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, 2e
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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
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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, 2e
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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
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, 2e
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Chapter 5 Externalities: Problems and Solutions
5.4
Distinctions Between Price and Quantity
Approaches to Addressing Externalities
Basic Model
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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
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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, 2e
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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, 2e
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Chapter 5 Externalities: Problems and Solutions
5.4
Distinctions Between Price and Quantity
Approaches to Addressing Externalities
Multiple Plants with Different Reduction Costs
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Chapter 5 Externalities: Problems and Solutions
Example with Multiple Firms
Ē1, Ē2;
X1, X2;
E1 = Ē1 - X1;
E2 = Ē2 - X2
Pollution damage = D(E1+E2) =D(Ē1 + Ē2 - X1 - X2)
© 2007 Worth Publishers Public Finance and Public Policy, Jonathan Gruber, 2e
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Chapter 5 Externalities: Problems and Solutions
* Optimal abatement:
Minimize C1(X1) + C2(X2) + D(Ē1 + Ē2 - X1 - X2)
 C1’ (X1) = C2’(X2) = 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, 2e
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Chapter 5 Externalities: Problems and Solutions
Example
Assume:
D(E) =10 E
=>
D’(E) =10
C1(X1)=F + 1/10 (X1)2
=>
C1’(X1) =1/5 (X1)
C2(X2)=F + 1/30 (X2)2
=>
C2’(X2) =1/15 (X2)
Setting C1’(X1) = C2’(X2) = D’(E)
=> X1=50; X2=150
© 2007 Worth Publishers Public Finance and Public Policy, Jonathan Gruber, 2e
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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:
C1 = F + 1/10 (100)2
C2 = F + 1/30 (100)2
© 2007 Worth Publishers Public Finance and Public Policy, Jonathan Gruber, 2e
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Chapter 5 Externalities: Problems and Solutions
Total cost of abatement=
C1 + C2 = 2F + (100)2 [1/10 + 1/30] = 2F + 4000/3
Versus the total cost for the Pigouvian solution:
C1 = F + 1/10 (50)2
C2 = F + 1/30 (150)2
=> C1 +C2 = 2F + 1000.
© 2007 Worth Publishers Public Finance and Public Policy, Jonathan Gruber, 2e
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Chapter 5 Externalities: Problems and Solutions
Market for Permits
•
•
•
•
•
•
Suppose Ē1 + Ē2 = 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, 2e
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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.
•
C1’(X1)= C2’(X2)
If p=t, we will have the Pigouvian solution.
© 2007 Worth Publishers Public Finance and Public Policy, Jonathan Gruber, 2e
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