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ch13_solutions_solved edit

ch13_solutions_solved edit - Solutions to Chapter 13...

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Solutions to Chapter 13 Exercises SOLVED EXERCISES S1. (a) The payoff table for the two types of travelers is: High Low High 100, 100 30, 70 Low 70, 30 50, 50 (b) The graph is: (c) There are three possible equilibria: a stable monomorphic equilibrium of all Low types ( h = 0), a stable monomorphic equilibrium of all High types ( h = 1), and an unstable polymorphic equilibrium where two-fifths of the population are High types ( h = 0.4). S2. Throughout the answers for this exercise, we let x represent the population proportion of the invading type. Solutions to Chapter 13 Solved Exercises 1 of 10
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(a) In a population primarily consisting of A types with only a small proportion ( x ) of invading T types, the A-type fitness is F (A) = 864(1 – x ) + 936 x = 864 + 72 x and the T-type fitness is F (T) = 792(1 – x ) + 972 x = 792 + 180 x. F (A) > F (T) as long as 864 + 72 x > 792 + 180 x, or 72 > 108 x, or x < 72/108 = 2/3. An all-A population can’t be invaded by T types unless the T types are more than two- thirds of the population, so a small number of mutant Ts cannot successfully invade. Similarly for a small proportion ( x ) of invading Ns: F (A) = 864(1 – x ) + 1,080 x = 864 + 216 x and F (N) = 648(1 – x ) + 972 x = 648 + 324 x. F (A) > F (N) as long as 864 + 216 x > 648 + 324 x, or 216 > 108 x, or x < 216/108 = 2. This is always true because x must be between 0 and 1. Therefore, the A types are always fitter than the N types, so an all-A population can’t be invaded by Ns. (b) In a primarily N population, mutant Ts have fitness F (T) = 972(1 – x ) + 972 x = 972 and Ns have fitness F (N) = 972(1 – x ) + 972 x = 972. The fitnesses are equal, so Ts and Ns do equally well in the population and Ns cannot prevent Ts from invading. A population of Ns invaded by Ts thus exhibits neutral stability, where both the primary and secondary criteria for an ESS give ties. Since neither type is more fit than the other, their proportions in the population will persist, only slightly adjusting as mutations occur. Against a group of mutant As, the N types have fitness F (N) = 972(1 – x ) + 648 x = 972 – 324 x and the A types have fitness F (A) = 1,080(1 – x ) + 864 x = 1,080 – 216 x. F (N) > F (A) when 972 – 324 x > 1,080 - 216 x, or 108 + 108 x < 0. This condition never holds, so As can invade an all-N population. An all- N population is unstable when an A mutation is possible. (c) In a primarily T population, mutant As have fitness F (A) = 936(1 – x ) + 864 x = 936 – 72 x , and the T-type fitness is F (T) = 972(1 – x ) + 792 x = 972 – 180 x. F (T) > F (A) when 972 – 180 x > 936 – 72 x, or 36 > 108 x, or x < 36/108 = 1/3. An all-T population can’t be invaded by A types unless the A types are more than one-third of the population, so a small number of mutant As cannot successfully invade. When the mutants are type N, F (T) = 972(1 – x ) + 972 x = 972, and the mutant Ns also have fitness F (N) = 972(1 – x ) + 972 x = 972. Again, the fitnesses are equal, so Ts and Ns do equally well in the population, and Ts cannot prevent Ns from invading. S3. (a) The payoff table is at right. Solutions to Chapter 13 Solved Exercises 2 of 10 Column A T Row A 20, 20 11, 35 T 35, 11 6, 6
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(b) Let x be the population proportion of T players. Then the As’ expected years in jail are 20(1 – x ) + 11 x = 20 – 9 x , and the Ts’ expected years in jail are 35(1 – x ) + 6 x = 35 – 29 x . Then the T
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