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 =
InitialCost  $ 


SavingsPerYear  $/yr 
[1]
Example:
Calculate the simple payback of a lighting retrofit that will cost
$1,000 to implement and will save $250 per year.
SP =
$1,000
= 4 years
$250/year
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 =
SavingsPerYear  $/year 


InitialCost  $ 
[2]
Example:
Calculate the ROR of a lighting retrofit that will cost $1,000 to
implement and will save $250 per year.
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Rate of Return =
$250/year
= 25% per year
$1,000
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
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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.
A
0 1 2 3
n
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 =
A
(1  i)
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P2 =
A
A

2
(1  i)
(1  i)
 1
A
1 
Pn = A 

 ... 

n
n1
(1  i) 
 (1  i) (1  i)
To find closed-form solution, do a little algebra...
1)

Pn (1  i)n = A 1  (1  i)  ...  (1  i)n1
n1

n1
2)
Pn (1  i)
2-1)
Pn (1  i)n1  (1  i)n = A (1  i)n  1

Pn =

= A (1  i)  ...  (1  i)

 

 (1  i)n


 1  (1  i)n 
A (1  i)n  1

A


i
(1  i)n (1  i)  1


Thus, the present value of a series of n payments of amount A is:
 1  (1  i)n 
Pn = A 

i


[5]
 1  (1  i)n 
The factor 
 is sometimes called the series present worth factor, SPWF(i,n). The
i


reciprocal of the series present worth factor is sometimes called the capital recovery factor,
CRF(i,n). Thus,
 1  (1  i)n 
Pn = A 
 = A SPWF(i,n)
i


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{SPWF(i,n) = 1 / CRF(i,n)}
4
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
 1  (1  .10) 10 
Pstd costs = $100 + $40 
 = $346
.10


10
 1  (1  .10) 
Phigh costs = $200 + $20 
 = $323
.10


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’.
$20
-$100
$100
 1  (1  .10) 10 
Phigh savings = -$100 + $20 
 = $23
.10


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.
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Example:
Find the annualized savings from investing in the high efficiency
furnace over the standard efficiency furnace:
 1  (1  .10) 10 
Ahigh savings = P / SPWF(.10,10) = $23 / 

.10


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.
A(1+e)n
A
0 1 2 3
n
Using a method similar to the previous derivation, the present value of an escalating series is:
A
n
1  1 e 
1  
  if e  i
(i  e)   1  i  
A
n
(1  e)
[6a]
P=
The factor
if e = i
[6b]
n
1  1 e 
n
(depending on whether i = e) is called the escalating
1  
  or
(1  e)
(i  e)   1  i  
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).
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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:
0 1 2 3
10
$50(1+.05)10
Pcurrent equip costs
= A ESPWF(.05, .05,10)
n
10
= A
= $50
= $476
(1  e)
(1  .05)
New Equipment:
0 1 2 3
10
$10
$400
Pnew equip costs
= $400 + A SPWF(.05,10)
 1  (1  .05) 10 
= $400 + $10 
 = $477
.05


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 =
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

A (1  i)n  1
i
[7]
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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:


 2,000 1  0.05 20  1 
F= 
  $66,132
0.05


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.
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P = F (1+i)-n = F PWF(i,n)
 1  (1  i)n 
P = A
 = A SPWF(i,n)
i


Fn =


A (1  i)n  1
i
A
n
1  1 e 
1

 
  if e  i
(i  e)   1  i  
P=
= A ESPWF(i,e,n)
A
n
(1  e)
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.
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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
$20
-$100
$100
1 2 3
10
If the discount rate is 10%, then:
 1  (1  .10) 10 
Phigh-eff savings = -$100 + $20 
 = $23
.10


The positive savings indicate that the high-efficiency furnace is
the best investment.
If the discount rate is 30%, then:
 1  (1  .30) 10 
Phigh-eff savings = -$100 + $20 
 = $-38
.30


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
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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.
A ESPWF(i=.05, e=.02, n=20)
-$16,000
$100=.
1 2 3
20
PV_IC = - 2 kW x $8,000 /kW = -$16,000
A = 3,000 kWh/yr x $0.10 /kWh = $300 /yr
NPV = PV_IC + A ESPWF(i=.05, e=.02, n=20)
NPV = -$16,000 + $300 14.665
NPV = -11,600
Thus, this is not a cost-effective investment compared to
alternative investments which return 5% per year.
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Return on Investment
Because of the discount rate’s large influence on the results of time value of money
calculations, the discount rate is sometimes solved for, rather than input, in time value of
money calculations. The discount rate when the present value of an investment is zero
represents the return on an equivalent investment, and is called the return on investment, ROI.
To calculate ROI, set NPV = 0 and solve for i.
Example:
Reconsider the furnace example. Find the return on investment
for upgrading to a high-efficiency furnace if the enatural gas =
0.84%.
Savings with Real Fuel Price Escalation
$20 ESPWF(i, e=.0084, n=10)
-$100
$100
1 2 3
10
NPV = -$100 + $20 ESPWF(i,e=.0084,n=10)
  1  .0084  10 
1
0 = -$100 +$20
1  
 
(i  .0084)   1  i  
  1  .0084 
1
$100/$20 = 5 =
1  

(i  .0084)   1  i 
10



By iteration: i = ROI = 16%
In Microsoft Excel, return on investment (ROI) is called internal rate of return (IRR). The IRR is
calculated by an internal function: IRR(range), where the range is made up of all cash flows.
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Example:
A high efficiency furnace costs $100 more than a low efficiency
furnace and saves $20 per year over a ten year period. Use
Microsoft Excel to calculate the return on investment.
A cash flow diagram of savings is shown below.
$20
-$100
$100
In Excel the cash flows below should be entered in separate
cells such as A1: A11.
-100, 20, 20, 20, 20, 20, 20, 20, 20, 20, 20.
Use the IRR function, “=IRR(A1:A11)” to calculate return on
investment.
Evaluating Energy Efficiency Options: The Problem with Simple Payback
In the industrial sector, the economic criterion used to evaluate energy saving opportunities is
frequently simple payback. Moreover, many companies demand very short simple paybacks on
the order of 2 years as the criteria for implementation. On the surface, this appears to be
unrational economic decision making, since a simple payback of 2 years is a rate-of-return of
50%. Thus, demanding a 2-year simple payback before a project will be funded appears to
indicate that a company has alternative investments that return 50% per year. Since this is
seldom the case, the wisdom of demanding very short simple paybacks appears to be
questionable.
However, the real problem is that simple payback, and hence rate-of-return, are poor metrics
for evaluating cost effectiveness because they do not take into account project lifetime.
Consider for example a project with a simple payback of 4 years, and hence a rate-of-return of
25% per year. This appears to be a highly profitable project. However, if the project lifetime is
3 years, then the initial investment will never be paid off. In this case, the 25% per year return is
completely misleading. If the project lifetime is 5 years, the investment is barely cost effective,
since it positive revenue is generated for only one year. If, however, the project lifetime is 20
years, then the project is highly cost effective since the project will generate positive revenue
for 16 years after it has paid back the initial investment. Thus, to properly evaluate energy
saving investments, the economic analysis must include project lifetime.
The economic criteria return on investment (ROI) includes project lifetime, and is thus a much
better measure of cost effectiveness. The return on investment, in contrast to rate of return,
can be directly compared to alternative investments in order to evaluate economic merit.
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Example:
A project has a simple payback of 3 years. Calculate the rate of return, ROR,
and return on investment, ROI, if the project has a lifetime of 3, 6, 12, 15 and 18
years.
Rate of Return = Annual Savings / Initial Cost
Rate of Return = 1 / Simple Payback = 1 / 3 years = 33% /year
To evaluate return on investment for a project with a 33% rate of return, let the
initial cost be $1,000. If so the annual savings would be $333. The net present
value of the investment, assuming that the initial investment has no value at
the end of the project life, is:
 1  (1  i)n 
NPV = -$1,000 + $333 SPWF(i,n) = -$1,000 + $333 

i


To solve for return on investment ROI, set NPV = 0 and solve for i. If the project
lifetime is 3 years, then:
 1  (1  i)3 
0 = -$1,000 +$333 
 By iteration: i = ROI  0%
i


If the project lifetime is 6 years, then:
 1  (1  i)6 
0 = -$1,000 +$333 
 By iteration: i = ROI  24.2%
i


Graphing the rate of return and return on investment as functions of project life
gives:
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Note that ROI calculated in the preceding example is for the case in which the initial investment
has no value at the end of the project life. If the initial investment had value at the end of the
project life, then that value should be added to the NPV equation explicitly. However, once ROI
is calculated using this method, the ROI represents the value of an alternative investment with
both capital and interest accumulation. For example, if an alternative investment returned
24.2% per year for six years, the value of a $1,000 investment after six years would be:
F = P (1 + i)n = $1,000 (1 + .242)6 = $3,671
The ROI of this alternative investment would be calculated as:
NPV = 0 = -$1,000 + $3,671 / (1+i)6
and solving for i gives i = ROI = 24.2%.
Thus, ROI represents the value of an alternative investment with both capital and interest.
A comparison of rate-of-return and return-on-investment, shows that return-on- investment is:



negative when the project life is less than the simple payback
much less than rate-of-return for short project lifetimes
approaches rate-of-return as project life increases.
This shows the importance of project lifetime in determining project cost effectiveness. It may
also explain why some companies require very short simple paybacks to fund a project. In
these cases, the company may believe that the project lifetime is very short. In our view,
however, the use of return-on-investment makes these assumptions explicit and is therefore a
much more transparent metric of economic fitness than simple payback.
Inflation and the ‘Real’ Discount Rate
Inflation causes the price of goods and services to rise or, equivalently, the value of money to
decrease. A widely used metric of the overall rate of inflation in the United States is the
“Implicit Price Deflator”, IPD. Implicit Price Deflator is determined by the U.S. Department of
Commerce. IPDs using year 2000 as the base year, are shown in the table below (Annual Energy
Review 2007, Energy Information Agency, U.S. Department of Energy, Table D1).
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IPD, 1949-2007, 2000 = 1.000
Year
1949
1950
1951
1952
1953
1954
1955
1956
1957
1958
1959
1960
1961
1962
1963
1964
1965
1966
1967
1968
1969
IPD
0.16352
0.16531
0.17718
0.18022
0.18243
0.18417
0.18743
0.19393
0.20038
0.20498
0.20751
0.21041
0.21278
0.21569
0.21798
0.22131
0.22535
0.23176
0.23893
0.24913
0.26149
Year
1970
1971
1972
1973
1974
1975
1976
1977
1978
1979
1980
1981
1982
1983
1984
1985
1986
1987
1988
1989
1990
IPD
0.27534
0.28911
0.30166
0.31849
0.34725
0.38002
0.40196
0.42752
0.45757
0.49548
0.54043
0.59119
0.62726
0.65207
0.67655
0.69713
0.71250
0.73196
0.75694
0.78556
0.81590
Year
1991
1992
1993
1994
1995
1996
1997
1998
1999
2000
2001
2002
2003
2004
2005
2006
2007
IPD
0.84444
0.86385
0.88381
0.90259
0.92106
0.93852
0.95414
0.96472
0.97868
1.00000
1.02399
1.04187
1.06404
1.09462
1.13000
1.16567
1.19664
The annual rate of inflation, j, over any period of n years can be calculated by applying Equation
3.
Example:
Find the general rate of inflation, j, from 1996 to 2006 using
IPDs.
Solution:
F = P (1+j)n
1.16042 = 0.93852 (1 + j)10
j = 0.02145
General inflation rates for various periods are shown in the table below.
1967-2006
Input
n
39
P
0.23893
F
1.16042
Calculations
j
0.041354
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1996-2006
Input
n
10
P
0.93852
F
1.16042
Calculations
j
0.02145
1996-2001
Input
n
5
P
1.02399
F
1.16042
Calculations
j
0.025331
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The rate of inflation is important, since inflation erodes the value of money. For example, if an
investment appreciates at an interest rate of 3% per year and the rate of inflation is also 3% per
year, then the investment does not deliver ‘real’ value.
The discount rate with the effect of inflation removed is called the “real” discount rate and
indicates the “real” return from an investment. To calculate the real discount rate, i’, one needs
to explicitly remove the effect of inflation, j. To do so, consider the following derivation.
Interest Adds Value to Money
Inflation Devalues Money
P
F=
(1  j)n
F = P(1 i)n
If both are active:
P(1  i)n
(1  j)n
Find i’, the “effective” or “real” discount rate, such that:
F=
F = P(1 i')n =
i’ =
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i j
1 j
P(1  i)n
(1  j)n
[8]
17
Example:
Returns from an S&P index fund indicate that a single share of
the fund was worth $14.90 in 1976 and $124.56 in 1994. What
is the market” rate of return from this investment?
Using F = P (1+i)n, we can solve for i, the average annual rate of
return. We assume that the investment began with a present
value P = $14.90 and ended with a future value F = $124.56
after 18 years.
i = market discount rate
i= e
 1  F 
 ln  
 n  P 
1= e
 1  124.56  

 ln 
 18  14.90  
 1 = 12.52 %
If the GDP implicit price deflator (an indication of general
inflation) was 52.3 in 1976 and 126.1 in 1994, what is the
annual rate of inflation over this period?
j = annual rate of inflation
j= e
 1  F 
 ln  
 n  P 
1= e
 1  126.1  

 ln 
 18  52.3  
 1 = 5.01%
What is the real annual rate of return (discount rate) over the
period?
i’ = real discount rate =
.1252  .0501
i j
=
= 7.15%
1 j
1  .0501
Energy Cost Escalation
In energy economics, many calculations include cost of energy, which frequently changes over
time. In most cases, the rate of escalation of energy prices is different than the general rate of
inflation. One way to estimate energy price escalation rates in the future is to consider past
rates of energy price escalation. The tables below show historical U.S. electricity and natural
gas prices (Annual Energy Review 2006, Energy Information Agency, U.S. Department of
Energy).
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Table 8.10 Average Retail Prices of Electricity, 1960-2006
(Cents per Kilowatthour, Including Taxes)
Year
Residential
Commercial 1
Industrial 2
Nominal Real
Nominal Real
Nominal Real
1960
2.6
12.4
2.4
11.4
1.1
1961
2.6
12.2
2.4
11.3
1.1
1962
2.6
12.1
2.4
11.1
1.1
1963
2.5
11.5
2.3
10.6
1
1964
2.5
11.3
2.2
9.9
1
1965
2.4
10.7
2.2
9.8
1
1966
2.3
9.9
2.1
9.1
1
1967
2.3
9.6
2.1
8.8
1
1968
2.3
9.2
2.1
8.4
1
1969
2.2
8.4
2.1
8
1
1970
2.2
8
2.1
7.6
1
1971
2.3
8
2.2
7.6
1.1
1972
2.4
8
2.3
7.6
1.2
1973
2.5
7.9
2.4
7.5
1.3
1974
3.1
8.9
3
8.6
1.7
1975
3.5
9.2
3.5
9.2
2.1
1976
3.7
9.2
3.7
9.2
2.2
1977
4.1
9.6
4.1
9.6
2.5
1978
4.3
9.4
4.4
9.6
2.8
1979
4.6
9.3
4.7
9.5
3.1
1980
5.4
10
5.5
10.2
3.7
1981
6.2
10.5
6.3
10.7
4.3
1982
6.9
11
6.9
11
5
1983
7.2
11
7
10.7
5
1984
7.15
10.57
7.13
10.54
4.83
1985
7.39
10.6
7.27
10.43
4.97
1986
7.42
10.41
7.2
10.11
4.93
1987
7.45
10.18
7.08
9.67
4.77
1988
7.48
9.88
7.04
9.3
4.7
1989
7.65
9.74
7.2
9.17
4.72
1990
7.83
9.6
7.34
9
4.74
1991
8.04
9.52
7.53
8.92
4.83
1992
8.21
9.5
7.66
8.87
4.83
1993
8.32
9.41
7.74
8.76
4.85
1994
8.38
9.28
7.73
8.56
4.77
1995
8.4
9.12
7.69
8.35
4.66
1996
8.36
8.91
7.64
8.14
4.6
1997
8.43
8.84
7.59
7.95
4.53
1998
8.26
8.56
7.41
7.68
4.48
1999
8.16
8.34
7.26
7.42
4.43
2000
8.24
8.24
7.43
7.43
4.64
2001
8.58
8.38
7.92
7.73
5.05
2002
8.44
8.1
7.89
7.57
4.88
2003
8.72
8.2
8.03
7.55
5.11
2004
8.95
8.18
8.17
7.47
5.25
2005
9.45
8.38
8.67
7.69
5.73
2006
10.4
8.96
9.36
8.07
6.09
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5.2
5.2
5.1
4.6
4.5
4.4
4.3
4.2
4
3.8
3.6
3.8
4
4.1
4.9
5.5
5.5
5.9
6.1
6.3
6.9
7.3
8
7.7
7.14
7.13
6.92
6.52
6.21
6.01
5.81
5.72
5.59
5.49
5.28
5.06
4.9
4.75
4.64
4.53
4.64
4.93
4.68
4.8
4.8
5.08
5.25
19
Table 6.8 Natural Gas Prices by Sector, 1967-2006
(Dollars per Thousand Cubic Feet)
Year
Residential Sector
Commercial Sector 1 Industrial Sector 2
Prices
Prices
Prices
1967
1968
1969
1970
1971
1972
1973
1974
1975
1976
1977
1978
1979
1980
1981
1982
1983
1984
1985
1986
1987
1988
1989
1990
1991
1992
1993
1994
1995
1996
1997
1998
1999
2000
2001
2002
2003
2004
2005
2006
Nominal Real
Nominal Real
Nominal Real
1.04
4.35
0.74
3.10
0.34
1.42
1.04
4.17
0.73
2.93
0.34
1.36
1.05
4.02
0.74
2.83
0.35
1.34
1.09
3.96
0.77
2.80
0.37
1.34
1.15
3.98
0.82
2.84
0.41
1.42
1.21
4.01
0.88
2.92
0.45
1.49
1.29
4.05
0.94
2.95
0.50
1.57
1.43
4.12
1.07
3.08
0.67
1.93
1.71
4.50
1.35
3.55
0.96
2.53
1.98
4.93
1.64
4.08
1.24
3.08
2.35
5.50
2.04
4.77
1.50
3.51
2.56
5.59
2.23
4.87
1.70
3.72
2.98
6.01
2.73
5.51
1.99
4.02
3.68
6.81
3.39
6.27
2.56
4.74
4.29
7.26
4.00
6.77
3.14
5.31
5.17
8.24
4.82
7.68
3.87
6.17
6.06
9.29
5.59
8.57
4.18
6.41
6.12
9.05
5.55
8.20
4.22
6.24
6.12
8.78
5.50
7.89
3.95
5.67
5.83
8.18
5.08
7.13
3.23
4.53
5.54
7.57
4.77
6.52
2.94
4.02
5.47
7.23
4.63
6.12
2.95
3.90
5.64
7.18
4.74
6.03
2.96
3.77
5.80
7.11
4.83
5.92
2.93
3.59
5.82
6.89
4.81
5.70
2.69
3.19
5.89
6.82
4.88
5.65
2.84
3.29
6.16
6.97
5.22
5.91
3.07
3.47
6.41
7.10
5.44
6.03
3.05
3.38
6.06
6.58
5.05
5.48
2.71
2.94
6.34
6.76
5.40
5.75
3.42
3.64
6.94
7.27
5.80
6.08
3.59
3.76
6.82
7.07
5.48
5.68
3.14
3.25
6.69
6.84
5.33
5.45
3.12
3.19
7.76
7.76
6.59
6.59
4.45
4.45
9.63
9.40
8.43
8.23
5.24
5.12
7.89
7.57
6.63
6.36
4.02
3.86
9.63
9.05
8.40
7.89
5.89
5.54
10.75
9.82
9.43
8.62
6.53
5.97
12.84
11.39
11.59
10.28
8.56
7.59
13.76
11.86
11.97
10.32
7.89
6.80
Using this data, the energy price escalation rate, e, over any period can be calculated by
applying Equation 3.
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Example:
Find the energy price escalation rate, e, for natural gas for the
residential sector from 1996 to 2006 using nominal cost data
from the Annual Energy Review.
Solution:
F = P (1+e)n
13.76 = 6.34 (1 + e)10
e = 0.081
Residential natural gas price escalation rates for various periods are shown in the table below.
Residential
Nominal
1967-2006
Input
n
P
F
39
1.04
13.76
Calculations
e
0.068461
Residential
Nominal
1996-2006
Input
N
P
F
10
6.34
13.76
Calculations
e
0.08057
Residential
Nominal
2001-2006
Input
n
P
F
5
9.63
13.76
Calculations
e
0.073986
Residential electricity price escalation rates for various periods are shown in the table below.
Residential
1967-2006
Input
n
39
P
2.3
F
10.4
Calculations
e
0.039448
Residential
1996-2006
Input
n
10
P
8.36
F
10.4
Calculations
e
0.022075
Residential
2001-2006
Input
n
5
P
8.58
F
10.4
Calculations
e
0.039224
The ‘real’ energy price escalation rate, e’, which represents the energy price escalation rate
with the effect of inflation removed, can be determined in two ways. The first is by removing
the effect of inflation, j, from the nominal energy escalation rate e using Equation 9. The
derivation for Equation 9 is analogous to the derivation for ‘real’ discount rate in Equation 8.
e’ =
ej
1 j
[9]
Example: Find the real energy price escalation rate, e, for
residential natural gas from 1996 to 2006.
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Solution:
The nominal annual energy escalation rate, e, from 1996 to
2006 was e = 0.081. The annual rate of inflation over this
period was 0.021. The real annual energy escalation rate e’ is:
ej
e’ =
1 j
0.081  0.021
e’ =
= 0.058
1  0.021
Alternately, ‘real’ energy escalation rates, e’, could be determined by applying Equation 3 to
energy costs reported in “constant dollars”.
Example:
Find the real energy escalation rate, e’, for natural gas for the
residential sector from 1996 to 2006 using constant dollar cost
data from the Annual Energy Review.
Solution:
F = P (1+e)n
11.86 = 6.76 (1 + e)10
e = 0.058
Choice of Real or Nominal Discount and Escalation Rates
It is important to use either real rates or nominal rates in time-value of money calculations and
to avoid mixing real and nominal rates in the same calculation. For example, either use nominal
discount rate, i, and nominal energy price escalation rate, e, or use real discount rate, i’, and
real energy price escalation rates e’. When money is borrowed or invested at interest, the
interest is typically represents a nominal rate; thus, to avoid mixing nominal and real rates, it is
common practice to use nominal rates in time value of money calculations.
Tax Deductions for Fuel, Interest and Equipment Depreciation
Tax laws sometimes allow businesses to deduct fuel, interest and equipment depreciation
expenses for tax purposes. Because combined local, state and federal corporate income taxes
are often nearly 50% of profits, it is often necessary to consider these deductions when
evaluating investment options.
In many cases, taxes have the effect of equaling reducing both initial costs and operating
savings. Initial costs are reduced since the initial cost can depreciated over the lifetime of the
equipment and deducted from income. Similarly, annual savings are reduced by the tax rate
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applied to profit. Hence, when considering taxes, the initial cost and annual savings are both
reduced by the tax rate, however the simple payback and ROI remain unchanged.
Example:
Find the NPV and annualized NPV for an energy investment which costs $1,000
and returns $500 per year for 10 years if the discount rate is 5% per year and
the energy cost escalation rate is 2% per year. Do so without considering taxes
and with considering taxes assuming the total tax rate is 40%.
Solution Without Taxes:
Input Data
AS ($/yr)
IC ($)
n (yrs)
i
e
Calculations
ESPWF = (1-((1+e)/(1+i))^n)/(i-e)
Ps ($)= AS x ESPWF
NPV ($) = Ps - IC
SPWF = (1-(1-i)^(-n))/i
A_NPV ($/yr) = NPV / SPWF
500
1,000
10
0.05
0.02
8.388106
4,194
3,194
7.721735
414
Solution With Taxes:
Input Data
AS ($/yr)
IC ($)
n (yrs)
i
e
tr
Calculations
ASAT ($/yr) = AS x (1-tr)
ICAT ($) = IC x (1-tr)
ESPWF = (1-((1+e)/(1+i))^n)/(i-e)
Ps ($) = ASAT x ESPWF
NPV ($) = Ps - ICAT
SPWF = (1-(1-i)^(-n))/i
A_NPV ($/yr) = NPV / SPWF
500
1000
10
0.05
0.02
0.5
250
500
8.388106
2,097
1,597
7.721735
207
Interest Expenses
If equipment is purchased with a mortgage, the interest is usually deductible. Because energy
saving equipment usually has a higher first cost than traditional equipment, interest deductions
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usually enhance the cost effectiveness of energy saving investments. Following the fuel savings
example:
Tax Savings = Tax1 - Tax2 = TR(Interest2 - Interest1) = TR(I2 - I1)
The example below shows how to calculate the interest and principle components of a
mortgage payment.
Example:
Show the interest and principle components of loan payments for a $10
loan borrowed at an interest rate of 10% for 5 years.
P = A SPWF(.10,5)
A = P / SPWF(.10,5) = $10 / 3.79 = $2.64 per year
The components are shown in the table below.
At end
of year
1
2
3
4
5
Payment
amount
2.64
2.64
2.64
2.64
2.64
Interest part
of payment
.1(10) = 1.00
.1(8.36) = .84
.1(6.56) = .66
.1(4.58) = .46
.1(2.40) = .24
Principle part
of payment
2.64 - 1.00 = 1.64
2.64 - .84 = 1.80
2.64 - .66 = 1.98
2.64 - .46 = 2.18
2.64 - .24 = 2.40
Principle
remaining
10 - 1.64 = 8.36
8.36 - 1.80 = 6.56
6.56 - 1.98 = 4.58
4.58 - 2.18 = 2.40
2.40 - 2.40 = 0
Depreciation Expenses
Because equipment wears out over time, many tax laws allow businesses to deduct the cost of
equipment wear from income taxes. The annual amount of wear is called the depreciation, D.
Several methods are usually allowed by the tax laws to calculate the depreciation. Straight-line
depreciation calculates D as:
D = (Purchase Cost - Salvage Value) / Equipment Lifetime
As in the following examples,
Tax Savings = TR(D1 - D2)
The higher costs of energy conserving equipment usually increases the depreciation deduction
and make energy saving investments more attractive.
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Uncertainty in Economic Analyses
Studies have shown that the results of economic analyses are most sensitive to i, e and n.
Because of this, it is recommended that one determine and report the sensitivity of the
economic analyses results using analytical or substitutional methods.
Example:
Determine the sensitivity of the present value of a future
amount of $100 in year 10 if the discount rate is 6%  2%.
By calculus...
P = F (1+i) -n
dP =
P
P
P
di 
dn 
dF 
i
n
F
if F and n are known exactly, then dn = dF = 0 and
dP =
P
di 
i
= F (-n) (1+i)-n-1 di = $100 (-10) (1+.06)-10-1 (.02)
= 
Or by substitution...
P = F (1+i) -n at limits of i
Plow = $100 (1+.08)-10 = $46.3
Phigh = $100 (1+.04)-10 = $67.5
Corporations and Energy Investments
The Congressional Office of Technology Assessment concluded that the biggest factor affecting
industrial energy efficiency is the will to invest in new technology. In normal course of business,
worn out and obsolete technologies are replaced with better and usually more energy efficient
technologies.
Will to invest influenced by:
 Maturity of industry: Young, high-growth industries tend to invest heavily, while
mature industries with price based competition and low profit margins have little
incentive.
 Business climate: Growth and competition encourage investment.
 Corporate climate: “If it ain’t broke don’t fix it” versus
“continual improvement”
 Shortage of technical personnel, especially in “lean” companies.
 Regulations (mainly environmental)
 Energy’s fraction of production costs.
Many corporations demand a simple payback of 2 or 3 years for energy efficiency investments.
This extremely high investment hurdle causes the “efficiency gap”, which economists have
identified as “chronic underinvestment” in energy efficiency. Some corporations report that
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the reason for the high economic hurdle is that energy efficiency projects must compete for inhouse capital and management time. Some analysts have observed that mandatory projects
(such as regulatory compliance, replacement of essential equipment, and maintenance of
product quality) and strategic projects (such as those which increase market share or new
product development) have higher corporate priority than discretionary (energy efficiency)
projects. Another commonly cited reason for high discount rates for energy efficiency projects
is the risk associated with those projects. Although risk means different things to different
organizations, when risk is measured in terms of volatility, the risk of energy efficiency
investments is about the same as U.S. T-bills, while the return on investment is greater than
that of small company stocks.
Source: Laitner, J., Ehrhardt-Martinez, K. and Prindle, W., 2007, “American Energy Efficiency
Investment Market”, Energy Efficiency Finance Forum, American Council or and Energy Efficient
Economy .
Some of these barriers can be lessened by energy service companies offering shared savings
and the linking of energy efficiency to pollution prevention and a good corporate image.
Some Good References
Bartlett, A.A., “Forgotten Fundamentals of the Energy Crisis", The American Journal of Physics,
Volume 46, September 1978, pages 876 to 888.)
Duffie, J. and Beckman, W., 1991. Solar Engineering of Thermal Processes, John Wiley and Sons,
Inc., New York, NY.
Stoecker, W., Design of Thermal Systems, 1989. Design of Thermal Systems, McGraw-Hill, Inc.,
New York, NY.
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Thuesen, G. and Fabrycky, W., 1993. Engineering Economy, Prentice Hall, Englewood Cliffs, NJ.
U.S. Congress, Office of Technology Assessment, Industrial Energy Efficiency, OTA-E-560, 1993.
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