Deck 5: Present Worth Analysis
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Deck 5: Present Worth Analysis
1
Jim, the owner of Computers-on-Wheels, an on-site repair service, just purchased a $0,000 diagnostics system and software for PCs, laptops, and LANs.During the expected 5-year life, he determines that the annual recovery amount for this investment is $0,000/5 = $000.( a ) Explain why Jim has mistakenly underestimated the annual increase needed in revenue.( b ) Determine the required annual revenue increase for a 15% per year return.
The formula for calculating annual payment is as follows:
Here,
P is present value,
n is number of years, and
i is annual rate of interest.
The standard notation equation is
where the value of the factor is seen in the compound interest factor table.
(a)Here, capital of $20,000 is invested by J, and he expects to recover back the investment in 5 years period.J has simply divided the amount by number of years and got the annual revenue amount to recover investment.But it is wrong because J has not considered the implications of interest rate i.e.opportunity cost of the interest rate that he is losing on the amount.
(b)Here,
P is $20,000,
n is 5 years, and
i is 15% annual rate of interest.
Calculate the annual worth, as follows:
Calculate the additional amount needed in annual revenue calculated by J, as follows:
Thus, the additional amount needed in annual revenue calculated by J is
.

P is present value,
n is number of years, and
i is annual rate of interest.
The standard notation equation is

(a)Here, capital of $20,000 is invested by J, and he expects to recover back the investment in 5 years period.J has simply divided the amount by number of years and got the annual revenue amount to recover investment.But it is wrong because J has not considered the implications of interest rate i.e.opportunity cost of the interest rate that he is losing on the amount.
(b)Here,
P is $20,000,
n is 5 years, and
i is 15% annual rate of interest.
Calculate the annual worth, as follows:



2
A 600-ton press used to produce composite- material fuel cell components for automobiles using proton exchange membrane (PEM) technology can reduce the weight of enclosure parts up to 75%.At MARR = 12% per year, calculate ( a ) capital recovery and ( b ) annual revenue required.
Installed cost = $? 3.8 million n = 12 years Salvage value = $50,000 Annual operating costs = $50,000 to start increasing by $5,000 per year
Installed cost = $? 3.8 million n = 12 years Salvage value = $50,000 Annual operating costs = $50,000 to start increasing by $5,000 per year
The following information is given:
a.
Calculate capital recovery ( CR ) as follows:
The capital recovery is -$603,112.
b.
The annual revenue required is equal to annual worth of the project, therefore is calculated as follows:
The annual revenue required is -$1,057,854

Calculate capital recovery ( CR ) as follows:

b.
The annual revenue required is equal to annual worth of the project, therefore is calculated as follows:

3
Five years ago, Diamond Electronics, a division of De Beers, paid $,150,000 for new diamond cutting tools.At that time, the company estimated an added revenue need of $00,000 to recover the investment at 10% per year.If there is an estimated 8 more years of service with a salvage value of $00,000, compare the revenue needed over the entire life with that estimated 5 years ago.
The formula for calculating annual payment is as follows:
Here,
P is present value,
n is number of years, and
i is annual rate of interest.
The standard notation equation is
where the value of the factor is seen in the compound interest factor table.
Here,
P is -$3,150,000,
S is $300,000
n is 13 years, and
i is 10% annual rate of interest.
Calculate the annual worth, as follows:
Calculate the difference in annual revenue amount calculated before, as follows:
Thus, the difference in annual revenue amount calculated before is
.

P is present value,
n is number of years, and
i is annual rate of interest.
The standard notation equation is

Here,
P is -$3,150,000,
S is $300,000
n is 13 years, and
i is 10% annual rate of interest.
Calculate the annual worth, as follows:



4
Brent owns Beck Trucking.Seven years ago, he purchased a large-capacity dump truck for $15,000 to provide short-haul earth moving services.He sold it today for $5,000.Operating and maintenance costs averaged $500 per year.A complete overhaul at the end of year 4 cost an extra $200.( a ) Calculate the annual cost of the truck at 7% per year.( b ) If Brent estimates that Beck cleared at most $0,000 per year added revenue from using the truck, was the purchase economically advantageous?
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5
You have the estimates for two alternatives, A and B, from which to select the cheaper one using an AW evaluation.Your boss selects a study period of 5 years, but, A has an expected life of 3 years and B has a 10-year life.Concerning the estimates, what is necessary to correctly perform the AW calculations over the 5-year period?
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6
BP Oil is in the process of replacing sections of its Prudhoe Bay, Alaska oil transit pipeline.This will reduce corrosion problems, while allowing higher line pressures and flow rates to downstream processing facilities.The installed cost is expected to be about $70 million.Alaska imposes a 22.5% tax on annual profits (net revenue over costs), which are estimated to average $5 million per year for a 20 year period.( a ) At a corporate MARR of 10% per year, does the project AW indicate it will make at least the MARR? ( b ) Recalculate the AW at MARR values increasing by 10% per year, that is, 20%, 30%, etc.At what required return does the project become financially unacceptable?
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7
Equipment needed at a Valero Corporation refinery for the conversion of corn stock to ethanol, a cleaner- burning gasoline additive, will cost $75,000 and have net cash flows of $5,000 the first year, increasing by $0,000 per year over the life of 5 years.( a ) Determine the AW amounts at different MARR values to determine when the project switches from financially justified to unjustified.(Hint: Start at 12% per year and increase by 1% per year.) ( b ) If a spreadsheet is used, plot a graph of AW versus interest rate on the spreadsheet.
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8
Your office mate performed an 8% per year PW evaluation of two alternatives before he left on vacation.Now, your supervisor wants you to provide the results in AW terms.The PW results are:
PW A = $517,510 for a life of 4 years
PW B = $- 812,100 for a life of 8 years
PW A = $517,510 for a life of 4 years
PW B = $- 812,100 for a life of 8 years
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9
The TT Racing and Performance Motor Corporation wishes to evaluate two alternative machines for NASCAR motor tuneups.Use the AW method at 9% per year to select the better alternative.


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10
Rework Problem 5.9 if there is a major overhaul cost every 2 years that was not included in the original estimates.The cost is $0,000 for R and $5,000 for S.
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11
The Haber Process, developed in the early 1900s, is commonly used to produce ammonia for industrial and consumer applications.Higher yields take place as temperature and pressure (in atmospheres or atm) are varied.New pressure chamber equipment is necessary at KGC Industries in India.Three alternatives are available, the third being contracting all services.Use the annual worth method for a study period of 4 years and i = 6% per year to select the lowest annual cost alternative.There are no positive salvage values expected, but the market values are expected to decrease by 20% of the first cost for each year of use.


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12
Holly Farms is considering two environmental chambers to accomplish detailed laboratory confirmations of on-line bacteria tests in chicken meat for the presence of E.coli 0157:H7 and Listeria monocytogenes.
a.If this project will last for 6 years and i = 10% per year, perform an AW method evaluation todetermine which chamber is more economical.
b.Chamber D103 can be purchased with different options and, therefore, at different installedcosts.They range from $00,000 to $00,000.Will the selection change if one of these othermodels is installed?

a.If this project will last for 6 years and i = 10% per year, perform an AW method evaluation todetermine which chamber is more economical.
b.Chamber D103 can be purchased with different options and, therefore, at different installedcosts.They range from $00,000 to $00,000.Will the selection change if one of these othermodels is installed?
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13
Blue Whale Moving and Storage recently purchased a warehouse building in Santiago.The manager has two good options for moving pallets of stored goods in and around the facility.Alternative 1 includes a 4000-pound capacity, electric forklift ( P = $30,000; n = 12 years; AOC = $?1000 per year; S = $000) and 500 new pallets at $0 each.The forklift operator's annual salary and indirect benefits are estimated at $2,000.
Alternative 2 involves the use of two electric pallet movers ("walkies") each with a 3000-pound capacity (for each mover, P = $2,000; n = 4 years; AOC = $ 150 per year; no salvage) and 800 pallets at $0 each.The two operators' salaries and benefits will total $5,000 per year.For both options, new pallets are purchased now and every two years that the equipment is in use.If the MARR is 8% per year, which alternative is better?
Alternative 2 involves the use of two electric pallet movers ("walkies") each with a 3000-pound capacity (for each mover, P = $2,000; n = 4 years; AOC = $ 150 per year; no salvage) and 800 pallets at $0 each.The two operators' salaries and benefits will total $5,000 per year.For both options, new pallets are purchased now and every two years that the equipment is in use.If the MARR is 8% per year, which alternative is better?
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14
Work Example 4.2 in Chapter 4 using the annual worth method and compare your conclusion with that reached using the PW method.
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15
McLaughlin Services is on contract to Tuscany County, Florida.One of the McLaughlin engineers is evaluating alternatives that use a robotic, liquid- propelled "pig" to periodically inspect the interior of buried potable water pipes for leakage, corrosion, weld strength, movement over time, and a variety of other parameters.Two equivalent robot instruments are available.
Robot Joeboy: P = $85,000;
annual M O costs = $30,000; S =$0,000
Robot Watcheye: P =$97,000;
annual M O costs = $27,000; S =$2,000
a.Select one using an AW comparison, i =10% per year, and a 3-year study period.
b.If the first cost of Watcheye is somewhat negotiable, by how many dollars must it increase or decrease to make the two instruments equally attractive financially? Use the 3-year study period.
Robot Joeboy: P = $85,000;
annual M O costs = $30,000; S =$0,000
Robot Watcheye: P =$97,000;
annual M O costs = $27,000; S =$2,000
a.Select one using an AW comparison, i =10% per year, and a 3-year study period.
b.If the first cost of Watcheye is somewhat negotiable, by how many dollars must it increase or decrease to make the two instruments equally attractive financially? Use the 3-year study period.
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16
Cheryl and Gunther wish to place into a retirement fund an equal amount each year for 20 consecutive years to accumulate just enough to withdraw $4,000 per year starting exactly one year after the last deposit is made.The fund has a reliable return of 8% per year.Determine the annual deposit for two withdrawal plans: ( a ) forever (years 21 to infinity); ( b ) 30 years (years 21 through 50).
( c ) How much less per year is needed when the withdrawal horizon decreases from infinity to 30 years?
( c ) How much less per year is needed when the withdrawal horizon decreases from infinity to 30 years?
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17
Baker|Trimline owned a specialized tools company for a total of 12 years when it was sold for $8 million cash.During the ownership, annual net cash flow (revenues over all costs) varied significantly.
The company made 12% per year on its positive cash flows and paid the same rate on short-term loans to cover the lean years.The president wants to use the cash accumulated after 12 years to improve capital investments starting in year 13 and forward.If an 8% per year return is expected after the sale, what annual amount can Baker|Trimline invest forever?

The company made 12% per year on its positive cash flows and paid the same rate on short-term loans to cover the lean years.The president wants to use the cash accumulated after 12 years to improve capital investments starting in year 13 and forward.If an 8% per year return is expected after the sale, what annual amount can Baker|Trimline invest forever?
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18
A major repair on the suspension system of Jane's 3-year old car cost her $,000 because the warranty expired after 2 years of ownership.Based on this experience, she will plan on additional $000 expenses every 3 years henceforth.Also, she spends $00 every 2 years for maintenance now that the warranty is over.This is for years 2, 4, 6, 8, and 10 when she plans to donate the car to charity.Use these costs to determine Jane's equivalent perpetual annual cost (for years 1 through infinity) at i = 5% per year, if cars she owns in the future have the same cost pattern.
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19
A West Virginia coal mining operation has installed an in-shaft monitoring system for oxygen tank and gear readiness for emergencies.Based on maintenance patterns for previous systems, costs are minimal for the first few years, increase for a time period, and then level off.Maintenance costs are expected to be $50,000 in year 3, $75,000 in year 4, and amounts increasing by $5,000 per year through year 6 and remain constant thereafter for the expected 10-year life of this system.If similar systems will replace the current one, determine the perpetual equivalent annual maintenance cost at i = 10% per year.
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20
Harmony Auto Group sells and services imported and domestic cars.The owner is considering outsourcing all of its new auto warranty service work to Winslow, Inc.a private repair service that works on any make and year car.Either a 5-year contract basis or 10-year license agreement are available from Winslow.Revenue from the manufacturer will be shared with no added cost incurred by the car/warranty owner.Alternatively, Harmony can continue to do warranty work in-house.Use the estimates made by the Harmony owner to perform an annual worth evaluation at 12% per year to select the best option.All dollar values are in millions.


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21
ABC Drinks purchases its 355 ml cans in large bulk from Wald-China Can Corporation.The finish on the anodized aluminum surface is produced by mechanical finishing technology called brushing or bead blasting.Engineers at Wald are switching to more efficient, faster, and cheaper machines to supply ABC.Use the estimates and MARR = 8% per year to select between two alternatives.
Brush alternative: P = $400,000; n = 10 years; S = $0,000; nonlabor AOC = $50,000 in year 1, decreasing by $000 annually starting in year 2.
Bead blasting alternative: P = $300,000; n is large, assume permanent; no salvage; nonlabor AOC = $50,000 per year.
Brush alternative: P = $400,000; n = 10 years; S = $0,000; nonlabor AOC = $50,000 in year 1, decreasing by $000 annually starting in year 2.
Bead blasting alternative: P = $300,000; n is large, assume permanent; no salvage; nonlabor AOC = $50,000 per year.
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22
You are an engineer with Yorkshire Shipping in Singapore.Your boss, Zul, asks you to recommend one of two methods to reduce or eliminate rodent damage to silo-stored grain as it awaits shipment.Perform an AW analysis at 10% per year compounded quarterly.Dollar values are in millions.


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23
The least common multiple (LCM) of lives need not be used when evaluating two alternatives by the annual worth method because, if inflation and deflation effects are neglected,
A)the assumptions for annual worth analysis are different from those for the present worth method.
B)cost and revenue estimates never remain the same over more than one life cycle.
C)the annual worth value used to evaluate the alternatives is determined over a large number of life cycles.
D)the annual worth over one life cycle is assumed to be the same for all succeeding life cycles.
A)the assumptions for annual worth analysis are different from those for the present worth method.
B)cost and revenue estimates never remain the same over more than one life cycle.
C)the annual worth value used to evaluate the alternatives is determined over a large number of life cycles.
D)the annual worth over one life cycle is assumed to be the same for all succeeding life cycles.
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24
An automation asset with a high first cost of $0 million has a capital recovery (CR) of $,985,000 per year.The correct interpretation of this CR value is that
a.the owner must pay an additional $,985,000 each year to retain the asset.
b.each year of its expected life, a net revenue of $,985,000 must be realized to recover the $0 million first cost and the required rate of return on this investment.
c.each year of its expected life, a net revenue of $,985,000 must be realized to recover the $0 million first cost.
d.the services provided by the asset will stop if less than $,985,000 in net revenue is reported in any year.
Problems 5.25 through 5.27 refer to the following estimates for three mutually exclusive alternatives.The MARR is 6% per year.

a.the owner must pay an additional $,985,000 each year to retain the asset.
b.each year of its expected life, a net revenue of $,985,000 must be realized to recover the $0 million first cost and the required rate of return on this investment.
c.each year of its expected life, a net revenue of $,985,000 must be realized to recover the $0 million first cost.
d.the services provided by the asset will stop if less than $,985,000 in net revenue is reported in any year.
Problems 5.25 through 5.27 refer to the following estimates for three mutually exclusive alternatives.The MARR is 6% per year.

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25
The annual worth of vendor 2 cash flow estimates is closest to
a.$7,000.
b.$3,370.
c.$?43,370.
d.$3,370.
a.$7,000.
b.$3,370.
c.$?43,370.
d.$3,370.
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26
Of the following three relations, the correct one(s) to calculate the annual worth of vendor 1 cash flow estimates is (note: All dollar values are in thousands)
Relation 1: AW 1 = ?200( A / P ,6%,10) + 70 +25( A / F ,6%,10)
Relation 2: AW 1 = [?200 ? 50( P / A ,6%,10) + 120( P / A ,6%,10) + 25( P / F ,6%,10)]( A / P ,6%,10)
Relation 3: AW 1 = ?200( F / P ,6%,10) + 25 + (?50 + 120)( A / P ,6%,10).
a.1 and 3.
b.1.
c.1 and 2.
d.3.
Relation 1: AW 1 = ?200( A / P ,6%,10) + 70 +25( A / F ,6%,10)
Relation 2: AW 1 = [?200 ? 50( P / A ,6%,10) + 120( P / A ,6%,10) + 25( P / F ,6%,10)]( A / P ,6%,10)
Relation 3: AW 1 = ?200( F / P ,6%,10) + 25 + (?50 + 120)( A / P ,6%,10).
a.1 and 3.
b.1.
c.1 and 2.
d.3.
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27
The AW values for the alternatives are as listed.The vendor(s) that should be recommended are
AW 1 = $4,723
AW 2 = $3,370
AW 3 = $0,000
a.1 and 2.
b.3.
c.2.
d.1.
AW 1 = $4,723
AW 2 = $3,370
AW 3 = $0,000
a.1 and 2.
b.3.
c.2.
d.1.
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28
If a revenue alternative has a negative AW value and it was correctly calculated, it means one or more of the following:
1.? The equivalent annual worth of revenues does not exceed that of the costs.
2.? The estimates are wrong somewhere.
3.? A minus or plus sign of a cash flow was entered incorrectly into the Excel PMT function.
4.? The alternative should have a longer life so revenues will exceed costs.
a.1
b.2
c.3
d.4
1.? The equivalent annual worth of revenues does not exceed that of the costs.
2.? The estimates are wrong somewhere.
3.? A minus or plus sign of a cash flow was entered incorrectly into the Excel PMT function.
4.? The alternative should have a longer life so revenues will exceed costs.
a.1
b.2
c.3
d.4
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29
Estimates for one of two process upgrades are: first cost of $0,000; annual costs of $000 per year; market value that decreases 5% per year to the salvage value of $0,000 after the expected life of 10 years.If a 4-year study period is used for AW
analysis at 15% per year, the correct AW value is closest to
a.$15,000.
b.$11,985.
c.$7,600.
d.$12,600.
analysis at 15% per year, the correct AW value is closest to
a.$15,000.
b.$11,985.
c.$7,600.
d.$12,600.
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30
The perpetual annual worth of investing $0,000 now and $0,000 per year starting in year 16 and continuing forever at 12% per year is closest to
a.$200.
b.$650.
c.$655.
d.$0,655.
a.$200.
b.$650.
c.$655.
d.$0,655.
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31
You graduated with an MS degree in engineering and have a great job.Next month you start an automatic draft from your paycheck of $000 each month, placing it into a mutual fund retirement account.You plan to continue at this rate for 20 years then stop making deposits.Assume a return of 6% per year compounded monthly is a good expectation for all years.If you retire 10 years after the last deposit and the first retirement benefit is withdrawn at the end of the first month of year 31, the equivalent monthly worth of each retirement payment to you and your heirs forever is closest to
a.$290 per month.
b.$820 per month.
c.$200 per month.
d.$180 per month.
a.$290 per month.
b.$820 per month.
c.$200 per month.
d.$180 per month.
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32
If the present worth PW of a perpetual investment and the interest rate are known, the AW value is determined using the relation(s)
Relation 1: AW = PW divided by i
Relation 2: AW = PW times i
Relation 3: AW = PW times the ( A / P ,i%,?) factor
A)1 only.
B)2 and 3.
C)2 only.
D)1 and 3.
Relation 1: AW = PW divided by i
Relation 2: AW = PW times i
Relation 3: AW = PW times the ( A / P ,i%,?) factor
A)1 only.
B)2 and 3.
C)2 only.
D)1 and 3.
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