Unformatted text preview: 1 / 3 . d. With a fee of 60, farm 1 sets 60 = 20 + 2 / 3 R , and calculates R 1 = 60. Farm 2 calculates 60 = 20 + 2 R 2 or R 2 = 20. Again, the average reduction is 40. Total costs now for farm 1 are 20 × 60 + 1 / 2 (60 – 20)(60) = 1,200 + 1,200 = 2,400 and for farm 2 20 × 20 + 1 / 2 (60 – 20)(20) = 400 + 400 = 800 so total costs are 3,200. The fee achieves the same average reduction at a much lower total cost. e. The fee procedure recognizes the differential costs of methane reduction. By having the low cost farm (farm 1) make relatively greater reductions, overall costs are reduced....
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 Fall '09
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 Economics, Fee, average reduction

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