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IEEN 5329 Homework 3 4

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IEEN 5329

Homework 3

4.42, 4.47, 4.54

5.4, 5.8, 5.14 (Calculate by hand and computer), 5.20, 5.25, 5.33, 5.39 (Calculate by hand and computer), 5.44, 5.51

Nominal and Effective Rates

4.1 Identify the compounding period for the following interest statements: (a) 1% per month; (b) 2.5% per quarter, and (c) 9.3% per year compounded semiannually.

4.2 Identify the compounding period for the following interest statements: (a) Nominal 7% per year compounded quarterly: (b) effective 6.8% per year compounded monthly; and (c) effective 3.4% per quarter compounded weekly.

4.3 Determine the number of times interest would be compounded in 1 year for the following interest statements: (a) 1% per month; (6)2% per quarter, and (c) 8% per year compounded semiannually.

4.4 For an interest rate of 10% per year com-pounded quarterly, determine the number of times interest would be compounded (a) per quarter. (b) Per year, and (c) per 3 years.

4.5 For an interest rate of 0.50% per quarto, determine the nominal interest rate per (a) semiannual period. (b) year, and (c) 2 years.

4.6 For an interest rate of 12% pa year com-pounded every 2 months, determine the nominal interest rate per (a) 4 months, (6) 6 months, and (c) 2 years.

4.7 For an interest rate of 10% per year, com-pounded quarterly, determine the nominal rate per (a) 6 months and (6) 2 years.

4.8 Identify the following interest rate statements as either nominal or effective: (0)1.3% per month; (b) 1% per week, compounded weekly; (c)nominal 15% per year, compounded monthly; (c) effective 1.5% per month, compounded daily; and (r)15% per year, compounded semiannually.

4.9 What effective interest rate per 6 months is equivalent to 14% per year, compounded semiannually? 4.10 An interest rate of 16% per year, com-pounded quarterly, is equivalent to what effective interest rate per year?

4.11 What nominal interest rate per year is equivalent to an effective 16% per year, compounded semiannually?

4.12 What effective interest rate per year is equivalent to an effective 18% per year, compounded semiannually?

4.13 What compounding period is associated with nominal and effective rates of 18% and 18.81% per year, respectively?

4.14 An interest rate of 1% per month is equivalent to what effective rate per 2 months?

4.15 An interest rate of 12% per year, com-pounded monthly, is equivalent to what nominal and effective interest rates per 6 months?

4.16 (a) an interest rate of 6.8% per semi-annual period compounded weekly, is equivalent to what weekly interest rate?

       (b) Is the weekly rate a nominal or effective rate? Assume 26 weeks per 6 months.

Payment and Compounding Periods

4.17 Deposits of $100 per week are made into a savings account that pays interest of 6% per year, compounded quarterly. Identify the payment and compounding periods.

4.18 A certain national bank advertises quarterly compounding for business checking accounts. What payment and compounding periods are associated with deposits of daily receipts?

4.19 Determine the F/P factor for 3 years at an interest rate of 8% per year, compounded quarterly.

4.20 Determine the P/G factor for 5 years at an effective interest rate of 6% per year, compounded semiannually.

Equivalence for Single Amounts and Series

4.21 A company that specializes in online security software development wants to have $85 million available in 3 years to pay stock dividends. How much money must the company set aside now in an account that earns interest at a rate of 8% per year, compounded quarterly?

4.22 Because testing of nuclear bombs was halted in 1992, the U.S. Department of Energy has been developing a laser project that will allow engineers to simulate (in a laboratory) conditions in a thermonuclear reaction. Due to soaring cost overruns, a congressional committee undertook an investigation and discovered that the estimated development coat of the project increased at an average rate of 1% per month over a 5-year period. If the original cost was estimated to be $2.7 billion 5 years ago, what is the expected cost today?

4.23 A present sum of $5000 at an interest ram of 8% per year, compounded semiannually, is equivalent to how much money 8 years ago?

4.24 In an effort to ensure the safety of all phone users, the Federal Communications Commission (FCC) requires cell phases to have a specific absorbed radiation (SAR) number of 1.6 watts per kilogram (W/Kg) of tissue or less. A new cell phone company estimates that by advertising its favorable 1.2 SAR number. It will increase sales by $1.2 million 3 months from now when its phones go on sale. At an interest rate of 20% per year, compounded quarterly, what is the maximum amount the company can afford to spend now for advertising in order to break even?

4.25 Radio Frequency Identification (RFID) is technology that is used by drivers with speed passes at toll booths and ranchers who track livestock from farm to fork. Wal-Mart expects to begin using the technology to track products within its stores. If RFID-tagged products will result in better inventory control that will save the company $1.3 million per month beginning 3 months from now, how much could the company afford to spend now to implement the technology at an interest rate of 12% per year, compounded monthly, if it wants to recover its investment in 2½ years?

4.35 Magnetek Instrument and Controls, a manufacturer of liquid-level sensors, expects sales for one of its models to increase by 20% every 6 months into the foreseeable future. If the sales 6 months from now are expected to be $I50.000, determine the equivalent semiannual worth of sales for a 5-year period at an interest rate of 14% per year, compounded semiannually.

4.36 Metal fab Pump and Filter projects that the cost of steel bodies for certain valves will increase by $2 every 3 months. If the coat for the first quarter is expected to be $80, what is the present worth of the costs for a 3-year period at an interest rate of 3% per quarter?

4.37 Fieldsaver Technologies, a manufacturer of precision laboratory equipment, borrowed $2 million to renovate one of its testing labs. The loan was repaid in 2 years through quarterly payments that increased by $50,000 each time. At an interest rate of 3% per quarter, what was the size of the first quarterly payment?

4.38 For the cash flows shown below, deter-mine the present worth (time 0), using an interest rate of 18% pa year, compounded monthly.

 

 

Month	Cash Flow, S/Month
0 1-12 13-28	1000 2000 3000

 

4.39 The cash flows (in thousands) associated with Fisher-Price's Touch learning system are shown below. Determine the uniform quarterly series in quarters 0 through 8 that would be equivalent to the cash flows shown at an interest rate of 16% per year, compounded quarterly.

Quarter	Cash Flow, S/ Quarter
1 2-3 5-8	1000 2000 3000

 

Equivalence When PP < CP

4.40 An engineer deposits $300 per month into a savings account that pays interest at a rate of 6% per year, compounded semiannually. How much will be in the account at the end of 15 years? Assume no interperiod compounding.

4.41 At time t= 0, an engineer deposited %10,000 into an account that pays interest at 8% per year compounded semiannually. If she withdrew $1000 in months 2, 11, and 23, what was the total value of the account at the end of 3 years? Assume no interperiod compounding.

4.42 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 semiannually. Assume no interperiod compounding.

End of Quarter	Amount of Deposit, s/Quarter	Amount of Withdrawal. S/Quarter
1 2-4 7 11	900 700 1000 -	2600 1000

 

4.43 The New Mexico State Police and Public Safety Department owns a helicopter that it uses to provide transportation and logistical support for high-level state officials. The $495 hourly rate coven operating expenses and the pilot's salary. If the governor uses the helicopter an average of 2 days per month for 6 hours each day, what is the equivalent future worth of the costs for 1 year at an interest rate of 6% per year, compounded quarterly (treat the costs as deposits)?

Continuous Compounding

4.44 What effective interest rate per year, compounded continuously, is equivalent to a nominal rate of 13% per year?

4.45 What effective interest rate per 6 months is equal to a nominal 2% per month, compounded continuously?

4.46 What nominal rate per stunner is equivalent to an effective rate of 12.7% per year, compounded continuously?

4.47 Corrosion problems and manufacturing defects rendered a gasoline pipeline be-tween El Paso and Phoenix subject to longitudinal weld seam failures. Therefore, the pressure was reduced to 80% of the de-sign value. If the reduced pressure results in the 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?

4.48 Because of a chronic water shortage in Santa Fe, new athletic fields must use artificial turf or xeriscape landscaping. If the value of the water saved each month is $6000, how much can a private developer afford to spend on artificial turf it he wants to recover his investment in 5 years at an interest rate of 18% per year, compounded continuously?

4.49 A Taiwan-based chemical company had to file for bankruptcy because of a nation- wide phase-out of methyl tertiary butyl ether (MTBE). If the company reorganizes and invests $50 million in a new ethanol production facility, how much money must it make each month if it wants to re-cover its investment in 3 years at an interest rate of 2% per month, compounded continuously?

4.50 In order to have $85,000 four years from now for equipment replacement, a construction company plans to set aside money today in government-insured bonds. If the bonds earn interest at a rate of 6% per year, compounded continuously, how much money must the company invest?

4.51 How long would it take for a lump-sum investment to double in value at an interest rate of 1.5% per month, compounded con-tenuously?

4.52 What effective interest rate per month, compounded continuously, would be required for a single deposit to triple in value in 5 years?

Varying Interest Rates

4.53 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% in years 1 through 3 and 12% in years 4 and 5?

4.54 What is the future worth in year 8 of a present sum of $50, 000 if the interest rate is 10% per year in years 1 through 4 and 1% per month in years 5 through 8?

 

Price $90,000 show the payback period to be between 3 and 4 years, while the NPV results in column F (selling price $120.000) indicate PW switching from positive to negative between 6 and 7 years. The NPV function reflect the relation presented in the hand solution except the cost gradient of $100 has been incorporated into the cost in column B.

If more exact payback values are needed, interpolate between the PW results on the spreadsheet. The values will be the same as in the solution by hand, namely, 3.96 and 6.90 years.

 CHAPTER SUMMARY

The present worth method of comparing alternatives involves convening all cash flows to present dollars at the MARR. The alternative with the numerically larger (or largest) PW value is selected. When the alternative have different lives, the comparison must be made for equal-service periods. This is done by performing the comparison over either the LCM of lives or specific study period. Both approaches compare alternatives in accordance with the equal service requirement. When a study period is used, any remaining value in an alternative is recognized through the estimated future market value.

Life-cycle cost analysis is an extension of PW analysis performed for systems that have relatively long lives and a large percentage of their lifetime costs in the form of operating expenses. If the life of the alternatives is considered to be infinite, capitalized cost is the comparison method. The CC value A/I, because the P/A factor reduces to 1/i in the limit ∞

Payback analysis estimates the number of years necessary in recover the initial investment plus a stated rate of return (MARR).This is a supplemental analysis technique used primarily for initial screening of proposed project prior to a full economic evaluation by PW or some other method. The technique has some drawbacks, especially for non-return payback analysis. Where i=0%, is used as MARR.

Finally, we learned about bonds. Present worth analysis determines if the MARR will he obtained over the life of a bond, given specific values for the bond's face value, term and interest rate.

PROBLEMS.

Types of Projects

5.1 What is meant by service alternative?

5.2 When you tee evaluating projects by the present worth method, how do you know which one(s) to select if the projects are (a) independent and (b) mutually exclusive?

a revenue or a service project: (a) Problem 2.12. (b) Problem 2.31, (c) Problem 2.51. (d) Problem 3.6. (e) Problem 3.10, and (∫) 1 Problem 3.14.

5.4 A rapidly growing city is dedicated to neighborhood integrity. However, increasing traffic and speed on a through street are of concern to residents. The city manager has proposed live independent options to slow traffic:

1. Stop sign at corner A.

2. Stop sign at corner B

3. Low-profile speed bump at point C.

4. Low-profile speed bump at point D.

5. Speed dip at point E.

These cannot be any of the following combinations in the final alternatives:

• No combination of dip and one or two bumps

• Not two bumps

• Not two stop signs

Use the five independent options and the restrictions to determine (a) the total number or mutually exclusive alternatives possible and (b) the acceptable mutually exclusive alternatives.

5.5 What is meant by the term equal service?

5.6 What two approaches can be used to satisfy the equal service requirement?

5.7 Define the term capitalized cost and give a real-world example of something that might he analyzed using that technique.

Alternative Comparison—Equal Lives

5.8 Lennon hearth Products manufactures glass-door fireplace screens that have two types of mounting brackets for the frame. An L-shaped bracket is used for relatively small fireplace openings, and a U-shaped bracket is used for all others. The cam box with the product, and the purchaser discards the one not needed. The cost of these two brackets with screws and other parts is $3.50. If the frame of the fireplace screen is redesigned, a single universal bracket can he used that will cost $1.20 to make. However, retooling will cost $6000, In addition, inventory write-downs will amount to another $8000. If the company sells 1200 fireplace units per year, should the company keep the old brackets or go with the new ones, assuming the company uses an interest rate of 15% per year and it wants to recover its investment in 5 years? Use the present worth method.

5.9 Two methods can be used for producing expansion anchors. Method A costs $80,000 initially and will have a $15.000 salvage value alter 3 years. The operating cost with this method will be $30,000 per year. Method B will have a first cost of $120,000 an operating cost of $8000 per year and a $40.000 salvage value after its 3-year life. At an interest rate of 12% per year. Which method should be used on the basis of a present worth analysis?

5.10 Sales of bottled water in the UNITED States totaled 16.3 gallons per person in 2014. Evian Natural Spring Water costs 40¢ per bottle. A municipal water utility provides tap water for $2.10 per 1000 gallons. If the average person drinks 2 bottles of water per day or uses 5 gallons per day in getting that amount of water from the tap. What are the present worth values of drinking bottled water or tap water per person for 1 year? Use an interest rate of 6% per year. Compounded monthly and 30 days per month.

5.11 A software package created by Navarro & Associates can be used for analyzing and designing three-sided guyed lowers and three and four-sided self-supporting $400 per year. A site license has a one-time cost of $15,000. A structural engineering consulting company is trying to decide between two alternatives: first, to buy one single-user license now and one each year for the next 4 years (which will provide 5 years of servicer: or second, to buy a site license now. Determine which strategy should be adopted at an interest rate of 12% per year for a 5-year planning period, using the present worth method of evaluation.

5.12 A company that manufactures amplified pressure transducers is trying to decide between the machines shown below. Compare them on the basis of their present worth values, using an interest rate of 15% per year.

Variable Speed		Dual Speed
First Cost, $	-250,000	-224,000
Annual operating cust. $/year	231,000	-235,000
Overhaul in year 3,$	-	-26,000
Overhead in Year 4 ,$	140,000	-
Salvage value	50,000	10,000
Life, years	6	6

 

Alternative Comparison over Different Time Periods

5.13 NASA is considering two materials for use in n space vehicle. The costs are shown below. Which should be selected on the basis of a present worth comparison at an interest rate of 10% per year?

	Material JX	Material KZ
First cost, $	-205,000	-235,000
Maintenance cost, $/year	29,000	27,000
Salvage Value	2,000	20,000
Life, Years	2	4

 

 5.14 Two processes can be used for producing polymer that reduces friction loss in engines. Process K will have a first cost of $160.000, an operating cost of $7000 per quarter and a salvage value of $40,000 after its 2-year life. Process L will have a first cost of $210.000 an operating cost or $5000 per quarter. and a $26,000 salvage Value after its 4-year life. Which process should be selected on the basis of a present worth analysis at an interest rate of 8% per year. Compounded quarterly?

5.15 Two methods are under consideration for producing the case for a portable hazardous material photoionization monitor. A plastic case will require an initial investment of $75,000 and will have an annual operating cost of $27,000) with no salvage after 2 years. An aluminum case will require an investment of $125.000 and will have annual costs of $12.000. Some of the equipment can he sold For $30,000 after its 3-year life. At an interest rate of 10% per year, which case should be used on the basis of a present worth analysis?

5.16 Three different plans were presented to the GAO by at high-technology facilities manager for operating a small weapons production facility. Plan A would involve renewable 1-year contracts with payments of $1 million at the beginning of each year. Plan B would be a 3-year contract and it would require four payments of $600,000 each. With the first one to be made now and the other three at 6-month intervals. Plan C would he a 3-year contract and it would entail a payment of $1 .5 million now and another payment of $0.5 million 2 year from now. Assuming that GAO could renew any of the plans under the same conditions if it wants to do so. Which plan is better on the basis of a present worth analysis at an interest rate of 6% per year, compounded semiannually?

Future worth Comparison

5.17 A remotely located air sampling station can be powered by solar cells or by running an electric line to the site and using conventional power. Solar cells will cost $12,600 to install and will have a useful life of 4 years with no salvage value. Annual costs for inspection, cleaning, etc.. Are expected to be $1400. A new power line will cost $11,000 to install, with power costs expected to be $800 per year. Since the air sampling project will end in 4 years, the salvage value of the line is considered to be zero. At an interest rate of 10% per year. Which alternative should be selected on the basis of a future worth analysis!

5.18 The Department of Energy is proposing new rules mandating a 20% increase in clothes washer efficiency by 2005 and a 35% increase by 2008. The 20% increase is expected to add $100 to the current price of a washer, while the 35% increase will add $240 to the price. If the cost for energy is $80 per year with the 20% increase in efficiency and $65 per year with the 35% increase. Which one of the two proposed standards is more economical on the basis of a future worth analysis at an interest rule of 10% per year? Assume a 15.yuat life for all washer lintels.

5.19 A small strip-mining coal company is trying to decide whether it should purchase of tense a new clamshell. If purchased, the shell will cost $150,000) and is expected to have a $65,000 salvage value in 6 years. Alternatively the company can lease a clamshell for $30,000 per year, but the lease payment will have to he made at the beginning of each year. If the clamshell is purchased it will he leased to other strip mining companies whenever possible an activity that is expected to yield revenues of $12,000 per year. if the company's minimum attractive rate of return is 15% per year, should the clamshell be purchased or leased on the basis of a future worth analysis?

5.20 Three types of drill bits can he used in a certain manufacturing operation. A bright high-speed steel (ASS) bit is the least eV pensive to buy, but it has a shorter life than either gold oxide or titanium nitride bits. The HSS bits will cost $3500 to buy and will last for 3 months under the conditions in which they will be used. The operating cost for these bits will be $2000 per month. The gold oxide bits will cost $6500 to buy and will last for 6 months with an operating cost of $1500 per month. The titanium nitride hits will cost $7000 to buy and will last 6 months with an operating cost $1200 per month at an interest rate of 12% per year, compounded monthly, which type of drill hit should be used on the basis of a future worth analysis?

5.21   EL Paso Electric is considering two alternatives for satisfying state regulations regarding pollution control for one of its generating stations. This particular station is located at the outskirts of the city and a short distance from Juarez, Mexico. The station is currently producing excess VOC.s and oxides of nitrogen. Two plans have been proposed for satisfying the regulators. Plan A involves replacing the burners and switching from fuel oil to natural gas. The cost or the option will he $300,000 initially and an extra $900,000 per year in fuel costs. Plan B involves going to Mexico and running gas lines to many of the "backyard" brickmaking sites nun now use wood tires, and other combustible waste materials for firing the bricks. The idea behind plan B is that by reducing the particulate pollution responsible for smog in EL Paso, there would be greater benefit in U.S. citizens than would be achieved through plan A. The initial cost of plan B will be $1.2 million for installation of the lines. Additionally the electric company would subsidize the cost of gas for the brick makers to the extent of $200,000) per year. For a 10 year project period and no salvage value for either plan, which one should be selected on the basis of a future worth analysis at an interest rate of 12% per year?

Capitalized Costs

5.22 The cost of painting the Golden Gate Bridge is $400,000. If the bridge is painted now and every 2 years hereafter. What is the capitalized cost of painting at an interest rate at 6% per year?

5.23 The cost of extending a certain road at Yellowstone National Park is $1.7 million resurfacing and other maintenance arc expected to cost $350,000 every 3 year. What is the capitalized cost of the road at an interest rate of 6% per year?

5.24 Determine the capitalized cost of an expenditure of $200,000 at time 0. $25,000 in year 2 through 5, and $40,000 per year from year 6. Use an interest rate of 12% per year.

5.25 A city that is attempting to attract a professional football team is planning to build a new stadium costing $2.50 million. Annual upkeep is expected to amount to $800,000 per year. The artificial will have to be replaced every 10 years at a cost of $950,000. Painting every 5 years will cost $75,000. If the city expects to maintain the facility indefinitely. What will be its capitalized cost at an interest rate of 8% per year?

5.26 A certain manufacturing alternative has a first cost of $82,000 an annual maintenance cost of $9000, and a salvage value of $15000 after its 4 year life. What is it’s capitalized cost of an interest rate of 12% per year?

5.27 If you want to be able to withdraw $80,000 per year forever beginning 30 years from now, how much will you have to have in your retirement account that 8% per year interest in (a) year 29 and (b) year 0?

5.28 What is the capitalized cost absolute value of the difference between the following two plans at an interest rate of 10% per year? Plan A will require an expenditure of $50,000 every 5 years forever (be-ginning in year 5 Plan B will require an expenditure of $100,000 every 10 years forever (beginning in year 10).

5.29 What is the capitalized cost of expenditures of $3,000,000 now $50,000 in months 1 through 12. $100,000 in months 13 through 25, and $50,000 in months 26 through infinity if the interest rate is 12% per year, compounded monthly?

5.30 Compare the following alternative’s on the basis of their capitalized cost and an interest rate of 10% per year.

	Petroleum Based Feedstock	Inorganic Based Feedstock
First cost, $	-250,000	-110,000
Annual Operating cost, $/year	-130,000	65,000
Annual revenues $/year	400,000	270,000
Salvage value, $ life, years	6	4

 

 5.31 An alumna of Ohio State University wanted to set up an endowment fund that would award scholarships to female engineering students totaling $100,000 per year forever. The first scholarship are to be granted now and continue each year for-ever. How much must the alumna donate now, If the endowment fund is expected to earn interest at a rate of 8% per year?

5.32 Two large-scale conduits ate under consideration by a large municipal utility district (MUD) The firm involves construction of a steel pipeline at a cost of $225 million. Portions of the pipeline will have to be replaced every 40 years at a cost of $50 million. The pumping and other operating costs are expected to be $10 million per year. Alternatively, a gravity flow canal can be constructed at a cost of $350 million. The M&O costs for the annul arc expected to be $0.5 million per year. If both conduits are expected to last forever, which should he hulk at an interest rate of 10% per year?

5.33 Compare the alternatives shown below on the basis of their capitalized casts, using an interest rate 12% per year, compounded quarterly?

	Alternative E	Alternative F	Alternative G
First Cost $	200,000	-300,000	-900,000
Quarterly income S/quarter	30,000	10,000	40,000
Salvage Value, $	50,000	70,000	100,000
Life, Years	2	4

 

Payback Analysis

5.34 What is meant by no-return payback or simple payback?

PROBLEMS 207

5.35 Explain why the alternative that recovers its initial investment at a specified rate of return in the shortest time is not necessary the most economically attractive one.

5.36 Determine the payback period for an asset that has a first cost of $40.000. a salvage value of 58(YJ0 anytime within 10 years of its purchase, and generates income of $6000 per year. The required return is 8% per year.

5.37 Accusoft Systems is offering business owners a software package that keeps track of many accounting functions front the company's bank transactions sales in voices. The site license will met $22,000 to install Rad will involve a quarterly fee of $2000. If a certain small company can save $3500 every quarter and have the security of managing its books in hose, how long will it take for the company to recover its investment at an interest rate of 4% per quarter?

5.38 Damell Enterprises constructed in addition to its building as a cast of $70,000. extra annual expenses are expected to by $18.50, but extra income will be $14.000 per year. How long will it take for the company to recover its investment at an interest rate of 10% per year?

5.39 A new process for manufacturing laser levels will have a first cost of $35,000 with annual cost of $ 17,000. Extra income associated with the new process is expected to be $22,000 per year. What is the pay-back period at (a) i=0% and (b)i=10% per year?

5.40 A multinational engineering consulting firm that wants to provide resort accommodations to certain clients is considering the purchase of a three bedroom lodge in upper Montana that will cost $250,000. The property in that area is rapidly appreciating in value because people anxious to get away from urban developments are bidding up the prices, if the company spends an average of $500 per month for utilities and the investment increases at a rate of 2% per month, how long would it be before the company could sell the property for $100,000 more than it has invested in it?

 5.41 A window frame manufacturer is searching for ways to improve it’s triple-insulated sliding windows, sold primarily in the far northern areas of the United States. Alternative A is an increase in TV and radio marketing. A total of $300,000 spent now is expected w increase revenue by $60,000 per year. Alternative requires the same investment for enhancements to the in-plant manufacturing process that will improve the temperature retention properties of the seals around each glass pane. New revenues start slowly for Mix alternative at an estimated $10,000 the first year, with growth of $15,000 per year as the improved product gains reputation among builders. The MARR is 8% per year and the maximum evaluation period is 10 year for either alternative. Use both payback analysis and present worth analysis at 8% (for 10 years) to select the more economical alternative state the reasons for any difference in the alternative chosen between the two analyses.

Life-Cycle Costs

5.42 A high technology defense contractor has been asked by the pentagon to estimate the life cycle cost (LCC) for a proposed light duty support vehicle. Its list of items included the following general categories. R&D costs (R&D), nonrecurring investment (NRI) costs, recurring investment (RI) costs, scheduled and un-scheduled maintenance costs (Maint) equipment usages costs (Equip) and disposal cost (Disp). The cost (in millions) for the 20 year life cycle are as indicated. Calculate the LCC at an interest rate of 7% per year.

Year	R&D	NRI	RI	Maint	Equip	Disp
0	5.5	1.1
1	3.5
2	2.5
3	0.5	5.2	1.3	0.6	1.5
4		10.5	3.1	1.4	3.0
5		10.5	4.2	1.6	5.3
6-10			6.5	2.7	7.8
11 on			2.2	3.5	8.5
18-20						2.7

 

5.43 A manufacturing software engineer at a major aerospace corporation has been as-signed the management responsibility of a project to design, build, test and implement AREMSS. a new-generation auto-mated scheduling system for routine Mid expedited maintenance. Reports on the disposition of each service will also be entered by field personnel, then filed and archived by the system. The initial application will be on existing Air Force in-flight refueling aircraft. The system is expected to be widely used over time for other aircraft maintenance scheduling. Out it is fully implemented, enhancement will have to be made, but the system is expected to serve as a worldwide scheduler for up to 15,000 separate aircraft. The engineer, who must make a presentation next week of the best estimates of costs over a 20-year Ills period, has decided to use the life-cycle cost approach of cost estimations. Use the following in formation to determine the current LCC at 6% per year for the AREMSS scheduling system.

Cost Category	1	2	3	4	5	6	10	18
Field Study	0.5
Design of System	2.1	1.2	0.5
Software design		0.6	0.9
Hardware Purchase			5.1
Beta Testing		0.1	0.2
User’s manual development		0.1	0.1	0.2	0.2	0.06
system implementation				1.3	0.7
Field Hardware		0.4	6.0	2.9
Training trainees			0.3	2.5	2.5	0.7
Software upgrades						0.6	3.0	3.7

 

5.44 The U.S. Army received two proposals for a turnkey design-build project for barracks for infantry unit soldiers in training. Proposal A involves an off the shelf barebones design and standard-grade construction of walls, windows, doors and other Features. With this option heating and cooling will he greater, maintenance costs will be higher and replacement will be sooner than for proposal B. The initial cost for A will he $750.000. Heating and cooling costs will average $6000 per month with maintenance costs averaging $2000 per month. Minor remodeling will be required in years 5, 10, and 15 at a cost of $150.000 each time in order to render the units usable tor 20 years. They will have no salvage value.

Proposal B will include tailored design and construction cost of $1.1 million initially with estimated heating and cooling cost of $3000 per month and maintenance cost of $1000 per months. There will be no salvage value at the end or the 20-year life.

Which proposal should he accepted on the basis of a life-cycle cost analysis. If the interest rate is 0.5% per month?

5.45 A medium-size municipality plans to develop a software system to assist in project selection during the next 10 years. A life cycle cast approach has been used to categorize cash into development, programming, operating, and support cost of each alternative under consideration, identified as a Oai-lured system). B (adapted system) and C (current system. The cost are summarized below. Use a life-cycle cost approach to identify the best alternative at 8% per year.

Alternative	Cost Component	Cost
A	Development	$250,000 now $150,000 years 1 through 4
	Programming	$45,000 now $35,000 years 1,2
	Operation	$50,000 years 1 through 10
B	Development Programming	$10,000 now &45,000 year 0
	Operation	$80,000 years 1 through to
	Support	$40,000 years 1 through 10
C	Operation	$175,000 years 1 through 10

 

Bonds

5.46 A mortgage bond with a face value or $10,000) has n bond interest rate of 6% per year payable quarterly. What are the amount and frequency of the interest payments?

5.47 What is the face value of a municipal bond that has a bond interest rate or 4% per year with semiannual interest payments of $800?

 

5.48 What is the bond interest rate on a $20.000 bond that has semiannual interest payments of $1500 and a 20-year maturity date?

5.49 What is the present worth of a $50,000 bond that has interest of 10% per year, payable quarterly? The bond matures in 20 years. The interest rate in the marketplace is 10%, per year. Compounded quarterly.

5.50 What is the present worth of a $50,000 municipal bond that has an interest rate of 4% per year? Payable quarterly? The bond matures in 15 years and the market interest rate is 8% per year, compounded quarterly.

5.51 General Electric issued 1000 debenture bonds 3 years ago with a face value of $5000 each and a bond interest rate of 8% per year payable semiannually, the bonds have a maturity date of 20 years from the date they were issued. If the interest rate in the market place is 10% per year. Compounded semiannually, what is the present worth of one bond to an investor who wishes to purchase it today?

552 Charleston Independent School District needs to raise $200 million to refurbish

FE REVIEW PROBLEMS

5.54 For the mutually exclusive alternative’s shown below, determine which one should be selected..

 

 

Alternative	Present Worth, $
A	-25,000
B	-12,000
C	10,000
D	15,000

 

its existing schools and build new ones. The bonds will pay interest semiannually at a rate of 7% per year and they will mature in 30 year. Brokerage fees associated with the sale of the bonds will be $1 million. If the interest rate in the marketplace rises to 8% per year, Compounded semiannually, before the bonds are issued. What will the face value of the bonds have to be for the school district to net $200 million?

5.53 An engineer planning for his retirement thinks that the interest rates in the market place will decrease before he retires. Therefore, he plans to invest in corporate bonds. He plans to buy a $50.000 bond that has a bond interest rate of 12% per year, payable quarterly with a maturity date 21 years, from now.

(a) How much should he be able to sell the bond for in 5 years if the market interest rate is 8% per year. Compounded quarterly?

(b) If he invested the interest he received at an interest rate of 12% per year. Compounded quarterly, how much will he have (total) immediately after he sells the bond 5 years from now?

(a) Only A

(B) Only D

(C) Only A and B

(D) Only C and D

5.55 The present worth of $50,000 now. $10,000 per year in years 1 through 15, and $20,000 per year 16 through infinity at 10% per year is closed to

(a) Less than $169,000

(b) $169,580

(c)  $173,940

(d) $195,730

5.56 A certain donor wishes to start an endowment at her alma matter that will provide scholarship money of $40,000 per year beginning in year 5 and continuing indefinitely. If the university earns 10% per year on the endowment, the amount she must donate now is closest to

 (a) $225,470

 (b) $248,360

(c)  $273,200

(d) $293,820

5.57 At an interest rate of 10% per year, the amount must deposit in your retirement account each year in years 0through 9 (i.e...10 deposits) if you want to withdraw $50,000 per year forever beginning 30 years from now is closest to

(a) $4239

(b) $4662

(c) $4974

(d) $5471

Problems 5.58 through 5.60 are based on the following estimates. The cost of money is 10% per year.

	Machine X	Machine Y
Initial Cost, $	-66,000	46,000
Annual Cost, $ per Year	10,000	15,000
Salvage Value, $	10,000	24,000
Life, Years	6	3

 

5.58 The present worth of machine X is closed to

(a) $65,270

(b) $87,840

(c) $103,910

(d) $-114,340

5.59 In comparing the machines on a present worth basis, the present worth of machine Y is closest to

(a) $- 65.270 (b) $-97.840 (c) $-103.910 (4) $-114.310

5.60 The capitalized cost of machine X is closest to

(a) $ —103.910

(b) $-114,310

(c)   $-235.990

 (d) $-233.580

5.61 The cost of maintaining a public monument in Washington D.C. Occurs as periodic outlays of $10.000 every 5 years. It the first outlay is now, the capitalized cost of the maintenance at an interest rate of I QC per year is closest to

(a) $-16.380 (b) $-26.380 (c) $-29.360 (d) S-41.050

5.62 The alternatives shown below are to be compared on the basis of their capitalized costs. At an interest rate of 10% per year. Compounded continuously the equation that represents the capitalized cost of alternative A is

	Alternative A	Alternative B
First Cost, $	-50,000	90,000
Annual cost, $/Year	-10,000	-4,000
Salvage value, $	13,000	15,000
Life, Years	3	6

 

1. PW_a =-50,000     10,000p/a

10.52%-37,000 p/F. 10.52%,3+13,000(p/f, 10.52 %,6)

 

2. PW_a =-50,000-10,000 p/A

(10.52%,3+13,000p/F10.52%,3)

(c) PW_a = 1-50,000(A/P,10.52%,3)-10000+13,000(A/F,10.52%,3)1/0.1052

(D) PW_a=1-50,000(A/P,10%,3)-10,000+13,000A/F10%,3)1/0.10

5.63 A corporate bond has a face value of $10,000 a bond interest rate of 6% per year payable semiannually and a maturity date of 20 years from now. If a person purchases the bond for $9000 when the interest rate in the marketplace is 8% per year. Compounded semiannually the size and frequency of the interest payments the person will receive are closest to

(a) $270 every 6 months

(b) $300 every 6 months

(c) $360 every 6 months

(d) $ $400 every 6 months

5.64 A municipal bond Mot was issued 3 years ago has a face value of $5000 and a bond interest rate of 4% per year payable semi-annually. The bond has a maturity date of 20 years from the date it was issued. If the interest rate in the marketplace is 8% per year, compounded quarterly the value of n that must be used in the PIA equation to calculate the present worth of the bond is

(A) 34

(B) 40

(C) 68

(D) 80

5.65 A $10,000 bond has an interest rate of 6% per year payable quarterly. The bond matures 15 years from now. At an interest rate of 8% per year, compounded quarterly, the present worth of the bond is represented by which of the equations below

 (a) PW = 150(P/A , 1.5%,60)+10,000

  (p/f,1.5%,60)

(b) PW = 150(P/A , 2%,60)+10,000

                (p/f,2%,60)

 

(c) PW = 600(P/A , 8%,15)+10,000

                (p/f,8%,15)

(d) PW = 600(P/A , 2%,60)+10,000

             p/f,2%,15)

           

EXTENDED EXERCISE

EVALUATION OF SOCIAL SECURITY RETIREMENT ESTIMATES

 Charles is a senior engineer who has worked for18 years since he graduated from college. Yesterday in the mail he received a report from the U.S. social Security Administration. In short, it stated that if he continues to earn at the same rate social security will provide him with the following estimated monthly retirement benefits:

• Normal retirement at age 66: full benefit of $1500 per month starting at age 66.

• Early retirement at age 62: benefit reduced by 25% starting at age 62.

• Extended retirement at age 70, benefit increased by 30% starting an age 70.

Charles never thought much about social security: he usually thought of it as a monthly deduction from his payback that helped pay for his parent’s retirement benefits from social security. But this time he decided an analysis should be per-formed. Charles decided to neglect the effect of the following over time income taxes, cost-of-living increases, and inflation. Also, he assumed the retirement benefits are all received at the end of each year, that is, no compounding effect occurs during the year. Using an expected rate of return on investments of 8% per year and an anticipated death just after his 85th birthday, use a spreadsheet to do the following for Charles

1. Calculate the total future worth of each benefit scenario through the age of 85.

2. Plot the annual accumulated future worth for each benefit scenario through the age of 85.

The report also mentioned that if Charles dies this year his spouse is eligible at full retirement age for a benefit of $1600 per month for the remainder of her life. If Charles and his wife are both 40 years old today, determine the following about his wiles survivor benefits, if she starts at age 66 and lives through her 85 birthday

3. Present worth now.

4. Future worth for his wife alter her 85th birthday,

CASE STUDY

PAYBACK EVALUATION OF ULTRALOW-FLUSH TOILET PROGRAM

Introduction

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PM BACK EVALUATION OF ULIRALOW-FLUSH TOILET PROGRAM

Introduction

 In many cities in the southwestern part of the United States, water is being withdrawn from subsurface aquifers faster than it is being replaced. The Attendant depletion of groundwater supplies has forced some of of these cities to take action ranging from restrictive pricing policies to mandatory conservation measures in residential, commercial, and industrial establishments. Beginning in the mid-1990s a city undertook a project encourage installation of ultralow-Flush toilets in existing louses. To evaluate the cost-effectiveness or the pregame, an economic analysis was conducted.

Background

The heart of the toilet replacement program involved a rebate of 75% of the cost of the fixture (up to $100 per. Unit) providing the toilet used to no more than 1.6 gallons of water per flush. There was no limit on the number of toilets any individual or business could have replaced.

Procedure

To evaluate the water saving achieved (if any) through the program, monthly water use second were searched for 325 of the household participants, representing a sample size of approximately 13%, water consumption data were obtained for 12 months before and 12 months after installation of the ultralow flush toilets. If the house changed ownership during the evolution, since water consumption increases dramatically during the hot summer months for lawn watering, evaporative cooling, ear washing. etc... only the winter months of December, January, and February were used to evaluate water consumption before and after installation of the toilet. Before any calculations were made high-volume water users (usually Business) were screened out by eliminating all records whose average monthly consumption ex-ceeded 50 CCF (1 CCF =100 cubic feet =N 748 gallons). Additionally accounts which had monthly averages of z CCF or less (either before or after installation were also eliminated because it was believed that such low consumption rates probably rep-resented an abnormal condition. such as a house for sale which was vacant during part of the study period. The 268 records that remained after the screening procedures were then used to quantify the effectiveness of the program.

Results

Water Consumption

Monthly consumption before and after installation of the ultralow-flush toilets was founded to be 11.2 and 9.1 CCF, respectively for an average reduction of 18.8%. When only the months of January and February were used in the before and after calculation, the respective value Were 11.0 and 8.7 CCF, resulting in a water saving rate 20.09%

Economic Analysis

The following table shows some of the program totals through the first 1½ years of program.

Program Summary

Number of household participating	2466
Number of toilets replaced	4096
Number of Persons	7981
Average cost of toilet	$115.83
Average rebate	$76.12

 

The result in the previous section indicated monthly water saving of 2.1 CCF. For the average program

Participant, the payback period n, in years with no interest considered is calculated using Equation {5.7}

N= net cost of toilets + installation cost/ net annual saving for water and sewer charges

The lowest rate block or water charges is $0.76 per CCF. The sewer surcharge is $0.62 per CCF. Using these values and a $50 cost for installation the payback period is

 (115.83 — 76.12) + 50/ (2.1 CCF/month x 12 months) x (0.76 +0.62)/CCF   =2.6 Years

 Less expensive toilets of lower installation costs would reduce the payback period accordingly. while consideration of the time value of money would lengthen it.

From the standpoint of the utility which supplies water, the cost or the program must be compared against the original cost of water delivery and wastewater treatment. The marginal cost c may be represented as c = Cost of rebates/ volume of water not delivered + volume of wastewater not treated

Theoretically, the reduction in water consumption would go on far an infinite period of time, since replacement will never be with a less efficient model. But for a worst-case condition, it is assumed the toilet would have a "productive" life of only 5 years after which it would leak and not to be repaired. The cost to the city for the water not delivered or wastewater not treated would he

C= $76.12/(2.1+2.1 CCF/month)(12 months)(5years)  = $0.302/CCF or $0.40/ 1000gallons

Thus unless the city can deliver water ad treat the resulting wastewater for less than $0.40 per 1000 gallons, the toilets replacement program would be considered economically attractive. For the city, the operating costs alone, that is, without the capital expenses, for water and wastewater services that were not expended were about $1.10 per 1000 gallons, which far exceeds $0.40 per 1000 gallons. Therefore, the toilets replacement program was clearly very cost-effective.

Case Study Exercises

1. For an interest raw of 8% and a toilet life of 5 years, what would the participant's payback period be? 2. Is the participant's payback period more sensitive to the interest rate used or to the life of the toilet? 3. What would the cost to the city he if an interest rate of 6% per year were used with a toilet life of 5 years? Compare the cost in $/CCF and $/1000 gallons to those determined at 0% interest

From the city's standpoint, is the success of the program sensitive to (a) the percentage of toilet cost rebated (b) the interest rate, if rates or 4% to 15% are used, or (c) the toilet life, if lives of 2 to 20 years are used?

What other factors might be important to (a) the participants and (b) the city in evaluating whether the program is a success?