Economics 431 Final Exam 200 Points. Answer each of the questions below. Round off values to one decimal place where necessary.

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1 Fall 009 Name KEY Economics 431 Final Exam 00 Points Answer each of the questions below. Round off values to one decimal place where necessary. Question 1. Think (30 points) In an ideal socialist system, natural resources are owned by all the people rather than by individual, profitmaximizing firms. Socialist governments claim to work for the welfare of the people. However, the record seems to show that such socialist systems use their natural resources very inefficiently. In fact, many socialist governments are now attempting to privatize many of their industries to avoid such inefficiencies. How does the material we covered in this course help you to better understand this situation? Think before you start and keep your answer short and to the point. When resources are owned by "all the people", this results in a common property problem. Individuals fail to account for the effect of their resource use on the well-being of others (as in the 'give-to-the-group' game we played in class). Common property resources are non-excludable, and this leads to free-riding behavior, much as in the case of public good problems. Privatization is one way to resolve the common property problem, because privatizing resources creates property rights. Question. (30 points) Environmental Policy and Firm Profitability Draw a diagram that shows a competitive equilibrium with a negative pollution externality. Define the competitive output level and price as (Q c,p c ). Show on your diagram that a pollution standard (or quota) that achieves the socially optimal resource allocation (Q*) can increase firm profits. $ D = MSB MSC Pc P* Gain in Rent Loss in rent MPC MEC Q* Qc Environmental policy that internalizes external costs raises consumer prices, which can benefit firms is the increase in revenue (the rise in price, P* - Pc times the equilibrium quantity under the standard, Q*) exceeds the loss in producer surplus on units prohibited under the standard (the triangle shown in the figure). Q

2 Question 3. (30 points) Controlling Pollution at the Widget Factory Suppose a firm uses two inputs to produce widgets, labor (L) and capital (K). Capital pollutes and labor does not. Suppose further that the output level of the firm at an optimal allocation is 0 percent lower than the current (unregulated) level. Draw a diagram that shows that a standard on output that requires the firm to produce 0 percent less output is suboptimal. Show the optimal solution on your diagram, explain how it is different than a standard requiring a 0 percent reduction of output, and describe types of alternative policies that could attain the optimum. L L* L 0 L 1 Q 0 Q 1 K* K 1 K 0 C/r* C 1 /r C 0 /r K The initial production level is shown as the isoquant Q 0. The cost of producing Q 0 is C 0, where C 0 = wl + rk is the isocost line. The K-intercept is therefore C 0 /r. If a production standard is implemented requiring a 0% reduction in output to Q 1, the firm scales back production proportionately by selecting (roughly) 0% less labor (L 1 L 0 ) and 0% less capital (K 1 K 0 ). This is not optimal, since input prices in the economy do not fully reflect negative external costs. Specifically, since K polluted and L does not, the rental rate on capital is not high enough (r < r*). The social price that reflects the marginal social cost of capital is r* and the optimal mix of inputs that achieves the isoguant line Q 1 at the social price level is L* and K*, which involves less capital and an additional input substitution effect away from capital and towards labor due to the relative price change. The optimal policy involves L* and K* at the optimal production level Q 1. Two policies that can achieve this result: (1) a standard on K* and L*; or () a tax of t*= E (K*) on polluting capital Question 4. (40 points) Transferable Permits

3 The U.S. Clean Air Act Amendments in 1990 introduced a permit trading system for SO emissions at electric utilities in the U.S. This regulation largely impacts coal-fired electric plants. Among coal-fired electric plants subject to regulation, there are two types of electric utilities: (1) Table A plants, which are the dirtiest electric utilities that burn high-sulfur coal from the Appalachian region; and () type B plants, which use low-sulfur coal from the Powder River Basin of Wyoming and Montana. Suppose a representative Table A plant has total benefit from SO emissions given by TB A = 300S A ½S A, and a representative type B plant has total benefit from SO emissions given by TB B = 300S B S B. Total emissions of SO (in tons) for the two plants combined is S = S A + S B, and the total social cost of SO emissions is given by C(S) = 10S + ½ S. A. (10 points) Calculate the marginal benefit of SO emissions for each electric plant (MB A and MB B ). How much SO do the plants emit in an unregulated market (S A C and S B C )? Marginal benefit is the derivative of total benefit with respect to emissions: for firm A: for firm B: MB A (S A ) = S A MB B (S B ) = S B In an unregulated market, each firm pollutes until MB(pollution) = 0, so S C A : 300 S A = 0 => S C A = 300. S C B : 300 S B = 0 => S C B = 150. B. (10 points) Draw a diagram that shows the marginal benefit of air pollution for each firm, the combined marginal benefit of air pollution for both firms, the marginal social cost of pollution and the social optimum. $/S P MSC P* P 1 MB T = Total pollution demand S B * S A * S* S*/ Pollution C. (10 points). What is the socially optimal level of air pollution for each firm, S A * and S B *, and what is the total amount of SO emissions at the social optimum, S*?

4 Total Social Cost of pollution is C(S) = 10S + ½ S. Marginal social cost is C (S) = 10 + S = 10 + S A + S B To calculate the socially optimal standard on total emissions, maximize welfare: W = TB 1 + TB C(S) = 300S A ½S A + 300S B S B 10(S A + S B ) ½(S A + S B ) Conditions for an optimum (FOC): (1) W/ S A = 300 S A 10 (S A + S B ) = 0 () W/ S B = 300 S B 10 (S A + S B ) = 0 Note: Optimality condition equate MB A = MSC and MB B = MSC Rearranging (1): 90 S A = S B Rearranging (): 90 3S B = S A Plugging () into (1): 90 (90 3S B ) = S B => 5S B = 90 => S B * = 58 Solving for S A in (): S A * = 90 3(58) = 116 => S A * = 116 S* = S A * + S B * = = 174 D. (10 points) Suppose transferable permits are distributed to firms, where each firm is given an equal share of ½S* as an initial allocation of permits. Derive the equilibrium permit price. Show the gains to permit trading for each firm from the initial transferable permit allocation on your diagram for part C. See diagram above. The permit price equates MB A = MB B = MSC => p* = 10 + S* = $184

5 Question 5. (30 points). Non-Renewable Resources and New Discoveries Draw a -panel diagram that shows demand for a non-renewable resource and (constant) marginal cost of extraction in one panel and the price of the resource over time in the second panel. Clearly label the initial price (P 0 ) and the terminal time in which the last unit of the resource is extracted (T*). Now suppose a new source of reserves is discovered, increasing the initial stock of the resource. Show on your graph the outcome of the new discovery for the initial price and terminal time of extraction. P(t) P 0 (t) P 1 (t) $/Q P c = choke price P c User cost P c MC T 0 T 1 Time Demand (MB) Q With a new discovery, the user cost must fall, because the opportunity cost of extracting a unit of the resource in the current period must go down with the rise in available resource stock. The user cost must still rise at the discount rate over time until the choke price is met, so the initial price of the resource must fall to extract the larger stock along the new time path, P 1 (t). The terminal time of extraction rises from T 0 to T 1. Question 6. (40 points). Optimal Harvest of a Steady-State Fishery Consider the population dynamics of a California halibut fishery resource with a growth function given by g(s) = 100S - S. Halibut have a catchability coefficient given by q = 1/. Consequently the yield-effort relationship is given by Y = (1/)E*S, where Y is the size of the catch (i.e., harvest) in tons and E is the level of effort exerted by fishers. The market price of Halibut is P = $0 (per ton) and the total costs of fishing per unit of effort is C(E) = 100E. A. Calculate the maximum sustainable yield (Y MSY ) and the stock associated with Y MSY for this fishery. dg ds = 100 S = 0 ==> S MSY = 50. We plug S MSY = 50 into the growth function to find the growth: Y = g = 100(50) (50 ) = g = 5,000 -,500 =,500. Since any harvest level equal to the growth of the stock results in a steady-state (or sustainable level of the catch), Y MSY =,500 tons.

6 B. Derive the yield-effort curve for a steady-state fishery. Noting that growth equals yield in a steady-state equilibrium, Y = g = 100S S. (1) We also have the yield-effort relationship: Y = 1 E S ==> S = Y. E Plugging this into equation (1), we get: Y = 100(Y / E) (Y / E) Y = 00( Y / E) 4Y / E YE = 00YE 4Y ==> Y = 50E 0.5E. This is the yield-effort curve for the steady-state fishery. C. What is the socially optimal level of effort for the fishing industry, E*? At this level of effort, what is the size of the catch, Y*, and the associated steady-state level of Halibut biomass, S*? Verify that an annual catch of Y* is indeed a sustainable equilibrium for the Halibut fishery. The socially optimal level of effort E* is derived by seeking the maximum profit of fishery industry. The industry total revenue is TR = py = 0(50E ¼E ) = 1,000 E 5E. The industry total cost of fishing is C(E) = 100E. The industry profit is therefore: π = TR TC ==> π = 1,000E 5E π = 900E 5E 100E By taking the derivative of profit and setting it equal to zero, we find the maximum: π ' = E = 0 ==> E* = 90 So, the optimal effort level is 90 units. Next, plug E* into the yield-effort function to get the size of the optimal catch: Y* = 50E *.5E * = 50(90).5(90) = 4, (8,100) => Y* =,475 tons The associated steady-state level of Halibut biomass is: S * = Y * / E * = (,475) / 90 = 70 => S* = 55 tons

7 D. What is the level of effort for the fishing industry under open access, E? Under open access, what is the size of the catch, Y, and the associated steady-state level of Halibut biomass, S? Under open access, the fishing industry realizes zero profit, that is π = TR TC = 0. So, π = 1,000E 5E 100E = 0. At his effort level, the size of catch is Y = 50E => 900 = 5E ==> E = E = 50(180).5(180) The associated steady-state level of Halibut biomass is S = Y / E = 9,000-8,100 ==> Y = 900 (tons) = (900) /180 ==> S = 10 (tons) Under open access, the fishing industry expends too much effort, keeps too small a level of stock of fish, and lands a smaller annual catch than the optimal level. ***The End. Have a great Winter Break***