Regulation and Antitrust Policy (Econ 180) Drake University, Spring 2011 William M. Boal Course page: www.drake.edu/cbpa/econ/boal/180 Blackboard: bb.drake.edu Email: william.boal@drake.edu

Oligopoly and Collusion

### Version A

I. Multiple choice [2 pts each: 16 pts total]

(1)b. (2)b. (3)c. (4)b. (5)b. (6)d. (7)a. (8)b.

II. Problems

(2) [Game theory: 12 pts]

1. low price.
2. low price.
3. There is only one Nash equilibrium: both firms play "low price."

(2) [Oligopolist's marginal revenue: 16 pts]

1. increase.
2. \$17 = MRA - MC = P + q (ΔP/ΔQ) - MC.
3. decrease.
4. \$3 = MRcartel - MC = P + Q (ΔP/ΔQ) - MC.

(3) [Collusion, joint profit maximization: 32 pts]

1. MR = 13 - Q.
2. MR curve is straight line with intercepts at P=13 and Q=13.
3. \$9.
4. 8 thousand.
5. \$8 thousand.
6. \$7.
7. 12 thousand.
8. \$0 thousand, because price = marginal cost.

(4) [Cournot duopoly: 20 pts]

1. P = (15 - qA/20) - qB/20.
2. MR = (15 - qA/20) - qB/10.
3. qB = 120 - qA/2.
4. qA* = qB* = 80.
5. P* = \$7.

III. Challenge question
Set MCA = MCB
12 + (qA/100) = 1 + (qB/50)
12 + (qA/100) = 1 + (1000-qA)/50
and solve to get qA* = 300 and qB* = 700.

### Version B

I. Multiple choice [2 pts each: 16 pts total]

(1)d. (2)d. (3)b. (4)d. (5)d. (6)b. (7)c. (8)c.

II. Problems

(2) [Game theory: 12 pts]

1. Standard #2.
2. Standard #1.
3. There are two Nash equilibria:
1. Both firms play "Standard #1."
2. Both firms play "Standard #2."

(2) [Oligopolist's marginal revenue: 16 pts]

1. increase.
2. \$9 = MRA - MC = P + q (ΔP/ΔQ) - MC.
3. decrease.
4. \$3 = MRcartel - MC = P + Q (ΔP/ΔQ) - MC.

(3) [Collusion, joint profit maximization: 32 pts]

1. MR = 14 - 2Q.
2. MR curve is straight line with intercepts at P=14 and Q=7.
3. \$10.
4. 4 thousand.
5. \$4 thousand.
6. \$8.
7. 6 thousand.
8. \$0 thousand, because price = marginal cost.

(4) [Cournot duopoly: 20 pts]

1. P = (18 - qA/10) - qB/10.
2. MR = (18 - qA/10) - qB/5.
3. qB = 75 - qA/2.
4. qA* = qB* = 50.
5. P* = \$8.

III. Challenge question
Set MCA = MCB
3 + (qA/100) = 1 + (qB/50)
3 + (qA/100) = 1 + (1000-qA)/50
and solve to get qA* = 600 and qB* = 400.