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Unformatted text preview: ENGINEERING ECONOMY
REVIEW FOR FUNDAMENTALS OF ENGINEERING EXAMINATION
Prepared by: Charles E. Dare, P.E., Professor Emeritus Department of Civil, Architectural, and Environmental Engineering, UMRolla A, Introduction and Deﬁnitions 1. Engineering Economy Study: An evaluation and comparison of alternatives in
which differences among those alternatives are expressed as monetary values.
Usually involves a decisionmaking process with these steps: a‘ Recognition and formulation of the problem b. Search for feasible alternatives 0. Quantification and analysis of alternatives
d. Selection of the best alternative (least cost) Time Value of Money: When making comparisons among alternatives, the
cash flows (transactions) are important in terms of their amounts and location in time. Cash flows that occur closer to the present are generally worth more to
you than similar dollar cash flows occurring many years in the future, Equivalence: Cash flow amounts that differ in terms of magnitude may have the
same value to you depending on their timing and the interest rate involved. Interest: A return, or gain, that may be expected on funds that are productively
invested; or cost to borrow money from another source for your own use, Compound Interest: The interest calculation procedure which requires interest to
be earned (charged) against the interest already earned as well as the principal.
(As opposed to Simple Interest, which does not compute interest on interest) B. Basic Financial Mathematics 1. Definition of Symbols (P. F, A, G, C, i, and n): P = a single “Present” amount occurring at the start of an interest period F = a single “Future" amount occurring at the end of an interest period A = a series of equal (uniform) amounts occurring at the end of consecutive
interest periods G = a uniformly increasing (decreasing) series of endofperiod amounts, where successive amounts change by a constant increment from one interest
period to the next. C = a geometrically increasing (decreasing) series of endofperiod amounts, where successive amounts change by constant percentage muitiplier from
one interest period to the next. i : compound interest rate, per interest period
n = number of interest periods Cash Flow Diagram: A “time line” sketch showing when cash flows occur, their amount, and whether or not the cash is flowing in or out of a project. investment,
or alternative. 3. Standard formulas: Relationships among the previously defined cash flow
symbols have been derived for standard configurations, so equations and
interest tables are readily available. When using the equations and tables, the
cash flows must conform to the standard pattern of cash flows for which the
equations were developed. If a situation deviates from a standard pattern, then
extra steps are usually necessary in order to solve a problem. Several standard
configurations of cash flows for P, F, A, G, and C follow: F P: F(’£7€«t,7’t)
o / 2 3..__nr 77 {Cr/apr/P/C”)
\L l l l l
p
F FrAV/A/LW)
lO { 2 i ”it; 7? A =F(A/F,i,?’t)
MA A A
P .
P=A(F/Alv(x7’i)
0 x 2 3 77472 ,
: P A
Mi ii A (”3"“)
A A A A A
P
m 7t P= 6 (ID/6,472)
or 2 3 
i it i l 6: Mama”)
re
26 022% 07—06
F:
To I 2 3 .. ”4 n W/ICQ f" : I
F i i ‘13: C71.
C (HP)
C({f‘f‘} C(/+P)h_l 4. Examples of Standard Form Applications: a. What amount must be invested today in order to have $40,000 available 10
years from now, if interest is 10 percent? 1::
40K P:F(P/F//O%,/O) 0 ““0 :4OK (0.3855)
P = 75 :. V5, 420
b. If $5,000 is placed in an account that earns 6.7 33 interest, how much will that
account be worth in 28 years? 3?
F: 2? F = [30 + r)
.28
o h=28 [email protected]¢75)
= 3/, Be. 5_7
P: 5 K c. How much will be accumulated in a tund earning 8 percent interest it $1,000 is
deposited in that fund at the end of the year for the next 12 years? F=? F:A(F/A,8’/3,/2)
nzrz = 4000 (£83970
,5. A A A
A =i,ooo d. An individual needs to have $5,000 for a down payment to be placed on a
tract of land 15 months from now. If she has access to a tund that earns
1.00 percent interest per month. what uniform amount must be placed in that fund starting one month from now? F5=K A : F(A/F, mus)
lo I: ’2'" R=15 5 5K(0.oo2/Z)
A A A A = /8,997. P : 310.69
At? e. An account which earns 8 percent interest currently has a balance of $10,000.
How much can be withdrawn from this account at the end of each year for the
next 7 years and have the account depleted at the time of the iast withdrawal? A: PM/R 8%, 7) P: 0 t Z “I'_‘_‘ 1a.??? I /o/< (O'IQZO
A A A A: 3. _ Do
IOK _ 4592/. ~ f. What amount would you need to deposit at 8 percent interest on January 1,
2003, in order to be able to draw out $2,000 at the end of each year for the
next 8 years, leaving the fund depleted at the end of that time? P: A 0%, 8%, 8) o i 2""ﬂn=8
‘ L : Limo : 2:000 (5 74m) p2"? : M433. 2.3 g. Repair costs on a new pump are estimated to be $ 0 at the end of the first
year; $25 at the end of the second year; $50 at the end of the third year, $?5
at the end of the fourth year and continuing in this manner until the end ofthe
tenth year. If interest is 8 percent, find the present worth equivalent of these maintenance costs.
P: a (P/e, a %, t0) 6=25 =25(2s.<27¢e)
O ' 2 3 ““0 3643.43 25 5O ‘i‘x25
P‘? =225 h. A geometric gradient with r = 10 percent per year has an EOY 1 value
equal to 100. What is the present worth equivalent of this gradient if
interest is 10 percent and n = 5 years? my“ 100 NO P: (rt/(HF) = xoo (9/010)
: 45+ :5 C. Problems having the number of compounding periods “n" as the unknown. 1. Direct Solution with exponents or logarithms.
P(t+i)"n:F n=og(FfP)iiog(‘l+i) EXAMPLE: How many years must $2,000 be invested at 8 percent in order to
reach a value of $4,000? n = log (4000:2000) I log (1 +08) = 0.30103 r' 0.033424 = 9.006 = 9 years Search (interpolate) fortabled interest factor. When using the more complex
interest factors, the best method is to calculate the value of the interest factor and attempt to locate it in the appropriate interest table, thus determining the
number of interest periods. EXAMPLE: A deposit of $1 ,000 is now placed in an account earning 8 percent interest. How many successive endof—year withdrawals equal to $98 each may
be transacted before the account is depleted? Use: 1000 (NP, 8%, n = ?) 2 98 (NP, 8%, n = ?) = 0098 Look in 8.00% Interest Factor Table under (NP) factor for 0.098: n = 22 years D. Problems having the interest rate “i’ as the unknown. 1. Direct solution with exponents: P (t + i) "‘ n = F i = [(Fi’P) "‘ (1in)] — 1 EXAMPLE: A deposit of $1,000 increases in value to a sum of $2,500 in just 5
years. What compound interest rate was earned? i=[(2,500r1,000)“(1i5)]u1 =1.2011—1 = 0.2011 = 20.11% Search (interpolate) for tabled interest factor. May involve a search between interest tables. The interest factor must first be calculated; then attempt to locate
it in the tables using the pertinent “n" and correct column. EXAMPLE: A series of 10 annual endof—year deposits equal to $500 each were placed in an account. The account was worth $7,600 just after the last deposit
was made. What interest rate was earned in this situation? Use: 500 (FIA, i: ?, 10) = 7,600 Then find: (FIA, i =?, 10) = 15.2
i (FiA, i, 10)
8.00 % 14.4866 Linear Interpolation gives: i r 8.98%
'P 15.2 10.00 % 15.9374 E. Interest Compounding More Frequent than Annual (NonAnnual Compounding): if interest is compounded at a consistent interval, shorter than a year, the following
terms must be defined: The compounding interval and the number of intervals in one year, m. The “Nominal Interest Fiate" per year, r. {sometimes shown as in ). The “Effective Interest Rate" per year, i eff . ierr = {l1 + (rtmil’rni 1
The interest rate which applies to the interest interval: i : ri’m EXAMPLE: A deposit of $500 is placed in an account yielding 18 % nominal
interest compounded monthly. What is this account worth at the end of 6
months; and what was the effective interest rate which was earned? i = ri’m = 0.18i'12 = 0.015 F = 500 (1+0,015)r\s = 500(10934) : $546.72 ieff ={{1+{0.1si12)}A12}1 = (1.015)A12 1 = 0.1956 = 19.56% A special type of very frequent compounding is when the compounding interval
becomes so short that there are an infinite number of them within a year. Thus
“m” becomes infinitely large. This is referred to as "Continuous Compounding"
and we have a different set of equations using “e" the base of natural iogarithms
to be applied. The Continuous Compounding rate is given the symbol ”r“. EXAMPLE: Five endof year deposits each equal to $800 are placed in an account which yields 10 % compounded continuously. What is this account worth at the
end of 5 years, immediately after the last deposit? F = A (FM, r%, 5) = 800 [eA(rn) 1]/[e/\(r) 1]
= 800[e’0.10x5 1]/[e/~0.10 1] = 800 [1.64872— 1] r [ 1.10517— 1] H 800 (6.16825) = $ 4,934.60 F. Present Worth Evaluations and Comparisons 1. To evaluate any configuration of future costs in terms of their present worth equivalent, each component of those costs needs to be “discounted” to the point
in time chosen to be the present for the problem. EXAMPLE: Find the present worth equivalent of a series of $100 payments which starts at the end~ofyear 21 and terminates at the end of year 35. The
interest rate is 8 % compounded annually. P 2 100 (PIA, 8%, 15)(F’i’F, 8%, 20) = 800(8.5595)(O.2145) 2 1,468.81 2. To compare alternatives, whatever the configuration of costs andior revenues,
there must be consistency in the number of years of service offered by those
alternatives. Often it is necessary to assume an identical replacement is
available in order to make an alternative having a short service life be
compatible with an alternative having a longer service life. EXAMPLE: Identify the most economical pump to buy of the two candidates
listed below. Assume identical replacements will be available in the future if
needed. The interest rate is 10 percent. Tiger Pump Miner Pump
initial Cost, $ 7,000 10,000
Annual Op. Cost, $ 2.000 1,700
Salvage, S 1,500 3,000
Service Life, yrs. 4 8 PW (Two Tigers) if 7000 + (7000 1500)(PfF,10%,4) — 1500(PfF, 10%, 8)
+ 2000mm, 10%, 8) = 20,727
PW (One Miner) 10000  3000(PrF, 10%, a) + 1700(P/A, 10%, 8)
17,570 ll  Seiect a Miner Pump since it has the lowest Present Worth of Costs for 8 years! G. Capitalized Cost Capitalized Cost evaluation or comparison is a special case of present worth
analysis where the service life or project life is assumed to be infinitely long. The
time value of money calculations are base on the use of the following: CC:P=A(PJ’A,i,n= co)=A(1/i) if any configuration of costs is presented, it is necessary to discount all the single
value costs, the series costs, and intermittent costs to the present time in order to find the Capitalized Cost. lf alternatives are being compared, the assumption of perpetual service assures the same number of years of service will be obtained from
the alternatives being considered. Capitalized Costs: EXAMPLE: Find the Capitalized Cost for a project having an initial cost of
$500,000 and annual maintenance expenses of $10,000. Interest is 8%. CC = 500,000 + 10,000!(0.08) = $525,000 EXAMPLE: What is the Capitalized Cost associated with a speciﬁc major repair expense that will be $100,000 every 5 years, with the first expense occurring at
the endof—year 5. Interest is 8%. cc = 100,000rArF, 8%, 5)f(0.08) = $213,125 H. Equivalent Uniform Annual Cost (EUAC) Evaluations and Comparisons 1 . To evaluate any conﬁguration of amounts in terms of their uniform annual
equivalent, each amount must converted to an annual endof—year value, with all
of the resulting endofyear values being equal. This is like ﬁnding an “annual
average" cost, but it is done taking into account the time value of money. One
common situation is where an item is purchased for an initial amount (P value), but is ﬁnanced with payments (A value) that are uniform and at a regular interval. EXAMPLE: A vehicle costs $14,000 when new and has a tradein value of
$6,000 at the end of4 years. Its operating expense is $3,500 per year, with an
additional $1 ,000 for tires, tune up, and battery at the end of the 2nd year. What
is the EUAC for owning this vehicle 4 years, assuming interest at 10%. EUAC = 140000013, 10%,4) — 6000(A1F,10%,4) + 3,500
+1.000(Pi’F,10%,2)(A1P,10%,4) = $6,885 EXAMPLE: A prOperty is purchased for $100,000 now, with this amount being
ﬁnanced for 15 years at 8% interest. What is the annual endofyear payment? Annual Payment = 100,000 (NP, 8%, 15) = $11,680 When comparing alternatives using the EUAC method, for each alternative all of the pertinent costs and revenues must be converted to equal endofyear values. If the alternatives have different service lives, it is often assumed that identical
replacements will be available in the future thus the ﬁrst item in a series of
assumed identical replacements is all that need be evaluated. EXAMPLE: Compare the Tiger Pump and Miner Pump by EUAC EUAC (Tiger) = 7000(AlP, 10%, 4)— 1500(AfF, 10%, 4) +2000 = $3,885 EUAC (Miner) = 10000(A!P, 10%, 8) — 3000(NF, 10%, 8) + 1700 = $3,312 l. EUAC of Perpetual Service The Equivalent Uniform Annual Cost of Perpetual Service is a special case of annual
cost analysis where the service life or project life is assumed to be infinitely long. To
evaluate any conﬁguration of future costs in terms of their annual equivalent for an infinite time, it is often best to discount the set of costs to be a Capitalized Cost, then
use the relationship: A = CC x ( i) EXAMPLE: Find the EUAC of perpetual service for a railroad right of way which
has a present value of $1,800,000 if interest is 10 percent compounded annually. EUAC = 1,800,000 (0.10) = $ 180,000 per year EXAMPLE: Find the EUAC of perpetual service for a series of maintenance
expenses which is $100,000 every five years. Interest is 8 %. EUAC = {100,000 (NF, 8%, 5)i(0.08)] [0.08] = $17,050 EXAMPLE: Find the EUAC of perpetual service for a set of construction costs
which were: EOY 1 = 200,000; EOY 2 = 700,000; EOY 3 = 300,000. i = 8%. EUAC = [200K(Pi’F, 8%,1)+?00K(PIF,8%, 2)+300K(PIF, 8%,3)] [0.08] = $81,874 J. Future Worth Evaluations and Comparisons 1. To evaluate any configuration of costs in terms of their future worth equivalent, each component of those costs must be “compounded” to the future point in time
chosen for the problem. EXAMPLE: Find the future worth equivalent at the endofyear 25 of a series of $1,000 deposits made from endof~year 1 through 15. How much of that future
total is interest? The interest rate is 10%. FW = 1,000(FIA, 10%, 15)(F!P 10%, 10) = $ 82,408
Interest portion of this total: 82,408 H {15)(1,000) = $67,408 is interest To compare alternatives by Future Worth evaluation, there must be consistency
among the alternatives in the number of years of service offered by each of them. EXAMPLE: Compare the Tiger Pump and the Miner Pump by FW evaluation. FW (Two Tigers) = 7,000(FIP,10%,8) + (7,0001,500)(FIP, 10%,4) — 1,500
+ 2000(FI‘A, 10%, 8) = $44,430 FW (One Miner) 10,000(FIF’, 10%, 8) + (1,700)(F!A, 10%, 8) — 3,000 $ 37,877 Select a Miner Pump since it has the lowest Future Worth of Costs for 8 years. K. Benefit — Cost Analysis (Often noted as: BenetiUCost Ratio) The Benefit/Cost (BIC) Ratio identifies the benetits (often determined as savings
or cost reductions) to be realized per each dollar that is expended to achieve those
benefits. Several aSpects of setting up the BC Ratio are 1. All dollar values used in the BIG calculations must be expressed consistently,
such as all annual amounts, or all present worth amounts 2. Projects, or alternatives, must be tested incrementally, proceeding from the
one having the least cost to the one having the greatest cost to the public
agency (this assures the denominator will not be a negative value) 3. BIC Ratio less than 1.0 indicates the more costly project is unsatisfactory
BIC Ratio equal 1.0 indicates the more costly project is marginal
BIC Ratio greater than 1.0 indicates the more costly project is acceptabie EXAMPLE: The Annual Benefits and Annuat Costs have already been calculated for 4 alternatives shown below. Determine which is the best alternative using the
BIC Ratio procedures ALT. ANNUAL BENEFITS ANNUAL COSTS WITHIN ALT. B/C A1 182,000 91,500 1.99
A2 167,700 79,500 2.10
A3 115,000 78,500 1.46
A4 95,000 50,000 1.90 Although Alternative A2 has the highest within project BIC ratio (2.10), A2 must
be also be proven best in an incremental comparison involving all projects, as: BIC (A3 vs. A4) = (115.000  95,000) f (78,500 — 50,000): 0.70
A3 is not accepted, next incremental test is A2 vs. A4 BIC (A2 vs. A4) = (167,700 — 95,000) r (79,500 — 50,000) = 2.46
A2 is accepted, but must defend itself against A1 BIC (A1 vs. A2) = (182,000 — 167,700) i’( 91,500 ~ 79,500) = 1.19
A1 is accepted Thus, the Incremental BIC Ratio analysis indicates A1 is the best choice. L. Valuation and Depreciation From the perspective of economic analysis, depreciation is an accounting practice
which affects the amount of taxes a profitmaking organization pays. Depreciation
views the cost of an asset as a prepaid expense that is to be charged against profits
over a certain number of years. Allowed depreciation procedures are specified by
the IRS. The “Straight Line” method could be specified to simpiify computations, or
the Modiﬁed Accelerated Cost Recovery System may be Specified since it allows
larger depreciation deductions earlier in the life of the asset. The “Book Value” of an
asset is its original acquisition cost less its accumulated depreciation costs. (Note: The IRS does not allow LAND to be depreciated!) Straight Line Depreciation provides for the depreciation charge to be the same each
year for the lite of the asset and is based on the equation: D j 2 (C — S ) r (n) for the depreciation charge in year “j" EXAMPLE: Given C = $18,000 8 = $3,000 n = 3 years D i = (18,000w3,000)f(3) = $5,000 each year The depreciation charge and book value pattern : YEAR START OF YEAR DEPRECIATION END OF YEAR
BOOK VALUE CHARGE BOOK VALUE 1 18,000 5,000 13,000 2 13,000 5,000 8,000 3 8,000 5,000 8,000 Modified ACRS Depreciation is based on a set of IRS tabled percentage factors that
are applied to the initial cost of the asset. Salvage value is ignored when computing
annual ACFtS depreciation charges, however any subsequent gain on a sale of item
would be subject to a gains tax. The iRS sets up Recovery Period Classes for
assets, and those classes specify the duration of time for allowing depreciation, as wet] as the specific percentages to apply. For instance, “Special Toots" belong to a 3 Year Recovery Period, however the actual depreciation extends over 4 years, with
allowed annual percentages being: 33.3%; 44.5%; 14.8%; and 7.4% respectively. EXAMPLE: 3 year recovery period class, Initial Cost = $16,000 Salvage = $3,000 YEAR MACRS START OF YEAR DEPRECIATION END OF YEAR
% BOOK VALUE CHARGE BOOK VALUE 1 83.3 16.000 5,328 10,672 2 44.5 10,672 7,120 3,552 3 14.8 3,552 2,368 1,184 4 ?.4 1,184 1,184 0 M. BreakEven Analysis and Payback Period A large percent of the data used in an economy study are subject to uncertainty
since they are future estimates. A breakeven analysis may be performed to specify
a range of conditions for which one or the other alternative is preferred. If a break
even point can be determined, it may be possible to predict on which side of the
breakeven point operations are most likely to occur. A special type of breakeven
evaluation is the amount of time necessary to recapture the amount of an initial
investment. This is referred to as a payback period analysis. EXAMPLE (BreakEven on hours of operation) Two electric motors are being compared for possible purchase, with costs: Royal Elec.. Co. Cardinal Power, Inc.
initial Cost $1.600 $1 r250
Annual Maint. Cost $ 50 $ 60
Salvage...
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