Economics 357, Fall 2006
Assignment 2 Solutions
Total marks: 40
1.
Chapter 5, Question 2, 95-96 (5 marks)
1
and transportation costs are T = 2d
d2
($1 per kilometre, each way). Daily total costs are therefore given by
(a) We are given that “noise costs” are N =
Daily TC = T + N =
1
+ 2d
d2
(b) Where would Fritz live? He can choose d to minimize total costs from noise and
transportation. Differentiating TC with respect to d and setting this expression
equal to zero, the condition for minimization of the function in part (a) is
2
− 3 + 2 = 0 , d3 = 1, d* = 1km
d
Alternatively you could find the optimal distance by graphing the total cost function
over a reasonable ranges such as d=0.1 to d=5.
So daily total costs are minimized where d = 1km.
N = 1, T = 2, TC = 3
(c) Now Fritz is told he will be compensated for any noise damage he suffers. Total
costs are now given by
TC = 2d +
1
1
− 2 = 2d
2
d
d
The problem is therefore reduced to one of minimizing travel costs. Fritz would
move as close as possible to the airport, which in this case is d = 0.1km.
Compensation, in turn, is maximized at
Compensation = N =
1
1
=
= $100
2
d
(0.1) 2
This outcome demonstrates what is known as “moving to the nuisance”, and
represents a strong argument against compensation based on damages for victims of
externalities.
Ec 357, fall 2006, Assignment 2 solutions
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2. Chapter 5, Question 5, page 96 (12 marks)
a) We have the individual marginal damage functions for the two types of
residents:
MDW = 2e
MDr ==6e
Because pollution is a non-rival bad, the aggregate marginal damage function is
the vertical sum of these individual damage functions:
MDT = 8e
b) In the absence of any regulation, the firm choose the level of emissions where
the marginal abatement cost is zero. That is, the uncontrolled level of pollution,
e0, is found where:
MAC = 20 – 2e = 0, p0 = 10
The socially optimal level of pollution, p*, is defined by the level at which
MAC=MD:
20 – 2e = 8e; e* = 2
c)
Marginal abatement cost and marginal damages as a function of emissions
90
80
70
$/unit of emissions
60
50
MAC(e)
MD(e)
40
30
20
10
0
0
1
2
3
4
5
6
7
8
9
10
emissions
The total abatement cost (TAC) is represented by the triangle with yellow dots on its
corners. The TAC is (1/2)8(16)=$64
Ec 357, fall 2006, Assignment 2 solutions
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(d) Note that the level of emissions will equal the emissions level in the unregulated
state less the amount of abatement that has been carried out: e=e0-A. Substitute
for e in the expressions for MD and MAC.
MDT = 8e = 8(10-A) = 80-8A.
MAC = 20 – 2e = 20 – 2(10-A) = 2A.
e)
Marginal abatement cost and marginal damages as a function of abatement
(abatement is relative to e=10)
90
80
70
$/unit of abatement
60
50
MAC(A)
MD(A)
40
30
20
10
0
0
1
2
3
4
5
6
7
8
9
10
$/unit
The optimal level of abatement is at A=8 where MAC=MD.
The total cost of abatement is shown by the triangle with yellow dots on its corners.
Total abatement cost is (1/2)(8)(16) = $64. This agrees with the answer in part (c)
Ec 357, fall 2006, Assignment 2 solutions
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3. Chapter 6, Question 2, page 114 (5 marks)
(a)
The information given in the problem gives us the payoff matrix for the two firms
in Table S-6.2. π M are mill profits and π F are fishery profits. The mill earns $500 profit
where it undertakes no pollution control, and $300 when it pays the $200 for pollution
control. The fishery chooses to operate upstream or downstream.
The joint profit-maximizing outcome is where the mill installs the pollution control
and the fishery operates downstream. The sum of profits from this outcome is $800.
upstream
downstream
Control
π M = 300
πF = 300
No control
π M = 500
π F = 200
π M = 300
πF = 500
πM = 500
πF = 100
Table S-6.2
(b) The problem asks to identify the outcomes where no bargaining is possible. First,
where the fishery has rights to an unpolluted river it will operate downstream with profits
of 500. Here, the mill is forced to purchase the pollution control equipment or shut
down. Because $300 > 0, the mill purchases the equipment. The efficient outcome
results.
Where the mill has the right to pollute, they are not forced to purchase the pollution
control equipment, so do not. The fishery responds by operating upstream. Combined
profits are $700.
(c) When the fishery has the right to clean water, we have the efficient outcome at the
outset as shown in part b.
Will bargaining lead to the efficient outcome under the other set of property rights? The
initial equilibrium in the case where the mill has the right to pollute is the upper-right cell
in the payoff matrix. The mill is earning $500 and the fishery $200. The efficient
solution requires the installation of pollution control equipment, which costs $200. In
this case, the fishery could offer to purchase the equipment for the mill. The efficient
solution will still result because the increase in profit to the fishery where the pollution
control is installed (and they move downstream) is $300, which is greater than the $200
needed to cover the costs of the equipment.
Ec 357, fall 2006, Assignment 2 solutions
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Note that while the efficient outcome is achieved either way with bargaining, the
firms are not indifferent between property rights regimes. Where the fishery has the right
to a clean river, they earn $500 and the mill $300. When the mill has the right to pollute,
after bargaining the fishery earns $300 and the mill $500. What is equivalent in terms of
economic efficiency are the actions of the two firms: the installation of pollution control
equipment at the mill and the location (downstream) of the fishery.
4. Chapter 7, Question 1, page 132 (8 marks)
a) The aggregate marginal savings curve is given by the horizontal sum of the individual
firms marginal savings curves.
e1 = 5 – MS1
2e2 = 8 – MS2 ; e2 = 4 – ½ MS2
By the equimarginal principle: MS1 = MS2 = MS
e1 + e2 = eT = 5 – MS + 4 – ½ MS
eT = 9 –
3
MS
2
MS = 6 -
2
eT
3
This is the effective aggregate level marginal savings function when emissions are
distributed cost effectively. It is shown in Figure S-7.1 below.
b) It is straightforward to calculate aggregate marginal damage. Since pollution is nonrival, aggregate damage is the vertical sum of the individual marginal damage functions:
MDT = MD1 + MD2 = e + e = 2e
This is shown in Figure S-7.1
Ec 357, fall 2006, Assignment 2 solutions
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$
MD r
MS2(e 2)
MD 1 = MD2 = e
MS1(e 1)
7/4
1/2
4
e* = 2 1/4
MSr(e)
5
9
e
Figure S-7.1
c) To find the optimal level of pollution we equate MS and MD:
6-
6=
2
e = 2e
3
8
1
e ; e* = 2
3
4
The Pigouvian fee that supports this optimal level of emissions is equal to the level of
marginal savings (and marginal damage) at the optimum:
1
2 ⎛ 1⎞
⎛ 1⎞
⎜2 ⎟ = 2 ⎜2 ⎟ = 4
3 ⎝ 4⎠
2
⎝ 4⎠
fee* = 6 -
The last part of the solution is to determine the individual firms’ levels of emissions
at the optimum. These can be found by setting the original marginal savings functions
equal to the Pigouvian fee:
5 – e1 = 4
1
1
; e1* =
2
2
8 – 2e2 = 4
1
;
2
e2* =
7
4
Verify that these sum to the optimal total emissions:
Ec 357, fall 2006, Assignment 2 solutions
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e1* + e2* =
2 7 9
1
+ = =2
4 4 4
4
5. Chapter 7, Question 5, page 132-133 (10 marks)
(a) We are given that the competitive price is P = 60. Fireyear will produce where
marginal cost is equal to this market price:
4QF = 60 ; QF = 15
ΠF = 60(15) – 300 – 2(15)2 = 150
Goodstone will do the same:
2QG = 60 ; QG = 30
ΠG = 60(30) – 500 – (30)2 = 400
(b) The problem is simplified by the constant marginal damage function: MD = 12.
This, combined with the l:1 relationship between output and pollution, establishes the
proper Pigouvian fee at £12 per Q. Marginal costs increase to each firm by £12. The
new quantities and profits:
12 + 4QF = 60 ; QF(tax) = 12
ΠF(tax) = 60(12) – 300 – 2(12)2 – 12(12) = -12
12 + 2QG = 60 ; QG(tax) = 24
ΠG(tax) = 60(24) - 500 - (24)2 - 12(24) = 76
( c) The subsidy that achieves the efficient level of output and emissions from each
firm is also £12. This subsidy is the logical equivalent of a lump-sum transfer of £12•Q
to each firm, followed by a £12 tax per unit of output. That is, the firm can receive a
subsidy of £12•Q if it chooses to produce nothing, but this subsidy is reduced by £12 with
each unit produced. In effect, the firms’ cost functions become:
Fireyear:
Costs: 300 + 2 QF2 - 12(15) + 12QF
Goodstone:
Costs: 500 + QG2 - 12(30) + 12QG
Resulting in the same marginal cost functions applied in part b:
Fireyear: MC = 12 + 4QF
Goodstone:
MC = 12 + 2QG
Ec 357, fall 2006, Assignment 2 solutions
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The market price does not change, so the quantities under a subsidy are the same as
with a tax:
QF(subsidy) = 12
QG(subsidy) = 24
The profits of the two firms, however, are not the same:
ΠF (subsidy) = 60(12) – 300 – 2(12)2 – 12(12) + 12(15) = 168
ΠG (subsidy) = 60(24) – 500 – (24)2 – 12(24) + 12(30) = 436
Note that the total amount of subsidy paid is equal to the subsidy per unit times emissions
reduced compared to the no-regulation level. For Fireyear, for example, this would be:
subsidy = (15-QF)($12). This is shown in the last two terms in the above equation for πF
(d) The unregulated outcome in (a) is inefficient because of the pollution externality.
The objective of the Pigouvian fee applied in (b) was the socially optimal combination of
output and emissions. It would appear at first that the outcome in (c) is the same as in
(b), but this is not exactly right. In the long run, Fireyear would leave the industry where
an emissions fee is levied, but would remain in business under a subsidy program (since
ΠF (tax) < 0, ΠF (subsidy) > 0).
Which of these outcomes is optimal? The Pigouvian fee results in the socially
efficient level of emissions because it imposes on the firm the absent portion of social
costs associated with production of each unit of pollution. This process is also present
with a subsidy program, but provides an initial lump-sum transfer to all firms. This
transfer is distortionary in that it allows for the continued presence of firms that cannot
operate profitably when all social costs are internalized. In short, the solution in (b) is
optimal, the solution in (c) is not.
Ec 357, fall 2006, Assignment 2 solutions
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