short discussion responses

Description

View attached document. Response to each post.THE TOPIC: Why would engineers be involved in cost analysis?? Isn’t this better left to
accountants or CEO type people??
POST1:
Engineers are involved in cost analysis because there are many things in engineering that
require knowing specifics. Accountants may look at cost as one simple blanket cost per sq ft.
As I have understood, many general contractors will do this to get a rough ballpark type price.
While this can work out okay in typical simple projects, it will likely not provide you with a very
accurate number for other projects. An engineer would be able to get a more detailed bid with
labor, materials such as pipe lengths, type of pipe, fittings, specialty sprinklers, fire pumps etc.
A building with a dry system will not cost the same as if it was a wet system. More intricate
ceilings and building designs can increase the number of sprinklers significantly on a project.
Specialty sprinklers like window sprinklers or attic sprinklers that are a lot more costly may be
needed..
RESPONSE EXAMPLE:
Exactly my thoughts Jessica. Accountants might have a good knowlegde of costs calculations,
but not how different dimensions of a construction project impact the costs as well as quality. Besides
maintaining the costs, engineers monitor and safeguard the equipment used on a construction project.
They are responsible for the integrity and safety of this equipment. Thus, they can provide a better picture
of costs associated with a project.
.
YOU RESPONSE TO POST1:
2-3 SENTENCES
POST2:
Cost analysis helps determine a project’s feasibility and the anticipated profits. The
managerial and accounting departments do a general overview of the involved costs of
production but lack a deeper understanding of material type, labor, and workforce for
successful completion. When analyzing the price of a project, engineers would be involved
because they will give a more comprehensive look into the operating costs and find the bestsuited financial option within the stipulated budget. Engineers also have a more accurate
estimate of when a project is due given the suitable materials and human resources; this saves
cost and prevents last-minute risks from financial strain (Project Management; Cost Estimation,
2023). Also, the company executives are more focused on increasing company income hence
involving the engineer in cost breakdown will improve efficiency in delivering the top benefits
of a project on time, attracting more clientele.
References
Project Management for Construction: Cost Estimation. (2023). Cmu.edu.
https://www.cmu.edu/cee/projects/PMbook/05_Cost_Estimation.html
RESPONSE EXAMPLE:
Great post! One thing I would like to add is calculated losses. There are many projects out there
that end up losing money because of no proper planning and some mistakes along the way! I think a
good engineer can utilize his/her experience to validate why one should use a certain manufacturer over
another by past experience. I think a lot of companies would like to profit but some would just like to break
even because too many issues during the process. Well done!
YOUR RESPONSE TO POST2:
2-3 SENTENCES
THE TOPIC: You are going to buy a new car (an awesome one – I might add). Review some
websites and go over their various options (buy, lease, 0% APR vs. cash back, etc.). When would
these options make a difference? When would they be more beneficial? Why?
POST 2.1:
The economy in whichever way you look at it has its own pros anfd cons. Therefore I
would say that whether you lease a car, buy a car, have a 0% APR, or get cash back on your
purchases, all depend on your situation.
When you lease a car, some of the issues you deal with are having to deal with the sale price, a
certain lenght of the period of lease, maximum amaount of mileage allowed, no ownership,
fees and other costs, lack of control, but you also get away with low monthly costs, a new car
every few years, worry-free maintenance, no resale worries, and you may even enjoy some tax
deductions.
When you buy and own a car, it is the other way round leasing but in this case you will also
have to deal with rapid depreciation that will affect resale, and the immense amount of driving
and maintenance costs.
So, one cap does not fit all when it comes to economical benefit because every situation is
different.
RESPONSE EXAMPLE:
HiI 100% agree with you! I think a lot of it also depends on the economy too! I think, at the
moment, cash is king and if someone is capable of getting a loan for less than 5%, they are in the green
because of current inflation and interest rates. I have always been a cash buyer, but I definitely rethought
that in our current economy! Great post!
YOUR RESPONSE TO POST 2.1:
2-3 SENTENCES.
POST 2.2:
Today, cars and other automobiles are one of the most common necessities. The
discussion of purchasing a new car can be quite unsettling due to current prices as well as
increased demand. Understanding your options, the market, and other factors will in theory
make your next purchase easier. Most dealerships allow three forms of payment those being
leasing, financing, and purchasing. Leasing a car, like a long-term rental, requires an upfront
and monthly payment system. At the end of the lease, you have the option to decide to
continue that lease, purchase, or even change cars. Financing your car involves taking out a
loan to purchase the vehicle and then paying back that loan over time. Applied interest rates
and fees are involved. Finally, purchasing your vehicle simply means paying full price for the
vehicle on the lot. At the end of the day, all options have their unique pros and cons but the
most important factor to consider is what can you and your budget afford and whether is it
within your parameters.
RESPONSE EXAMPLE:
You make a good point here! Because cars are more of a necessity than luxury to many,
the control of which acquisition joyce is limited even though the determining factors are many.
In the investment perspective, the markets really determine which way is more beneficial than
the other. When interest rates are low borrowing is not a big problem as compared to when
interest rates are high. And again individual situations would determine the best beneficial
options for them.
Thanks for your post.
YOUR RESPONSE TO POST 2.2:
2-3 SENTENCES.
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Page 93
SECTION FIVE
CHAPTER 7
Engineering Economics
John M. Watts, Jr., and Robert E. Chapman
Introduction
Engineering economics is the application of economic
techniques to the evaluation of design and engineering alternatives.1 The role of engineering economics is to assess
the appropriateness of a given project, estimate its value,
and justify it from an engineering standpoint.
This chapter discusses the time value of money and
other cash-flow concepts, such as compound and continuous interest. It continues with economic practices and
techniques used to evaluate and optimize decisions on selection of fire safety strategies. The final section expands
on the principles of benefit-cost analysis.
An in-depth treatment of the practices and techniques
covered in this compilation is available in the ASTM
compilation of standards on building economics.2 The
ASTM compilation also includes case illustrations showing how to apply the practices and techniques to investment decisions.
A broader perspective on the application of engineering economics to fire protection engineering can be
found in The Economics of Fire Protection by Ramachandran.3 This work is intended as a textbook for fire protection engineers and includes material and references that
expand on several other chapters of this section of the
SFPE handbook.
Cash-Flow Concepts
Cash flow is the stream of monetary (dollar) values—
costs (inputs) and benefits (outputs)—resulting from a
project investment.
Dr. John M. Watts, Jr., holds degrees in fire protection engineering,
industrial engineering, and operations research. He is director of the
Fire Safety Institute, a not-for-profit information, research, and educational corporation located in Middlebury, Vermont. Dr. Watts also
serves as editor of NFPA’s Fire Technology.
Dr. Robert E. Chapman is an economist in the Office of Applied Economics, Building and Fire Research Laboratory, National Institute of
Standards and Technology.
Time Value of Money
The following are reasons why $1000 today is
“worth” more than $1000 one year from today:
1. Inflation
2. Risk
3. Cost of money
Of these, the cost of money is the most predictable,
and, hence, it is the essential component of economic
analysis. Cost of money is represented by (1) money paid
for the use of borrowed money, or (2) return on investment. Cost of money is determined by an interest rate.
Time value of money is defined as the time-dependent
value of money stemming both from changes in the purchasing power of money (inflation or deflation) and from
the real earning potential of alternative investments over
time.
Cash-Flow Diagrams
It is difficult to solve a problem if you cannot see it. The
easiest way to approach problems in economic analysis is
to draw a picture. The picture should show three things:
1. A time interval divided into an appropriate number of
equal periods
2. All cash outflows (deposits, expenditures, etc.) in each
period
3. All cash inflows (withdrawals, income, etc.) for each
period
Unless otherwise indicated, all such cash flows are considered to occur at the end of their respective periods.
Figure 5-7.1 is a cash-flow diagram showing an outflow or disbursement of $1000 at the beginning of year 1
and an inflow or return of $2000 at the end of year 5.
Notation
To simplify the subject of economic analysis, symbols are introduced to represent types of cash flows and
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$2000
F(2) C F(1) = F(1)(i)
Interest is applied to the new sum:
C (F)(1)(1 = i) C P(1 = i)2
F (3) C F (2)(1 = i) C P(1 = i)3
0
2
1
3
4
5
and by mathematical induction,
F(N) C P(1 = i)N
$1000
Figure 5-7.1.
Cash-flow diagram.
interest factors. The symbols used in this chapter conform
to ANSI Z94;4 however, not all practitioners follow this
standard convention, and care must be taken to avoid
confusion when reading the literature. The following
symbols will be used here:
P C Present sum of money ($)
F C Future sum of money ($)
N C Number of interest periods
i C Interest rate per period (%)
A complete list of the ANSI Z94 symbols is given in Appendix A to this chapter.
Interest Calculations
Interest is the money paid for the use of borrowed
money or the return on invested capital. The economic
cost of construction, installation, ownership, or operation
can be estimated correctly only by including a factor for
the economic cost of money.
Simple interest: To illustrate the basic concepts of interest, an additional notation will be used:
F(N) C Future sum of money after N periods
Then, for simple interest,
F(1) C P = (P)(i) C P(1 = i)
EXAMPLE:
$100 at 10 percent per year for 5 yr yields
F(5) C 100(1 = 0.1)5
C 100(1.1)5
C 100(1.61051)
C $161.05
which is over 7 percent greater than with simple interest.
EXAMPLE:
In 1626 Willem Verhulst bought Manhattan Island
from the Canarsie Indians for 60 florins ($24) worth of
merchandise (a price of about 0.5 cents per hectare [0.2
cents per acre]). At an average interest rate of 6 percent,
what is the present value (2001) of the Canarsies’ $24?
F C P(1 = i)N
C $24(1 = 0.06)375
C $7.4 ? 1010
Seventy-four billion dollars is a reasonable approximation of the present land value of the island of Manhattan.
Interest Factors
Interest factors are multiplicative numbers calculated
from interest formulas for given interest rates and periods. They are used to convert cash flows occurring at different times to a common time. The functional formats
used to represent these factors are taken from ANSI Z94,
and they are summarized in Appendix B to this chapter.
and
F(N) C P = (N)(P)(i) C P(1 = Ni)
For example: $100 at 10 percent per year for 5 yr yields
F(5) C 100[1 = (5)(0.1)]
C 100(1.5)
C $150
However, interest is almost universally compounded to
include interest on the interest.
Compound interest
F(1) C P = (P)(i) C P(1 = i)
is the same as simple interest,
Compound Amount Factor
In the formula for finding the future value of a sum of
money with compound interest, the mathematical expression (1 = i)N is referred to as the compound amount factor,
represented by the functional format (F/P, i, N). Thus,
F C P(F/P, i, N)
Interest tables: Values of the compound amount, present worth, and other factors that will be discussed
shortly, are tabulated for a variety of interest rates and
number of periods in most texts on engineering economy.
Example tables are presented in Appendix C to this chapter. Although calculators and computers have greatly reduced the need for such tables, they are often still useful
for interpolations.
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Engineering Economics
Present Worth
Present worth is the value found by discounting future cash flows to the present or base time.
Discounting: The inverse of compounding is determining a present amount which will yield a specified future
sum. This process is referred to as discounting. The equation for discounting is found readily by using the compounding equation to solve for P in terms of F:
Nominal versus effective interest: It is generally assumed that interest is compounded annually. However,
interest may be compounded more frequently. When this
occurs, there is a nominal interest or annual percentage rate
and an effective interest, which is the figure used in calculations. For example, a savings bank may offer 5 percent
interest compounded quarterly, which is not the same as
5 percent per year. A nominal rate of 5 percent compounded quarterly is the same as 1.25 percent every three
months or an effective rate of 5.1 percent per year. If
P C F(1 = i)>N
EXAMPLE:
What present sum will yield $1000 in 5 yr at 10 percent?
P C 1000(1.1)>5
C 1000(0.62092)
C $620.92
This result means that $620.92 “deposited” today at 10
percent compounded annually will yield $1000 in 5 yr.
Present worth factor: In the discounting equation, the
expression (1 = i)>N is called the present worth factor and is
represented by the symbol (P/F, i, N). Thus, for the present
worth of a future sum at i percent interest for N periods,
r C Nominal interest rate,
and
M C Number of subperiods per year
then the effective interest rate is
‹
M
r
>1
iC 1=
M
EXAMPLE:
Credit cards usually charge interest at a rate of 1.5
percent per month. This amount is a nominal rate of 18
percent. What is the effective rate?
i C (1 = 0.015)12 > 1
C 1.1956 > 1
C 19.56%
P C F(P/F, i, N)
Note that the present worth factor is the reciprocal of the
compound amount factor. Also note that
(P/F, i, N) C
1
(F/P, i, N)
EXAMPLE:
What interest rate is required to triple $1000 in 10 years?
PC
F
C (P/F, i, 10)
3
therefore,
(P/F, i, 10) C
1
3
From Appendix C,
(P/F, 10%, 10) C 0.3855
and
Continuous interest: A special case of effective interest
occurs when the number of periods per year is infinite.
This represents a situation of continuous interest, also referred to as continuous compounding. Formulas for continuous interest can be derived by examining limits as M
approaches infinity. Formulas for interest factors using
continuous compounding are included in Appendix B.
Continuous interest is compared to monthly interest in
Table 5-7.1.
EXAMPLE:
Compare the future amounts obtained under various
compounding periods at a nominal interest rate of 12 percent for 5 yr, if P C $10,000. (See Table 5-7.2.)
Series Payments
Life would be simpler if all financial transactions were
in single lump-sum payments, now or at some time in the
(P/F, 12%, 10) C 0.3220
Table 5-7.1
By linear interpolation,
Continuous Interest (%)
Effective
i C 11.6%
Interest Periods
Normally, but not always, the interest period is taken
as 1 yr. There may be subperiods of quarters, months,
weeks, and so forth.
Nominal %
Monthly
Continuous
5
10
15
20
5.1
10.5
16.1
21.9
5.1
10.5
16.2
22.1
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Table 5-7.2
Example of Continuous Interest N C 5 yr, r C 12%
Compounding
M
i
NM
F/P
F
Annual
Semi-annual
Quarterly
Monthly
Weekly
Daily
Hourly
Instantaneously
1
2
4
12
52
365
8760
ã
12.000
12.360
12.551
12.683
12.734
12.747
12.749
12.750
5
10
20
60
260
1825
43,800
ã
1.76234
1.79085
1.80611
1.81670
1.820860
1.821938
1.822061
1.822119a
17,623.40
17,908.50
18,061.10
18,167.00
18,208.60
18,219.38
18,220.61
18,221.19
aF/P (instantaneous) C e Ni C e 5(0.12) C e 0.6.
Series compound amount factor: Given a series of regular payments, what will they be worth at some future time?
Let
Capital recovery factor: It is also important to be able to
relate regular periodic payments to their present worth;
for example, what monthly installments will pay for a
$10,000 car in 3 yr at 15 percent?
Substituting the compounding equation F C P(F/P, i, N)
in the sinking fund equation, A C F(A/F, i, N), yields
A C the amount of a regular end-of-period payment
A C P(F/P, i, N)(A/F, i, N)
Then, note that each payment, A, is compounded for a
different period of time. The first payment will be compounded for N > 1 periods (yr):
And, substituting the corresponding interest factors gives
future. However, most situations involve a series of regular payments, for example, car loans and mortgages.
F C A(1 = i)N>1
and the second payment for N > 2 periods:
F C A(1 = i)N>2
AC P
In this equation, the interest expression is known as the
capital recovery factor, since the equation defines a regular
income necessary to recover a capital investment. The
symbolic equation is
A C P(A/P, i, N)
and so forth. Thus, the total future value is
F C A(1 = i)N>1 = A(1 = i)N>2 = ß = A(1 = i) = A
or
FC
A[(1 = i)N > 1]
i
The interest expression in this equation is known as the
series compound amount factor, (F/A, i, N), thus
F C A(F/A, i, N)
Sinking fund factor: The process corresponding to the
inverse of series compounding is referred to as a sinking
fund; that is, what size regular series payments are necessary to acquire a given future amount?
Solving the series compound amount equation for A,
8
4
i
AC F
[(1 = i)N > 1]
Or, using the symbol (A/F, i, N) for the sinking fund factor
A C F(A/F, i, N)
Here, note that the sinking fund factor is the reciprocal of
the series compound amount factor, that is, (A/F, i, N) C
1/(F/A, i, N).
[i(1 = i)N ]
[(1 = i)N > 1]
Series present worth factor: As with the other factors,
there is a corresponding inverse to the capital recovery
factor. The series present worth factor is found by solving
the capital recovery equation for P.
PCA
[(1 = i)N > 1]
[i(1 = i)N ]
or, symbolically
P C A(P/A, i, N)
Other Interest Calculation Concepts
Additional concepts involved in interest calculations
include continuous cash flow, capitalized costs, beginning of period payments, and gradients.
Continuous cash flow: Perhaps the most useful function of continuous interest is its application to situations
where the flow of money is of a continuous nature. Continuous cash flow is representative for
1. A series of regular payments for which the interval between payments is very short
2. A disbursement at some unknown time (which is then
considered to be spread out over the economic period)
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Engineering Economics
Factors for calculating present or future worth of a
series of annual amounts, representing the total of a continuous cash flow throughout the year, may be derived
by integrating corresponding continuous interest factors
over the number of years the flow is maintained.
Continuous cash flow is an appropriate way to handle economic evaluations of risk, for example, the present
value of an annual expected loss.
Formulas for interest factors representing continuous, uniform cash flows are included in Appendix B.
Fire protection engineering economic analysis is primarily concerned with cost-reduction decisions, finding
the least expensive way to fulfill certain requirements, or
minimizing the sum of expected fire losses plus investment in fire protection.
There are four common methods of comparing alternative investments: (1) present worth, (2) annual cost,
(3) rate of return, and (4) benefit-cost analysis. Each of
these is dependent on a selected interest rate or discount
rate to adjust cash flows at different points in time.
Capitalized costs: Sometimes there are considerations,
such as some public works projects, which are considered
to last indefinitely and thereby provide perpetual service.
For example, how much should a community be willing
to invest in a reservoir which will reduce fire insurance
costs by some annual amount, A? Taking the limit of the
series present worth factor as the number of periods goes
to infinity gives the reciprocal of the interest rate. Thus,
capitalized costs are just the annual amount divided by the
interest rate. When expressed as an amount required to
produce a fixed yield in perpetuity, it is sometimes referred to as an annuity.
Discount Rate
Beginning-of-period payments: Most returns on investment (cash inflows) occur at the end of the period during
which they accrued. For example, a bank computes and
pays interest at the end of the interest period. Accordingly,
interest tables, such as those in Appendix C, are computed
for end-of-year payments. For example, the values of the
capital recovery factor (A/P, i, N) assume that the regular
payments, A, occur at the end of each period.
On the other hand, most disbursements (cash outflows) occur at the beginning of the period (e.g., insurance
premiums). When dealing with beginning-of-period payments, it is necessary to make adjustments. One method
of adjustment for beginning-of-period payments is to calculate a separate set of factors. Another way is to logically
interpret the effect of beginning-of-period payments for a
particular problem, for example, treating the first payment as a present value. The important thing is to recognize that such variations can affect economic analysis.
Gradients: It occasionally becomes necessary to treat
the case of a cash flow which regularly increases or decreases at each period. Such patterned changes in cash
flow are called gradients. They may be a constant amount
(linear or arithmetic progression), or they may be a constant percentage (exponential or geometric progression).
Various equations for dealing with gradient series may be
found in Appendix B.
The term discount rate is often used for the interest
rate when comparing alternative projects or strategies.
Selection of discount rate: If costs and benefits accrue
equally over the life of a project or strategy, the selection
of discount rate will have little impact on the estimated
benefit-cost ratios. However, most benefits and costs occur at different times over the project life cycle. Thus,
costs of constructing a fire-resistive building will be incurred early in contrast to benefits, which will accrue over
the life of the building. The discount rate then has a significant impact on measures such as benefit-cost ratios,
since the higher the discount rate, the lower the present
value of future benefits.
In view of the uncertainty concerning appropriate discount rate, analysts frequently use a range of discount
rates. This procedure indicates the sensitivity of the analysis to variations in the discount rate. In some instances,
project rankings based on present values may be affected
by the discount rate as shown in Figure 5-7.2. Project A is
preferred to project B for discount rates below 15 percent,
while the converse is true for discount rates greater than
15 percent. In this instance, the decision to adopt project A
in preference to project B will reflect the belief that the appropriate discount rate is less than or equal to 15 percent.
A
Net present value
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B
Comparison of Alternatives
Most decisions are based on economic criteria. Investments are unattractive, unless it seems likely they will
be recovered with interest. Economic decisions can be divided into two classes:
1. Income-expansion—that is, the objective of capitalism
2. Cost-reduction—the basis of profitability
0
5
10
15
20
Discount rate (%)
Figure 5-7.2. Impact of discount rate on project selection.
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A comparison of benefits and costs may also be used
to determine the payback period for a particular project or
strategy. However, it is important to discount future costs
or benefits in such analyses. For example, an analysis of
the Beverly Hills Supper Club fire compared annual savings from a reduction in insurance premiums to the costs
of sprinkler installation. Annual savings were estimated at
$11,000, while costs of sprinkler installation ranged from
$42,000 to $68,000. It was concluded that the installation
would have been paid back in four to seven years (depending on the cost of the sprinklers). However, this
analysis did not discount future benefits, so that $11,000
received at the end of four years was deemed equivalent
to $11,000 received in the first year. Once future benefits
are discounted, the payback period ranges from five to
eleven years with a discount rate of 10 percent.
Inflation and the discount rate: Provision for inflation
may be made in two ways: (1) estimate all future costs
and benefits in constant prices, and use a discount rate
which represents the opportunity cost of capital in the absence of inflation; or (2) estimate all future benefits and
costs in current or inflated prices, and use a discount rate
which includes an allowance for inflation. The discount
rate in the first instance may be considered the real discount rate, while the discount rate in the second instance
is the nominal discount rate. The use of current or inflated
prices with the real discount rate, or constant prices with
the nominal discount rate, will result in serious distortions in economic analysis.
Present Worth
In a present worth comparison of alternatives, the
costs associated with each alternative investment are all
converted to a present sum of money, and the least of
these values represents the best alternative. Annual costs,
future payments, and gradients must be brought to the
present. Converting all cash flows to present worth is often referred to as discounting.
EXAMPLE:
Two alternate plans are available for increasing the capacity of existing water transmission lines between an unlimited source and a reservoir. The unlimited source is at a
higher elevation than the reservoir. Plan A calls for the construction of a parallel pipeline and for flow by gravity. Plan
B specifies construction of a booster pumping station. Estimated cost data for the two plans are as follows:
Plan A
Pipeline
Plan B
Pumping Station
Construction cost
Life
$1,000,000
40 years
$200,000
40 years (structure)
20 years (equipment)
Cost of replacing
equipment at the
end of 20 yr
Operating costs
0
$1000/yr
$75,000
$50,000/yr
If money is worth 12 percent, which plan is more economical? (Assume annual compounding, zero salvage
value, and all other costs equal for both plans.)
Present worth (Plan A) C P = A(P/A, 12%, 40)
C $1,000,000 = $1000(8.24378)
C $1,008,244
Present worth (Plan B) C P = A(P/A, 12%, 40)
= F(P/F, 12%, 20)
C $200,000 = $50,000(8.24378)
= $75,000(0.10367)
C $619,964
Thus, plan B is the least-cost alternative.
A significant limitation of present worth analysis is
that it cannot be used to compare alternatives with unequal economic lives. That is, a ten-year plan and a
twenty-year plan should not be compared by discounting
their costs to a present worth. A better method of comparison is annual cost.
Annual Cost
To compare alternatives by annual cost, all cash flows
are changed to a series of uniform payments. Current expenditures, future costs or receipts, and gradients must be
converted to annual costs. If a lump-sum cash flow occurs
at some time other than the beginning or end of the economic life, it must be converted in a two-step process: first
moving it to the present and then spreading it uniformly
over the life of the project.
Alternatives with unequal economic lives may be
compared by assuming replacement in kind at the end of
the shorter life, thus maintaining the same level of uniform payment.
Partial system
Full system
System Cost
Insurance
Premium
Life
$ 8000
$15,000
$1000
$250
15 yr
20 yr
EXAMPLE:
Compare the value of a partial or full sprinkler system purchased at 10 percent interest.
Annual cost (partial system) C A = P(A/P, 10%, 15)
C $1000 = $8000(0.13147)
C $2051.75
Annual cost (full system) C A = P(A/P, 10%, 20)
C $250 = $15,000(0.11746)
C $2011.90
The full system is slightly more economically desirable.
When costs are this comparable, it is especially important
to consider other relevant decision criteria, for example,
uninsured losses.
Rate of Return
Rate of return is, by definition, the interest rate at
which the present worth of the net cash flow is zero. Computationally, this method is the most complex method of
comparison. If more than one interest factor is involved,
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the solution is by trial and error. Microcomputer programs
are most useful with this method.
The calculated interest rate may be compared to a
discount rate identified as the “minimum attractive rate
of return” or to the interest rate yielded by alternatives.
Rate-of-return analysis is useful when the selection of a
number of projects is to be undertaken within a fixed or
limited capital budget.
EXAMPLE:
An industrial fire fighting truck costs $100,000. Savings in insurance premiums and uninsured losses from
the acquisition and operation of this equipment is estimated at $60,000/yr. Salvage value of the apparatus after
5 yr is expected to be $20,000. A full-time driver during
operating hours will accrue an added cost of $10,000/yr.
What would the rate of return be on this investment?
@ 40% present worth
C P = F(P/F , 40%, 5) = A(P/A, 40%, 5)
C >$100,000 = $20,000(0.18593)
= ($60,000 > 10,000)(2.0352)
C $5,478.60
@ 50% present worth
C P = F(P/F, 50%, 5) = A(P/A, 50%, 5)
C >$100,000 = $20,000(0.13169)
= ($60,000 > $10,000)(1.7366)
C >$10,536.40
By linear interpolation, the rate of return is 43 percent.
Benefit-Cost Analysis
Benefit-cost analysis, also referred to as cost-benefit
analysis, is a method of comparison in which the consequences of an investment are evaluated in monetary
terms and divided into the separate categories of benefits
and costs. The amounts are then converted to annual
equivalents or present worths for comparison.
The important steps of a benefit-cost analysis are
1. Identification of relevant benefits and costs
2. Measurement of these benefits and costs
3. Selection of best alternative
4. Treatment of uncertainty
Identification of Relevant Benefits and Costs
The identification of benefits and costs depends on
the particular project under consideration. Thus, in the
case of fire prevention or control activities, the benefits
are based on fire losses prior to such activities. Fire losses
may be classified as direct or indirect. Direct economic
losses are property and contents losses. Indirect losses include such things as the costs of injuries and deaths, costs
incurred by business or industry due to business interruption, losses to the community from interruption of services, loss of payroll or taxes, loss of market share, and
loss of reputation. The direct costs of fire protection activities include the costs of constructing fire-resistive buildings, installation costs of fire protection systems, and the
5–99
costs of operating fire departments. Indirect costs are
more difficult to measure. They include items such as the
constraints on choice due to fire protection requirements
by state and local agencies.
A major factor in the identification of relevant benefits and costs pertains to the decision unit involved. Thus,
if the decision maker is a property owner, the relevant
benefits from fire protection are likely to be the reduction
in fire insurance premiums and fire damage or business
interruption losses not covered by insurance. In the case
of a municipality, relevant benefits are the protection of
members of the community, avoidance of tax and payroll losses, and costs associated with assisting fire victims.
Potential benefits, in these instances, are considerably
greater than those faced by a property owner. However,
the community may ignore some external effects of fire
incidents. For example, the 1954 automobile transmission
plant fire in Livonia, Michigan, affected the automobile
industry in Detroit and various automobile dealers
throughout the United States. However, there was little
incentive for the community to consider such potential
losses in their evaluation of fire strategies, since they
would pertain to persons outside the community. It might
be concluded, therefore, that the more comprehensive the
decision unit, the more likely the inclusion of all relevant
costs and benefits, in particular, social costs and benefits.
Measurement of Benefits and Costs
Direct losses are measured or estimated statistically
or by a priori judgment. Actuarial fire-loss data collected
nationally or for a particular industry may be used, providing it is adequately specific and the collection mechanism is reliable. More often, an experienced judgment of
potential losses is made, sometimes referred to as the maximum probable loss (MPL).
Indirect losses, if considered, are much more difficult
to appraise. A percentage or multiple of direct losses is
sometimes used. However, when indirect loss is an important decision parameter, a great deal of research into
monetary evaluation may be necessary. Procedures for
valuing a human life and other indirect losses are discussed in Ramachandran.3
In the measurement of benefits, it is appropriate to
adjust for utility or disutility which may be associated
with a fire loss.
Costs may be divided into two major categories:
(1) costs of private fire protection services, and (2) costs of
public fire protection services. In either case, cost estimates will reflect the opportunity cost of providing the
service. For example, the cost of building a fire-resistive
structure is the production foregone due to the diversion
of labor and resources to make such a structure. Similarly,
the cost of a fire department is the loss of other community services which might have been provided with the
resources allocated to the fire department.
Selection of Best Alternative
There are two considerations in determining benefitcost criteria. The first pertains to project acceptability,
while the second pertains to project selection.
Project acceptability may be based on benefit-cost difference or benefit-cost ratio. Benefit-cost ratio is a measure
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of project worth in which the monetary equivalent benefits are divided by the monetary equivalent costs. The first
criterion requires that the value of benefits less costs be
greater than zero, while the second criterion requires that
the benefit-cost ratio be greater than one.
The issue is more complicated in the case of project
selection, since several alternatives are involved. It is no
longer a question of determining the acceptability of a
single project, but rather selecting from among alternative
projects. Consideration should be given to changes in
costs and benefits as various strategies are considered.
Project selection decisions are illustrated in Figure 5-7.3.
The degree of fire protection is given on the horizontal
axis, while the marginal costs and benefits associated
with various levels of fire safety are given on the vertical
axis. As the diagram indicates, marginal costs are low initially and then increase. Less information is available concerning the marginal benefit curve, and it may, in fact, be
horizontal. The economically optimum level of fire protection is given by the intersection of the marginal cost
and marginal benefit curves. Beyond this point, benefits
from increasing fire protection are exceeded by the costs
of providing the additional safety.
A numerical example is given in Table 5-7.3. There are
five possible strategies or programs possible. The first
strategy, A, represents the initial situation, while the remaining four strategies represent various fire loss reduction activities, each with various costs. Strategies are
arranged in ascending order of costs. Fire losses under
each of the five strategies are given in the second row,
while the sum of fire losses and fire reduction costs for
each strategy is given in the third row. The sum of fire
losses and fire reduction costs of each strategy is equivalent to the life-cycle cost of that strategy. Life-cycle cost
analysis is an alternative to benefit-cost analysis when the
outcomes of the investment decision are cost savings
rather than benefits per se. Additional information on
life-cycle cost analysis is found in Fuller and Petersen.5
Marginal costs
Table 5-7.3
Use of Benefit-Cost Analyses in Strategy
Selection
Strategy
Category
A
B
C
D
E
Fire reduction costs
Fire losses
Sum of fire reduction
costs and fire losses
0
100
10
70
25
50
45
40
70
35
100
80
75
85
105
0
0
30
10
20
15
10
20
5
25
0
—
20
3.0
5
1.33
–10
0.5
–20
0.2
Marginal benefits
Marginal costs
Marginal benefits minus
marginal costs
Marginal benefit-cost ratio
Data in the first two rows may then be used to determine the marginal costs or marginal benefits from the replacement of one strategy by another. Thus, strategy B
has a fire loss of $70 compared to $100 for strategy A, so
the marginal benefit is $30. Similarly, the marginal benefit
from strategy C is the reduction in fire losses from B to C
or $20. The associated marginal cost of strategy C is $15.
Declining marginal benefits and rising marginal costs result in the selection of strategy C as the optimum strategy.
At this point, the difference between marginal benefits
and marginal costs is still positive.
Marginal benefit-cost ratios are given in the last row.
It is worth noting that, while the highest marginal benefitcost ratio is reached at activity level B (as is the highest
marginal benefit-cost difference), project C is still optimum, since it yields an additional net benefit of $5. This
finding is reinforced by examining changes in the sum of
fire losses and fire reduction costs (i.e., life-cycle costs).
Total cost plus loss first declines, reaching a minimum at
point C, and then increases. This pattern is not surprising,
since as long as marginal benefits exceed marginal costs,
total losses should decrease. Thus, the two criteria—
equating marginal costs and benefits, and minimizing the
sum of fire losses and fire reduction costs—yield identical
outcomes.
Treatment of Uncertainty
$
Marginal benefits
A final issue concerns the treatment of uncertainty.
One method for explicitly introducing risk considerations
is to treat benefits and costs as random variables which
may be described by probability distributions. For example, an estimate of fire losses might consider the following
events: no fire, minor fire, intermediate fire, and major fire.
Each event has a probability of occurrence and an associated damage loss. The total expected loss (EL) is given by
EL C
0
100%
Degree of fire safety
Figure 5-7.3.
Project selection.
3
}
piDi
iC0
where
p0 C probability of no fire
p1 C probability of a minor fire
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Engineering Economics
p2 C probability of an intermediate fire
p3 C probability of a major fire
Dn C associated damage loss, n C 0,1,2,3
Expected losses may be computed for different fire
protection strategies. Thus, a fire protection strategy that
costs C3 and reduces damage losses of a major fire from D3
to D3 will result in an expected loss
EL C p0D0 = p1D1 = p2D2 = p3D3 = C3
Similarly, a fire control strategy that costs C2 and reduces the probability of an intermediate fire from p2 to p 2
has an expected loss
EL C p0D0 = p1D1 = p 2D2 = p3D3 = C2
A comparison of expected losses from alternative
strategies may then be used to determine the optimal
strategy.
Use of expected value has a limitation in that only the
average value of the probability distribution is considered. Discussion of other procedures for evaluating uncertain outcomes is given by Anderson and Settle.6
References Cited
1. ASTM E833, Definitions of Terms Relating to Building Economics, American Society for Testing and Materials, West Conshohocken, PA (1999).
2. ASTM Standards on Building Economics, 4th ed., American Society for Testing and Materials, West Conshohocken, PA (1999).
3. G. Ramachandran, The Economics of Fire Protection, E & FN
Spon, London (1998).
4. American National Standards Institute Standard Z94.0-1982,
“Industrial Engineering Terminology,” Chapter 5, Engineering Economy, Industrial Engineering and Management Press,
Atlanta, GA (1983).
5. S.K. Fuller and S.R. Petersen, “Life-Cycle Costing Manual
for the Federal Energy Management Program,” NIST Handbook 135, National Institute of Standards and Technology,
Gaithersburg, MD (1996).
6. L.G. Anderson and R.E. Settle, Benefit-Cost Analysis: A Practical Guide, Lexington Books, Lexington, MA (1977).
Additional Readings
R.E. Chapman, “A Cost-Conscious Guide to Fire Safety in Health
Care Facilities,” NBSIR 82-2600, National Bureau of Standards, Washington, DC (1982).
R.E. Chapman and W.G. Hall, “Code Compliance at Lower
Costs: A Mathematical Programming Approach,” Fire Technology, 18, 1, pp. 77–89 (1982).
L.P. Clark, “A Life-Cycle Cost Analysis Methodology for Fire Protection Systems in New Health Care Facilities,” NBSIR 822558, National Bureau of Standards, Washington, DC (1982).
W.J. Fabrycky, G.J. Thuesen, and D. Verma, Economic Decision
Analysis, 3rd ed., Prentice Hall International, London (1998).
E.L. Grant, W.G. Ireson, and R.S. Leavenworth, Principles of Engineering Economy, 8th ed., John Wiley and Sons, New York
(1990).
J.S. McConnaughey, “An Economic Analysis of Building Code
Impacts: A Suggested Approach,” NBSIR 78-1528, National
Bureau of Standards, Washington, DC (1978).
D.G. Newnan and J.P. Lavelle, Engineering Economic Analysis, 7th
ed., Engineering Press, Austin, TX (1998).
C.S. Park, Contemporary Engineering Economics, 2nd ed., AddisonWesley, Menlo Park, CA (1997).
J.L. Riggs, D.D. Bedworth, and S.U. Randhawa, Engineering Economics, 4th ed., McGraw-Hill, New York (1996).
R.T. Ruegg and H.E. Marshall, Building Economics: Theory and
Practice, Van Nostrand Reinhold, New York (1990).
R.T. Ruegg and S.K. Fuller, “A Benefit-Cost Model of Residential
Fire Sprinkler Systems,” NBS Technical Note 1203, National
Bureau of Standards, Washington, DC (1984).
W.G. Sullivan, J.A. Bontadelli, and E.M. Wicks, Engineering Economy, 11th ed., Prentice Hall, Upper Saddle River, NJ (2000).
Appendix A: Symbols and Definitions of Economic Parameters
Symbol
Definition of Parameter
Aj
A
–
A
Cash flow at end of period j
End-of-period cash flows (or equivalent end-of-period values) in a uniform series continuing for a specified number of periods
Amount of money (or equivalent value) flowing continuously and uniformly during each period, continuing for a specified number
of periods
Future sum of money—the letter F implies future (or equivalent future value)
Uniform period-by-period increase or decrease in cash flows (or equivalent values); the arithmetic gradient
Number of compounding periods per interest perioda
Number of compounding periods
Present sum of money—the letter “P” implies present (or equivalent present value). Sometimes used to indicate initial capital
investment.
Salvage (residual) value of capital investment
Rate of price level increase or decrease per period; an “inflation” of “escalation” rate
Uniform rate of cash flow increase or decrease from period to period; the geometric gradient
Effective interest rate per interest perioda (discount rate), expressed as a percent or decimal fraction
Nominal interest rate per interest period,a expressed as a percent or decimal fraction
F
G
M
N
P
S
f
g
i
r
aNormally, but not always, the interest period is taken as 1 yr.
Subperiods, then, would be quarters, months, weeks, and so forth.
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Fire Risk Analysis
Appendix B: Functional Forms of Compound Interest Factorsa
Name of Factor
Algebraic
Formulation
Functional
Format
Group A. All cash flows discrete: end-of-period compounding
Compound amount (single payment)
(1 = i)N
(F/P, i, N)
Present worth (single payment)
(1 = i)–N
(P/F, i, N)
Sinking fund
i
(1 = i )N – 1
(A/F, i, N)
Capital recovery
i(1 = i)N
(1 = i )N – 1
(A/P, i, N)
Compound amount (uniform series)
(1 = i )N – 1
i
(F/A, i, N)
Present worth (uniform series)
(1 = i )N – 1
i(1 = i )N
(P/A, i, N)
Arithmetic gradient to uniform series
(1 = i )N – iN – 1
i (1 = i )N – i
(A/G, i, N)
Arithmetic gradient to present worth
(1 = i )N – iN – 1
i 2(1 = i )N
(P/G, i, N)
Continuous compounding compound amount
(single payment)
erN
(F/P, r, N)
Continuous compounding present worth (single payment)
e–rN
(P/F, r, N)
Continuous compounding present worth (single payment)
erN – 1
erN(er – 1)
(P/A, r, N)
Continuous compounding sinking fund
er – 1
erN – 1
(A/F, r, N)
Continuous compounding capital recovery
erN(er – 1)
erN – 1
(A/P, r, N)
Continuous compounding compound amount
(uniform series)
erN – 1
er – 1
(F/A, r, N)
Continuous compounding sinking fund (continuous,
uniform payments)
r
erN – 1
–
A /F, r, N
Continuous compounding capital recovery (continuous,
uniform payments)
rerN
erN – 1
–
A /P, r,N
Continuous compounding compound amount (continuous,
uniform payments)
erN – 1
r
–
F/A , r, N)
Continuous compounding present worth (continuous,
uniform payments)
erN – 1
rerN
–
P/A , r, N)
Group B. All cash flows discrete: continuous compounding
at nominal rate r per period
Group C. Continuous, uniform cash flows: continuous
compounding
aSee Appendix A for definitions of symbols used in this table.
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Engineering Economics
Appendix C—Interest Tables
Table C-7.1
Present Worth Factor (Changes F to P)
Yr.
2%
4%
6%
8%
10%
12%
15%
20%
25%
30%
40%
50%
1
2
3
4
5
.9804
.9612
.9423
.9238
.9057
.9615
.9246
.8890
.8548
.8219
.9434
.8900
.8396
.7921
.7473
.9259
.8573
.7938
.7350
.6806
.9091
.8264
.7513
.6830
.6209
.8929
.7972
.7118
.6355
.5674
.8696
.7561
.6575
.5718
.4972
.8333
.6944
.5787
.4823
.4019
.8000
.6400
.5120
.4096
.3277
.7692
.5917
.4552
.3501
.2693
.7143
.5102
.3644
.2603
.1859
.6667
.4444
.2963
.1975
.1317
6
7
8
9
10
.8880
.8706
.8535
.8368
.8203
.7903
.7599
.7307
.7026
.6756
.7050
.6651
.6274
.5919
.5584
.6302
.5835
.5403
.5002
.4632
.5645
.5132
.4665
.4241
.3855
.5066
.4523
.4039
.3606
.3220
.4323
.3759
.3269
.2843
.2472
.3349
.2791
.2326
.1938
.1615
.2621
.2097
.1678
.1342
.1074
.2072
.1594
.1226
.0943
.0725
.1328
.0949
.0678
.0484
.0346
.0878
.0585
.0390
.0260
.0173
11
12
13
14
15
.8043
.7885
.7730
.7579
.7430
.6496
.6246
.6006
.5775
.5553
.5268
.4970
.4688
.4423
.4173
.4289
.3971
.3677
.3405
.3152
.3505
.3186
.2897
.2633
.2394
.2875
.2567
.2292
.2046
.1827
.2149
.1869
.1625
.1413
.1229
.1346
.1122
.0935
.0779
.0649
.0859
.0687
.0550
.0440
.0352
.0558
.0429
.0330
.0254
.0195
.0247
.0176
.0126
.0090
.0064
.0116
.0077
.0051
.0034
.0023
16
17
18
19
20
.7284
.7142
.7002
.6864
.6730
.5339
.5134
.4936
.4746
.4564
.3936
.3714
.3503
.3305
.3118
.2919
.2703
.2502
.2317
.2145
.2176
.1978
.1799
.1635
.1486
.1631
.1456
.1300
.1161
.1037
.1069
.0929
.0808
.0703
.0611
.0541
.0451
.0376
.0313
.0261
.0281
.0225
.0180
.0144
.0115
.0150
.0116
.0089
.0068
.0053
.0046
.0033
.0023
.0017
.0012
.0015
.0010
.0007
.0005
.0003
21
22
23
24
25
.6698
.6468
.6342
.6217
.6095
.4388
.4220
.4057
.3901
.3751
.2942
.2775
.2618
.2470
.2330
.1987
.1839
.1703
.1577
.1460
.1351
.1228
.1117
.1015
.0923
.0926
.0826
.0738
.0659
.0588
.0531
.0462
.0402
.0349
.0304
.0217
.0181
.0151
.0126
.0105
.0092
.0074
.0059
.0047
.0038
.0040
.0031
.0024
.0018
.0014
.0009
.0006
.0004
.0003
.0002
30
35
40
45
50
.5521
.5000
.4529
.4102
.3715
.3083
.2534
.2083
.1712
.1407
.1741
.1301
.0972
.0727
.0543
.0994
.0676
.0460
.0313
.0213
.0573
.0356
.0221
.0137
.0085
.0334
.0189
.0107
.0061
.0035
.0151
.0075
.0037
.0019
.0009
.0042
.0017
.0007
.0003
.0001
.0012
.0004
.0001
60
70
80
90
100
.3048
.2500
.2051
.1683
.1380
.0951
.0642
.0434
.0293
.0198
.0303
.0169
.0095
.0053
.0030
.0099
.0046
.0021
.0010
.0005
.0033
.0013
.0005
.0002
.0011
.0004
.0002
1
(1 = i )y
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Fire Risk Analysis
Table C-7.2
Capital Recovery Factor (Changes P to A)
Yr.
2%
4%
6%
8%
10%
12%
15%
20%
25%
30%
40%
50%
1
2
3
4
5
1.020
.5150
.3468
.2626
.2122
1.040
.5302
.3603
.2755
.2246
1.060
.5454
.3741
.2886
.2374
1.080
.5608
.3880
.3019
.2505
1.100
.5762
.4021
.3155
.2638
1.120
.5917
.4163
.3292
.2774
1.150
.6151
.4380
.3503
.2983
1.200
.6545
.4747
.3863
.3344
1.250
.6944
.5123
.4234
.3719
1.300
.7348
.5506
.4616
.4106
1.400
.8167
.6294
.5408
.4914
1.500
.9000
.7105
.6231
.5758
6
7
8
9
10
.1785
.1545
.1365
.1225
.1113
.1908
.1666
.1485
.1345
.1233
.2034
.1791
.1610
.1470
.1359
.2163
.1921
.1740
.1601
.1490
.2296
.2054
.1874
.1736
.1627
.2432
.2191
.2013
.1877
.1770
.2642
.2404
.2229
.2096
.1993
.3007
.2774
.2606
.2481
.2385
.3388
.3163
.3004
.2888
.2801
.3784
.3569
.3419
.3312
.3235
.4613
.4419
.4291
.4203
.4143
.5481
.5311
.5203
.5134
.5088
11
12
13
14
15
.1022
.0946
.0881
.0826
.0778
.1141
.1066
.1001
.0947
.0899
.1268
.1193
.1130
.1076
.1030
.1401
.1327
.1265
.1213
.1168
.1540
.1468
.1408
.1357
.1315
.1684
.1614
.1557
.1509
.1469
.1911
.1845
.1791
.1747
.1710
.2311
.2253
.2206
.2169
.2139
.2735
.2685
.2645
.2615
.2591
.3177
.3135
.3102
.3078
.3060
.4101
.4072
.4051
.4036
.4026
.5059
.5039
.5026
.5017
.5011
16
17
18
19
20
.0737
.0700
.0667
.0638
.0611
.0858
.0822
.0790
.0761
.0736
.0990
.0954
.0924
.0896
.0872
.1130
.1096
.1067
.1041
.1019
.1278
.1247
.1219
.1195
.1175
.1434
.1405
.1379
.1358
.1339
.1679
.1654
.1632
.1613
.1598
.2114
.2094
.2078
.2065
.2054
.2572
.2558
.2546
.2537
.2529
.3046
.3035
.3027
.3021
.3016
.4019
.4013
.4009
.4007
.4005
.5008
.5005
.5003
.5002
.5002
21
22
23
24
25
.0588
.0566
.0547
.0529
.0512
.0713
.0692
.0673
.0656
.0640
.0850
.0830
.0813
.0797
.0782
.0998
.0980
.0964
.0950
.0937
.1156
.1140
.1126
.1113
.1102
.1322
.1308
.1296
.1285
.1275
.1584
.1573
.1563
.1554
.1547
.2044
.2037
.2031
.2025
.2021
.2523
.2519
.2515
.2512
.2510
.3012
.3009
.3007
.3006
.3004
.4003
.4002
.4002
.4001
.4001
.5000
30
35
40
45
50
.0446
.0400
.0366
.0339
.0318
.0578
.0536
.0505
.0483
.0466
.0726
.0690
.0664
.0647
.0634
.0888
.0858
.0839
.0826
.0817
.1061
.1037
.1023
.1014
.1009
.1241
.1223
.1213
.1207
.1204
.1523
.1511
.1506
.1503
.1501
.2008
.2003
.2001
.2001
.2000
.2503
.2501
.2500
.3001
.4000
60
70
80
90
100
.0288
.0267
.0252
.0241
.0232
.0442
.0428
.0418
.0412
.0408
.0619
.0610
.0606
.0603
.0602
.0808
.0804
.0802
.0801
.0800
.1003
.1001
.1000
.1200
.1500
i (1 = i )y
(1 = i )y – 1

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