Engineering Economics
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Topics
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Motivation
Types of Costs
Two Typical Scenarios for Analysis
“Simple” Methods
The Time Value of Money
The Present Value Method
The Project Timetable
Accounting for Taxes and Depreciation
Accounting for Inflation
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Motivation
 The objective is to introduce methods of
economic analysis for energy engineering
decision making in a corporate, institutional or
governmental setting.
 Legal, environmental, public relations, energy
efficiency, safety and ethical considerations
are important, but the most important
consideration for engineering design
decisions is economics-- “dollars and cents”.
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More Motivation
 If factors can be valued in $, they should be
included in the economic analysis.
 The analysis methods here do not require
tables. The use of computer spreadsheets
(like EXCEL) will be emphasized.
 A spreadsheet is a computerized
accountant’s ledger- ideal for economic
analysis.
 Spreadsheets give greater flexibility
than table-based methods.
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Types of Costs
There are usually two types of costs associated
with an engineering project, one-time costs,
which include first costs and salvage costs, and
annual costs
(or benefits) that occur every
year or several
years of the
project.
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First Costs
First Costs or Initial Costs are the costs
necessary to implement a project, including:
• Costs of new equipment
• Costs of shipping and installation
• Costs of renovations needed to install
equipment
• Cost of engineering
• Cost of permits,
licenses, etc.
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Items that Reduce First Cost
Some costs of starting projects may be offset by
immediate savings:
• Deduct gains from sale of replaced equipment
• Deduct investment tax credits. These are
government tax incentives to purchase certain
equipment, and reduce first costs by reducing
taxes.
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Salvage Value
We are attempting to estimate the total cost
of doing a project. Cost is reduced if we can
sell the equipment at end of project.
Salvage value is the money that can be
obtained at the end of the project by selling
equipment. Salvage value is a benefit rather
than a cost.
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Annual Costs or Benefits
Annual Costs and Benefits are costs and
benefits of the project that accrue over two or
more years of the project, including:
 Direct operating costs such as labor,
supervision, janitorial, supplies, maintenance,
material, electricity, fuel, etc.
 Indirect operating costs sometimes included,
such as a portion of building rent, a portion of
secretarial expenses, etc.
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Annual Costs or Benefits (Cont’d)
 Sometimes include tax costs or benefits. If the
project increases profits, it also increases taxes,
which are an additional cost. There is a tax
benefit if project reduces profits.
 Depreciation of equipment for tax purposes is
also a tax benefit.
 Finally, need to include savings or profits from
the project.
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Two Analysis Scenarios
New Project- selecting from two or more
alternative solutions. The objective in this
scenario is to find the lowest cost solution that
accomplishes some objective.
Example- Your client, a school district is
building a new school. You are designing the
heating and cooling system. Should you
select: (1) a gas boiler plus an air conditioner,
(2) an electric air-to-air heat pump, or (3) a
ground-source heat pump?
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Second Analysis Scenario
Replacement Project- a method for
accomplishing the goal is already in place, but
a new alternative solution can accomplish the
goal more cheaply. Do the future savings from
the new method justify spending money now to
cover the first costs?
Example- Should 2 existing machines with an
operator each be replaced with a single new
machine with the same output and a single
operator?
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“Simple” Methods
 Simple Payback Period (SPP)- The time
required for savings to offset first costs.
 Simple Return on Investment (ROI)- The
simple percent return the project pays over its
life.
 These methods are “simple” because they do not
consider the time value of money.
 Simple methods are OK for investments that are
very good and pay off over short time periods.
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Simple Payback Period
For a replacement project:
First
Cost
SPP 
AnnualSavings
For comparing two projects A and B, where
the first cost of A is greater than B, but the
annual costs of A are lower than B:
First Cost A  First Cost B
SPP 
AnnualCost A  AnnualCost B
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Simple Return on Investment
The simple rate of return on investment is:
First
Cost
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Annual Savings  
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Lifetime 
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ROI 
First Cost
Lifetime is the life of the project.
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Simple Methods Example
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The Acme Threaded Products Corp. now
uses 2 machinists to operate its 4 screw
machines. Each screw machine costs
$12,000/yr for electricity, maintenance and
wasted materials. Each machinist costs
$35,000/yr. Find SPP for replacing 4 old
machines with 2 new machines @ $60,000
each that produce same output, cost
$14,000/yr each to operate, and can be run
by only one machinist? The old machines
can be sold for $4000 each.
Solution
 First cost is 2 * $60,000 - 4 * $4000, or:
First cost = $104,000.
 Savings/yr is $35,000 (labor), + 4 * $12,000 - 2 *
$14,000 (operation), or:
Savings/yr = $55,000/yr
 SPP = First cost/(Savings/yr)
= $104,000/($55,000/yr)
or SPP = 1.89 yr
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Follow-On Example
For the previous example, which had a first cost
of $104,000 and an annual savings of
$55,000/yr, the expected useful life of the
equipment is 4 years. Acme Threaded Products
Corp. requires a minimum simple return on
investment, ROI, of 10%. (This minimum simple
ROI is often called the corporate hurdle rate.)
Does this investment exceed the corporate
hurdle rate of 10%?
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Solution
ROI =
[Savings/yr - (First Cost/Lifetime)]/First Cost
=[$55,000/yr - ($104,000/4 yr)]/$104,000/yr
ROI = 0.279 = 27.9% (per year)
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Time Value of Money
 “Would you prefer $100 now or a year from
now?” Most would prefer “cash up front”
because $1 today is worth more than $1
some time in the future. You could invest $1
today and have more than $1 in a year.
 A project with a life of several years has
cash flows at various times. To consider
dollar amounts at different times, we need to
put all amounts on an equal basis taking the
time value of money into account.
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Present and Future Value
 Present Value is the value now of an amount of
money F received n years in the future.
 Future Value is value n years in the future of an
amount of money P received now.
 If we can earn interest rate i on investments, the
relationship between P and F is:
F = P(1 + i)n
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or
P = F/(1 + i)n
Example
 Problem- If a savings bond with a yield of
6% matures with a value of $1000 in 8 years,
what does it cost now?
 Solution- This is equivalent to asking what is
the present value P of F = $1000 received n =
8 years in the future at an interest rate of i =
6%, so P =:
F/(1 + i)n = $1000/(1 +0.06)8 = $627.41
 A savings bond with these terms would cost
this amount today.
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Example
 Problem- If you invest $500 today at 9%
interest, how much time is required for the
investment to be worth $1200?
 Solution- Here P = $500, F = $1200, i =
9% and n is the unknown, so:
F = P(1 + i)n
or
F/P = (1 + i)n
ln F/P = n ln(1 + I)
or:
n = [ln F/P]/[ln(1 + i)]
n = [ln 1200/500]/[ln(1 + 0.09)] = 10.2 yr
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The Present Value Method
 Method determines the value in today’s dollars
of a set of cash inflows and outflows at various
times from now into the future.
 Set up a table (spreadsheet!) of all costs and
savings associated with the project, making
costs negative and income/savings positive.
 Sum up all cash flows from each year to get the
net cash flow (NCF) for that year.
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Present Value Method (Cont’d)
Convert each NCF from all years n to present
value by multiplying NCF * 1/(1 + i)n. This
process is called discounting the NCF to account
for present value. Most large companies have a
“discount rate” or “rate of return” i that they use for
all projects.
Add up discounted cash
flows for all years. This
is the present value of
the investment.
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Present Value Method (Cont’d)
 Replacement projects: If present value is
greater than zero, then the accumulated,
discounted savings exceed costs, and the
investment should be made.
 Using a spreadsheet, i can be adjusted so that
P is exactly zero. The value of i for which P = 0
is the discounted rate of return. This means
that the return on the proposed new investment
is the same as putting the money in the bank at
an interest rate of i.
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Present Value Method (Cont’d)
 New projects: The present value will normally
be negative for each alternative. The most
economically attractive alternative is the one
having the smallest (negative) value of P.
 The smallest P corresponds to the alternative
that will cost the least- cheapest.
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The Project Timetable
 One potentially confusing aspect of the present
value method is in which year to place various
expenses.
 The lifetime of the project must first be
determined. The economic analysis covers all
years of the project life.
 For analysis, the project starts on the very last
day (Dec. 31) of Year 0. All first costs plus
investment tax credits are considered to occur in
Year 0 for most projects.
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Project Timetable (Cont’d)
 All operating costs and savings plus any tax and
depreciation effects that occur at any time in
Year 1 are assumed to take place on the final
day of Year 1 , that is, one year after project
start. Similarly, all costs or savings occurring in
any year are considered to occur at the end of
that year.
 The salvage value is a positive cash flow taken
at the end of the final project year.
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Present Value Example
Given: CrimsonCorp lights its factory with
fluorescent lights with a power bill of $12,000/yr
and a lamp replacement cost of $5200/year.
CrimsonCorp is considering high pressure
sodium lights that could be operated for a power
cost of $5000/yr and a replacement cost of
$2600/yr. The cost of removing the old fixtures
and installing new fixtures is $35,000. The life of
the new lamp fixtures is estimated to be at least
8 years.
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Present Value Example
 Find- (a) For a 10% minimum rate of return,
should lights be replaced? (b) What is the actual
rate of return on this investment?
 Solution- The First Cost is $35,000. The
annual savings are the cost of operating old
lights less the cost of operating new lights. So,
the annual savings are ($12,000 + $5200) ($5000 + $2600) = $9600/yr. A spreadsheet
solution is used for this replacement problem.
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Solution Spreadsheet
Part a
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Year 1st Cost Savings NCF Discount
P
0 -35000
-35000
1
-35000
1
9600
9600 0.9091 8727.3
2
9600
9600 0.8264 7933.9
3
9600
9600 0.7513 7212.6
4
9600
9600
0.683 6556.9
5
9600
9600 0.6209 5960.8
6
9600
9600 0.5645 5418.9
7
9600
9600 0.5132 4926.3
8
9600
9600 0.4665 4478.5
9
9600
9600 0.4241 4071.3
10
9600
9600 0.3855 3701.2
Present Value Is: 23988
Discount Rate Is: 0.1
Solution, Part b
 The actual rate of return is the value of i that
causes the present value to exactly equal zero.
 This is easily determined once the Excel
spreadsheet is set up by using Goal Seek...
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Taxes
 Taxes, local, state
and federal, are a
fact of life
 Taxes are applied to
profits, therefore:
• an expense that decreases profit also decreases
taxes owed by the company.
• a cash inflow that increases profit also increases
taxes owed.
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Taxes (Cont’d)
 To account for taxes on the cash flow in a
particular year, add to the NCF the quantity: -1
* NCF * tax rate. The negative sign shows that
a “+” cash flow has a “-” tax effect.
 The federal corporate tax rate varies over time
(with politics and circumstances). Consult your
accountant for details!!
 The primary tax effect described here affects
the net cash flow any time there is any cash
flow in a year-- positive or negative.
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Taxes (Cont’d)
 In addition to the primary tax effect, in some
years there can be a tax effect from investment
tax credits or depreciation.
 To encourage investment in some goods (like
energy efficiency), the government some-times
legislates an investment tax credit.
 An investment tax credit is a specified
percentage of the purchase price that can be
subtracted directly from tax bill in the year of the
purchase (Year 0 for our purposes).
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Tax Credit vs. Deduction
 A tax credit is an amount subtracted directly
from the taxes owed. The taxpayer’s taxes are
reduced by the entire amount of a credit.
 A tax deduction is an amount subtracted from
income. The company benefits only by the tax
rate times the deduction for a tax deduction.
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Depreciation
 A company cannot deduct the cost of capital
purchases from its income as a “cost of doing
business.” The IRS regards capital purchases
as investments rather than expenses.
 The company can “write off,” or depreciate
capital purchases over a number of years.
 Depending on type of equipment, a company can
claim a depreciation deduction of a specified
amount for a specified number of years, following
a “depreciation schedule.”
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Depreciation (Cont’d)
 The tax effect of a depreciation deduction can be
calculated as:
(-1) * First Cost * Deprec. Rate * Tax Rate
 Note that First Cost is a negative quantity, so the
depreciation effect is a positive cash flow.
 The most commonly used gov’t approved
depreciation schedule is the Accelerated Cost
Recovery System (ACRS):
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ACRS Depreciation Schedule
Year
1
2
3
4
5
6
7
8
9
10
11
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3-Year
33.3
44.5
14.8
7.4
5-Year
20
32
19.2
11.5
11.5
5.8
7-Year
14.3
24.5
17.5
12.5
8.9
8.9
8.9
4.5
10-Year
10
18
14.4
11.5
9.2
7.4
6.6
6.6
6.5
6.5
3
Depreciation (Cont’d)
 Depending on what the government decides is
the lifetime of a type of capital equipment, a 3, 5,
7 or 10-year depreciation schedule is followed.
 Note: the fact that an item has been depreciated
100% for tax purposes has no relation to its
actual value or salvage value.
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Tax and Depreciation Example
 Given- Previously, CrimsonCorp was
considering an investment with a first cost of
$35,000 and annual savings of $9600/yr.
Assume that CrimsonCorp is entitled to a 5%
investment tax credit on the first cost, that pays
a 34% tax on profits, and that the federal gov’t
classifies the lighting equipment in the “5-year”
property class.
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Tax and Depreciation Example
 Find- (a) the present value of the investment
assuming a 10% discount rate, and (b) the
actual rate of return for the investment.
 Solution- Again, a computer spreadsheet
solution is recommended. The 5% investment
credit is the only Year 0 tax effect, there is a
primary tax effect for Years 1 - 10, and the 5year class depreciation schedule is followed in
Years 1 - 6...
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Measures of Inflation
 Prices tend to increase over time
due to inflation in money’s value.
 Although some goods inflate faster
than others (like health care and
education), the average rate of inflation is
reflected by the Consumer Price Index (CPI).
 The ratio of the CPI for one year to that of
another is inversely proportional to the ratio of
the buying power of a dollar for the same two
years.
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Inflation/CPI Example
 Given- The average CPI (based on 1982-84 =
100) for 1970 was 38.8 and the CPI for 1991 was
136.2. How much could a dollar buy in 1991
compared to 1970.
 Sol’n- The ratio of buying power is inversely
proportional to the ratio of CPI:
CPI-1970/CPI-1991 = 38.8/136.2 = 0.285
 Therefore, the 1991 dollar is worth only 28.5% of
the 1970 dollar for the average consumer
purchase.
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Inflation
 There is a simple relationship between the
purchasing power of a dollar now compared to
the value of a dollar n years in the future
assuming that the inflation rate is I:
$now = $future/(1 + I)n
 Note that i is used for interest rate, I for inflation
rate.
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Example
 Given- If it costs you $15,000/year to live now,
how much will it cost in 10 years if the inflation
rate is 4.5%?
 Sol’n-
$Future = $Now * (1 + I)n
so $15,000 * (1 + 0.045)10 = $23,300
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Example
 Given- For medical care, the 1970 CPI was 34.0
and the 1991 CPI was 177.
 Find- the average inflation rate for medical care
over the 21 year period.
 Sol’n- $34 in 1970 = $177 in 1992, so
$177 = $34 * (1 + I)21
ln(177/34) = 21 ln(1 + I)
exp[(1/21) ln(177/34)] = 1 + I
I = exp[(1/21) ln(177/34)] - 1= 0.0817= 8.17%
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Inflation
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Note that a dollar in the future is worth less than
a dollar now for two different reasons:
(1) You could invest your dollar today and have
more than a dollar at the future time, so it takes
more than a dollar in the future to have the
present value of a dollar today (time value of
money).
(2) Your dollar in the future won’t be able to buy as
much as it could today because of inflation.
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Accounting for Inflation
(1) Inflate all costs and savings using the
available information, so that net cash flow
before taxes is accurate for the year in
question. Lacking better information, use the
average inflation rate.
(2) Calculate NCFAT the usual way.
(3) Calculate the present value discount factor
the same as before, using discount rate i.
PV Discount = 1/(1 + i)n
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Accounting for Inflation (Cont’d)
(4) Calculate an inflation discount factor using
inflation rate I:
Inflation Discount = 1/(1 + I)n
(5) Calculate an overall discount factor:
Overall Disc. = Inflation Disc. * PV Disc.
(6) The Present Value for a year is the NCFAT
times the overall discount factor.
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Inflation Example
Given: Acme Axles needs a heat treating oven
for its account with Deutschmobile AG. An
electric oven and natural-gas fired oven are
considered. For both options assume a tax rate
of 34%, 7-year depreciation, a 10-year life, no
tax credits, salvage value of 10% of first cost,
discount rate of 10%, inflation rate of 4%,
electricity inflation rate of 5%, and natural gas
inflation rate of 7%, All costs of the two ovens
are identical except for the following:
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Inflation Example (Cont’d)
Electric Oven- First cost is $125,000, electricity
cost is $35,000/yr, maintenance cost is
$6000/yr.
Gas Oven- First cost is $185,000, fuel cost is
$10,000/yr, maintenance cost is $9000/yr.
Find: which option should Acme select?
Sol’n: Use spreadsheet solution method, as
follows:
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