Problem Set II Answers

1) For this problem, you need to assume that prey size is proportional to energy, which is reasonable if insects have similar composition. Then, size becomes a measure of energy (E). The time it takes to consume the prey item is the handling time (h), and the search time (S) is the inverse of the given encounter rate.

            a) Profitability of Juniper Cricket = E/h = 2000/1 = 2000

Profitability of Scorpion = E/h = 4000/4 = 1000

Profitability of June Beetle = E/h = 1000/3 = 333.3

Profitability of Grasshoppers = E/h = 2500/2 = 1250

Juniper cricket > Grasshopper > Scorpion > June beetle

b) Scorpions should be included if the energy per unit time gained from eating a scorpion is greater than passing the scorpion by and waiting for more profitable prey, i.e. Grasshoppers and Juniper Crickets

                       The formula to use in answering this question is the multiple prey choice formula.

                        E_s/h_s > (E_GH + E_JC)/(h_GH + S_GH + h_JC + S_JC)

1000 > (2500 + 2000)/ (0.2 + 2 + 1 + 1)

1000 > (4500)/(4.2)

1000 > 1071.43?

1000 is not greater than 1071.43

NO.  Scorpions should not be included.

c)  Yes.  Should always eat the most profitable food item if encountered.

2. a. The following graph indicates that a group size of 3 is optimal because at that group size the amount of energy / time is greatest for each lion.

b. Many possible answers: Lions might be in larger groups than 3 because of

Kin selection, e.g. sisters band together to protect against male infanticide

Enhanced protection against theft by competitors, e.g. hyenas, crocodiles

Information transfer

Imperfect knowledge about optimal group size

Age structure: e.g. young or old lions do not affect intake rate as much

The only other option for an individual is to be solitary, which is worse than being in a group of four.

3)  Dove-Bully Game

a.Payoff Matrix

	Bully	Dove
Bully	V/2	V
Dove	0	V/2

b. ESS if V=2

Black dots indicate strategy with highest payoff in a population playing a given strategy

	Bully	Dove
Bully	1	2
Dove	0	1

If population of Bullies, highest payoff to play Bully

If population of Doves, highest payoff to play Bully

Therefore always play Bully, meaning Bully is a pure ESS

c. Could Hawk invade a population of Bully if V=2, C=4

	Bully	Hawk
Bully	V/2	0
Hawk	V	(V-C)/2

	Bully	Hawk
Bully	1	0
Hawk	2	-1

Hawk can invade a population of Bully.

d. ESS if V=2, C=4

	Bully	Dove	Hawk
Bully	1	2	0
Dove	0	1	0
Hawk	2	2	-1

If population of Bullies, highest payoffs to play Hawk

If population of Dove, highest payoffs to play Bully or Hawk

If population of Hawk, highest payoff to play Bully or Dove.

Therefore this is a mixed ESS, because each strategy is invasible when pure.

Note that you won't be responsible for figuring out the ESS on three-way games, as their dynamics can be pretty complicated.

4) In this problem the payoff is the number of offspring produced by the actor. Since the Watchers actively scan for predators, they lose no offspring to predators but they do not produce as many young as do NonWatchers. Watchers produce 4 young. Nonwatchers do not watch for predators so they lose 40% of their young to predators if no one else is watching. However when there is a Watcher present, a NonWatcher can produce 5 young.

NonWatchers can invade a world of Watchers. Watchers can invade a world of NonWatchers. So the ESS is mixed and stable.

If you assume that p is the fraction of watchers in the population, the point at which the fitness of the Watcher equals the fitness of the NonWatcher is:

4p + 4(1-p) = 5p + 3(1-p)

4p + 4 – 4p = 5p + 3 -3p

4 = 2p + 3

1 = 2p

p = 1/2

At equilibrium, the proportion of Watchers in the population is 1/2.