NUMERICAL_EXAMPLE

# NUMERICAL_EXAMPLE - 10-2 x = 1 2 10-2 S-x 29 29 x M = 5 S 3...

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Numerical Examples Used in Class for Nonrenewable Resources Optimal Solutions vs. Monopoly Suppose demand = B / x = 10 - x , Extraction cost = 0 r = 1, interest rate of 100% We know that x 0 and x 1 are constrained to be smaller than 10 since, at x = 0, B x x = 10 ( 29 = 0. The Optimal Solution as Function of S 0 if the Constraints Are Binding B x x 0 ( 29 = 1 1 + r B x x 1 ( 29 = λ 10 - x 0 = 1 2 10 - S 0 - x 0 ( 29 [ ] which implies x 0 = 10 + S 0 3 x 1 = S 0 - x 0 If S 0 = 6, x 0 = 16 3 , x 1 = 2 3 S 0 = 8, x 0 = 6, x 1 = 2 S 0 = 10, x 0 = 6 2 3 , x 1 = 3 1 3 S 0 = 12, x 0 = 7 1 3 , x 1 = 4 2 3 Monopoly Solutions Note that x cannot exceed 5 since MR x ( 29 = 10 - 2 x

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2 At x = 5, marginal revenue = 0; therefore, production will not exceed 5. The optimality conditions of monopoly are λ 0 M = MR x 0 ( 29 = 1 1 + r MR x 1 ( 29 = 1 1 + r MR S 0 - x 0 ( 29 when the resource constraints are binding
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Unformatted text preview: 10-2 x = 1 2 10-2 S-x ( 29 ( 29 x M = 5 + S 3 . For S = 6, x M = 11 3 , x 1 M = 7 3 when S = 8, x M = 13 3 , x 1 M = 11 3 . In both cases, consumption in period 0 is lower under monopolistic solutions than under optimal solutions. For S = 10, x M = x 1 M = 5 For S = 12, x M = x 1 M = 5. For cases with S 10, the monopoly solutions does not change. However, optimal solutions will utilize all the resources for S ≤ 20. 3...
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NUMERICAL_EXAMPLE - 10-2 x = 1 2 10-2 S-x 29 29 x M = 5 S 3...

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