Stackelberg - 1 : q & 1 = 280 = 3 = 93(1 = 3) (2) Also...

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Stackelberg Solution: a numerical example For comparison with the Cournot example, consider the same model with p = 100 (0 : 5)[ q 1 + q 2 ] C 1 = 5 q 1 C 2= = (0 : 5) q 2 2 The second duopolist is a follower. So, as in the Cournot model, derive her reaction function as q 2 = 50 (0 : 25) q 1 (1) (1) into 1 = pq 1 C 1 ( q 1 ) = pq 1 5 q 1 = [ 100 (0 : 5)[ q 1 + q 2 ] q 1 5 q 1 (using (1) to eliminate q 2 = { 100 (0 : 5)[ q 1 + (50 (0 : 25) q 1 )] g q 1 5 q 1 Simplifying: = 70 q 1 ( : 0375) q 2 1 Setting 1 =dq 1 = 0
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Unformatted text preview: 1 : q & 1 = 280 = 3 = 93(1 = 3) (2) Also check that the second order condition is satis&ed. & & 1 = 3266(2 = 3) From (1) and (2) we get q & 2 = 80 = 3 = 26(2 = 3) & & 2 = 155(5 = 9) Finally, p & = 100 & (0 : 5)[ q & 1 + q & 2 ] = 100 & (1 = 2)[(280 + 80) = 3] = 100 & 60 = 40 Compare the numbers with those obtained in the Cournot model: q & 1 = 80; q & 2 = 30; p & = 45 ; & & 1 = 3200; & & 2 = 900 : 1...
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This note was uploaded on 12/23/2009 for the course ECON 3130 at Cornell University (Engineering School).

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