Question

Question 1: An incumbent sells steel and faces a potential entrant. Inverse demand curve for steel sales is given by by P = 400-Q, where Q is the total amount sold by both firms. The marginal cost to produce one unit of steel is MC = 100 The entrant incurs fixed costs of F. (a) What are the Cournot prices and quantities in the market, assuming the entrant enters? (b) What are the Stackelberg prices and quantities in the market, if that the incumbent accommodates entry? (c) Briefly explain why the profits for the incumbent are higher in (b) than (a). (d) What happens if you take the Stackelberg quantity for the entrant and substitute it into the Cournot BRF for the incumbent? Does this exercise tell us anything about whether or not it is important that the incumbent commits to the Stackelberg quantity? (e) Now suppose that the entrant incurs fixed costs to operate. Suppose the incumbent is thinking about using a "limit price" strategy, by committing to a quantity of 200. Calculate how large would the fixed costs have to be in order for the limit strategy to deter the entrant. (f) Is limit pricing a good choice for the incumbent here, relative to accommodating entry in Stackelberg competition? (g) (challenge problem) Now suppose that the firms incur fixed costs to operate. As we discussed in class, if F = 0, limit pricing doesn't make sense. What is the lowest level of fixed costs for the entrant where limit pricing provides higher profits for the incumbent than accommodating entry and playing the Stackelberg quantity?

Answer 1

Page No Date Salution: Accordling to the Cournot_malel firm 1 LTncumbent) P= 400- ( Ginen) to Now, Total Rewenue = eX px9 (t00-19+92)] x qu → (400 - q1-2 2.400gi-912 => ¿(TR) Marginal Revenue MR = {tog=qi= q192] MR = 400- 2gı -92 [Given] Marginal lost MC = 10 for Equilibnium, MR = MC 400-291-92 100 ) 300 = 241+9,2 = 30০

firm 2 (Entrant) Total rewenue pxq2 [400-(q1+q 71400 - q1- q2-)92 Now diferentiation ef Tatal Revenue gives Marginal Revenia MR = 400 2-292 MC=100 की MR MC 400-91-29,2=100 =300 +242 Now fom O22 we gets -300 2:300 246+92 z300 =600 -3 (-) -) -39,2 = -300 42 Substituting g2= 100 in O

2q1+92=300 7291 + 100 = 300 21=200 - 100 Now P becomes, 400 - (qi+q22 = 400 - (100 +100) 200 Now p91 - [f+10q] E-X1ρ 1001200-1007-F 200 X 100 10000 T2 = pq2-[F + 100g2] 200X100-f (100x100) 10000-F

) According to the Stackelbarg madel, fom eg" e we have = 300 300 -91 2. FIRM1 x= TR-TC 1009- 400g1-912-9q2-f-10091 300q1- q2 -q 300-91 300-ag- 150 tqp-0 =D0 150-91=0 १|= |50. Now 42= 300-150 442=7 P = 400 -75-150 175

Page No. Date : = TV. 176X 150 – E -(100X150) 1250-F -(100x75) 75 x 175 -f 56२5- £ The fts in inuumbent are fa) beiause aftex soeing the quuantity set by FIRM 2, the firM1 moves higher în (b) than

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