(a) Consider the following Table: Cows in field Yield per cow Total Yield 1 8 8 2 5 10 3 3 9 4 2 8
To maximize total milk production (social optimum) the men should graze 2 cows total. But consider the following payoff table: Mr. Two x2=1 0,8 5,5 6,3
x1=0 x1=1 x1=2
x2=0 0,0 8,0 10,0
x2=2 0,10 3,6 4,4
Grazing two cows is a dominant strategy for each farmer. In equilibrium they each receive 4 quarts, even though it is possible to obtain 5. If Mr. One grazes two cows he can pay Mr. Two not to graze any, but he must pay him at least 4 quarts per day, that being what Mr. Two could get by grazing his own cows. Paying 5 or 6 quarts will work as well. But if the price for Mr. Two to graze no cows were 7 quarts, then Mr. One would prefer to make no deal, since he can be assured of getting 4 quarts with 2 cows rather than just getting 3 quarts. Alternatively, Mr. Two can graze the cows and pay Mr. One not to. Finally, each can agree to graze only one cow. This is probably the most sensible outcome. Each gets 5 quarts of milk and no transfers are necessary. (b) There are two possible kinds of equilibria, cooperative and non-cooperative. If the men make no deal, then each will graze two cows and get 4 quarts per day, since grazing 2 cows is a strongly dominant strategy. But this equilibrium is unlikely, since it is easy to do better. Transfer payments of milk are possible. Communication is easy (since they are on the same field). It is easy to enforce a contract by withholding milk and easy to observe when the contract is broken (when the other guy grazes cows even though he promised not to.) So a cooperative equilibrium where one farmer grazes two cattle and pays the other to graze none or where each promises only to graze one cow is most likely, particularly in a situation where the men are repeating the game day in and day out.
(c) Mr. i gets Qi=(250 - X)xi = (250 - X -i - xi )xi , where X -i is the total...