ERE HW 3 Solution

# ERE HW 3 Solution - AEM 2500 Fall 2009 Homework #3: Market...

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1 AEM 2500 Fall 2009 Homework #3: Market Failure – Externalities: SOLUTIONS 1) a. Step 1 : Using the information in the table, find the MB and MC curves (or you can solve in terms of S and D functions): MB = -Q d + 340 (or Q d = 340 – P) MC = 1/2 Q s + 10 (or Q s =2P – 20) Set MB = MC (the mantra!!!) to find Q* (private optimum quantity) -Q + 340 = 1/2 Q + 10 (or 340-P = 2P-20) Q* = 220 (or P*=120) Plug Q* into either MB or MC to obtain P* P* = MC = (1/2) * (220) + 10 P* = 120 Or you can plug P* into either the supply or demand function to obtain Q* Q* = Q s = 2P*-20= 2(120)-20 Q* = 220 Private (or Market) equilibrium is at (Q*, P*) = (220, 120) → e M on graph Step 2 : Find the Social MC curve and solve for the social optimum (from here on out in this homework we will use MC and MB only – the solutions following this route are much more direct): SMC = MC + external costs (this is what we refer to as marginal external social costs in the lectures) SMC = 1/2 Q s + 10 + 60 = 1/2 Q s + 70 Note: that in contrast to the lecture which used an increasing damage function, we are using a damage function in which the damages per unit of pollution are constant across emission levels. Set SMC = MB (the mantra again!!!) and follow the same process as above to obtain: Social equilibrium at (Q*, P*) = (180, 160) e S on graph

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2 b. Net Economic Surplus at e M : You can approach this using a producer and consumer surplus approach, or the C, P, E [G] accounting framework we used in class to identify the NES at both points. At e M the NES = A + B + C – D using the areas in the graph above. Area A+B+C = ½[(340-70) *180] = 24,300 - Area D = -½(60*40) = - 1,200 = 23,100 (note: times 1 million) At e S the NES = A + B + C Area A+B+C = ½[(340-70) *180] = 24,300 The reason for the difference is that for every unit of Q in excess of the social optimal level (Q=180) the MSC = MC + MESC exceeds the marginal benefits of consumption. Therefore for each unit of Q in excess of Q=180, steel production makes society worse off. c. i.) A Pigouvian tax is a tax levied on an agent who is creating a negative externality with the intention of reducing the level of the negative externality. It is a per unit tax set equal to the Marginal External Social Cost at the social optimum. \$ 340 160 120 70 10 180 220 Q (mil. tons) SMC MC e S MB e M \$60 per ton external cost A B C D
3 ii) The optimal level of the Pigouvian tax is equal to the marginal external social cost at the social optimum. In this case we have constant external cost of \$60 per unit for all levels of output, including the socially optimal level. So the optimal tax is \$60 per unit. iii) When the Pigouvian tax is implemented, the total tax revenue collected will be equal to the number of units produced times the amount of the tax per unit. So the amount will be: \$60 * 100 mil. tons = \$10,800 mil iv) The NES with the Pigouvian tax is the same as the NES at the social optimum. The difference is that now tax revenue is collected. Remember,

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## This note was uploaded on 12/16/2009 for the course AEM 2500 taught by Professor Poe,g. during the Fall '07 term at Cornell University (Engineering School).

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ERE HW 3 Solution - AEM 2500 Fall 2009 Homework #3: Market...

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