Question

1. Consider the following Cournot game between two firms. In each cell the first number gives the profit of firm 1 and the second number the profit of firm 2. Firm 2 q2 = 1 q2 = 2 q1 = 1 12, 12 0,50 Firm 1 q1 = 2 50,0 2,2 Firm 2 tells firm 1: "today, let's call it date 0, I will produce 1 unit. At every date t I will produce 1 unit if and only if you produced 1 unit up to date t-1. If at any date t you produce 2 units, then from date t+1 onwards I will also produce 2 units". Suppose that firm 1 believes that firm 2 will follow this strategy. Suppose that the rate of interest is 10%. A What is the value of the discount factor 8? B. Assuming an infinite horizon, is it better for firm 1 to always produce 1 unit or to produce 1 unit up to date 9 and then switch to 2 units from date 10 onwards?

Answer 1

A).

Consider the given problem here the interest rate is “r=10%=0.1”, => the discount factor is “d=1/1+r=1/1.1”.

B).

Now, given the strategy in the question the payoff of “firm1” if he always produces “1 unit” is given by.

=> (12 + 12*d + 12*d^2 + …..) = 12*(1 + d + d^2 + …..) = 12*(1 + d + d^2 + …..) = 12/1-d = 12/0.0909 = 132.0132.

Now, let’s assume that “firm 1” will produce 1 unit or to produce 1 unit up to date 9 and then switch to 2 units from date 10 onwards, => the payoff of “firm1” is given by.

=> (12 + 12*d + 12*d^2 + 12*d^3 + 12*d^4 + 12*d^5 + 12*d^6 + 12*d^7 + 12*d^8 + 12*d^9 + 50*d^10 + 2*d^11 + 2*d^12 + 2*d^13 + …..).

=> 12*(1-d^10)/(1-d) + 50*d^10 + 2*d^11(1 + d + d^2 + …..).

=> 12*(1-d^10)/(1-d) + 50*d^10 + 2*d^11/(1 - d).

=> 12*0.614457/0.090909 + 19.2772 + 0.700988/0.090909.

=> 81.108405 + 19.2772 + 7.710876 = 108.0965 < 132.0132.

So, here the payoff of producing “1 unit” is more than produce “1 unit” up to date 9 and then switch to 2 units from date 10 onwards, => “firm 1” will produce only “1 units”.