Lecture 6

Lecture 6: Strategic commitment &
applications to entry and exit (II)
Entry, Capacity and Price Competition: Gelman & Salop, 1983,
Bell Journal of Economics, 14: 315-325. Simplified version in
section 8.4.4, Shy, 198-200.
Paper examines the role of precommitment in strategic
interaction between a dominant incumbent firm and an entrant
•A “small” competitor partially offsets its demand disadvanatge
by engaging in capacity limitation and discount pricing.
•Strategy credibly reduces the threat posed to the dominant
incumbent firm and makes retaliation more costly
•This is an example of “Judo” economics in which a small firm
uses its rival’s large size to its own advantage
Model in General:
•Homogeneous Products
•Incumbent is at least as efficient as entrant, i.e., c1 c2.
•Two-stage game
•Stage 1: Entrant chooses capacity (k2) & price (p2).
•Stage 2: Incumbent chooses price (p1).
•If incumbent matches price, it takes the whole market (and it has
the capacity to serve the whole market.)
•Without capacity or if k2=, incumbent takes the whole market.
Incumbent has two strategies: (I) Deter entry & (II)
accommodate entry.
•Deterrence profits: D (p1=p2) = (p2-c1)D(p2)
•Accommodation Profits: The incumbent chooses p1 > p2, and
allows the entrant to sell all of its output. The incumbent is
then a monopolist on the residual demand curve. (They assume
reservation-price rationing so that the entrant gets the k2
consumers with the highest valuation.)
A = (p*1-c1)[D(p*1)-k2] (where p*1 is the monopoly price)
•Hence the entrant chooses so that D  A.
•As p2 increases, D increases. As k2 increases, A decreases.
•Thus there is a tradeoff between p2 and k2. If the entrant
chooses to large a capacity or too high a price, the incument
will respond by deterring entry.
•Hence the entrant underinvests in order not to provoke a
response from the incumbent. This is an example of a
“puppy-dog” strategy.
Numerical Example:
•Let P=100-Q, and let c1=c2=0.
•Hence D(p1)-k2 =100- p1-k2
•Hence to find A, maximize p1 (100-p1-k2)
•FOC imply that p1 =(100- k2)/2.
•Hence A = (100- k2)2/4.
•D = p2D(p2)= p2(100- p2).
•A = D implies that (100- k2)2/4= p2(100- p2).
•Solving yields p2={ 100-[(200- k2)k2].5} /2.
•Clearly, there is an inverse relationship between p2 and k2 for the
•As increases, profits increase, but must fall to insure that the
constraint (A = D) holds. Otherwise the incumbent will deter
•In the first stage, entrant maximizes E =p2k2, where p2={ 100[(200- k2)k2].5} /2. This needs to be solved numerically.
Finally we need to check that p2<p1. Otherwise, the entrant
will not make any sales.
•p2={ 100-[(200- k2)k2].5} /2 < p1 =(100- k2)/2 implies
[(200- k2)k2].5 > k2, which implies
[(200- k2)k2] > k22, which implies
200k2 > 2k22, which implies
100 > k2.
•This is true because demand function is P=100-Q.

similar documents