A question about Nash Equilibrium

Question

I have some trouble with Nash Equilibrium. The specific question as follows.

Suppose that there are $2N$ people in the village, of which $N$ residents live in the first district, and each person chooses to raise $q_1$ sheep, and the cost of each sheep is $c_1$. In addition, n people live in the second district, and each person chooses to raise $q_2$ sheep. The cost of a sheep is $c_2$. The income brought by each sheep is $200-q$, where $q$ is the total number of sheep in the village.
Q: Find the number of sheep raised by each resident in the two regions under the Nash equilibrium of this game.

Here is my thought. For simplicity, let $I$ denotes the index set of individuals in the first district, similarly, $II$ for the second district.

For $\forall i\in I$, the profit maximization problem is \begin{equation} \max_{q_i} \pi_i=(200-q_i-\sum_{k\neq i,\\i\in I}q_k-\sum_{l\in II}q_l)q_i-c_1q_i\\ F.O.C\qquad \frac{\partial \pi_i}{\partial q_i}=200-2q_i-\sum_{k\neq i\\k\in I}q_k-\sum_{l\in II}q_l-c_1=0\\ BR_{q_i}(q_{-i})=q_i=100-\frac{1}{2}\sum_{k\neq i\\k\in I}q_k-\frac{1}{2}\sum_{l\in II}q_l-\frac{1}{2}c_1 \end{equation}

Similarly, for $\forall j\in II$, we have best reaction function \begin{equation} BR_{q_j}(q_{-j})=q_j=100-\frac{1}{2}\sum_{k\in I}q_k-\frac{1}{2}\sum_{l\neq j\\l\in II}q_l-\frac{1}{2}c_2 \end{equation}

Since every individual belonging to the same area is in the same situation, we can assume that \begin{align} \begin{cases} q_i^*=a, &\forall i\in I\\ q_j^*=b, &\forall j\in II \end{cases} \end{align} and then substitute it into the best reaction function, we have \begin{align} \begin{cases} (N+1)a+Nb=200-c_1\\ Na+(N+1)b=200-c_2 \end{cases} \end{align} \begin{align} \begin{cases} a=\frac{ \begin{vmatrix} 200-c_1 & N\\ 200-c_2 & N+1 \end{vmatrix}}{\begin{vmatrix} N+1 & N\\ N & N+1 \end{vmatrix}}\\ b=\frac{ \begin{vmatrix} N+1 & 200-c_1\\ N & 200-c_2 \end{vmatrix}}{\begin{vmatrix} N+1 & N\\ N & N+1 \end{vmatrix}} \end{cases} \end{align}

I don’t know if I’m doing it right, and I’m going to solve it under the assumption that there is an interior point solution. I think if it is strictly proved, the cost $c_1,c_2$ may be considered.

Answer 1

From symmetry we can say that for each player $i$ in $I$, $q_r=q_k$ $\forall r,k \in I$. Similarly for $II$. Let this be $q_I$ for producers in $I$ and $q_{II}$ for $II$.

So profit function for individual $i$ in $I$: \begin{align} \max_{q_i} \pi_i&=(200-(N-1)q_I-Nq_{II}-q_i))q_i-c_1q_i\\ \text{FOC: }\,\,\frac{\partial \pi_i}{\partial q_i}&=200-(N-1)q_I-Nq_{II}-q_i-q_i-c_1=0 \\ \implies q_i^*(q_I,q_{II})&=\frac{200-Nq_{II}-c_1-(N-1)q_I}{2} \end{align}

Similarly, we will get for individual $j$ in $II$:

$$q_j^*(q_I,q_{II})=\frac{200-Nq_I-c_2 - (N-1)q_{II}}{2N}$$

At NE, we know that $q_i^*=q_I^*$ and $q_j^*=q_{II}$. Using this:

\begin{align} q_I^*&=\frac{200-Nq_{II}^*-c_1-(N-1)q_I^*}{2} \\ &=\frac{200-Nq_{II}^*-c_1}{N+1} \end{align}

And so, $$q_{II}^*=\frac{200-Nq_{I}^*-c_2}{N+1}$$

Solving for $q_I^*$:

\begin{align} (N+1)q_I^*&=200-N\frac{200-Nq_I^*-c_2}{N+1}-c_1 \\ \implies q_I^* &=\frac{200-c_1+N(c_2-c_1)}{2N+1} \end{align} Similarly, $$q_{II}^* =\frac{200-c_2+N(c_1-c_2)}{2N+1}$$