( n + 1 – n ) + (100 – n – 1)(4 + 3 n + 3 – 4 – 3 n ) = –3 – 2 n + n + 3(100 – n – 1) = 294 – 4 n. This formula is positive for all n up to n = 73, but turns negative for n = 74 and higher. Thus, the last time the change in total social payoff expression is positive is for T (74) – T (73). Thus T is maximized at n = 74. Using calculus, you can find the same answer by differentiating T ( n ) from part b. T´( n ) = 2 n + 3(100 – n ) – (4 + 3 n ) = 296 – 4 n . Set this derivative equal to zero to find that T ( n ) is maximized at n = 74.
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ch12_solutions_solved edit - Solutions to Chapter 12...

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