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Engineering Economics
Cash Flow
Cash flow is the sum of money recorded as receipts or disbursements in a project’s financial records.
A cash flow diagram presents the flow of cash as arrows on a time line scaled to the magnitude of the cash
flow, where
expenses are down arrows and receipts are up arrows.
Year-end convention ~ expenses occurring during the year are assumed to occur at the end of the year.
Example (FEIM):
A mechanical device will cost $20,000 when purchased. Maintenance will cost $1000 per year. The device will
generate
revenues of $5000 per year for 5 years.
The salvage value is $7000.
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Discount Factors and Equivalence
Present Worth (P): present amount at t = 0
Future Worth (F): equivalent future amount at t = n of any present amount at t = 0
Annual Amount (A): uniform amount that repeats at the end of each year for n years
Uniform Gradient Amount (G): uniform gradient amount that repeats at the end of each year, starting at the
end of the second year and stopping at the end of year n.
NOTE: To save time, use the calculated factor table provided in the NCEES FE Handbook.
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Example (FEIM):
How much should be put in an investment with a 10% effective annual rate today to have $10,000 in five
years?
Using the formula in the factor conversion table,
P = F(1 + i) –n = ($10,000)(1 + 0.1) –5 = $6209
Or using the factor table for 10%,
P = F(P/F, i%, n) = ($10,000)(0.6209) = $6209
Example (FEIM):
What factor will convert a gradient cash flow ending at t = 8 to a future value? The effective interest rate is
10%.
The F/G conversion is not given in the factor table. However, there are different ways to get the factor using
the factors that are in the table.
For example,
NOTE: The answers arrived at using the formula versus the factor table turn out to be slightly different. On
economics problems, one should not worry about getting the exact answer.
Nonannual Compounding
Effective Annual Interest Rate
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An interest rate that is compounded more than once in a year is converted from a compound nominal rate to
an annual effective rate.
Effective Interest Rate Per Period
Example (FEIM):
A savings and loan offers a 5.25% rate per annum compound daily over
365 days per year. What is the effective annual rate?
Discount Factors for Continuous Compounding
The formulas for continuous compounding are the same formulas in the factor conversion table with the limit
taken as the number of periods, n, goes to infinity.
Comparison of Alternatives
Present Worth
When alternatives do the same job and have the same lifetimes, compare them by converting each to its cash
value today. The superior alternative will have the highest present worth.
Example (EIT8):
Capitalized Costs
Used for a project with infinite life that has repeating expenses every year.
Compare alternatives by calculating the capitalized costs (i.e., the amount of money needed to pay the start-up
cost and to yield enough interest to pay the annual cost without touching the principal).
NOTE: The factor conversion for a project with no end is the limit of the P/A factor as the number of periods,
n, goes to infinity.
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Annual Cost
When alternatives do the same job but have different lives, compare the cost per year of each alternative.
The alternatives are assumed to be replaced at the end of their lives by identical alternatives. The initial costs
are assumed to be borrowed at the start and repaid evenly during the life of the alternative.
Cost-Benefit Analysis
Project is considered acceptable if B – C ≥ 0 or B/C ≥ 1.
Example (FEIM):
The initial cost of a proposed project is $40M, the capitalized perpetual
annual cost is $12M, the capitalized benefit is $49M, and the residual
value is $0. Should the project be undertaken?
B = $49M, C = $40M + $12M + $0
B – C = $49M – $52M = –$3M < 0
Rate of Return on an Investment (ROI)
The ROI must exceed the minimum attractive rate of return (MARR).
The rate of return is calculated by finding an interest rate that makes the present worth zero. Often this must
be done by trial and error.
The project should not be undertaken.
Depreciation
Straight Line Depreciation
The depreciation per year is the cost minus the salvage value divided by the years of life.
Depreciation
Accelerated Cost Recovery System (ACRS)
The depreciation per year is the cost times the ACRS factor (see the table in the NCEES Handbook). Salvage
value is not considered.
4 of 12 11-12-2018, 02:09 PM

Depreciation
Example (FEIM):
An asset is purchased that costs $9000. It has a 10-year life and a salvage value of $200. Find the straight-line
depreciation and ACRS depreciation for 3 years.
Straight-line depreciation/year
ACRS depreciation
First year ($9000)(0.1) = $ 900
Second year ($9000)(0.18) = $1620
Third year ($9000)(0.144) = $1296
Depreciation
Book Value
The assumed value of the asset after j years. The book value (BVj) is the initial cost minus the sum of the
depreciations out to the j th year.
Example (FEIM):
What is the book value of the asset in the previous example after 3 years using straight-line depreciation?
Using ACRS depreciation?
Straight-line depreciation
$9000 – (3)($800) = $6360
ACRS depreciation
$9000 – $900 – $1620 – $1296 = $5184
Tax Considerations
Expenses and depreciation are deductible, revenues are taxed.
Example (EIT8):
5 of 12 11-12-2018, 02:09 PM

Tax Considerations
Gain or loss on the sale of an asset:
If an asset has been depreciated and then is sold for more than the book value, the difference is taxed.
Bonds
Bond value is the present worth of payments over the life of the bond.
Bond yield is the equivalent interest rate of the bond compared to the bond cost.
Example (EIT8):
Break-Even Analysis
Calculating when revenue is equal to cost, or when one alternative is equal to another if both depend on some
variable.
Example (FEIM):
How many kilometers must a car be driven per year for leasing and buying to cost the same? Use 10% interest
and year-end cost.
6 of 12 11-12-2018, 02:09 PM

Leasing: $0.15 per kilometer
Buying: $5000 purchase cost, 3-year life, salvage $1200,
$0.04 per kilometer for gas and oil, $500 per year for insurance
EUAC (leasing) = $0.15x, where x is kilometers driven
EUAC (buying) = $0.04x + $500 + ($5k)(A/P,10%,3) – ($1.2k)(A/F,10%,3)
= $0.04x + $500 + ($5k)(0.4021) – ($1.2k)(0.3021)
= $0.04x + $2148
Setting EUAC (leasing) = EUAC (buying) and solving for x
$0.15x = $0.04x + $2148
x = 19,527 km must be driven to break even
Inflation
Inflation-Adjusted Interest Rate
Additional Examples
Example 1 (FEIM):
What is the uninflated present worth of $2000 in 2 years if the average inflation rate is 6% and i is 10%?
d = i + f + if = 0.06 + 0.10 + (0.06)(0.10) = 0.166
P = ($2000)(P/F,16.6%, 2) = ($2000)(1 + d)–n
= ($2000)(1 + 0.166)–2 = $1471
Additional Examples
Example 2 (FEIM):
It costs $75 per year to maintain a cemetery plot. If the interest rate is 6.0%, how much must be set aside to
pay for maintenance on each plot without touching the principal?
(A) $1150
(B) $1200
(C) $1250
(D) $1300
P = ($75)(P/A,6%,∞) = ($75)(1/0.06) = $1250
Therefore, (C) is correct.
Additional Examples
Example 3 (FEIM):
It costs $1000 for hand tools and $1.50 labor per unit to manufacture a product. Another alternative is to
manufacture the product by an automated process that costs $15,000, with a $0.50 per-unit cost. With an
7 of 12 11-12-2018, 02:09 PM

annual production rate of 5000 units, how long will it take to reach the break-even point?
(A) 2.0 yr
(B) 2.8 yr
(C) 3.6 yr
(D) never
Cumulative cost (hand tools) = $1000 + $1.50x, where x is the number of units.
Cumulative cost (automated) = $15,000 + $0.50x
Set cumulative costs equal and solve for x.
$1000 + $1.50x = $15,000 + $0.50x
$1x = $14,000
x = 14,000 units
tbreak-even = x/production rate = 14,000/5000 = 2.8 yr
Therefore, (B) is correct.
Additional Examples
Example 4 (FEIM):
A loan of $10,000 is made today at an interest rate of 15%, and the first payment of $3000 is made 4 years
later. The amount that is still due on the loan after the first payment is most nearly
(A) $7000
(B) $8050
(C) $8500
(D) $14,500
loan due = ($10k)(F/P,15%,4) – $3000
= ($10k)(1 + 0.15)4 – $3000
= ($10k)(1.7490) – $3000
= $14,490 ($14,500)
Therefore, (D) is correct.
Additional Examples
Example 5 (FEIM):
A machine is purchased for $1000 and has a useful life of 12 years. At the end of 12 years, the salvage value is
$130. By straight-line depreciation, what is the book value of the machine at the end of 8 years?
(A) $290
(B) $330
(C) $420
(D) $580
BV = $1000 – ($1000 – $130)(8/12) = $1000 – $580 = $420
8 of 12 11-12-2018, 02:09 PM

Additional Examples
Example 6 (FEIM):
The maintenance cost for an investment is $2000 per year for the first 10 years and $1000 per year thereafter.
The investment has infinite life.
With a 10% interest rate, the present worth of the annual disbursement is most nearly
(A) $10,000
(B) $16,000
(C) $20,000
(D) $24,000
The costs or benefits for a cash flow that repeat should be broken into different benefits and costs that all start
or finish at the time of interest.
Take the $2000 cost that repeats for 10 years and break it into two $1000 costs to have one $1000 cost that
goes on infinitely and one $1000 cost that goes on for 10 years.
P = ($1000)(P/A,10%,10) + ($1000)(P/A,10%,∞)
= ($1000)(6.1446) + ($1000)(1/0.10)
= $6144.6 + $10,000 = $16,144.6 ($16,000)
Therefore, (B) is correct.
Additional Examples
Example 7 (FEIM):
With an interest rate of 8% compounded semiannually, the value of a $1000 investment after 5 years is most
nearly
(A) $1400
(B) $1470
(C) $1480
(D) $1800
ie = (1 + r/m)m – 1= (1 + 0.08/2)2 – 1 = 0.0816
F = ($1000)(F/P,8.16%,5) = ($1000)(1 + 0.0816)5
= ($1000)(1.480) = $1480
Additional Examples
Example 8 (FEIM):
The following data applies for example problems 8.1 through 8.3. A company is considering the purchase of
either machine A or machine B.
machine A machine B
9 of 12 11-12-2018, 02:09 PM

initial cost $80,000 $100,000
estimated life 20 years 25 years
salvage value $20,000 $25,000
other costs $18,000 per year $15,000 per year for the first 15 years
$20,000 per year for the next 10 years
Example 8.1 (FEIM):
The interest rate is 10%, and all cash flows may be treated as end-of-year cash flows. Assume that equivalent
annual cost is the value of the constant annuity equal to the total cost of a project. The equivalent annual cost
of machine B is most nearly
(A) $21,000
(B) $21,500
(C) $23,000
(D) $26,500
Additional Examples
The $15,000 cost for 15 years and the $20,000 cost for the next 10 years can be broken into a $20,000 cost
for the full 25 years and a $5000 benefit that is present for the first 15 years.
The present worth of $20k for 25 years is
P($20,25) = ($20k)(P/A,10%,25) = ($20k)(9.0770) = $181.54k
The present worth of $5k for 15 years is
P($5,15) = ($5k)(P/A,10%,15) = ($5k)(7.6061) = 38.03k
Pother costs = $181.54 k + $38.03k = $143.51k
Aother costs = Pother costs(A/P, 10%, 25) = ($143.51k)(0.1102) = $15,815
EUAC = ($100k)(A/P,10%,25) – ($25k)(A/F,10%,25) + $15,815
= ($100k)(0.1102) – ($25k)(0.0102) + $15,815
= $11,020 – $255 + $15,815 = $26,610 ($26,500)
Additional Examples
Example 8.2 (FEIM):
If funds equal to the present worth of the cost of purchasing and using machine A over 20 years were invested
at 10% per annum, the value of the investment at the end of 20 years would be most nearly:
(A) $548,000
(B) $676,000
(C) $880,000
(D) $1,550,000
P(A) = $80k + ($20k)(P/F,10%,20) – ($18k)(P/A,10%,20)
= $80k + ($20k)(0.1486) – ($18k)(8.5136)
= $80k + $2.972k – $153.245k
= $230,273
10 of 12 11-12-2018, 02:09 PM

F(A,10%,20) = ($230,273)(F/P,10%,20) = ($230,273)(6.7275)
= $1,549,162 ($1,550,000)
Additional Examples
Example 8.3 (FEIM):
How much money would have to be placed in a sinking fund each year to replace machine B at the end of 25
years if the fund yields 10% annual compound interest and if the first cost of the machine is assumed
to increase at a 6% annual compound rate? (Assume the salvage value does not change.)
(A) $2030
(B) $2510
(C) $2540
(D) $4110
F = P(F/P,6%,25) – salvage value = ($100,000)(4.2919) – $25,000
= $404,190
A = F(A/F,10%,25) = ($404,190)(0.0102)
= $4123 ($4110)
Related Document: Chapter - Viscoelasticity, PPT, Economics, Semester, Engineering (https://edurev.in
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