OLIGOPOLY Intermediate Microeconomics 1. π. ππππ£π πππ ππ 1000000 ππ = q − 10ππ ππ + ππ a ππ 1000000ππ = − 10ππ πππ ππ + ππ ππ 1000000(ππ + ππ ) − (1000000ππ )(1) = − 10 (ππ + ππ )2 πππ 1000000ππ = 10 (ππ + ππ )2 (ππ + ππ )2 ππ = 100,000 (ππ + ππ )2 πππππππππ¦, ππ = 100,000 πππππ ππ = ππ (2ππ )2 ππ = 100,000 4ππ2 ππ = 100,000 ππ2 ππ = 25000 ππ∗ = 25000 ππ∗ = 25000 ππππ£πππ πππ πππππ 1000000 π= 25000(2) ∗ π = ππ 2. π. ππππ£π πππ π 1 (ππ )πππ π 2 (ππ ) ππππππ‘ = [320 − 4(ππ + ππ )]ππ − 20ππ π = 320 − 8ππ − 4ππ − 20 πππ 8ππ = 300 − ππ ππ ππ (ππ ) = 37.5 − 2 Since both firms face the same cost and demand function then π 1 (ππ ) = π 2 (ππ ) ππ = 37.5 − 37.5 − ππ 2 2 37.5 ππ ππ = 37.5 − + 2 4 3ππ = 18.75 4 ππ = 25 ππ = 25 π = 320 − 4(25 + 25) π = 320 − 200 π = 120 3. πππππ π = ππ§π + πππ’ π(π) = 900 − 2πππππ π π(π) = 900 − 2(ππ§π + πππ’ ) ππ§π = ππ§π (900 − 2(ππ§π + πππ’ )) 2 ππ§π = 900ππ§π − 2ππ§π − 2πππ’ ππ§π πππ§π = 900 − 4ππ§π − 2πππ’ ππ1 900 − 4ππ§π − 2πππ’ = 0 πππ’ ππ§π = 225 − 2 πππππππππ¦, ππ§π πππ’ = 225 − 2 Substituting pcu in pzn π 225 − π§π 2 ππ§π = 225 − 2 225 ππ§π ππ§π = 225 − + 2 4 3 225 π = 4 π§π 2 ππ§π = 150 πππ’ = 150 πππππ π = ππ§π + πππ’ πππππ π = 150 + 150 πππππ π = 300 4. ππ = [160 − 2(ππ + ππ )]ππ − 10ππ ππ = 160ππ − 2ππ2 − 2ππ ππ − 10ππ πππ = 160 − 4ππ − 2ππ − 10 πππ 4ππ = 150 − 2ππ ππ ππ (ππ ) = 37.5 − 2 Since both firms face the same cost and demand function then π 1 (ππ ) = π 2 (ππ ) ππ = 37.5 − 37.5 − ππ 2 2 37.5 ππ ππ = 37.5 − + 2 4 3ππ = 18.75 4 ππ = 25 ππ = 25 5. ππ = [180 − 3(ππ + ππ )]ππ ππ = 180ππ − 3ππ2 − 3ππ ππ πππ = 180 − 6ππ − 3ππ πππ 6ππ = 180 − 3ππ ππ ππ (ππ ) = 30 − 2 6. Solve for Firm B’s reaction function ππ = (110 − 0.5ππ − 0.5ππ )ππ − 10ππ ππ = 110ππ − 0.5ππ2 − 0.5ππ ππ − 10ππ πππ = 110 − 0.5ππ − ππ − 10 πππ ππ = 100 − 0.5ππ Compute for Leader’s output using Follower’s reaction function ππ = (110 − 0.5ππ − 0.5ππ )ππ − 10ππ ππ = (110 − 0.5ππ − 0.5(100 − 0.5ππ ))ππ − 10ππ ππ = (110 − 0.5ππ − 50 + 0.25ππ )ππ − 10ππ ππ = (60 − 0.25ππ )ππ − 10ππ πππ = 60 − 0.5ππ − 10 πππ 0.5ππ = 50 ππ = 100 Compute for Follower’s output 100 ππ = 100 − 2 ππ = 50 7. πππππ πΆπππΈπΆ πππππ ππ‘ πππ‘β ππππππ‘ππππ π‘π π πππ ππ‘ π‘βπ πππ£ππ ππππππ , π‘βππ ππ 1 = ππ 2 1 ππ 1 = ππ 2 = π2 (1 + ) π 1 100 = 150 (1 + ) π 150 100 = 150 + π 150 100 − 150 = π −50π = 150 π = −3 8. D. the price elasticity of demand is equal to 1. Since firms in a cartel behave like a monopoly, they will produce where market demand is inelastic. 9. Monopoly Output π = (20 − π)π − 8π π = 20π − 8π − π 2 ππ = 12 − 2π ππ πππππππππ¦ = 6 Cournot Output ππ = [20 − (ππ + ππ )]ππ − 8ππ ππ = 20ππ − ππ2 − ππ ππ − 8ππ πππ = 12 − 2ππ − ππ πππ 2ππ = 12 − 2ππ ππ ππ (ππ ) = 6 − 2 ππ (π ) ππ π = 6 − 2 π 6 − 2π ππ = 6 − 2 6 ππ ππ = 6 − + 2 4 3 π =3 4 π ππ = 4 ππ = 4 πππ‘ππ ππππ’π π‘ππ¦ ππ’π‘ππ’π‘ ππππ’ππππ‘ = 4 + 4 = 8 Stackelberg output ππ ππ = [20 − (ππ + 6 − )] ππ − 8ππ 2 ππ ππ = (14 − ) ππ − 8ππ 2 ππ = 14ππ − ππ = 6ππ − ππ2 − 8ππ 2 ππ2 2 πππ = 6 − ππ πππ ππ = 6 Type equation here. 6 ππ = 6 − 2 ππ = 3 The Monopoly output is 6. The total Cournot output for a duopoly is 8. A Stackelberg leader will produce 6 and a follower will produce 3. π 10. ππΆ1 = ππ¦ 2π¦ + 500 = 2 π 2π¦ + 400 = 2 ππ¦ πππππ ππΆ1 = ππΆ2 πππ π‘βπ ππππ πππππ π‘βπ π πππ ππππππ‘ ππππππ π(π), π‘βππ ππ π πππ’ππππ‘ πππ’πππππππ’π π 1 (π2 ) = π 2 (π1 ) πβπ’π , π1 = π2 πππ πππ‘β πππππ π€πππ πππππ’ππ π‘βπ π πππ πππ£ππ ππ ππ’π‘ππ’π‘. ππΆ2 =