978-0393919684 Chapter 5 Solution Manual Part 2

Solutions to Chapter 5 Exercisess

UNSOLVED EXERCISES

U1. (a) DTC’s profit is YDTC = PQDTC – 12QDTC = (120 – QDTC)QDTC – 12QDTC = –QDTC2 + 108QDTC.

The profit-maximizing quantity for DTC is QDTC = 54.

(b) Now DTC’s profit is given by YDTC = PQDTC – 12QDTC = (120 – QDTC – QADA)QDTC –

Similarly, since Adamantia faces the same cost and wholesale price as DTC, Adamantia’s profit is

To find the solution for the equilibrium quantities analytically, substitute Adamantia’s

(c) When Adamantia enters the market, the total quantity of diamonds supplied in the market

per week increases from 54 to 72. At the same time, the market price of diamonds decreases from $6,600

U2. (a) The demand for Modern Multiplex is QMM = 14 – PMM + PSS and the demand for Sticky

(b) Modern Multiplex’s profit is given by YMM = QMMPMM – 4QMM = (14 – PMM + PSS)PMM –

Similarly, Sticky Shoe’s profit is given by YSS = QSSPSS – 2QSS = (8 – 2PSS + PMM)PSS – 2(8 – 2PSS

(c) To find the solution for the equilibrium quantities analytically, substitute Sticky Shoe’s

Given these equilibrium prices, the demand for Modern Multiplex is QMM = 14 – PMM + PSS = 14 –

(d) Colluding to set prices to maximize the sum of profits means the firms maximize the joint

To find the solution for the Nash equilibrium prices analytically, substitute La Fromagerie’s

best-response function for P2 into La Boulangerie’s best-response function. This yields P1 = 4.5 – 0.25(9 –

(b) Colluding to set prices to maximize the sum of profits means the individual divisions of

L’Épicerie maximize the joint profit function: Y = Y1 + Y2 = 10P1 + 18.5P2 – P12 – P22 – P1P2 – 40. Using

At these prices, the quantities produced are Q1 = 8 – P1 – 0.5P2 = 8 – (0.5) – 0.5(9) = 3, or 3,000

(c) When firms produce complements, a drop in the price of one good leads to an increase in

U4. (a) Cart 0 serves x customers and Cart 1 serves (1 – x), where x is defined by the equation

(b) Profits for Cart 0 are [(5/6)p1 – (5/6)p0 + 0.5](p0 – 0.25). Profits for Cart 1 are symmetric:

(c) The Nash equilibrium prices are the values of p0 and p1 that solve simultaneously the two

best-response rules found in part (b). Substituting Cart 1’s best-response rule into that for Cart 0, we find

U5. No matter what beliefs B may hold about what A is playing, Right is never B’s best response.

Therefore neither (Up, Right) nor (Down, Right) is rationalizable. A recognizes this, and Up is A’s best

U6. To find rationalizable strategies, use iterated elimination of never-best responses. In the Evens or

Odds game, each player has the option of showing one or two fingers. For Anne, one is a best response

U7. Graphically, L’s best-response function, x = 10y – y, is shown below:

Note that any value of x greater than 25 is never a best response for L. Thus all values of x greater than 25

The Nash equilibrium strategies, x = 25, y = 25 and x = 0, y = 0, are clearly rationalizable. Any

The set of L’s rationalizable strategies is 0 ≤ x ≤ 25, and the set of R’s rationalizable strategies is 0

≤ y ≤ 25.

(b) The Nash equilibrium occurs when each firm is best responding to the other, so the

equilibrium quantities are found by solving the best-response funtions simultaneously:

The Nash equilibrium price is thus P = 210 – 46 – 58 = 106. The profits of each firm in equilibrium are

(c) As the hint warns, forging through the math alone yields a nonsensical answer. The

Upon reflection, this maximization problem has an intuitive corner solution. Once the firms have

merged, mid-range chips should only be produced by the division with the lower per-chip cost. Thus qIntel

U9. (a) The profit functions for the three countries are

The best-response functions are thus:

qK = 80 – 0.5qJ – 0.5qC

Substituting for qK in Japan’s best-response function gives

Substituting for qK and qJ in China’s best-response function gives

thus:

Market shares are

Korea: 55/(55 + 35 + 15) ≈ 52.4%

Profits are

YKorea = (P – 20)qK = (75 – 20) * 55 = 3,025, or $3,025 million

(b) The profit functions for the three countries are

YKorea = (P – 20)qK = –(qK)2 + (160 – qJ – qC)qK

The best-response functions are thus:

qK = 80 – 0.5qJ – 0.5qC

Substituting for qK in Japan’s best-response function gives

Substituting for qK and qJ in China’s best-response function gives

thus:

Market shares are

Korea: 44/(44 + 24 + 48) ≈ 37.9%

Profits are

YKorea = (P – 20)qK = (64 – 20) * 44 = 1,936, or $1,936 million

U10. (a) Monica’s payoff is 0.5(5m + 4n + mn) – m2, so her best-response function is m = 1.25 +

0.25n. When she expects Nancy to put in an effort of n = 4/3, Monica’s best response is to put in an effort

(b) Nancy’s payoff is 0.5(5m + 4n + mn) – n2, so her best-response function is n = 1 + 0.25m,

as in Exercise S10. Substituting Nancy’s best-response function into Monica’s from part (a):

(c) Compared with the equilibrium found in part (c) of Exercise S10, Monica puts in more

effort: 1.6 > 1.333. Nancy also puts in more effort: 1.4 > 1.333.

(d) Monica’s payoff is 0.5[5(1.6) + 4(1.4) + (1.6)(1.4)] – (1.6)2 = 5.36, while Nancy’s payoff

In the end, Nancy receives more benefit from Monica’s additional training. This is due to the

U11. (a) In a symmetric Nash equilibrium all students submit the same number, so the mean is

identical to each student’s submission. Since the number in question is its own best response, the mean X

must be equal to (2/3) * (X + 9):

(b) For the target number to be 5, it would have to be the case that

That is, the average of the submissions would have to be –1.5. Since all of the submissions are required to

(c) For the target number to be 90, it would have to be the case that

But the average of the submissions can be at most 100, so submitting 90 is dominated, in particular by

(d) From the reasoning in parts (b) and (c), any number less than 6 is a dominated strategy,

and any number greater than 218/3 is also a dominated strategy.

(e) If Elsa believes that none of her classmates will play the dominated strategies in part (d),

then the smallest the class mean could possibly be is 6 and the largest it could possibly be is 218/3.

(f) Iterating the logic of parts (b), (c), (d), and (e) through multiple rounds will whittle the set

U12. (a) R’s best-response function is derived by differentiating the payoff function with respect to R’s

(c) When the effectiveness of advertising dollars is increased, both parties benefit and the new

(d) Square L’s best-response function, x = √(yk) – yk, and find (x + yk)2 = yk. Now, take R’s

best-response function, y = √[x/(kc)] – x/k, rewrite it as y + x/k = √[x/(kc)], and then multiply through by k

(e) For k = 1 and c changing, y = 1/(c + 1)2 increases as c decreases, while x = c/(c + 1)2 is

,