Energy Economics
Introduction
Most investments involve an initial payment in return for future income. This is especially true of investments in energy efficiency and renewable energy systems, both of which typically require an up-front investment in equipment in order to derive future savings or future income. In order to evaluate these investments, it is necessary to understand how the value of money changes over time. Energy Economics describes the methods used to evaluate investments which contain cash flows at different times.
Engineers who can clearly and correctly communicate the financial impacts of energy saving ideas have more influence in the decision-making process. Those who do not have these skills are less able to judge the economic merit of an idea or to advocate for ideas they believe in.
Although economics is important, many important factors are difficult to translate into dollars. For this reason, economic analysis should not be the only criteria in accepting or rejecting a design or investment option.
Simple Payback and Rate of Return
The simplest index of economic feasibility, and one that is very widely used, is simple payback. Simple payback, SP, is the time period required for an investment to create a positive cash flow. Simple payback is:
SP = [1]
Example:
Calculate the simple payback of a lighting retrofit that will cost $1,000 to implement and will save $250 per year.
SP = = 4 years
The rate of return, ROR, is the reciprocal of the simple payback. Rate of return represents the annual return on the investment.
ROR = 1 / SP = [2]
Example:
Calculate the ROR of a lighting retrofit that will cost $1,000 to implement and will save $250 per year.
Rate of Return = = 25% per year
One of the strengths of the simple payback and rate of return methods for evaluating investments is that the results are independent of assumptions about the time-value of money. Over short time periods, the value of money does not change much with time. Thus, simple payback and rate of return are appropriate methods to analyze investments with short paybacks.
Time Value of Money
The notion of economic growth, of investing in capital to generate future profit is a central concept in capitalism. Entrepreneurs and growing companies are interested in acquiring money today to make a profit with it tomorrow. Thus, in the right hands $100 today is worth more than $100 tomorrow; money has a time-value component.
Example:
Would you rather have $100 now or $100 next year?
In a growing economy, I’d rather have $100 now, because I could put it in the bank at 5% interest and have $105 next year.
To compare investment options with cash flows that occur at different times, it is useful to covert all cash flows to a common time. The most common way to do so is to covert all cash flows into their “present value”, and then compare the present values to evaluate alternative investments. Cash flows involving future amounts of money can be converted to their present values using three important equations:
· Present Value of a Future Amount
· Present Value of a Series of Annuities
· Present Value of an Escalating Series of Annuities
Present Value of a Future Amount
Consider the common situation of investing a present amount, P, in an account that pays a rate of interest, i, over n compounding periods. The future amount, F, can be determined from the following example. Start with a present amount, P = $100 at a rate of interest, i = 5% per year. The future amount after n years, Fn, is:
F0 = 100
F1 = 100 + 100(.05) = P + Pi = P (1+i)
F2 = [100 + 100(.05)] + [100 + 100(.05)](.05) = P(1+i) + P(1+i)i = P(1+i)2
Fn = P (1+i)n
Equation 3 is the fundamental equation of exponential growth and can be applied whenever growth is a fixed percentage of the current quantity.
F = P (1+i)n [3]
Equation 3 can be rearranged to show the present value of a future amount, as in Equation 4.
P = F (1+i)-n [4]
The factor (1+i)-n is sometimes called the present worth factor, PWF(i,n). Thus,
P = F(1+i)-n = F PWF(i,n)
Example:
Someone promises to pay you $1,000 in 5 years. If the interest rate is 10% per year, what amount would you take today that is equivalent to $1,000 in 5 years (i.e. what is the present value of $1,000 5 years from now?) ?
P = F(1+i)-n = $1,000 (1+.10)-5 = $621
Present Value of a Series of Annuities
An annuity is a regular payment of income made at the end of a fixed period. Consider investing an annuity of amount, A, during each of n compounding periods with an interest rate i. This situation can be shown graphically in a cash flow diagram. In cash flow diagrams, income is shown as line extending upward and payments are shown as lines extending downward. The cash flow diagram for a series of n investments of amount A is shown below.
The following derivation shows how to calculate the present value, P, of this series of payments.
Pn = present value of n payments of amount A
= present amount that is equal to a series of payments, A, for n years
P0= 0
P1 =
P2 =
Pn =
To find closed-form solution, do a little algebra...
1) Pn =
2) Pn =
2-1) Pn =
Pn =
Thus, the present value of a series of n payments of amount A is:
Pn = [5]
The factor is sometimes called the series present worth factor, SPWF(i,n). The reciprocal of the series present worth factor is sometimes called the capital recovery factor, CRF(i,n). Thus,
Pn = = A SPWF(i,n) {SPWF(i,n) = 1 / CRF(i,n)}
Example:
A standard-efficiency furnace costs $100 and consumes $40 per year in fuel over its 10 year lifetime. A high-efficiency furnace costs $200 and consumes $20 per year in fuel over its 10 year lifetime. If interest rates are 10% per year, which is the better investment?
Standard efficiency High-efficiency
$40 $20
$100
$200
Pstd costs = $100 + $40= $346
Phigh costs = $200 + $20= $323
and
Phigh savings = $346 - $323 = $23
Or you could solve directly for the present value of savings by setting up your cash flow diagram to reflect ‘savings’ instead of ‘costs’.
Phigh savings = -$100 + $20= $23
It is sometimes easier to understand “annualized savings” than the “present value” of savings”. To find annual savings, find the present value of savings, and then annualize that amount by solving P = A SPWF(i,n) for A.
Example:
Find the annualized savings from investing in the high efficiency furnace over the standard efficiency furnace:
Ahigh savings = P / SPWF(.10,10) = $23 /
Ahigh savings = $4
Present Value of an Escalating Series of Annuities
Sometimes a recurring annuity, A, is expected to increase over time at some escalation rate e. For example, as equipment gets older it often requires more and more maintenance. A cash flow diagram of an escalating series is shown below.
Using a method similar to the previous derivation, the present value of an escalating series is:
if e ¹ i [6a]
P =
if e = i [6b]
The factor or (depending on whether i = e) is called the escalating series present worth factor, ESPWF(i,e,n). Thus,
P = A ESPWF(i,e,n)
Note that when e = 0, escalating series present worth factor ESPWF(i,0,n) is identical to the series present worth factor SPWF(i,n).
Example:
The maintenance director says that it will cost $50 this year to maintain an aging piece of equipment and estimates that the equipment will require 5% more maintenance every year for the next 10 years. New replacement equipment would cost $400 and would require only $10 of maintenance per year with no expected escalation over the next 10 years. Which is a better investment if interest rates are 5%?
Current Equipment:
Pcurrent equip costs = A ESPWF(.05, .05,10)
= = = $476
New Equipment:
Pnew equip costs = $400 + A SPWF(.05,10)
= $400 + $10 = $477
Hence, the two options are expected to cost about the same amount.
Future Value of a Series of Payments
In addition, it is sometimes useful to calculate the future value of a series of annuities. Using a derivation similar to that for the present value of a series of annuities, the future value, F, of a series of equal annuities, A, that accrue interest at a rate, i, over n periods is:
Fn = [7]
Example:
The future value of an annual investment of $2,000 per year for 20 years in an IRA that accrues interest at 5% per year is:
F =
Compounding Periods
Typically, interest is paid or payments are due on fixed intervals rather than continuously. These intervals are called compounding intervals. Interest rates are typically reported for an annual compounding period. If interest is paid or payments are due at other compounding intervals, simply divide the annual interest rate, i, in the time value of money equations by the number of compounding periods per year, m, and multiply the number of years, n, by m.
Example:
Calculate the future value of $100 earning 8% annual interest compounded quarterly for 10 years.
F = P(1 + i/m)nm = $100 (1 + .08/4)(10*4) = $220
Example:
Calculate the monthly house payment if $100,000 is borrowed at 8% on a 30 year mortgage.
P = A SPWF(i,n)
A = P / SPWF(.08/12, 30*12) = $734 / month
Summary of Time Value of Money Equations
The figure below shows the four time value of money equations developed so far.
P = F (1+i)-n = F PWF(i,n)
P = = A SPWF(i,n)
Fn =
if e ¹ i
P = = A ESPWF(i,e,n)
if e = i
The Discount Rate
So far, we have referred to the rate of growth i as the rate of interest. More formally, i is the discount rate. The discount rate is the expected rate of return from an alternative investment. The alternative investment could be interest from a bank, stock market appreciation or expected profits from one’s own company. High discount rates reflect the belief that a large profit can be made from an alternative investment; thus, money today is very valuable and future money is less valuable. High discount rates have the effect of discounting future sums of money or “discounting the future”. Hence the name “discount” rate. To get a feeling for how the discount rate affects time value of money, reconsider a previous example, but compare solutions with two discount rates.
Example:
A standard-efficiency furnace costs $100 and consumes $40 per year in fuel over its 10 year lifetime. A high-efficiency furnace costs $200 and consumes $20 per year in fuel over its 10 year lifetime. If the discount rate is 10% per year, which is the better investment? What if the discount rate is 30%?
Cash flow diagrams are:
Standard Efficiency High Efficiency
$40 $20
$100
$200
Savings from high-efficiency furnace
If the discount rate is 10%, then:
Phigh-eff savings = -$100 + $20= $23
The positive savings indicate that the high-efficiency furnace is the best investment.
If the discount rate is 30%, then:
Phigh-eff savings = -$100 + $20= $-38
The negative savings indicate that when the discount rate is 30%, the traditional furnace is the better investment.
In summary, a high discount rate reduces the value of future cash flows, including savings from energy efficiency measures.
In general, renewable energy and energy conservation technologies have high first capital costs and low future fuel costs. Thus, high discount rates (which value the present and ‘discount’ the future) work against these technologies. Some people argue that a discount rate of zero ought to be used in non-renewable resource decisions. A zero discount rate implies that the future is equally as important as the present.
Lifecycle Cost and Net Present Value
The most comprehensive way to make investment decisions is to consider the total cost of a system over its entire life. Because the costs or revenues during these phases occur at different times, time-value of money equations can be used to calculate the present value of all costs and revenues over the lifecycle of a product. The net present value is the sum of the present values of the costs and revenues of a system over its lifetime. A positive net present value indicates that an investment is more cost-effective than investing the money at the discount rate used in the calculations. When the net present value is calculated for all costs and revenues over a product’s lifetime, including manufacturing, operating and post-use phases, it is called the lifecycle cost.
Example:
Determine the net present value (i.e. lifecycle cost) of a 2 kW photovoltaic system that costs $8,000 per kW, generates 3,000 kWh per year that displaces electricity purchased from the utility at $0.10 /kWh with a projected cost escalation of 2% per year. The system lifetime is 20 years and the discount rate is 5% per year. System recycle income or removal costs at the end of the systems life are negligible.