Question

2. Consider a two-team league in which the teams play 100 games. The large market team gets gate revenue 12w – (w²/20) if it wins w games while the small market team gets 8w – (w²/20) if it wins w games. There are no other sources of revenue.

i) Find the equilibrium number of wins for each team and the marginal cost of a win.

ii) Suppose that there is revenue sharing, with each team keeping 50% of its revenue and receiving 50% of the other team’s revenue. Determine the equilibrium number of wins for each team and the marginal cost of a win.

iii) Now suppose that there is a payroll cap of 150 for each team (and no revenue sharing). Find the equilibrium number of wins for each team and marginal cost of a win. How do the results change if the cap is 100 for each team?

Answer 1

i) Total Revenue of large market team(R) = 12w-w² /20

Taking derivative of either side with respect to w, we get,

R/w = 12-2w/20 or, R/w = 12-w/10 which is marginal revenue of the large market team.

Under profit maximization, marginal revenue = 0

therefore R/w =0 or, 12-w/10 = 0 or w =120. But the maximum value of w =100

Therefore total revenue of the large market team when w=100, is 12*100-(100)² /20=700

Therefore, average revenue of a win for large market team = 700/100 = 7

under equilibrium condition,

                           Average Revenue or P = Marginal Cost

Therefore marginal cost of a win for the large market team is 7

Total revenue of small market team is R = 8w--w² /20

therefore, marginal revenue or R/w = 8-w/10

Under profit maximization rule, marginal revenue = 0 or 8-w/10 =0 or w=80

Therefore total revenue of winning 80 games is R = 8.80- (80)²/20= 640-320=$320

Therefore Average Revenue or marginal cost of a win for small market team is $320/80 = $4 per win.

ii) If each team shares 50% of its total revenue with other then total revenue of each team will be R= 1/2{ (12w -w² /20) +(8w-w²/20)} = 1/2(20w-w² /10) = 10w-w²/20

Therefore, marginal revenue of each team will be R/w = 10-w/10

Under profit maximization rule, 10-w/10=0 or w= 100 which is equilibrium number of wins of each team

Therefore, total revenue of each team = 10w-w²/20 = 10.100 - (10)²/20= 1000-500= $500

Therefore average revenue or marginal cost of each team = $500/100 = $5 per win.

iii) If there is payroll cap of 150 for each team, then total revenue of large market team = 12w-w² /20-150

marginal revenue of large market team = 12-w/10 or w=120 (under profit maximiization rule from (i))

Also total revenue of the large market team under equilibrium condition will be 700 (From (i))

Therefore, total revenue of large market team = 700-150 = 550

and Average revenue or marginal cost for a win for large market team = 550/100= 5.5

Equilibrium number of wins for small market team is 80

Total revenue of small market team = 8w-w²/20 -150 = 320-150 = 170

and marginal cost of a win = 170/80= 2.125

If payroll cap is made 100 for each team,

Then total revenue of large market team for winning 100 games = 700-100= 600

and marginal cost of a win = 600/100 = 6

Also, total revenue of small market team = 320-100= 220

And marginal cost of a win = 220/80= 2.75