978-0393919684 Chapter 12 Solution Manual Part 3

U7. (a) When t = 0.5, the fitness of tortoises is FT = 0.5c + 0.5(–2) = 0.5c – 1, and the fitness of

hares is FH = 0.5(1) + 0.5(0) = 0.5. Tortoises are more fit than hares when 0.5c – 1 > 0.5, that is, when c >

(b) When t = 0.1, the fitness of tortoises is FT = 0.1c + 0.9(–2) = 0.1c – 1.8, and the fitness of

(c) When c = 1, and a single hare invades a population of pure tortoises,

(e) In a polymorphic equilibrium, the fitness of tortoises must equal the fitness of hares. That

(f) All of the polymorphic equilibria described in part (e) are unstable. For a given value of

c, tortoises are exactly as fit as hares when (2 + c)t – 2 = t. If the value of t is perturbed up from the

U8. Creating a spreadsheet for this exercise is most helpful and strongly recommended. Consider the

following example in Excel.

Enter Period in cell A1, xt in cell A2, FX in cell A3, and FY in cell A4. In cell B1 enter 0, and in

cell C1 enter the formula =B1+1. The symmetric game used in this exercise has the following payoff

structure:

In cells A6, A7, A8, and A9, enter a =, b =, c =, and d =, respectively. Given the payoffs of the

game used in this exercise (and Exercise S8 as well), enter 2 in cell B6, 5 in cell B7, 3 in cell B8, and 1 in

cell B9.

Select cells B3 and B4, copy them, select cells C3 and C4 and paste. Now select the range C1:C4,

X[x] Y[1 – x]

(a) x1 ≈ 0.63636, FX1 ≈ 3.09091, and FY1 ≈ 2.27273. (Enter 0.5 in cell B2 and see cells C2,

C3, and C4.)

(b) x20 ≈ 0.79991, FX20 ≈ 2.60028, and FY20 ≈ 2.59982. (See cells V2, V3, and V4.)

(g) As seen in part (e), monomorphic equilibria are possible. If a population consists of only

type X, the next generation—barring mutations—will also consist of only type X, and likewise for type Y.

U9. Note from the payoffs that the game is symmetric:

Green Purple

Let g be the proportion of Green types in the population. The fitness of a Green type is then FG = ag + b(1

Case 1: d > b

0 0, since 0

1 0

d b a c d b d b

a c d b

d b d b a c d b a c a c

a c d b

–

< Þ – + – > – >

– + –

–< Þ – < – + – Þ < – Þ >

– + –

Case 2: d < b

0 0, since 0

1 0

d b a c d b d b

a c d b

d b d b a c d b a c a c

a c d b

–

< Þ – + – < – <

– + –

–< Þ – > – + – Þ > – Þ <

– + –

Note that when d = b, g* = 0, implying a monomorphic rather than a polymorphic equilibrium.

For there to be a polymorphic equilibrium, it must be the case either that a > c when d > b or that

A stable polymorphic equilibrium must have the property that if the proportion of greens is

slightly increased from g* to gʹ = g* + ε, where ε > 0, the purple types will be more fit, so that in future

Consider first when gʹ = g* + ε. The fitness of the two types will be

( ) ( )

1 * 1 * ( )

G

P

F ag b g ag b g a b

F cg d g cg d g c d

e

¢ ¢

= + – = + – + –

¢ ¢

= + – = + – + –

The purple types must be more fit when the proportion of Greens is above g* for the polymorphic

equilibrium to be stable, implying:

( ) ( )

P G

F F c d a b c d a b b d a ce e> Þ – > – Þ – > – Þ – > –

Now consider the implications when gʹʹ = g* – ε and Green types need to be more fit for stability:

( ) ( )

1 * 1 * ( )

( ) ( )

G

P

G P

F ag b g ag b g a b

F cg d g cg d g c d

F F a b c d a b c d b d a c

e

e e

¢¢ ¢¢

= + – = + – – –

¢¢ ¢¢

= + – = + – – –

> Þ – – >- – Þ – + >- + Þ – > –

For a polymorphic equilibrium to be stable, then, the condition b – d > a – c must hold. Look back at the

cases under which a polymorphic equilibrium exists in the first place. The conditions of Case 1 (d > b and

For a symmetric game with the above payoff table, a stable polymorphic equilibrium will be

guaranteed when d < b and a < c.

U10. (a) To verify that the derivatives conform to the statements regarding population dynamics,

note first that q1 increases if and only if dq1/dt is positive. Because dq1/dt = –q2 + q3, it follows that q1

(c) In three-dimensional space, q1 + q2 + q3 = 1 represents a plane, passing through the point

1 on each of the three axes. We also know that along the path of motion of the three variables, q12 + q22 +

q32 is constant. The value of the constant depends on the initial condition; suppose that the constant is

(d) It only remains to determine whether the point moves clockwise or counterclockwise

along the circle. Consider the point shown by a bullet on the figure, near the bottom of the circle where q3

The two-dimensional Figure 12.15 in the text is simply the projection of this three-dimensional