BIO calculus problem

Question

How cooperative behaviors can evolve is an open problem in biology. A cooperative behavior is one in which each organism performs a costly task, but the reward from performing this task is shared somewhat indiscriminately among organisms. For example, emperor penguins huddle together to keep warm during the cold Antarctic winter. The penguins at the edge of the flock endure the worst wind, but their bodies shield the penguins inside the flock. If penguins take it in turns to be on the outside, the cost (shield other penguins with your body) and the reward (being warmed and shielded from the wind) are shared quite evenly between penguins. Although cooperation benefits the population as a whole, it is also vulnerable to an effect called defection. If a single organism decides not to cooperate then it will still share in the rewards from cooperation, but will not need to pay the cost. For example, a penguin could choose to not spend any time at the edge of the flock, so it is shielded from the wind by the other penguins but does not take its turn shielding the others.

We will analyze a simple example of a cooperative system. Two organisms must choose whether to cooperate or to defect (not cooperate). If either organism cooperates then both organisms get a reward b. If 1 neither organism cooperates then there is no reward. If both organisms cooperate then each must pay a cost c/2, but if only one organism cooperates, that organism pays a cost c, and the defecting organism pays no cost. If organism A interacts with organism B then the net reward to organism A can be summarized using a pay-off matrix

Net Benefit to A B Cooperates B Defects

A Cooperates b − c/2 b − c

A Defects b 0

Lets assume that a fraction p of the time B decides to cooperate, and a fraction 1−p of the time it defects. Then if A cooperates the average net reward that it receives is

r¯C = p(b − c/2) + (1 − p)(b − c)

Whereas if A defects, the average net reward that it receives is:

r¯D = pb + (1 − p) · 0 = pb

So if A cooperates a fraction of the time, q ,and defects a fraction of the time, 1−q, its average net reward is:

r(q) = q · r¯C + (1 − q) · r¯D = q · (p(b − c/2) + (1 − p)(b − c)) + (1 − q) · pb

(a) What should A do? One possible goal for A may be to choose the value of q (that is, the fraction of times that it cooperates) that maximizes r(q).

(i) Show that if p < b−c/(b−c/2) then the optimal value of q (which optimizes r(q)) is q = 1: that is, A should always cooperate.

(ii) Show that if p > b−c/(b−c/2) then the optimal value of q (which optimizes r(q)) is q = 0: that is, A should always defect.

(b) Lets suppose now that A and B are from the same species and that they follow exactly the same strategy (that is, p = q). Show that the average reward to A is then given by:

r(p) = (b − c/2)(2 − p)p

If b > c/2 , show that the optimal strategy for each organism is p = 1 (always cooperate).

(c) What happens if b < c/2 . Explain in words why b > c/2 and b < c/2 might lead to different optimal strategies.

Answer 1

The problem statement really should say that b > c. Otherwise, no organism would have incentive to cooperate.

We're told that r(q) = q · r¯C + (1 − q) · r¯D = q · (p(b − c/2) + (1 − p)(b − c)) + (1 − q) · pb,

where p = Prob(A cooperates) and q = Prob(B cooperates).

Collecting like terms and simplifying,

r(q) = { p[b - c/2) - (b - c)] + b - c - pb }q + pb

This is a line, and the part in the curly brackets is the slope.So, simplifying the part in the curly brackets, we have

slope = { p[b - c/2) - (b - c)] + b - c - pb } = { c/2 * p -pb + b - c} = { p (c/2 - b) + b - c }

It's important to note that c/2 - b is negative (follows from b > c > c/2. So the term p (c/2 - b) in the last formula above is negative.

(a) (i) if p < (b - c) / (b - c/2), then | p (c/2 - b) | < b - c, so the slope is positive. The graph of r(q) is line segment with positive slope, so it is maximized at the right-hand endpoint of domain, i.e., when q = 1. But that means A always cooperates.

(ii) if p > (b - c) / (b - c/2), then | p (c/2 - b) | > b - c, so the slope is negative. (Remember that p (c/2 - b) is negative.) In this case the graph of r(q) is a line segment with negative slope and is maximized at the left-hand endpoint of domain, i.e., when q =0. But that means A always defects.

(b) In the case that p=q, we substitute p for q:

r(p) = { p (c/2 - b) + b - c }p + pb = { p (c/2 - b) + b - c + b } p. Now notice that the part in the curly brackets is equal to (b - c/2)(2 - p), so r(p) = (b - c/2)(2 - p)p

Using the product rule, r'(p) = (b - c/2) (-1)p + (b - c/2)(2-p)1 = (b - c/2)(1-p). If b > c/2, then r(p) is increasing on [0,1) and is therefore maximized when p = 1.

(c) If b < c/2, then r'(p) is negative on [0, 1) and so r(p) is decreasing. In this case r is maximized when p = 0. The reason for different outcomes boils down to increasing versus decreasing.