1. (Arbitrage) A supplier is selling hammers in two cities, Pleasantville and Ha

Question

1. (Arbitrage) A supplier is selling hammers in two cities, Pleasantville and Happy Valley. It costs him $5.00 per hammer delievered in each city. Let p1 be the price of hammers in Pleasantville and p2 be the price of hammers in Happy Valley. The price-response curves for each city are:

Pleasantville: d1(p1) = 10,000-800p1

Happy Valley: d2(p2) = 8,000-500p2

a.) Assuming the supplier can charge any prices he likes, what should he charge for hammers in Pleasantville and Happy Valley to maximize total contribution? What the corresponding demands and total contributions?

b) An enterprising arbitrageur discovers a way to transport hammers from Pleasantville to Happy Valley for $0.50 each. He begins buying hammers in Pleasantville and shipping htem to Happy Valley to sell. Assuming the supplier does not change his proces from those given in part a, what will be the optimal price for the arbitrageur to sell hammers in Happy Valley? How many will he sell? What will his total contribution be? (Assume Happy Valley customers will buy hammers from the cheapest vendor.) What will happen to the total sales and contribution for the supplier? (Remember that he is now selling to the arbitageur too.)

c.) The supplier decides to eliminate the arbitrage opportunity by ensuring that his selling price in Happy Valley is no more than $0.50 more than the selling price in Pleasantville (and vice versa). What is his new selling price in each city? What are his corresponding sales and total contribution?

d.) From among the Pleasantville buyers, the Happy Valley buyers, and the sellers, who wins and who loses from the threat of arbitrage?

Explanation / Answer

a) the revenue at pleasentville R1= (10000-800*p1)*p1 = 10000p1 - 800p1^2 it will be maximum when dR1/dp1 = 0

so 10000 - 1600p1 = 0 so p1 = $6.25 d1 = 10000-800*6.25 = 5000 and total revenue = $31250

the revenue at Happy valley R2= (8000-500*p2)*p2 = 8000p2 - 500p2^2 it will be maximum when dR2/dp2 = 0

so 8000 - 1000p2 = 0 so p 2= $8 d2 = 8000-500*8 = 4000 and revenue = $32000 so total revenue = 63250

b) As arbitrageur has found a way for to sell to Happy valley(H) at $0.5 higher than pleasentville(P) so now p1 = 6.25 and p2 = 6.75 So demand for P will remain 5000 but demand at H will rise to =8000-6.75*500 = 4625. His revenue will be $31218.75 But people of H will not buy from supplier but they will buy from arbitrageur so supplier will sell 5000+4625 = 9625 to pleasentville (including arbitrageur) at $6.25 and total revenue will be 9625*6.25 = $60156.25

c) Now p2 = p1 + 0.5 and best condition will be dR1/dp1 + dR2/dp2 = 0 so 10000 - 1600p1 + 8000 - 1000p2 = 0

this is 18000 = 1600p1 + 1000(p1+0.5) solving it we get p1 = $6.73 and p2 = $7.23 d1 = 4616 and d2 = 4385 total revenue of supplier is $62769.23

d) so we see that customers in Pleasentville and the supplier loses in arbitration while customer of Happy valley were benefited.

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