Engineering Economics
Energy Economics
IntroductionMost 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 ReturnThe 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 MoneyThe 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, Id 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 AmountConsider 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 = 100F1 = 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)2Fn =
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 AnnuitiesAn 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 =
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= $323and
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
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 AnnuitiesSometimes 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:
if e 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:
0 1 2 3 10
$50(1+.05)10
Pcurrent equip costs = A ESPWF(.05, .05,10)
= = = $476
New Equipment:
0 1 2 3 10
$10
$400
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 PaymentsIn 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 PeriodsTypically, 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 iP = = A ESPWF(i,e,n)
if e = i
The Discount RateSo 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 ones 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
$20
1 2 3 10-$100$100
If the discount rate is 10%, then:
Phigh-eff savings = -$100 + $20= $23The 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 ValueThe 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 products
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) 1 2 3 20
-$16,000$100=.
PV_IC = - 2 kW x $8,000 /kW = -$16,000A = 3,000 kWh/yr x $0.10
/kWh = $300 /yrNPV = PV_IC + A ESPWF(i=.05, e=.02, n=20)NPV =
-$16,000 + $300 14.665NPV = -11,600
Thus, this is not a cost-effective investment compared to
alternative investments which return 5% per year.
Return on InvestmentBecause of the discount rates 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
-$100$100$20 ESPWF(i, e=.0084, n=10) 1 2 3 10
NPV = -$100 + $20 ESPWF(i,e=.0084,n=10)
0 = -$100 +$20
$100/$20 = 5 =
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.
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
PaybackIn 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.
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:
NPV = -$1,000 + $333 SPWF(i,n) = -$1,000 + $333
To solve for return on investment ROI, set NPV = 0 and solve for
i. If the project lifetime is 3 years, then:
0 = -$1,000 +$333 By iteration: i = ROI 0%
If the project lifetime is 6 years, then:
0 = -$1,000 +$333 By iteration: i = ROI 24.2%
Graphing the rate of return and return on investment as
functions of project life gives:
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 RateInflation 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).
IPD, 1949-2007, 2000 = 1.000
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)n1.16042 = 0.93852 (1 + j)10j = 0.02145
General inflation rates for various periods are shown in the
table below.
1967-20061996-20061996-2001
InputInputInput
n39n10n5
P0.23893P0.93852P1.02399
F1.16042F1.16042F1.16042
CalculationsCalculationsCalculations
j0.041354j0.02145j0.025331
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 MoneyInflation Devalues Money
F = F = If both are active:
F = Find i, the effective or real discount rate, such that:
F = =
i = [8]
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 = = = 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 = = = 5.01%
What is the real annual rate of return (discount rate) over the
period?
i = real discount rate = = = 7.15%
Energy Cost EscalationIn 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).
Using this data, the energy price escalation rate, e, over any
period can be calculated by applying Equation 3.
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)n13.76 = 6.34 (1 + e)10e = 0.081
Residential natural gas price escalation rates for various
periods are shown in the table below.
Residential NominalResidential NominalResidential Nominal
1967-20061996-20062001-2006
InputInputInput
n39N10n5
P1.04P6.34P9.63
F13.76F13.76F13.76
CalculationsCalculationsCalculations
e0.068461e0.08057e0.073986
Residential electricity price escalation rates for various
periods are shown in the table below.
ResidentialResidentialResidential
1967-20061996-20062001-2006
InputInputInput
n39n10n5
P2.3P8.36P8.58
F10.4F10.4F10.4
CalculationsCalculationsCalculations
e0.039448e0.022075e0.039224
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 = [9]
Example: Find the real energy price escalation rate, e, for
residential natural gas from 1996 to 2006.
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 =
e = = 0.058
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)n11.86 = 6.76 (1 + e)10e = 0.058
Choice of Real or Nominal Discount and Escalation RatesIt 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 DepreciationTax 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 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:
Solution With Taxes:
Interest ExpensesIf 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 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 yearPaymentamountInterest partof paymentPrinciple
partof paymentPrincipleremaining
12.64.1(10) = 1.002.64 - 1.00 = 1.6410 - 1.64 = 8.36
22.64.1(8.36) = .842.64 - .84 = 1.808.36 - 1.80 = 6.56
32.64.1(6.56) = .662.64 - .66 = 1.986.56 - 1.98 = 4.58
42.64.1(4.58) = .462.64 - .46 = 2.184.58 - 2.18 = 2.40
52.64.1(2.40) = .242.64 - .24 = 2.402.40 - 2.40 = 0
Depreciation ExpensesBecause 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.Uncertainty in Economic AnalysesStudies
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 = if F and n are known exactly, then dn = dF = 0 and
dP = = 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 iPlow = $100
(1+.08)-10 = $46.3Phigh = $100 (1+.04)-10 = $67.5
Corporations and Energy InvestmentsThe 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 aint broke dont fix
it versus continual improvement Shortage of technical personnel,
especially in lean companies. Regulations (mainly environmental)
Energys 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 the reason for the
high economic hurdle is that energy efficiency projects must
compete for in-house 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 ReferencesBartlett, 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.
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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