Imagine that there are 3 firms in a monopolistic market, F1, F2 and F3. Firms 1 and 2 are incumbent firms and act simultaneously whereas Firm 3 observes the actions of both firms before deciding whether to enter.

All firms face a market price of $P(Q) = 16-q_1-q_2-q_3$. Incumbent firms face a production cost of $C_i (q_i) = 4q_i$ for $i= 1,2$. Firm 3 faces a production cost of $C_3(q_3) = 4 + 4q_3$. Find both firm 3 and firm 1's payoff.

I understand that I need to take firm 3's best response function which is $$0.5 (12-q_2-q_1)$$ but I am not sure where to go from there. Do I substitute this back into the price equation? I tried doing this but it wouldn't yield any definitive answers. Thanks!

  • $\begingroup$ I just posted a hint because we generally don't do homework questions here. $\endgroup$ – Bayesian May 22 '19 at 10:53
  • $\begingroup$ Your best response is not right. you should have $q_3$ as a function of $q_1$ and $q_2$. I'll just give you the road map. Player 3 observes $q_1$ and $q_2$ so take these two as parameters. Then, his profit function is $\pi(q_3)=(16-q_1-q_2-q_3)*q_3-(4+4q_3)$. To obtain the best response you should take the derivative $\pi'(q_3)$ and set it equal to zero. You solve that equation for $q_3$ and that is F3's best response. You plug your result in F1's and F2's profit function. Lastly, you take FOC's of these two and solve for $q_1^*, q_2^*$, and plug these into F3's best response to obtain $q_3^*$ $\endgroup$ – Regio May 23 '19 at 5:21
  • $\begingroup$ Sorry i miss-wrote the function I have fixed it now, thanks $\endgroup$ – Oliver May 25 '19 at 2:49

You are right that you first have to find F3's best-response function. F1 and F2 take as given this reaction of F3 to whatever they produce. Hence, you plug this best-response function into the incumbents' profit maximization problem. In that way, you take care of the fact that the incumbents anticipate F3's reaction, indirectly determining $q_3$.

You should also check if the resulting reaction quantity $q_3$ is non-negative! It can be optimal to deter the entry of F3 completely.

  • $\begingroup$ I tried this, but i just get: Max: (16-q1-(6-0.5q1-0.5q2)q1-4q1--> 0.5q3q1-0.5q2q1 = 0 --> FOC: -0.5q3-0.5q2 = 0 --> How do i determine the actual value of q3? $\endgroup$ – Oliver May 22 '19 at 23:59
  • $\begingroup$ The reaction function of F3 is some $q_3(q_1,q_2)$. Plugged into the maximization problem of F1, you get some $q_1(q_2)$. Analoguously, you can find $q_2(q_1)$. Plug them into each other for some $q_1^*$ and $q_2^*$, which then solve $q_3^*=q_3(q_1^*,q_2^*)$. $\endgroup$ – Bayesian May 23 '19 at 8:33

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