Engineering Economy Problems 1
1. What is meant by the term time value of money?
2. List three intangible factors
3. What is meant by evaluation criterion?
4. What is the primary evaluation criterion used in economic analysis?
5. List three evaluation criteria besides the economic one for selecting the best
restaurant.
6. Discuss the importance of identifying alternatives in the engineering economic
process.
7. What is the difference between simple and compound interest?
8. Trucking giant Yellow Corp agreed to purchase rival Roadway for $966 million
in order to reduce so-called back-office costs (e.g., payroll and insurance) by $45
million per year. If the savings were realized as planned, what would be the rate
of return on investment?
9. If Astra International’s profits increased from 22 rupiahs per share to 29 rupiahs
per share in the April-June quarter compared to the previous quarter, what was the
rate of increase in profits for that quarter?
10. At an interest rate of 8% per year, $10,000 today is equivalent to how much (a) 1
year from now (b) 1 year ago?
11. A medium-size consulting engineering firm is trying to decide whether it should
replace its office furniture now or wait and do it 1 year from now. If it waits 1
year, the cost is expected to be $16,000. At an interest rate of 10% per year, what
would be the equivalent cost now?
12. Certain certificates of deposit accumulate interest at 10% per year simple interest.
If a company invests $240,000 now in these certificates for the purchase of a new
machine 3 years from now, how much will the company have at the end of the 3year period?
13. A local bank is offering to pay compound interest of 7% per year on new savings
accounts. An e-bank is offering 7.5% per year simple interest on a 5-year
certificate of deposit. Which offer is more attractive to a company that wants to
set aside $1,000,000 now for a plant expansion 5 years from now?
14. A company that manufactures in-line mixers for bulk manufacturing is company
borrow now or 1 year from now? Assume the total amount due will be
considering borrowing $1.75 million to update a production line. If it borrows the
money now, it can do so at an interest rate of 7.5% per year simple interest for 5
years. If it borrows next year, the interest rate will be 8% per year compound
interest, but it will be for only 4 years. (a) How much interest (total) will be paid
under each scenario, and (b) Should the paid when the loan is due in either case.
Engineering Economy Problems 2
1. Define the symbols involved when a construction company wants to know how
much money it can spend 3 years from now in lieu of spending $50,000 now to
purchase a new truck, when the compound interest rate is 15% per year?
2. State the purpose for each of the following built-in Excel functions:
a. FV(i%,n,A,P)
b. IRR(first_cell:last_cell)
c. PMT(i%,n,P,F)
d. PV(i%,n,A,F)
3. Identify the following as cash inflows or cash outflows to Daimler-Chrysler:
income taxes, loan interest, salvage value, rebates to dealers, sales revenues,
accounting services, cost reductions.
4. Construct a cash flow diagram for the following cash flows: $10,000 outflow at
time zero, $3000 per year outflow in years 1 through 3 and $9000 inflow in years
4 through 8 at an interest rate of 10% per year, and an unknown future amount in
year 8.
5. Use the rule of 72 to estimate the time it would take for an initial investment of
$10,000 to accumulate to 20,000 at a compound rate of 8% per year.
6. If you now have $62,500 in your retirement when the account is worth $2 million,
estimate the rate of return that the account must earn if you want retire in 20 years
without adding any more money to the account.
From the Case Study
1. Problem Formulation. The president of Innovations Plastics wants a
recommendation on whether the company should plan on offering the new
technology to the major manufacturers and an estimate of the necessary capital
investment to enter this market early.
2. Information gathering.
a. The technology and equipment are expected to last about 10 years.
b. The inflation and income taxes are ignored for simplicity
c. The expected ROI were compound rates of 15%, 5% and 18%. And 5%
rate for enhancing an employee-safety
d. Equity capital financing beyond $5 millions is not possible. The amount of
debt financing and its cost are unknown
e. Annual operating cost 8% of first cost for major equipment
f. Increased annual training cost and salary range from $800,000 to $1.2
million
g. Another economic and non economic factors that may influence to each
alternative.
3.
a.
b.
c.
Seeking possible alternatives
Alternatives A
Alternative B
Alternative C
4. Alternatives Analysis
5.
Nominal and Effective Interest Rates
Nominal interest rate, r, is an interest rate that does not include any
consideration of compounding. By definition:
r = interest rate per period x number per period
Effective interest rate is actual rate that applies for a stated period of
time. The compounding of interest during the time period of the
corresponding nominal rate is accounted for by the effective interest
rate. It is commonly expressed on an annual basis as the effective
annual rate ia, but any time basis can be used.
Examples: 4% per year compounded monthly.
ieff = ( 1 + r/m)m – 1
where: ieff = effective rate for specified time period
r = nominal interest rate for same time period
m = number of times interest is compounded per state time
period
Equivalence Relations: Single Amounts with PP ≥ CP
1. A present sum of $ 5000 at an interest rate of 8% per year,
compounded semi-annually, is equivalent to how much
money 8 years ago?
2. The optical products division of Panasonic is planning a
$3.5 million building expansion for manufacturing its
powerful Lumix DMC digital zoom camera. If the company
uses an interest rate of 20% per year compounded
quarterly, for all new investments, what is the uniform
amount per quarter the company must make to recover its
investment in 3 years?
3. Northwest Iron and Steel is considering getting involved in
electronic commerce. A modest e-commerce package is
available for $20,000. If the company wants to recover the
cost in 2 years, what is the equivalent amount of new
income that must be realized every 6 months, if the interest
rate is 3% per quarter?
Equivalence When PP < CP
1. An engineer deposit $300 per month into a savings account that pays interest at a
rate of 6% per year, compounded semi-annually. How much will be in the
account at the end of 15 years? Assume no inter-period compounding.
2. For the transactions shown below, determine the amount of money in
the account at the end of year 3 if the interest rate is 8% per year,
compounded semi-annually. Assume no inter-period compounding.
End of Quarter
Amount of deposit,
$/Quarter
Amount of
Withdrawal,
$/Quarter
900
700
1000
-
2600
1000
1
2-4
7
11
Continuous Compounding
Effective Interest rate for Continues Compounding;
i% = er – 1
where : r is nominal interest rate
e = 2.71828
1. What effective interest rate per year, compounding continuously, is
equivalent to a nominal rate of 13% per year?
2. Corrosion problems and manufacturing defects
rendered a gasoline pipeline between El Paso and
Phoenix subjects to longitudinal weld seam failures.
Therefore, pressure was reduced to 80% of the design
value. If the reduced pressure results in delivery of
$100,000 per month less product, what will be the value
of the lost revenue after a 2-year period at an interest
rate of 15% per year, compounded continuously?
3. Because of a chronic water shortage in Santa Fee, new athletic fields must
use artificial turf or xeriscape landscaping. If the value of water saved
each month is $6000, how much can a private developer afford to spend
on artificial turf if he wants to recover his investment in 5 years at an
interest rate of 18% per year, compounded continuously?
4. An interest rate of 21% per year, compounded every 4 months, is
equivalent to what effective rate per year?
5. In ‘N Out Payday Loans advertises that for a fee of only $10, you can
immediately borrow up to $200 for one month. If a person accepts the
offer, what are (a) the nominal interest rate per year and (b) the effective
rate per year?
6. How much will be in a high-yield account at the National Bank of Arizona
12 years from now if you deposit $5000 now and $7000 five years from
now? The account earns interest at rate of 18% per year, compounded
quarterly.
Varying Interest Rate
1.
How much money could the maker of fluidized bed scrubbers afford to spend
now instead of spending $150,000 in year 5 if the interest rate is 10% per year in
years 1 through 4 and 1% per month in years 5 through 8?
2.
For the cash flows shown below, determine (a) the future worth in year 5 and (b)
the equivalent A value for years 0 through 5.
Year
0
1–4
5
Cash Flow, $/year
5000
6000
9000
Interest rate per year. %
12
12
20
PRESENT WORTH ANALYSIS (PW ANALYSIS)
o A FUTURE AMOUNT OF MONEY CONVERTED INTO ITS
EQUIVALENT VALUE NOW HAS A PW THAT ALWAYS LESS
THAN ACTUAL CASH FLOW, FOR ANY INTEREST RATE > 0.
o P/F factor < 1
o PW values are often referred to as DCF (Discounted Cash Flow)
o Interest Rate = Discounted Rate
o PW = PV = NPV
Purpose: To compare mutually exclusive alternatives on a PW basis.
PW ANALYSIS PW ANALYSIS will help us:
o
o
o
o
o
o
o
o
Identify mutually exclusive and independent projects
Define a service and a revenue alternative
Select the best equal-life alternatives using PWA
Select the best different-life alternatives using PWA
Select the best alternative using FWA
Select the best alternative using CC calculations
Determine the payback period at i = 0% and i > 0%
Perform a life-cycle cost analysis for the acquisition & operation phases
of the system
o Calculate the PW of a bond investment
o Develop spreadsheet that use PWA, including payback period
Mutually exclusive project: only one of the viable projects can be selected
Independent project: more than one viable project may be selected
Example.
Perform a PWA of equal-service machines, if MARR is 10%/year
Revenues for all three alternatives are expected to be the same
First Cost, $
AOC, $
Salvage Value, $
Life, Years
Electric Powered(E)
-2500
-900
200
5
Gas Powered (G)
-3500
-750
350
5
Solar Powered (S)
-6000
-50
100
5
Answer:
PWE = -2500 – 900(P/A,10%,5) + 200(P/F, 10%, 5) = 4 -5788
PWG = -3500 – 750(P/A, 10%, 5) + 350(P/F, 10%, 5) = $ -5936
PWS = -6000 – 50(P/A, 10%, 5) + 100(P/F, 10%,5) = $ -6127
PWE is selected since PW of its cost is lowest, or numerically largest PW value.
If there are different life alternatives
o Compare the alternatives over a period of time equal to the LCM (Least Common
Multiple) of their lives
The assumptions of a PWA of different-life alternatives:
o The service provided by alternatives will be needed for the LCM of years or more
o The selected alternative will be repeated over each life cycle of the LCM in
exactly the same manner
o The cash flow estimates will be the same in every life-cycle
Example.
A project engineer with EnvironCare is assigned to start up a new office in a city where a
6-year contract has been finalized to take and analyze ozone-level readings. There are
two options:
First Cost, $
Annual Cost, $
Deposit Return, $
Lease Time, years
Location A
-15,000
-3,500
1000
6
Location B
-18,000
-3,100
2000
9
(a) Determine which lease option should be selected on the basis of PW, if MARR is
15%/year
(b) If a study period of 5 years is used, which location should be selected ?
(c) If a study period of 6 years is used, and the deposit return of alternative B is
estimated to be $ 6000 after 6 years, which location should be selected?
Answer.
Since the lease have different term, the LCM(6,9) is 18
(a) PWA = -15,000 – 15,000(P/F, 15%,6) + 1000(P/F, 15%, 6) – 15,000(P/F,
15%,12) + 1000(P/F, 15%, 12) + 1000(P/F, 15%, 18) – 3,500(P/A, 15%,18) = $ 45,036
PWB = -18,000 – 18,000(P/F, 15%,9) + 2000(P/F, 15%, 9) + 2,000(P/F, 15%,18)
- 3100(P/A, 15%, 18) = $ -41,384
Location B is selected, because the PWB value is numerically larger than PWA
(b) For 5 year study period, no cycle repeats are necessary
o PWA = -15,000 – 3,500(P/F, 15%,5) + 1000(P/F, 15%, 5) = $ – 26,236
o PWB = -18,000(P/F, 15%, 12)–3,100(P/A,15%,5)+2,000(P/F,15%,5) = $ - 27,397
Location A is the better choice
(c) For 6 year study period,
o PWA = -15,000 – 3,500(P/F, 15%,6) + 1000(P/F, 15%, 6) = $ – 27,813
o PWB = -18,000(P/F, 15%, 6)–3,100(P/A,15%,6)+6,000(P/A,15%,6) = $ - 27,397
Location B is numerically larger than A
Future Worth Analysis
o FW of an alternative may be determined directly from the cash-flows by
determining the FW value, or multiplying the PW by the F/P factor, at the
established MARR.
o FW analysis is applicable to large capital investment decision when a prime goal
is to maximize the future wealth of a corporation’s stockholders.
o FW analysis is often utilized if the asset might be sold or traded at some time after
its start-up, but before the expected life is reached.
Capitalized Cost Calculation and Analysis
o CC is present worth of an alternative that will last “forever”
Projects fall into this category are:
Bridges (Suramadu)
Dams (Asahan)
Irrigation Systems (Jatiluhur)
Rail Road (Jabodetabek) and Toll Road (Cipularang)
CC is derived from the relation P = (A/P, i, n), where n = 
P = A[{(1+i)n – 1}/{i(1+i)n}] or P = A[{1-1/(1+i)n}/i}] = A/i so, CC = A/i = AW/i
Procedure to calculate CC for an infinite sequence of cash-flows.
 Step 1. Draw cash flow diagram showing all non-recurring (one-time) cash-flows
and at least two cycles of all recurring cash-flows.
 Step 2. Find the PW of all non-recurring amounts . This is their CC value
 Step 3. Find the equivalent uniform AW (annual worth/value) through one life
cycle of all recurring amounts.
 Step 4. Divide the AW in step 3 by i to obtain a CC value.
 Step 5. Add CC values obtained in step 2 and 4.
Payback Period Analysis
The payback period np is the estimated time (in years), it will take for the estimated
revenues and other economic benefits to recover the initial investment and a stated rate of
return.
np
If the stated rate i>0%, then
0 = -P +  NCFt (P/F, i, t)
t=1
Where: P
: is the initial investment (first cost)
NCFt : Net Cash Flow for each year t = receipt – disbursement
If NCF values are expected to be equal each year:
If the NCF series are uniform:
np = P/NCF
0 = - P + NCF (P/F, i, np)
Life Cycle Cost
Typical applications for LCC are
 Buildings
 New product lines
 Manufacturing plants
 Commercial Aircraft
 New Automobile models
 Defense Equipment Systems
Generally, LCC estimates may be categorized into simplified format for the major phases
of acquisition and operation as following:
A. Acquisition Phase
 Requirement definition Stage:
(a) Determination of user need
(b) Anticipated system
(c) Preparation and documentation
 Preliminary design
(a) Feasibility Study
(b) Conceptual
(c) Early Stage Plans
 Detail Design. Detail Plan & Resources
(a)
Capital, human, facilities, information
(b) Some acquisition of assets
system,
marketing
B. Operation Phase
 All activities are functioning, products & services are available
 Construction & implementation, testing and preparation
 Usage Stage: to generate products and services
 Phase Out & Disposal Stage
Present Worth of Bonds
Equity Financing: The corporation uses its own funds from cash on hand, stock sales,
or retained earning. The individuals can use their own cash, savings or investment.
Debt Financing: The corporation borrows from outside sources and repays the
principal and interest. Sources of dept capital may be bonds, loans, mortgages, venture
capita pools. Individuals can utilize debt sources, such as credit card or from a credit
union.
A time-tested method of raising capital is through the issuance of IOU (I owe you), which
is financing through debt. One form of IOU is bond.
A bond: is a long term note issued by a corporation (financing institution) or government
to finance major project.
Bonds are usually issued in face value amounts of $100, $1000, $5000 or $10,000.
Bond Interest I (also called bond dividend) is paid periodically
The bond interest is paid c times per year.
The stated interest rate is called the bond coupon rate b
I = (face value)(bond coupon rate)/number of payment periods per year = Vb/c
The steps to calculate the PW of a bond are as follow:
1.
Determine I, the interest per payment period
2.
Construct the cash flow diagram
3.
Establish the required MARR or rate of return
4.
Calculate the PW value of the bond interest payments and the face value at
i=MARR
Example.
Determine the purchase price you should be willing to pay now for a 4.5% $5000 10-year
bond with interest paid semiannually. Assume your MARR is 8% per year compounded
quarterly.
Solution..
I = 5000(0.045)/2 = $112.50 every 6 month
The nominal semiannual MARR is r = 8%/2 = 4%
Effective I = ( 1 + 0.04/2)2 – 1 = 4.04% per 6 months
The PW of the bond is determined for n = 2(10) = 20 semiannual period
PW = $112.50(P/A,4.04%,20) + 5000(P/F,4.04%,20) = $3788
Problems
5.9, 5.11, 5.14, 5.16, 5.26, 5.28, 5.38, 5.39, 5.42, 5.49, 5.50