978-0393919684 Chapter 10 Solution Manual Part 2

Solutions to Chapter 10 Exercises

UNSOLVED EXERCISES

U1. See the following payoff table:

Baker

10 100

Both players have dominant strategies, 90:10 for Cutler and 100 for Baker. Therefore, the equilibrium

consists of these two strategies, and the payoffs are 90 to Cutler, 10 to Baker, (a) regardless of whether the

U2. (a) Each pizza store has a dominant strategy to price medium, so the unique Nash

equilibrium of the game is (Low, Low). However, both stores prefer the outcome in which both price

high, so this is a prisoners’ dilemma:

Pierce’s Pizza Pies

High Low

(b) Pierce’s dominant strategy is still Low but Donna’s dominant strategy is now High. The

unique Nash equilibrium is (High, Low). There is no dilemma in this version of the game:

Pierce’s Pizza Pies

High Low

(c) Donna plays a leadership role here. The captive market for Donna’s is so large now that if

Donna’s lowers its price, what it loses on the captive market outweighs what is gained by garnering the

U3. The three-dimensional payoff table follows:

C votes

Yes No

B B

Yes No Yes No

A vote of No is a dominant strategy for each of the three council members; the Nash equilibrium is thus

for all three to vote No.

With the given assumptions, two members of the council face no cost from voting Yes. The only

gain from voting No is that such an action looks good to voters. When a council member doesn’t have to

face voters for over a year, however, the voters will forget about the vote, and so will impose no

U4. (a) Computing payoffs based on the stated rules yields the payoff table in the text.

(b) The Nash equilibrium in the one-stage game is (Low, Low). Low is a dominant strategy

(c) Consider the second stage of the two-stage game. Call the amount that Row puts in the

Column

Low High

Row

High 0 + c, 11 – c

c

To simplify the analysis, assume that a player always chooses High when both actions would give

her the same payoff. (Otherwise, c will have to be slightly greater than 2 and r slightly greater than 4, in

The first stage of the game can now be analyzed. Column has two strategies in this first stage: set

c = 0 or set c = 2; Row also has two strategies: set r = 0 or set r = 4:

Column

c = 2 c = 0

The rollback equilibrium is then (r = 4, c = 2) in the first stage, and (High, High) in the second stage. The

prisoners’ dilemma is resolved in this game; players attain the joint-payoff-maximizing outcome of (High,

U5. (a) The game whereby Glassworks and Clearsmooth compete for one year is depicted in the

following payoff table (in millions of dollars):

Clearsmooth

Advertise Don’t

Advertise is a dominant strategy for each firm, so the unique Nash equilibrium is (Advertise, Advertise)

with payoffs (2, 2).

(b) Starting with the last round and rolling back, the subgame-perfect equilibrium is for each

(c) Using a tit-for-tat strategy, in every round after the first a firm would take the action taken

(d) If the game is infinitely repeated and both firms use a grim-trigger strategy—that is, they

don’t advertise until the other firm does, after which they advertise every year forever—then reaching the

cooperative outcome (Don’t advertise, Don’t advertise) each year is subgame perfect if neither firm wants

(b).

U6. (a) There are two ways to find the PPierce that maximizes YPierce for each PDonna. Method 1:

multiply out and rearrange terms on the right-hand side of the expression for YPierce:

YPierce =- 3(12 +0.5P

Donna )+(3+12 +0.5P

Donna )P

Pierce –P

Pierce

2

=- 3(12 +0.5P

Donna )+15 +0.5P

Donna

2

æ

è

çö

ø

÷

2

–15 +0.5P

Donna

2–P

Pierce

æ

è

çö

ø

÷

2

.

Only the last term above involves PPierce. That term appears with a negative sign, so what appears after the

negative sign should be made as small as possible in order to make the whole expression as large as

Method 2: Use the first line of the above expression for YPierce and take its derivative with respect

to PPierce (holding PDonna fixed):

The first-order condition for PPierce to maximize YPierce is that this derivative should be zero. Setting it equal

to zero and solving for PPierce gives the same equation as above for the best-response rule. Donna’s

best-response rule can be found similarly, using either method. The costs and sales of the two stores are

entirely symmetric, so the equation is

The Nash equilibrium prices solve the two best-response rules simultaneously. Substitute the

expression for PPierce into the expression for PDonna and find

(b) The two stores choose P to maximize

(c) To determine the best cheating price, use one firm’s best-response rule:

To determine if collusion is sustainable, compare the stream of profits when collusion holds to the

stream of profits earned by a cheater. It is reasonable to assume that the “noncheating” firm is using a

“tit-for-tat” strategy (or, equivalently, a “grim” strategy in the “cheat forever” case). Then we look first at

the “cheat forever” strategy:

Discounting the infinite stream of payoffs, Pierce finds cheating to offer a higher payoff if

Considering the cheat-forever strategy, therefore, Pierce will cheat only if the monthly interest rate r >

Now, consider the cheat-once strategy. There are some ambiguities in this case that involve

describing the outcome that holds when Pierce signals his intention to reestablish cooperation after his

(onetime) cheat. Note first that this problem does not arise in the high price/low price version of the

pricing game. When Pierce cheats in that version, the outcome is (High, Low), and when he signals that

Collude: 55,125, 55,125, 55,125, . . .

Cheat once: 62,016, 36,750, 55,125, . . .

Pierce finds cheating to offer a higher payoff if

Considering the cheat-once strategy, therefore, Pierce will cheat only if the monthly interest rate r >

1.667. In discount rate terms, Pierce cheats only if the (monthly) discount factor d < 0.375. Different

assumptions about the pricing in the second period will, of course, produce different conclusions about

the critical values of p and d.

U7. (a) The joint profit of Japan, Korea, and China from producing VLCCs is

( ) ( )

2

*

30 180 30 150

150 2 75.

P Q Q Q Q Q

Q Q

Q

p

= – = – – = –

¶= – Þ =

¶

Joint profit is maximized when the three countries produce a total of 75 VLCCs.

(c) When one country defects knowing the others are producing their collusive quantities, its

profit function is

( )

2

180 25 25 30 100 .

d d d d d

q q q qp= – – – – = –

The defecting country’s profit is maximized when qd = 50, which yields a profit of (180 – 25 – 25 – 50 –

30) • 50 = 2,500 or $2,500 million.

(d) A defecting country will gain 2,500 – 1,875 = 625 ($625 million) in the year of defection

and lose 1,875 – 1,406.25 = 418.75 ($418.75 million) in every subsequent year.

(e) Collusion will be sustainable if the amount gained during the defection is less than the

418.75 418.75

625 0.67.

625

r

< Þ < =

That is, if the triopoly game is infinitely repeated and all three countries are using grim-trigger strategies,

collusion will be sustainable whenever the annual interest rate is less than 67%.

Compared with the duopoly case in Exercise S7 (where collusion could be sustained if the annual

interest rate were less than 88.9%), the set of interest rates that can sustain collusion is smaller in the