1 topic 2: production externalities examples of types of abatement activities analyzed: –output...

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Topic 2: Production Externalities

• Examples of types of abatement activities analyzed:

– Output reduction

– Cleaner production involving VC (ex: input-switching)

– Cleaner production involving FC (ex: pollution control equipment)

• Policies we have looked to date involve the regulator:

– Deciding which of the different abatement activities is the “best” (efficient); then

– Designing policy to induce firms to undertake this activity.

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• Simpler regulatory approach: target emissions (E) directly, and let firms decide which way in which to achieve this level of E.

– Especially true in the real world, where there might be many (more than 3) different ways in which to abate pollution.

• From now, we will focus on policies that target E directly, leaving the firm to decide which abatement activities are undertaken. Two questions:

– What is the “right” (efficient) level of emissions/abatement?

– What policies can we use to achieve this?

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• Additional questions we will explore with respect to policy:

1. What information does the regulator need in order to implement a particular policy?

2. What incentives to innovate (come up with new abatement technology) are associated with a particular policy?

3. Which policies create the right incentive for firms to remain operating within an industry?

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Topic 2: Production Externalities• Efficiency (as always) maximizing the NB.

– In this context, we are interested in maximizing the NB of abatement.

• Recall: the efficient level of output is where the MB of consuming output = MC of producing output.

• Similar logic can be used to show that the efficient level of abatement is where the MB of abatement = MC of abatement.

• First step is therefore to understand what are the benefits and costs of abatement.

• Benefit of abatement?

• Avoided environmental damages. Can be measured using the marginal damages from emissions function (MD).

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Topic 2: Production Externalities• MD = extra damages associated with an increase in

emissions (for instance, and extra ton of SO2).• Recall our discussion about the shape of MD function.

– In many cases, each extra unit of pollution is more damaging than the last MD as E.

$

E

MD

E1E2

Area under the MD curve gives TD.

If E from E1 to E2, TD by area A.

A area A is the benefit of reducing pollution from E1 to E2.

Key point to understand: the marginal damages from emissions = marginal benefit of abatement.

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Topic 2: Production Externalities• If E from left to right in the diagram then E from right to

left abatement (A) is increasing from right to left in the diagram.

$

E

MDTells us that when marginal damages from emissions are increasing in E…..

… the marginal benefit of abatement is decreasing in A.

That is, as we abate more emissions, the MD of emissions is smaller, and so the benefit we get from additional abatement is also smaller.

EA

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Topic 2: Production Externalities• Example: Suppose that MD = (1/2)E, and firm emits E =

100 without regulation abatement (A) = 0 & TD = area under MD curve up to E = 100; so TD = 2500

$

E

MD If E from 100 to 80, this would A= 20.

Total benefit to society given abatement of 20 = area under the MD curve as E from 100 to 80 i.e.TB of abatement

= 900

E = 100A = 0

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E = 80A = 20

. .

40

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Topic 2: Production Externalities• Cost of abatement? • We know that abatement costs will depend on abatement

activity.• Easy to imagine that, no matter what those abatement

activities are, it is likely the case that additional abatement gets more expensive as A.

• That is, the marginal abatement costs (MAC) are an increasing function of A.

MAC is a decreasing function of E.

$

E

MAC

A as we move from right to left.So if MAC is sloping with respect to A it is sloping with respect to E.

NB: for many pollutants, it is unlikely that MAC is linear in practice!

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Topic 2: Production Externalities• Example: Suppose MAC = 120 - E.

$

E

MAC

120

120

Just like every other MC curve, area under MAC gives total costs.

In this case, total abatement costs (TAC).

Major difference: we read the diagram from right to left.

Horizontal intercept = 120: Tells us that, in the absence of regulation, the firm will choose to emit E = 120. Define this Emax.

At E = 120, A = 0 and TAC = 0.

If E to 100, A would be 20, and TAC = 200.

100

20

TAC if A = 20

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Topic 2: Production Externalities• Putting together MD and MAC to find the efficient level of

abatement/emissions (for a single firm).$

E

MAC

If we abated less (emitted more than E*), MD > MAC MB of further abatement > MC of further abatement. We should be abating more.

If we abated more (emitted less than E*), MD < MAC MB of further abatement < MC of further abatement. We should be abating less.

E*

MD

Emax

Efficient to abate until MAC(E*) = MD (E*).As E from Emax to E*:

TDfrom X+Y+Z to Z TB of abatement = X+Y

TAC from 0 to Y TC of abatement = Y

So NB = TB - TC = X.

X

YZ

A*

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Topic 2: Production Externalities• Example: Suppose that MD = (1/2)E and MAC = 120 - E.

$

E

MAC

80

MD

120

MD = MAC (1/2)E = 120 - E

E = 80 & A = 40.

A = 40

120

Total Benefit of Abatement (TBA) = area under MD curve for A = 40.

TBA = $2,000.40

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Total Abatement Costs (TAC) = area under MAC curve for A = 40.

TAC = $800.

NB = TBA - TAC = $1,200.

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Topic 2: Production Externalities• The case of finding the efficient level of abatement is

(relatively) simple in the case of just one polluting firm: MAC = MD.

• What if more than one firm polluting in a particular region?

• Example: Suppose we have two firms:

– Firm 1: MAC1 = 180 - E1

– Firm 2: MAC2 = 90 - (1/2)E2

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• Marginal Damages given by MD = (2/3)E

– E is total emissions E = E1 + E2.

MD = (2/3)(E1 + E2)

• Extra dimension to this problem: no matter what level of abatement we want, efficiency requires that this abatement be achieved at least cost.

• How to ensure that the target level of abatement is being “produced” in the least cost manner?

– Cost minimization MAC1 = MAC2

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Topic 2: Production Externalities• Tells us that efficiency with two firms requires

MAC1 = MAC2 = MD.

• So in this example we need:– MAC1 = 180 - E1 = 90 - (1/2)E2 = MAC2; and– MAC1 = 180 - E1 = (2/3)(E1 + E2) = MD

• 2 equations and 2 unknowns (E1 & E2) which we can solve for– E1 = 100– E2 = 20

• Check that this is efficient. – E1 = 100 MAC1 = 80– E2 = 20 MAC2 = 80– E = 120 MD = 80

E = E1 + E2 =120

MAC1 = MAC2 = MD.

Ensures that, whatever total A is, the “right” firms are doing it.

Ensures that the total A is the “right” amount.

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180

180

80

90

20 120100

MAC1

MAC2

MAC1 = 180 - E1,

MAC2 = 90 - (1/2)E2,

MD = (2/3)E

MAC1 = 80 when E1 = 100

MAC2 = 80 when E2 = 20

MD = 80 when E1 + E2 = 120

MD

MAC1

MAC2MD

If E1 = 100, then A1 = 80.If E2 = 20, then A2 = 160.

Why is A2 > A1?

Firm 2 can abate cheaper than firm 1.

Graphically:

A1A2

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• Note the two-step nature of this problem:

1. Recognize that no matter what level of abatement we want, cost minimization requires that MAC1 = MAC2.

2. NB maximization requires that we pick the level of abatement where the (equalized) MAC = MD.

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• It is useful to think in terms of a two stage process because we can generalize to a large amount of sources:

1. choose the emissions for each source to achieve a given target level of aggregate emissions at minimum abatement cost;

2. choose the target level of aggregate emissions to maximize social surplus.

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• In the previous slides, we solved this problem in one step.

• An alternative approach is to solve this problem in two steps:

1. Derive an equation that tells us all of the possible cost-minimizing allocations of abatement activity across the two firms, for each and every possible level of total abatement.

• Aggregate MAC (aka supply of abatement) curve.

2. Find where the aggregate MAC curve cuts the MD curve to identify the efficient level of total abatement.

1 topic 2: production externalities examples of types of abatement activities analyzed: –output...

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