5 2 Chapter 5 Externalities Problems and Solutions

5. 2 Chapter 5 Externalities: Problems and Solutions Private-Sector Solutions to Negative Externalities The Solution Coase Theorem (Part I) When there are well-defined property rights and costless bargaining, then negotiations between the party creating the externality and the party affected by the externality can bring about the socially optimal market quantity. Coase Theorem (Part II) The efficient solution to an externality does not depend on which party is assigned the property rights, as long as someone is assigned those rights. © 2007 Worth Publishers Public Finance and Public Policy, Jonathan Gruber, 2 e 1

5. 2 Chapter 5 Externalities: Problems and Solutions Example I Net Benefit to the factory associated with marginal production = $1. 0 Net Cost to the Laundromat associated with the firm’s marginal production = $1. 20 *Efficient outcome? Case (i): Factory has the property right. Case (ii): Laundromat has the property right © 2007 Worth Publishers Public Finance and Public Policy, Jonathan Gruber, 2 e 2

5. 2 Chapter 5 Externalities: Problems and Solutions Example II Net Benefit to the factory associated with marginal production = $1. 20 Net Cost to the Laundromat associated with the firm’s marginal production = $1. 0 *Efficient outcome? Case (i): Factory has the property right Case (ii): Laundromat has the property right © 2007 Worth Publishers Public Finance and Public Policy, Jonathan Gruber, 2 e 3

5. 2 Chapter 5 Externalities: Problems and Solutions The problem of the Common Example: 1000 identical persons who can do nothing but fish. Each can catch 4 fish on shore. * ** No of Men Total Catch on Board MP (on board) AP (on board) Net Social MP (on board) Social Total 0 0 0 4000+0=4000 1 6 +6 6 2 3396+6=4002 2 16 +10 8 6 3392+16=4008 3 24 +8 8 4 4012 4 30 +6 7. 5 2 4014 5 34 +4 6. 8 0 4014 6 36 +2 6 -2 4012 7 36 0 5. 14 -4 4008 8 32 -4 4 -8 4000 9 27 -5 3 -9 3991 10 21 -6 21 -10 3981 © 2007 Worth Publishers Public Finance and Public Policy, Jonathan Gruber, 2 e 4

Chapter 5 Externalities: Problems and Solutions 5. 4 Distinctions Between Price and Quantity Approaches to Addressing Externalities Basic Model © 2007 Worth Publishers Public Finance and Public Policy, Jonathan Gruber, 2 e 5

Chapter 5 Externalities: Problems and Solutions Abatement: Algebraic Illustration Ē = firm’s pollution without abatement X = abatement E = Ē-X = pollution C(X) = abatement cost D(E) = D(Ē–X) = pollution damage C’(X) = marginal abatement cost D’(E) = marginal damage of pollution © 2007 Worth Publishers Public Finance and Public Policy, Jonathan Gruber, 2 e 6

Chapter 5 Externalities: Problems and Solutions 1. Optimal abatement: Choose X to Minimize C(X) + D(E) = C(X) + D(Ē-X) • => C’(X) - D’(Ē-X)=0. • Or, C’(X) = D’(E). © 2007 Worth Publishers Public Finance and Public Policy, Jonathan Gruber, 2 e 7

Chapter 5 Externalities: Problems and Solutions 2. Optimal solution for a firm in the presence of a tax: Minimize C(X) + t E = C(X) + t Ē – t X (x) • => t= C’(x) • To attain social optimum then, set t= D’(E). © 2007 Worth Publishers Public Finance and Public Policy, Jonathan Gruber, 2 e 8

Chapter 5 Externalities: Problems and Solutions 5. 4 Distinctions Between Price and Quantity Approaches to Addressing Externalities Multiple Plants with Different Reduction Costs © 2007 Worth Publishers Public Finance and Public Policy, Jonathan Gruber, 2 e 9

Chapter 5 Externalities: Problems and Solutions Example with Multiple Firms Ē 1, Ē 2; X 1, X 2; E 1 = Ē 1 - X 1; E 2 = Ē 2 - X 2 Pollution damage = D(E 1+E 2) =D(Ē 1 + Ē 2 - X 1 - X 2) © 2007 Worth Publishers Public Finance and Public Policy, Jonathan Gruber, 2 e 10

Chapter 5 Externalities: Problems and Solutions * Optimal abatement: Minimize C 1(X 1) + C 2(X 2) + D(Ē 1 + Ē 2 - X 1 - X 2) Þ C 1’ (X 1) = C 2’(X 2) = D’(E). * Firm’s solution: Minimizes Ci(Xi) + t (Ēi - Xi) => Ci’(Xi) = t. => Set: t = D’(E) © 2007 Worth Publishers Public Finance and Public Policy, Jonathan Gruber, 2 e 11

Chapter 5 Externalities: Problems and Solutions Example Assume: D(E) =10 E => D’(E) =10 C 1(X 1)=F + 1/10 (X 1)2 => C 1’(X 1) =1/5 (X 1) C 2(X 2)=F + 1/30 (X 2)2 => C 2’(X 2) =1/15 (X 2) Setting C 1’(X 1) = C 2’(X 2) = D’(E) => X 1=50; X 2=150 © 2007 Worth Publishers Public Finance and Public Policy, Jonathan Gruber, 2 e 12

Chapter 5 Externalities: Problems and Solutions Equal pollution Reduction: Ask each firm to reduce pollution by 100. • Same benefit of damage reduction as with the Pigouvian solution. • Costs: C 1 = F + 1/10 (100)2 C 2 = F + 1/30 (100)2 © 2007 Worth Publishers Public Finance and Public Policy, Jonathan Gruber, 2 e 13

Chapter 5 Externalities: Problems and Solutions Total cost of abatement= C 1 + C 2 = 2 F + (100)2 [1/10 + 1/30] = 2 F + 4000/3 Versus the total cost for the Pigouvian solution: C 1 = F + 1/10 (50)2 C 2 = F + 1/30 (150)2 => C 1 +C 2 = 2 F + 1000. © 2007 Worth Publishers Public Finance and Public Policy, Jonathan Gruber, 2 e 14

Chapter 5 Externalities: Problems and Solutions Market for Permits • • • Suppose Ē 1 + Ē 2 = 500. Want 200 reduction Issue 300 permits (150 each) Firm i’s pollution level is Ei = Ēi - Xi = 150 + ni ni denotes the number of extra permits purchased. If ni is negative, it will be the number of permits sold. © 2007 Worth Publishers Public Finance and Public Policy, Jonathan Gruber, 2 e 15

Chapter 5 Externalities: Problems and Solutions • • Price of a permit= p Cost of polluting Ei = Ci (Xi) + ni p Or Ci (Xi) + (Ē - Xi – 150) p Minimizing costs yields Ci’(Xi)=p. • C 1’(X 1)= C 2’(X 2) If p=t, we will have the Pigouvian solution. © 2007 Worth Publishers Public Finance and Public Policy, Jonathan Gruber, 2 e 16