HW-5 sol

HW-5 sol - ECON101 Homework 5 Suggested Solutions 3/10/2011...

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Unformatted text preview: ECON101 Homework 5 Suggested Solutions 3/10/2011 1 Question: Nancy lives in a gated community of 100 homes (99 + Nancy) that is separated from Stone Canyon Reservoir by 1,000 feet of fencing. The fencing is old, and coyotes are wandering in to the communityl this is not good. The community association wants to repair the fence to prevent coyotes from coming in. The cost of repairing the fence is $10/foot. All the residents are just like Nancy: if F feet of fence are repaired and a given resident pays c towards the repair, then that resident derives utility 1 , 000- (1 , 000- F ) 1 / 2- c . (Notice that the benefit comes from the total amount of fence repaired, that benefit is increasing in amount of fence repaired, and that repairing the last foot of fencing is more important than repairing the first foot.) a Suppose the community association can decide to repair some or all of the fence and divide the cost equally among all 100 homes. What is the optimal amount of fence repair? (Remember that the maximum is 1,000 feet.) b Suppose instead that the community association asks for voluntary contributions. If each resident acts perfectly selfishly, how much will each resident contribute? How much total fence will be repaired? Solution: a The community association wishes to maximize everybodys utility. We will assume that the association views each resident the same, and therefore wishes to solve: max 100 X j =1 (1000- (1000- F ) 1 / 2- c j ) Since the association is going to split the total cost of F , C = 10 F evenly among the 100 residents, the problem then becomes: max 100 X j =1 (1000- (1000- F ) 1 / 2- 10 F 100 ) = 100 , 000- 100(1000- F ) 1 / 2- 10 F Taking the first order condition with respect to F yields: F.O.C.: 100 1 2 (1000- F )- 1 / 2- 10 = 0 5 = (1000- F ) 1 / 2 25 = 1000- F F = 975 This implies that the optimal amount of fencing repair is 975 feet and that each consumer will pay 10 * 975 100 = 97 . 5 and derive a utility of 1 , 000- (1 , 000- 975) 1 / 2- 97 . 5 = 897 . 5. However, for the result of the first order condition to be a solution to the utility maximization problem, the second order condition must hold, or: 2 U F 2 F =975 < must hold. Computing the second derivative yields: 2 U F 2 = 100 1 4 (1000- F )- 3 / 2 > , F < 1000 1 so F = 975 as a result of the first order condition does not solve the maximization problem. We have instead found a minimizer of the utility function. Since the second order condition does not hold for the only solution that satisfies the first order condition, we look towards the boundaries. Checking the boundaries, we set F = 0 and therefore c i = 0 , i to find that the utility for any one resident is: U i (0) = 1000- 1000 968 . 38 > 897 . 5 If we set F = 1000, then c i = 100 and the utility derived is: U i (1000) = 1000- sqrt (1000- 1000)- 100 = 900 < 968 . 38 Therefore, the community utility is higher when F = 0, so the total amount of fencing that should be repaired is 0...
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This note was uploaded on 10/23/2011 for the course ECON 101 taught by Professor Buddin during the Winter '08 term at UCLA.

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HW-5 sol - ECON101 Homework 5 Suggested Solutions 3/10/2011...

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