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Topic 2: Production Externalities
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  1. Topic 2: Production Externalities Burning Coal • So far we have only considered policies that reduce pollution by reducing output. Indirect way to reduce environmental costs. • What if there are ways in which we can make production cleaner? i.e., (E/Q) • Example: SO2 emissions are a function of the sulfur content of coal. • Lower sulfur content  lower SO2 emissions. Electricity SO2   Benefits Environmental Costs

  2. Topic 2: Production Externalities • In practice, can be many possible levels of sulfur content in coal. • For simplicity, we will assume there are just two types of coal: low sulfur coal and high sulfur coal. • Continuing with coal-fired power plant example… Suppose: • (as before) each ton of SO2  in $300 of damage, irrespective of the level of SO2 emissions (MD = 300); and • each kwh of electricity  • 1/10,000th of a ton of SO2 emissions if high sulfur coal; • 1/30,000th of a ton of SO2 emissions if low sulfur coal;  High sulfur coal: MECH = $0.03.  Low sulfur coal: MECL = $0.01. • Assume initially that price of coal not a function of sulfur content. • MPC not a function of sulfur content Unrealistic: we will relax this assumption later

  3. Topic 2: Production Externalities MSC depending on the sulfur content of coal. If MPC not a function of H or L, then: c MSCH 12 MECH MSCL MSCH = 3 + (1/50)Q A MPC C MSCL = 1 + (1/50)Q B 3 MECL MB 1 Q (thousands kwh) 300 3662/3 1,200 We get the benefit of cleaner production at no cost. Maximized NB = area A if H is used = $13,500 = areas A+B+C if L is used = $20,167 Clearly better to use low sulfur coal

  4. Topic 2: Production Externalities • If L costs the same as H, then clearly efficiency requires that the cleaner input be used. • MSC is lower using L than H. • Will the firm want to use the cleaner input? • Inputs cost the same  the firm will be indifferent • No real regulatory problem in ensuring the right choice of input by the firm.

  5. Topic 2: Production Externalities • Problem still remains, however, of getting the firm to produce the right quantity of output, given the use of the clean input. • Even with low sulfur coal, MEC > 0  firm will produce too much output absent government intervention. • So there still is a role for government in ensuring that the firm produces where MSC = MB. • Taxes, quotas, etc, in the output market.

  6. Topic 2: Production Externalities • The more interesting (and realistic) case is where the clean and dirty inputs have different costs. • For instance, it is likely that MPCL > MPCH • Specifically, in our example assume: • MPCH = (1/50)Q; and • MPCL = 1 + (1/50)Q • That is, using low sulfur coal  per kwh cost of electricity by 1c.

  7. Topic 2: Production Externalities • Questions: • From the viewpoint of efficiency, do we want H or L? • Will the firm make the right choice of inputs? • If not, what policies will induce the firm to make the right choice? • Once we have induced the right input choice, do we still need government intervention to ensure the right output choice?

  8. Topic 2: Production Externalities • The answers to Qs 1 and 2 are fairly obvious: • From the viewpoint of efficiency, do we want H or L? • Cost of low sulfur coal = 1c extra per kwh produced • Benefit of low sulfur coal = 2c per kwh lower MEC.  We want the firm to use low sulfur coal.

  9. Topic 2: Production Externalities • Will the firm make the right choice of inputs? • Cost of low sulfur coal = 1c extra per kwh produced • Benefit to the firm of low sulfur coal = nothing, without some form of government intervention. • the firm doesn’t account for the MEC, so doesn’t care that it is lower using low sulfur coal.  Firm will not choose to to use low sulfur coal, absent regulatory inducement. • Can also analyze Qs 1 and 2 graphically.

  10. Topic 2: Production Externalities Low sulfur coal increases production costs by 1c/kwh but reduces MEC by 2c/kwh. MSCH = MPCH + MECH = (1/50)Q + 3 c MSCH MSCL MECL MPCL MPCH MSCL = MPCL + MECL = 1 + (1/50)Q + 1 = 2 + (1/50)Q MECH 3 2 1 Q (thousands kwh) At any level of output, using L rather than H  Lower social costs; and Higher private costs  L is efficient but not profit- maximizing. Exercise: Calculate the  in maximized NB that result from using L rather than H, given the D curve

  11. Topic 2: Production Externalities • How can we induce the firm to adopt the cleaner input? • The policies we have considered so far won’t do it (without modification). • They all target output, not emissions. • Output tax: if we set t = $0.03 per kwh then, including the tax, the firm’s costs are: • MPCH + t = (1/50)Q + 3, if H used • MPCL + t = 1 + (1/50) + 3 = (1/50)Q + 4, if L used H still cheaper for firm than L

  12. Topic 2: Production Externalities • Quota: Mandates a reduction in output, not pollution. • Subsidy on output reduction: Has same effect as tax. • Obvious point is that policy should address the target of interest, which is emissions here (not output).

  13. Topic 2: Production Externalities • BUT, doesn’t mean that we can’t ever achieve efficiency in this context by policies targeting Q. • For example: we could introduce an output tax as follows: • t = $0.03 per kwh for power plants using high sulfur coal; or • t = $0.01 per kwh for power plants using low sulfur coal. • This policy would ensure the firm internalized all EC no matter which input it used. • Would induce all firms choose L rather than H. • Potentially difficult and costly in terms of monitoring/ensuring enforcement. • How do we know whether on any given day a firm is using one type of coal or another?

  14. Topic 2: Production Externalities • Alternative policies to ensure the right input is used: • Require the use of low sulfur coal: • Simply mandate that all firms use the low emissions input. • An example of “command and control” policy. • Tax high sulfur coal: • Note: different from taxing output. • We need to figure out a tax on high sulfur coal that will mean that MPC of producing output is greater for H rather than L. • If H and L are equally productive in terms of electricity output, this simply means making H more expensive than L. • Think about the case where L is less productive.

  15. Topic 2: Production Externalities • Subsidize low sulfur coal: • Similar to the tax on high sulfur coal • We need to figure out a subsidy on L that will mean that MPC of producing output is lower for L rather than H. • So far we have looked at two ways in which emissions of a pollutant can be abated • Pollution abatement = pollution reduction • In the electricity example, abatement can be achieved through: • Reducing output • Switching to a cleaner input.

  16. Topic 2: Production Externalities • In reality pollution can also be reduced through the installation of pollution control equipment. • For instance, SO2 emissions can be reduced through the use of an electro-magnetic precipitator, or “scrubber.” • How does it work? Gaseous SO2 is piped through a “reaction tower.” A slurry of finely ground limestone is also sprayed into the reaction tower. The SO2 reacts with the limestone, 95% of the SO2 is removed and gypsum (CaS04) is formed as a by-product. • Installing equipment such as scrubbers  fixed cost for the firm.

  17. Topic 2: Production Externalities • Qualitatively different from other abatement activities/policies we have considered. • per unit output tax • input switching • The decision of whether or not to require the installation of pollution control equipment also introduces a time dimension to the problem. • If we install the equipment, • we incur a (fixed) cost now; and • receive (flow) benefits (in terms of reduced emissions) now and in the future. Both  VC (hence shift MC)

  18. Topic 2: Production Externalities • We need some way in which to compare costs today with future benefits. Example: Suppose that • Current emissions = 280 tons of SO2 per year. • Each ton of SO2  $100 of TD (i.e., MD = $100). • Cost of a scrubber = $200,000 • Lifetime of a scrubber = 10 years • Assume (i) VC of scrubber = 0 and (ii) E = 0 if scrubber used.

  19. Topic 2: Production Externalities • Question: Is it economically efficient to install the scrubber? • Tempting to say: • Benefit of scrubber (saved environmental costs) = 10  $28,000 = $280,000 > $200,000 = cost of installing scrubber • Tempting…. But incorrect!! • We can’t compare today’s $s with future $s in this way.

  20. Topic 2: Production Externalities • Dollars today are worth more than the same number of dollars at some time in the future. • This is because if I receive a certain number of dollars today I can invest them  more dollars in the future. Example: I have $100 today, which I can invest at 5%. • In 1 year I will have: $100  (1.05) = $105. • In 2 years I will have: $105  (1.05) = ($100  (1.05))  (1.05) = $100  (1.05)2 = $110.25. • In 3 years I will have: $110.25  (1.05) = $100  (1.05)3 = $115.76. Etc…..

  21. Topic 2: Production Externalities General formula: FV of a single payment of X in n years time = X(1+r)n. = 100(1+r) = 100(1+r)2 • Tells us about future value (FV) of $100. If the interest rate is 5%, the FV of $100 is: • $105 in one year • $110.25 in two years • $115.76 in three years • FV tells how much a certain amount of today’s money will be worth at some point in the future. • In this class, we will more often be interested in the present value (PV) of a sum of money. • PV tells us how much a certain amount of future money is worth today. • For instance, if installing a scrubber will generate an annual benefit of $28,000 ten years from now, what is that worth to us now? = 100(1+r)3

  22. Topic 2: Production Externalities Example: • Suppose r = 5%. What is $105 received in one year’s time worth to us now? • i.e., what is the PV of $105 in 1 years time? • We already know that if we invest $100 at 5%, then we will have $105 in one year.  having $100 now and $105 in one year are (in some sense) equivalent.  the PV of $105 in one years time is $100.

  23. Topic 2: Production Externalities • Similar logic  the PV of $110.25 in 2 years time is $100. etc. • General formula: • PV of a single payment of X in n years time = X/(1+r)n. • Often useful to think of PV as the amount of money you would need to invest now, in order to have X in n years.

  24. Topic 2: Production Externalities Note timing assumption here! • General PV formula can help us figure out the PV of the saved EC from installing scrubber. (assume in this ex that r = 5%) • Savings of $28,000 per year for 10 years. • Saved EC in the: • 1st year: PV = 28,000/1.05 = $26,666.67 • 2nd year: PV = 28,000/(1.05)2 = $25,396.83 …… • 10th year: PV = 28,000/(1.05)10 = $17,189.57 • Total PV of saved EC (= benefit of installing scrubber) = $26,666.67 + $25,396.83 + … + $17,189.57 = $216,298.58 < 10  $28,000 • Fixed cost of installing scrubber = $200,000 • Benefit of scrubber exceeds cost  efficient to install.

  25. Topic 2: Production Externalities The effect of discounting graphically: (finding the PV referred to as “discounting”) annual savings in EC PV of EC savings r = 5% Saving to EC that occur in ten years time are worth much less (measured in PV dollars) than savings to EC in one year.

  26. Higher interest rates  lower PV: Topic 2: Production Externalities r = 5% r = 10% Using a higher r  discounting the future more  placing less weight on the future. If r = 10% in this example, the PV of benefits from installing scrubber = $172,047.88  If correct to use r = 10%, then NOT efficient to install scrubber.

  27. Topic 2: Production Externalities • Using the general PV formula can be computationally inefficient if we are looking at long time horizons. • For instance, if the scrubber lasted for 50 years, or longer. • If the scrubber lasted forever, this is easy computationally. • Formula for the PV of an infinite stream of payments of X per year (beginning in 1 years time) at interest rate r: • PV = X/r

  28. Topic 2: Production Externalities • Example: What is the PV of $100 per year forever, beginning in one year, if r = 10%? • PV = $100/(0.10) = $1,000 • Intuition for formula? Suppose we took $1,000 invested at 10%. • After 1st year, interest income = $100. Can withdraw that, leaving the principal intact at $1,000. • After 2nd year, interest income = $100. Can withdraw that, leaving the principal intact at $1,000. • After 3rd year, interest income = $100. Can withdraw that, leaving the principal intact at $1,000. • And so on….

  29. Topic 2: Production Externalities • Tells us that - if r = 10% - having $1,000 now, is the same as having an income stream of $100 per year forever. • We can combine the • formula for the PV of 1 payment received in n years time • PV = X/(1+r)n; with • the formula for the PV of an infinite stream of payments • PV = X/r To solve the problem of finding the PV of a finite stream of payments that lasts for a large number of periods.

  30. Topic 2: Production Externalities • Suppose the scrubber has a lifetime of 50 years, and in each year it operates it reduced EC by $28,000. Still assume r = 5%. • This is a finite stream of benefits, which we can actually think of as being the difference between two infinite streams of benefits. PV of finite stream = PV of A - PV of B $28,000 Stream A Stream B years 0 1 2 3 …………. 50 51 52 53 …………………… ∞ Infinite stream A: $28,000 per year forever, beginning in year 1. Infinite stream B: $28,000 per year forever, beginning in year 51. PV of finite stream of $28,000 per year beginning in year 1 = PV of infinite stream A - PV of infinite stream B.

  31. Topic 2: Production Externalities Infinite stream A: $28,000 per year forever, beginning in year 1. • PV = $28,000/0.05 = $560,000 Infinite stream B: $28,000 per year forever, beginning in year51. • PV in year50 = $28,000/0.05 = $560,000. $28,000 Stream A Stream B years 0 1 2 3 …………. 50 51 52 53 …………………… ∞ PV of stream B at year 50 = $560,000 PV of stream A at year 0 = $560,000 Infinite stream B is worth $560,000 to us in year50 dollars.

  32. Topic 2: Production Externalities Infinite stream B: worth $560,000 to us in year50 dollars. • What is this worth to us now (in year0 dollars)? • PV of $560,000 received in 50 years = $560,000/(1.05)50 = $48,834.09. $28,000 Stream A Stream B years 0 1 2 3 …………. 50 51 52 53 …………………… ∞ PV of stream B at year 50 = $560,000 PV of stream A at year 0 = $560,000 PV of stream B at year 0 = $48,834.09

  33. Topic 2: Production Externalities • Recall: we are trying to find the PV of $28,000 of benefits received every year for 50 years. • PV of the 50 year stream = PV of stream A - PV of stream B. = $560,000 - $48,834.09 = $511,165,91 $28,000 Stream A Stream B years 0 1 2 3 …………. 50 51 52 53 …………………… ∞ Tell us that the PV of the saved environmental costs = $511,165.91, over the 50 year life of the pollution control equipment. PV of stream A at year 0 = $560,000 - PV of stream B at year 0 = $48,834.09

  34. Topic 2: Production Externalities • PV of benefits from activities such as installing pollution control equipment yield a maximum willingness to pay for this equipment (if efficiency is our criterion for decision-making). • Recall in example the cost of the scrubber was $200,000. • We have seen that, if the scrubber  EC by $28,000 per year for 50 years, the PV of benefits is $511,165.91. • Tells us that the PV of net benefits associated with installing the scrubber = present value TB - TC = $511,165.91 - $200,000 = $311,165.91

  35. Topic 2: Production Externalities • Net benefits in this context often referred to as the Net Present Value (NPV). • NPV = PV of Total Benefits - PV of Total Costs. • Note: in example, the costs were incurred today, so no need to find the PV of costs (not always the case). • Efficiency in a dynamic context: • A context in which benefits and costs fall at different times. • Maximizing NB  maximizing NPV.

  36. Topic 2: Production Externalities • Decision rules for dynamic choices: • Binary decision-making: Do we undertake a given project? • Example: do we install a scrubber or not? • Yes, if the NPV is positive. • More complex choice: If we are faced with a number of different projects, which (if any) do we undertake? • Example: if there is a choice among different pollution control technologies, which should we adopt? • Undertake the project with the highest NPV.