StatstriveISI intelligencePractice
Back to MSQE practice
PEAMCQModerate

2023 PEA Q3

p=100−qp=100-q, q=∑i=123qiq=\sum_{i=1}^{23}q_i, ci(qi)=qi22c_i(q_i)=\dfrac{q_i^{2}}{2}. Cournot--Nash equilibrium:

Reveal answer and solution

Answer

C

Solution

  1. 1

    Firm ii maximises πi=(100−∑jqj)qi−qi22\pi_i=(100-\sum_j q_j)q_i-\dfrac{q_i^2}{2}. FOC:

  2. 2
    100−∑j≠iqj−2qi−qi=0  ⟹  100−Q−i−3qi=0. 100-\sum_{j\ne i}q_j-2q_i-q_i=0 \;\Longrightarrow\; 100-Q_{-i}-3q_i=0.
  3. 3

    By symmetry qi=q∗q_i=q^* and Q−i=22q∗Q_{-i}=22 q^*:

  4. 4
    100−22q∗−3q∗=0  ⟹  25q∗=100  ⟹  q∗=4. 100-22 q^*-3 q^*=0\;\Longrightarrow\; 25 q^*=100\;\Longrightarrow\; q^*=4.

Answer structure / marking notes

Marginal cost is c′(qi)=qic'(q_i)=q_i (not zero); forgetting the quadratic cost gives q∗=100/24q^*=100/24.

%------------------------------------------------------------------

Content note

Imported from public/resources/isi/msqe/solutions/pea/2023/ISI_MSQE_PEA_2023_Solutions.tex. Question wording is retained from the available local TeX source; incomplete option blocks or ambiguous source status are flagged for review.