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 3-year 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. Seeking possible alternatives
a. Alternatives A
b. Alternative B
c. Alternative C
4. Alternatives Analysis
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, Amount of
Withdrawal,
$/Quarter $/Quarter
1 900
2-4 700
7 1000 2600
11 - 1000
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 Cash Flow, $/year Interest rate per year. %
0 5000 12
1 – 4 6000 12
5 9000 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 Identify mutually exclusive and independent projects
o Define a service and a revenue alternative
o Select the best equal-life alternatives using PWA
o Select the best different-life alternatives using PWA
o Select the best alternative using FWA
o Select the best alternative using CC calculations
o Determine the payback period at i = 0% and i > 0%
o 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
Electric Powered(E) / Gas Powered (G) / Solar Powered (S)First Cost, $
AOC, $
Salvage Value, $
Life, Years / -2500
-900
200
5 / -3500
-750
350
5 / -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:
Location A / Location BFirst Cost, $
Annual Cost, $
Deposit Return, $
Lease Time, years / -15,000
-3,500
1000
6 / -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