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Coase-rent/sell. Industriøkonomi, uge 6 Christian Schultz 3 år, 2004. No commitment. 2 periods, good lasts these 2 periods Zero interest rate, no cost Competitive resale market. (p = p m ) In each period, demand for service of good (for instance light, cooling, transport) is

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coase rent sell

Coase-rent/sell

Industriøkonomi, uge 6

Christian Schultz

3 år, 2004

no commitment
No commitment
  • 2 periods, good lasts these 2 periods
  • Zero interest rate, no cost
  • Competitive resale market. (p = pm)
  • In each period, demand for service of good (for instance light, cooling, transport) is
  • Q(R) = 20 – R
if monopolist rents
If monopolist rents
  • In each period: max R RQ(R)
  • = max R R(20-R)
  • Foc : 20 – 2R = 0 so R = 10, Q = 20-10 = 10
  • Profit per period 10*(20-10) = 100
  • For two periods 2* 100 = 200
if mon sells at start of period 1
If mon. sells at start of period 1
  • If he can commit not to lower price in period 2.
  • Set price = 20 sell 10 units earn 200.
  • In period 2, everybody with reservation price above 10 has bought, so demand in period 2 is
  • 10 – p
if mon cannot commit and sells
If mon cannot commit and sells
  • Ass: Consumers have rational expectations
  • Time line
  • ---- p1 ,Q1 ------ p2 , Q2
  • Solve backwards!
  • Look at period 2, Q1 given
  • Residual demand: Q2 (p2) = 20 - Q1 – p2
selling no commitment ii
Selling no commitment, II
  • Max p2p2 (20 - Q1 – p2) 
  • p2 = (20 - Q1)/2 , Q2 = (20 - Q1)/2 ,
  • 2 = (20 - Q1)2/4
  • Notice, second period profit depends on how much was sold in first period!
period 1
Period 1
  • Rat exp: consumers know they can buy (or sell if they wish) in next period for p2.
  • If consumer pays p1 in the first period, she is really paying R1 = (p1 - p2 ) for 1st period use and R2 = p2 for 2nd period use.
  • So equivalent to renting for R1 = (p1 - p2 ) in first period and for R2 = p2 in second period.
  • So we can analyze period 1 as if the monopolist sets rent R1
period 1 ii
Period 1 ,II
  • 1st period demand is therefore
  • Q1 = 20 - R1  Q1 = 20 - (p1 - p2 )
  • Remember p2 = (20 - Q1)/2
  • So Q1 = 20 - p1 + (20 - Q1)/2
  • Q1 = 20 - (2/3) p1
  • Total profit Q1p1 + 2 = Q1p1 + (20 - Q1)2/4
  • = (20- (2/3) p1) p1 + (20 -(20- (2/3) p1))2/4
period 1 iii
Period 1, III
  • (20- (2/3) p1) p1 + (20 -(20- (2/3)p1))2/4
  • Maximize wrt p1 . Foc yields
  • p1 = 18, Q1 = 20- (2/3) p1 = 20-(2/3)18 =8
  • p2 = (20 - Q1)/2 = (20-8)/2 = 6
  • Q2 = (20 - 8)/2 = 6
  • Total profit 18*8 + 6*6 = 180
  • < 200!!!!!
example end
Example end
  • Profit lower when monopolist sells than when he rents.
  • Problem: he is his own competitor.
  • Notice he seeks to mitigate the problem by setting p1 high. But not perfect solution.
  • Coase’s conjecture
  • When number of periods go to infinity and there is no discounting (like in ex), then price  MC
  • This has been verified in subsequent research
  • Examples: Store Danske Encyklopædi !
how to solve problem for mon
How to solve problem for mon
  • Commit not to lower price . DSDE
  • Make good non-durable
  • Fads, fashion
  • Make capacity constraints so expanding output costly
  • Most favored costumer clause (NB)
  • Buy back guarantee
  • Reputation (de Beers)
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