Problem Set 3: Solutions

Econ 502: Advanced Microeconomics

Part I: Benchmarks

Throughout: \(P = 150 - Q\), \(c = 30\), so \(a - c = 120\).

Part (a): Perfect competition

Under perfect competition, \(P = MC\):

\[150 - Q = 30 \quad \Longrightarrow \quad Q_c = 120, \quad P_c = \$30\]

Consumer surplus is the triangle between the demand curve and the price (see figure below):

\[CS_c = \frac{1}{2}(150 - 30)(120) = \$7{,}200\]

Producer surplus is zero (\(P = MC\) with constant marginal cost).

\[\boxed{P_c = \$30, \quad Q_c = 120, \quad CS_c = \$7{,}200, \quad TS_c = \$7{,}200}\]

Part (b): Monopoly

Total revenue: \(TR = (150 - Q)Q = 150Q - Q^2\). Marginal revenue:

\[MR = 150 - 2Q\]

Setting \(MR = MC\):

\[150 - 2Q = 30 \quad \Longrightarrow \quad Q_m = 60, \quad P_m = 90\]

\[\pi_m = (90 - 30)(60) = \$3{,}600\]

\[CS_m = \frac{1}{2}(150 - 90)(60) = \$1{,}800\]

\[TS_m = CS_m + \pi_m = 1{,}800 + 3{,}600 = \$5{,}400\]

\[DWL = TS_c - TS_m = 7{,}200 - 5{,}400 = \$1{,}800\]

This equals the triangle: \(\frac{1}{2}(P_m - c)(Q_c - Q_m) = \frac{1}{2}(60)(60) = \$1{,}800\).

\[\boxed{Q_m = 60, \quad P_m = \$90, \quad \pi_m = \$3{,}600, \quad CS_m = \$1{,}800, \quad DWL = \$1{,}800}\]

Part II: Price Competition

Part (c): Bertrand

With \(n \geq 2\) firms competing on price, the Nash equilibrium is:

\[\boxed{p_1 = p_2 = \cdots = p_n = c = 30}\]

This replicates the perfectly competitive outcome: \(P = 30\), \(Q = 120\), zero profits.

Why? Suppose all firms charge some \(p > c\). Any single firm can capture the entire market by cutting its price to \(p - \epsilon\), roughly multiplying its profit. This undercutting incentive continues until price is driven down to marginal cost. At \(p = c\), no firm can profitably deviate: raising the price loses all customers, and lowering it means selling below cost.

This is the Bertrand paradox: just two firms are enough to eliminate all market power and reproduce the competitive outcome.

Part III: Quantity Competition

Part (d): Cournot duopoly (\(n = 2\))

Firm 1’s profit:

\[\pi_1 = (150 - q_1 - q_2)q_1 - 30q_1 = (120 - q_1 - q_2)q_1\]

First-order condition:

\[\frac{\partial \pi_1}{\partial q_1} = 120 - 2q_1 - q_2 = 0\]

Best response function:

\[\boxed{q_1^*(q_2) = 60 - \frac{q_2}{2}}\]

By symmetry, \(q_2^*(q_1) = 60 - q_1/2\). In the symmetric equilibrium \(q_1 = q_2 = q^*\):

\[q^* = 60 - \frac{q^*}{2} \quad \Longrightarrow \quad q^* = 40\]

\[Q^* = 80, \qquad P^* = 70, \qquad \pi^* = (70 - 30)(40) = \$1{,}600\]

\[\boxed{q^* = 40, \quad P^* = \$70, \quad \pi^* = \$1{,}600 \text{ per firm}}\]

Part (e): Cournot with \(n\) firms

Firm \(i\) maximizes:

\[\pi_i = \left(150 - q_i - \sum_{j \neq i} q_j\right)q_i - 30q_i\]

First-order condition:

\[120 - 2q_i - \sum_{j \neq i} q_j = 0\]

In a symmetric equilibrium, \(q_j = q^*\) for all \(j\), so \(\sum_{j \neq i} q_j = (n-1)q^*\):

\[120 - 2q^* - (n-1)q^* = 0 \quad \Longrightarrow \quad 120 = (n+1)q^*\]

\[\boxed{q^* = \frac{120}{n+1}}\]

Part (f): Convergence

Remember \(c=30\) and: \[ q^* = \frac{120}{n+1}, \quad P^* = 150-nq^*, \quad \pi^* = (P^* - c)q^*, \quad \mu = \frac{P^*}{c}\]

\(n\)	Per-firm output (\(q^*\))	Price (\(P^*\))	Per-firm profit (\(\pi^*\))	Markup (\(\mu\))
1	60	\(\$90\)	\(\$3{,}600\)	\(3.00\)
2	40	\(\$70\)	\(\$1{,}600\)	\(2.33\)
5	20	\(\$50\)	\(\$400\)	\(1.67\)
10	\(10.9\)	\(\$40.9\)	\(\$119\)	\(1.36\)
50	\(2.35\)	\(\$32.9\)	\(\$6.8\)	\(1.09\)

As \(n\) gets large, price gets closer to marginal cost (\(P^* \to 30\)) and the markup \(\mu\) approaches 1, replicating the competitive outcome.

Part IV: Welfare

Part (g): Deadweight loss

DWL is the triangle between the demand curve and the price, from \(Q^*\) to \(Q_c\):

\[DWL = \frac{1}{2}(P^* - c)(Q_c - Q^*)\]

Note that \[ Q^* = nq^* = \frac{120n}{n+1} \]

Since \(P^* = 150 - Q^*\), we can write:

\[P^* = 150 - \frac{120n}{n+1} = \frac{150 + 30n}{n+1}\]

Therefore:

\[DWL = \frac{1}{2}\left(\frac{150 + 30n}{n+1} - 30\right)\left(120 - \frac{120n}{n+1}\right) = \frac{7{,}200}{(n+1)^2}\]

Comparison:

• Monopoly (\(n = 1\)): \(DWL = 7{,}200/4 = \$1{,}800\)
• Cournot duopoly (\(n = 2\)): \(DWL = 7{,}200/9 = \$800\)

Going from monopoly to duopoly cuts the DWL by more than half. The DWL shrinks as \(1/(n+1)^2\), so adding more firms reduces welfare loss rapidly. By \(n = 5\), DWL is only \(\$200\) (compared to \(\$1{,}800\) under monopoly).