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) For Pleasentville P revenue = p1d1 = 10000p1-800p1^2 it will be maximum when dR1/dp1 = 0 i.e. 10000-1600p1 = 0 so p1 =$6.25 amd quantity d1 = 10000-800*6.25 = 5000 sellers revenue R1 = 5000*6.25 = $31250

For Happy valley H revenue = p2d2 = 8000p2-500p2^2 it will be maximum when dR2/dp2 = 0 i.e. 8000-1000p2 = 0 so p2 =$8 amd quantity d2= 8000-500*8 = 4000 sellers revenue R2= 4000*5 = $32000 So total revenue TR1 is $63250

b) Now due to arbitrageur p1 will remain $6.25 but $p2 = 6.75 so d2 will be = 8000-500*6.75 = 4625 but these units will be purchased from arbitrageur not directly from seller and as he is from P to d2 = 0 and d1 = 5000+4625 = 9625. Sellers revenueTR2 = $60156.25 and arbitrageur's revenue = $31218.75

c) As given p2 = p1+0.5 now to maximize revenue R which is dR1/dp1 + dR2/dp2 = 0 so 18000-1600p1-1000p2 = substituting p2 we get 17500-2600p1=0 sp p1 = $6.73 p2 will be $7.23 now d1 = 4616 and d2 = 4385 and new total revenue TR3 = $62769.23

d) We see that as p1 increased so people of P were at loss, p2 decreased to people of H were at profit and as total revenue also decreased so seller is also at loss.

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