Answer: (b) - Washington State University

Report
CE 466 FE Exam Review
Engineering Economics
Spring 2013
Economic Equivalence
Two cash flows are economically equivalent if, when the time value of
money is considered, the two cash flows are identical.
Analysis Methods
Present and Future Worth Analysis
Equivalent Uniform Cash Flow Analysis
Rate of Return Analysis
Benefit Cost Ratio Analysis
Related Topics
Capitalized Cost
Bonds
Break-Even Analysis vs Payback Period
Depreciation and Taxes
Inflation
Material Provided in Reference Manual
BE FAMILIAR WITH THIS!
 Formulas for
 Cash Flow Manipulation
 Effective Interest Rate
 Capitalized Cost
 Depreciation
 Inflation
 Descriptions of Terms
 Break-Even analysis
 Inflation
 Taxation
 Capitalized Cost
 Bonds
 Rate of Return
 Benefit-Cost Analyisis
 Interest Tables
Cash Flow Formulas – from Reference Handbook, p. 114
NOT PROVIDED: Pictorial Representation
CAN YOU DRAW THESE?
Pictoral Representation – From Newnan, et. Al, front cover
Interest Rates
r
The nominal annual interest rate
Possible wording 12%, compounded monthly
i
The interest rate per interest period, i = r/m
where m=# of compounding per year
Possible wording 1% monthly interest
ie The effective interest rate, ie = (1 + i)n -1
where n = # of compounding in the desired period
Possible effective interest rates for r = 12%, compounded monthly (i = 1%)
Effective quarterly interest rate: n = 3
Effective annual interest rate: n = 12
Example 1 – Effective Annual Interest
A credit card company offers a credit card with an
interest rate of 18%, compounded monthly.
What is the effective annual interest rate for the
card?
(a) 6.3%
(b) 18.0%
(c) 19.6%
(d) 21.6%
Solution to example 1
ie = (1 + r/m)m – 1
= (1 + 0.18/12)12 – 1 = 0.1956 = 19.56%
Answer: (c)
Example 2 – Effective quarterly Interest
A bank is paying 0.25% monthly interest on a
special savings account. What is the
effective quarterly interest?
(a)0.750%
(b)0.752%
(c) 1.000%
(d)1.010%
Solution to example 2
iquarterly = (1 + 0.0025)3 – 1 = 0.00752 = 0.752%
Answer: (b)
Present, Future Worth, and
Equivalent Uniform Cash Flow Analyses
Three Possible Goals
 Maximize BENEFITS or
 Minimize COSTS or
 Maximize the NET WORTH (BENEFITS – COSTS)
Present Worth or Future Worth –
 Move each cash flow to a single point in time.
Equivalent Uniform Cash Flow –
 Convert each to an equivalent uniform cash flow
 over the SAME TIME INTERVAL.
Example 3 – Present Worth Analysis
An individual can afford monthly car payments of
$450 for the next 5 years. Assuming interest on
car loans is 6% compounded monthly,
determine the greatest price of car he can
afford.
(a) $20,000
(b) $23,000
(c) $25,000
(d) $27,000
Solution to example 3
r = 6%  i = 6/12 = 0.5% monthly interest
n = 5x12 = 60 months
P = A(P/A, i, n) = $450(P/A, 0.5%, 60) =$450(51.7256) = $23,277
Alternate solution: use formula
P = [(1+i)n -1] / [(i)(1+i)n ] = $23,277
Answer: (b)
Example 4 – Equivalent Uniform Cash Flow Analysis
A college savings account paying 6% annual
interest is established for a 5 year old boy, with
the objective of having $60,000 on his 18th
birthday. If uniform deposits are made on each
of the boy’s birthdays, starting with his 7th
birthday and ending on his 18th birthday, how
much must each deposit be?
(a) $3558
(b) $3988
(c) $4185
(d) $5000
Solution to example 4
There are 12 uniform deposits.
A = F(A/F, 6%, 12)
= 60,000 (0.0593) = $3,558
Answer: (a)
Example 5 – Net Present Worth
A new milling machine will cost $150,000. The
net benefits from the purchase are expected to be
$40,000 the first year, increasing by $2,000 per
year for the 8 year life of the machine. Interest is
10%. The net present worth of the investment is
closest to:
(a)
(b)
(c)
(d)
$65,000
$75,000
$85,000
$95,000
Solution to example 5
NPW = -150,000 +
(40,000+2000(A/G,10%,8))(P/A,10%,8)
= -150,000+(40,000*2000(3.0045))(5.3349)
= $95,453
Answer: (d)
Example 6 – Equivalent Uniform Cash Flow Analysis
Two alternatives are being considered:
Alternative A
Alternative B
1st Cost
$25,000
$32,000
Net Annual Benefits
$9,500
$10,300
Salvage Value
$2,500
$3,200
10
10
Useful Life, years
At 6% interest, the difference between the two
Equivalent Uniform Cash Flows is most nearly:
(a) $100
(b) $720
(c) $840
(d) $1000
Solution to example 6
Alternative A:
EUA(B-C) = $9,500 + $2,500(A/F,6%,10) - $25,000(A/P,6%,10)
= $6292
Alternative B:
EUA(B-C) = $10,300 + $3,200(A/F,6%,10) - $32,000(A/P,6%,10)
= $6194
Difference: $6292-$6194 = $98
Answer: (a)
Example 7 – Future Worth Analysis
Each quarter for 40 years an individual deposits
$600 into an IRA that pays 8% interest,
compounded monthly. Determine the value of
the retirement account at the end of the 40
years.
(a) $554,000
(b) $621,000
(c) $693,000
(d) $852,000
Solution to example 7
r = 8%  i = 0.6667%
iquarterly = (1.006667)3 -1 = 0.02013 o4 2.013%
n = 40*4 = 160
Using the formula,
F = $600[((1.02013)160 -1)/(.02013)] = $693,280.81
Answer: (c)
Example 8 – Present Worth Analysis
A food processing company is considering investing in new
food packaging equipment that will cost $250,000. The
equipment will save the company $60,000 the first year,
decreasing by $5000 each year thereafter to $55,000 the
second year, $50,000 the third year, and so on. At the end
of its useful life of 8 years, the equipment will have a
salvage value of $12,500. The company’s MARR is 8%.
Determine the net present worth of the investment.
(a) $10,550
(b) $12,520
(c) $16,580
(d) $21,470
Solution to example 8
NPW = -250,000
+ 60,000(P/A, 8%,8)
– 5,000(P/G, 8%,8)
+ 12,500(P/F, 8%,8)
= -250,000+60,000(5.7466)-5000(17.8061)+12,500(.5403)
= $12,520
Answer: (b)
Rate of Return Analysis
 Find the internal rate of return (IRR) and compare to the minimum




acceptable rate of return (MARR)
The MARR is the minimum interest rate or return on investment one is
willing to accept.
The IRR is the interest rate at which the present worth of costs equal the
present worth of benefits
To be acceptable the IRR must be ≥ the MARR
Incremental analysis required to compare two alternatives
Example 9 – Rate of Return Analysis
A company is considering investing in a piece of
equipment that will cost $100,000 and have no
salvage value. The company estimates the
equipment will produce uniform benefits of $25,250
per year for the equipment’s useful life.
The internal rate of return for this equipment purchase
is most nearly:
(a) 4%
(b) 6%
(c) 8%
(d) 10%
Solution to example 9
PW of C = PW of B
$100,000 = $25,250(P/A, irr, 5)
3.96 = (P/A, irr, 5)
Look at 4%, 6%, 8%, and 10% to find the interest rate that
gives the closest value.
6%: 4.212
8%: 3.993
10%: 3.791
Answer: (c)
Example 10 – Rate of Return Analysis
A firm will purchase a new milling machine for
$25,000 that will save the company $4,000 each year
for 10 years. At that time the machine will be
salvaged for $2,500. The company’s rate of return on
this investment is most nearly:
(a) 6.5%
(b) 8.4%
(c) 10.5%
(d) 12.6%
Solution to example 10
This is a trial and error solution.
Suggestion: Work backward from the interest rates given to find the one that
gives a NPW closest to ZERO.
NPW = -25,000+4000[(1+i)10 -1]/[i(1+i)10 ] + 2,500(1+i)-10
Try i = 6.5%: NPW = $5087
i too small
Try i = 8.4%: NPW = $2479
i too small
Try i = 10.5%: NPW = -$19.79 CLOSE
Answer: (c)
-
Benefit/Cost Analysis
 Comparison of present worth of benefits to present worth of costs
(or EUAB to EUAC)
 The benefit/cost ratio must be ≥ 1 for an option to be considered
viable (benefits greater than costs)
 Salvage value is considered a REDUCTION IN COST
 INCREMENTAL ANALYSIS needed if comparing two or more
options
Example 11 – Benefit/Cost Analysis
A county is considering the following project
Initial Cost
$22,500,000
Maintenance
$525,000 per year
Savings
$5,300,000 per year
Given a useful life of 12 years and an interest
rate of 8%, the benefit to cost ratio is closest to:
(a) 0.67
(b) 1.01
(c) 1.51
(d) 1.67
Solution to example 11
PWcost = 22,500,000 + 525,000 (P/A, 8%, 12)
= 22,500,000 + 525,000 (7.5361)
= $26,456,453
PWbenefit = 5,300,000 (P/A, 8%, 12)
= 5,300,000 (7.5361)
= $39,941,330
B/C = PWbenefit/ PWcost
= 39,941,330/26,456,453
= 1.51
Answer: (c)
Example 12 – Benefit Cost Ratio Analysis
A project will require an initial equipment cost of
$9500 that will result in annual benefits of $2200
per year over the 15 year life of the equipment. At
the end of the 15 years the equipment will be
salvaged for $6000. The minimum acceptable rate
of return is 12%.
The Benefit/Cost Ratio for the project is nearest to:
(a) 0.56
(b) 1.24
(c) 1.78
(d) 2.21
Solution to Example 12
EUAB = $2200/year for 15 years
EUAC = $9500(A/P,12%,15)-$6000(A/F,12%,15)
= 9500(0.1468)-6000(0.0268) = $1233.8
B/C Ratio = (2200/1233.8) = 1.78
Answer: (c)
Infinite Analysis Period:
Capitalized Cost
 Capitalized Cost – The PRESENT cash
amount that would need to be set aside
now to cover a service indefinitely.
Formula: P = A/i
 Infinite Uniform Cash Flow
Formulas
 A= Pi
 A forever = A/cycle
Example 13 – Capitalized Cost Analysis
The cleanup of an environmental disaster will
cost the state $50,000 annually for perpetuity.
Assuming interest = 4%, determine the state’s
capitalized cost of this cleanup.
(a) $1,250,000
(b) $3,575,000
(c) $5,000,000
(d) $8,250,000
Solution to example 13
For perpetuity  forever
P=A/i = $50,000/.04 = $1,250,000
Answer: (a)
Example 14 – Infinite Life
The initial cost of constructing a road is estimated
to be $50 million. Annual maintenance is
estimated to be $0.18 million per year. In
addition, every 10 years the road will need
resurfaced at a cost of $2 million. Interest is 6%.
The equivalent uniform cost is most nearly:
a) $6.33 million
b) $5.33 million
c) $4.33 million
d) $3.33 million
Solution to Example 14
A = 0.18 + 50*.06 + 2(A/F, 6%,10) = .18+3+2(.0759)
= 3.3318
Answer: (d)
Bonds
 Bond - A loan an investor makes to a corporation or government.
 Face Value (par) – The original purchase price of the bond. The
bond holder will receive this amount when the bond reaches
maturity.
 Coupon (stated interest) – The interest that the bond holder will
receive while holding the bond.
 Maturity – The number of years the interest will be paid.
 Bonds can be resold for more or less than the face value. The rate of
return you actually get is the yield.
Example 15 – Bonds
A $5000 bond is being offered for sale. It has a
stated interest rate of 7%, paid annually ($350 each
year). The $5000 debt will be repaid at 8 years
along with the last interest payment. If you want
an 8% return on this investment (the bond yield),
what is the most you would be willing to pay for
the bond (the bond value)?
(a) $3500
(b) $4700
(c) $5000
(d) $5200
Solution to example 15
Bond value = PW of all future benefits
= 350 (P/A,8%,8) + 5000 (P/F,8%,8)
= 350 (5.747) + 5000 (0.5403) = $4713
Answer: (b)
Example 16 – Bonds
If the $5000 bond in the previous example could
be purchased for $4200, what is the bond yield?
(a) 5%
(b) 8%
(c) 10%
(d) 12%
Solution to example 16
Bond value = PW of all future benefits
$4200 = 350 (P/A, i %,8) + 5000 (P/F, i %,8)
By trial and error, try i = 10%
350 (5.335) + 5000 (0.467) = $4202 – close enough
Answer: (c)
Example 17 - Bonds
An investor is considering purchasing a bond
with a face value of $20,000 and 10 years left
to mature. The bond pays 12% interest
payable quarterly. If he wishes to get a 4% per
quarter return, the most he should pay for the
bond is closest to:
(a)$15,400
(b)$16,000
(c)$16,400
(d)$16,800
Solution to example 17
Since the bond pays 12%, paid quarterly, its effective
interest rate is 3% per quarter (every 3 months).
Interest payment = i(Face value) = 0.03(20,000)
= $600/quarter
n = 10*4 = 40 quarters
P = 600(P/A, 4%, 40) + 20,000(P/F, 4%, 40)
= 600(19.7928) + 20,000(0.2083) = $16,042
Answer: (b)
Break-Even Analysis
 Considers the time value of money.
 Single project
Determines the value of a particular variable that makes
the benefits = costs.
 Two projects
Determines the value of a particular variable that makes
the two projects equivalent.
Example 18 – Break-Even Analysis
A company is planning to update its production equipment and
is considering two different options. Each is anticipated to have
a 15 year life. The initial costs and salvage values of each are
shown. In addition, the annual saving for Option A is predicted
to be $14,000 per year, but the company is unsure of the annual
savings for Option B.
Initial Cost
Annual Savings
Salvage Value
Option A
$84,000
$14,000
$8,000
Option B
$140,000
??????
$12,000
If interest is 10%, determine the breakeven point between the
two projects.
(a)
(b)
(c)
(d)
18,720
21,240
23,210
26,930
Solution to example 18
EUA(B-C)A = -84,000(A/P, 10%, 15)
+ 14,000 + 8,000(A/F, 10%, 15) = 3208
EUA(B-C)B = 3208
= -140,000(A/P, 10%, 15) + A + 12,000(A/F,10%,15)
A = 21,240
Answer: (b)
Example 19 – Break-Even and Benefit /Cost Analyses
A proposed change to highway design standards is expected to
reduce the number of vehicle crashes by 9,200 per year, but
have initial cost of $150,000,000 and annual costs of
$25,000,000. Given an interest rate of 10% and a study
period of 8 years, the average cost of each vehicle crash in
order that the benefit-to-cost ratio be 1.0 is closest to:
(a) $5700
(b) $6700
(c) $8700
(d) $9700
Solution to example 19
In order that B/C = 1.0
PWcost = PWbenefit
PWbenefit
= 150,000,000 + 25,000,000 (P/A, 10%, 8)
= 150,000,000 + 25,000,000 (5.3349)
= $283,372,500
EUACbenefit = $283,372,500 (A/P, 10%, 8)
= $283,372,500 (0.18744)
= $53,115,341
$equivalent/crash = 53,115,341/9,200 = $5,773
Answer: (a)
Example 20 –Break-Even Analysis
A company is considering two alternative forklifts with equal
useful lives and the following characteristics
Initial Cost
Total Annual Costs
A
B
25,000
32,000
4350
2500
Given an interest rate of 10%, the service life in years at which
both machines have the same equivalent uniform annual cost
(EUAC) is most nearly:
(a) 5
(b) 7
(c) 9
(d) 11
Solution to example 20
EUACA = 25,000 (A/P, 10%, n) + 4,350
EUACB = 32,000 (A/P, 10%, n) + 2,500
EUACA = EUACB
25,000 (A/P, 10%, n) + 4,350 = 32,000 (A/P, 10%, n) + 2,500
(A/P, 10%, n) = (4,350 – 2,500) / 7,000 = 0.2643
From table look-up, the value of n that most nearly makes the
above relation true is 5.
Answer: (a)
Example 21 – Break-Even Analysis
A company is considering purchasing a new machine for
$650,000 that will increase the firm’s net income by
$150,000 per year over the next 5 years. If the company
wishes to obtain a 12% return on its investment, the
minimum salvage value of the machine at the end of the 5year useful life should be closest to:
(a) $120,000
(b) $193,000
(c) $256,000
(d) $307,000
Solution to example 21
S = 650,000 (F/P, 12%, 5) – 150,000 (F/A, 12%, 5)
= 650,000(1.7623) – 150,000(6.3528) = $192,575
Answer: (b)
Payback Period
 Does not consider the time value of money.
 Determines the time required to recover the initial cost of
a project or an investment.
Example 22 – Payback Period
An engineering department is considering purchase of an
advanced computational fluid dynamics software system to
enhance productivity. The initial cost of the software is
$55,000 but is expected to result in efficiency savings of
$25,000 the first year, with this amount decreasing by
$5,000 per year thereafter. The payback period for the
software is closest to:
(a) 2
(b) 2.67
(c) 3
(d) 3.67
Solution to example 22
Costs = 55,000
The payback period is the time when total income to date is
equal to the total costs.
Costs - Income = 0
55,000-25,000-20,000=10,000 to pay back at end of year 2.
Since the income is stated as $25,000 per year, one can
assume that savings occur uniformly throughout the year.
15,000 comes in during year 3. Therefore:
Payback period =2 + (10,000/15,000) = 2.67
Answer: (b)
Depreciation
Costs of capital assets (e.g., major equipment or facilities) are allocated
(depreciated) over time for tax purposes
Taxable income is total income less depreciation and ordinary
expenses
Book value at a point in time is the original cost of an asset minus
depreciation to that point in time
Example 23 - Depreciation
A company purchases a plastic extrusion machine for
$95,000. If this machine has an estimated salvage
value of $10,000 at the end of its five-year useful life
and recovery period, the second year straight line
depreciation is closest to:
(a) $13,000
(b) $15,000
(c) $17,000
(d) $19,000
Solution to example 23
Dt = D2 = (95,000 – 10,000)/5
= $17,000
Answer: (c)
Example 24 – Depreciation
A depreciable asset costs $25,000 and has an
estimated salvage value of $2500. It has a
MACRS class life of 5 years. At the end of
three years its book value is closest to:
(a) $4520
(b) $4800
(c) $7200
(d) $8980
Solution to example 24
MACRS depreciation ignores salvage value.
Depreciation the 1st year = (20/100)(25,000) = $5,000
Depreciation the 2nd year = (32/100)(25,000) = $8,000
Depreciation the 3rd year = (19.2/100)(25,000) = $4,800
Book value = cost – sum of depreciations
= $25,000 - $5,000 - $8,000 - $4,800 = $7,200
Answer: (c)
Inflation
 Inflation - increase in the price of goods/services.
 Actual dollars
 Dollars in the actual market
 inflate over time by Dn = Do(1+f)n
 f = annual inflation rate
 Real Dollars
 Buying power dollars
 Move through time with real interest rate
Inflation - continued
Equation in reference manual: d=i +f + (i)(f)
 d = market interest rate, for actual dollars
 i = real interest rate, for buying power dollars
 f = annual inflation rate
Example 25 - Inflation
A compact car costs approximately $21,000 today. If a
comparable car cost $15,000 ten years ago, the
average annual inflation in compact car prices over
the past ten years is closest to:
(a) 2.6%
(b) 3.0%
(c) 3.4%
(d) 3.8%
Solution to example 25
21,000 = 15,000 (1 + f)10
f = (21,000/15,000)0.1 – 1 = 0.034 = 3.4%
Note the similarity to the equation F = P(1+i)n
Answer: (c)
Example 26 – Inflation
An annuity pays $2500 per year for the next 10
years. Inflation is expected to be 4% per year.
After inflation, a real interest rate of 5% is
desired. The amount that should be paid for the
annuity is nearest to:
(a) $10,200
(b) $15,900
(c) $18,600
(d) $25,000
Solution to example 26
Step #1: Find the desired market interest rate, d.
d = 0.05 + 0.04 + (0.04)(0.05) = 0.092 or 9.2%
Step #2: Find the present worth of the annuity, at market interest = 9.2%.
Answer: (b)
.
Sources
Fundamentals of Engineering Supplied-Reference Handbook, 8th
Edition, Revised – pg. 114-120
Engineering Economic Analysis, 11th Edition, D.G. Newnan, J.P
Lavelle, & T.G. Eschenbach, Oxford University Press, 2012

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