Problem 4.1
State two conditions under which the do-nothing alternative is not an option.
Problem 4.2
When evaluating projects by the present worth method, how do you know which one(s)
to select, if the (a) projects are independent, and (b) alternatives are mutually exclusive?
Problem 4.3
A biomedical engineer with Johnston Implants just received estimates for replacement
equipment to deliver online selected diagnostic results to doctors performing surgery who
need immediate information on the patient’s condition. The cost is $200,000, the annual
maintenance contract costs $5000, and the useful life (technologically) is 5 years.
a. What is the alternative if this equipment is not selected? What other information is
necessary to perform an economic evaluation of the two?
b. What type of cash flow series will these estimates form?
c. What additional information is needed to convert the cash flow estimates to the other
type of series?
Problem 4.4
The lead engineer at Bell Aerospace has $1 million in research funds to commit this year.
She is considering five separate R&D projects, identified as A through E. Upon
examination, she determines that three of these projects (A, B, and C) accomplish exactly
the same objective using different techniques.
a. Identify each project as mutually exclusive or independent.
b. If the selected alternative between A, B, and C is labeled X, list all viable options
(bundles) for the five projects.
Problem 4.5
List all possible bundles for the four independent projects 1, 2, 3, and 4. Projects 3 and 4
cannot both be included in the same bundle.
Problem 4.6
What is meant by the term equal service?
Problem 4.7
An irrigation return flow drain has sampling equipment that can be powered by solar cells
or by running an electric line to the site and using conventional power. Solar cells will
cost $14,000 to install with a useful life of 10 years. Annual costs for inspection,
cleaning, etc. are expected to be $1500. A new power line will cost $12,000 to install and
the power costs are estimated at $600 per year. The salvage value of the solar cells is
expected to be 25% of the first cost when the sampling project ends in 4 years. The
electric line will stay in place, so its salvage value is considered to be zero. At an interest
rate of 10% per year, which alternative should be selected?
Problem 4.8
Oil from a particular type of marine microalgae can be converted to biodiesel that can
serve as an alternate transportable fuel for automobiles and trucks. If lined ponds are used
to grow the algae, the construction cost will be $13 million and the maintenance &
operating (M&O) cost will be $2.1 million per year. If long plastic tubes are used for
growing the algae, the initial cost will be higher at $18 million, but less contamination
will render the M&O cost lower at $0.41 million per year. At an interest rate of 10% per
year and a 5-year project period, which system is better, ponds or tubes? Use a present
worth analysis.
Problem 4.9
American Electric Power agreed to spend $4.6 billion to clean up 46 coal-fired power
plants that are believed to be contributing to acid rain. The plan is to reduce nitrogen
oxide emissions by 69% by 2016 and sulfur dioxide emissions by 79% by 2018. Assume
Plan A is to spend $0.575 billion per year in years 1 through 4 and an additional
$0.575 billion per year in years 7 through 10. Plan B is to spend $0.46 billion in each of
years 1 through 10. At an interest rate of 8% per year, which plan is more economical to
the company based on a present worth analysis?
Problem 4.10
The costs associated with manufacturing a multifunction portable gas analyzer are
estimated. At an interest rate of 8% per year and a present worth analysis, which method
should be selected?
Manual
Robotic
First cost, $
-425,000
-850,000
M&O cost, year 1, $
-90,000
-10,000
Increase in M&O, $/year
7,000
1,000
Salvage value, $
80,000
300,000
Life, years
5
5
Problem 4.11
A pilot plant for conducting research related to reverse osmosis concentrate recovery via
lime softening can be leased for the 4-month project duration for $6,900 per month.
Electrical work at the site will cost $8500 now and the technician to install the electrical
work will charge $2000 now. Shipping will cost $2300 each way (months 0 and 4). A
technician for demobilization will cost $1300 in month 4. At an interest rate of 6% per
year compounded monthly, what is the present worth of the pilot plant project?
Problem 4.12
Biomet Implants is planning new online patient diagnostics for surgeons while they
operate. The new system will cost $300,000 to install in an operating room, $5000
annually for maintenance, and have an expected life of 4 years. The revenue per system is
estimated to be $80,000 in year 1 and to increase by $10,000 per year through year 4.
Determine if the project is economically justified using PW analysis and an MARR of
10% per year.
Problem 4.13
An undergraduate engineering student and her husband operate a pet-sitting service to
help make ends meet. They want to add a daily service of a photo placed online for pet
owners who are traveling. The estimates are: equipment and setup cost $950; net monthly
income over costs $70. Over a period of 3 years, will the service make at least 12% per
year compounded monthly?
Problem 4.14
The CFO of Marta Aaraña Cement Industries knows that many of the diesel-fueled
systems in its quarries must be replaced at an estimated cost of $20 million 10 years from
now. A fund for these replacements has been established with the commitment of
$1 million at the end of next year (year 1) with 10% increases through the 10th year. If
the fund earns at 5.25% per year, will the company have enough to pay for the
replacements? Solve using (a) tabulated factors, and (b) a spreadsheet.
Problem 4.15
Burling Water Cooperative currently contracts the removal of small amounts of hydrogen
sulfide from its well water using manganese dioxide filtration prior to the addition of
chlorine and fluoride. Contract renewal for 5 years will cost $75,000 annually for the next
2 years and $100,000 in years 3, 4, and 5. Assume payment is made at the end of each
contract year. Burling Coop can install the filtration equipment for $125,000 and perform
the process for $50,000 per year. At a discount rate of 6% per year, does the contract
service still save money?
Problem 4.16
Halogen-free liquid crystal polymers are used for lead-free soldering without corrosion
and maintenance issues. The polymers can be produced by either of two methods.
Equipment for method A costs $70,000 initially and has a $15,000 salvage value after
3 years. The operating cost with this method will be $20,000 per year. Method B will
have a first cost of $140,000, 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? Solve using (a) tabulated factors, and
(b) a calculator.
Problem 4.17
A software package created by Navarro & Associates can be used for analyzing and
designing three-sided guyed towers and three- and four-sided self-supporting towers. A
single-user license will cost $4000 per year. A site license has a onetime cost of $15,000.
A structural engineering consulting company is trying to decide between 2 alternatives:
first, to buy one single-user license now and one each year for the next four years (which
will provide five years of service), or second, to buy a site license now. Determine which
strategy is more economical at an interest rate of 12% per year for a 5-year planning
period. Apply the present worth method of evaluation.
Problem 4.18
The Bureau of Indian Affairs provides various services to American Indians and Alaskan
Natives. The Director of Indian Health Services is working with chief physicians at some
of the 230 clinics nationwide to select the better of two medical X-ray system alternatives
to be located at secondary-level clinics. At 5% per year, select the more economical
system. Solve using (a) tabulated factors, and (b) a spreadsheet.
Del Medical
First cost, $
-250,000
Annual operating cost,
$ per year
-231,000
Overhaul in year 3, $
—
Overhaul in year 4, $
-140,000
Salvage value, $
50,000
Expected life, years
6
Problem 4.19
The Briggs and Stratton Commercial Division designs and manufactures small engines
for golf turf maintenance equipment. A robotics-based testing system will ensure that
their new signature guarantee program entitled “Always Insta–Start” does indeed work
for every engine produced. Compare the two systems at MARR = 10% per year. Solve
using (a) tabulated factors, and (b) single-cell spreadsheet functions.
Pull System Push System
Robot and support
equipment first cost, $ -1,500,000 -2,250,000
Annual M&O cost,
$ per year -700,000 -600,000
Rebuild cost in year 3, $ 0 -500,000
Salvage value, $ 100,000 50,000
Estimated life, years 8 8
Problem 4.20
Chevron Corporation has a capital and exploratory budget for oil and gas production of
$19.6 billion in one year. The Upstream Division has a project in Angola for which three
offshore platform equipment alternatives are identified. Use the present worth method to
select the best alternative at 12% per year.
Problem 4.21
The TechEdge Corporation
offers two forms of 4-year service
contracts on its closed-loop
water purification
system used in the manufacture of semiconductor packages for microwave and highspeed
digital devices. The Professional Plan has an initial fee of $52,000 with annual fees
starting at $1000 in contract year 1 and increasing by $500 each year. Alternatively, the
Executive Plan costs $62,000 up front with annual fees starting at $5000 in contract year
1 and decreasing by $500 each year. The initial charge is considered a setup cost for
which there is no salvage value expected. Evaluate the plans at a MARR of 9% per year.
Solve using (a) factors, and (b) a spreadsheet. (c) How is the analysis performed using a
financial calculator?
Problem 4.22
A
B
C
First cost, $ million
-200
-350
-475
Annual cost, $ million
per year
-450
-275
-400
Salvage value, $ million
75
50
90
Estimated life, years
20
20
20
Allison and Joshua are engineers at Raytheon. Each has presented a proposal to track
fatigue development in composite materials installed on special-purpose aircraft. Which
is the better plan economically, if i = 12% per year compounded monthly?
Allison’s Plan Joshua’s Plan
Initial cost, $
Monthly M&O costs,
-40,000
-60,000
$ per month
-5,000
—
Semiannual M&O
cost,
$ per 6-month
—
-13,000
Salvage value, $
10,000
8,000
Life, years
5
5
Problem 4.23
What is the present worth of a $40,000 bond that has a bond interest rate of 6% per year,
payable semiannually? The bond matures in 20 years. The interest rate in the marketplace
is 8% per year compounded semiannually.
Problem 4.24
The present worth of a $10,000 municipal bond due 6 years from now is $11,000. If the
bond interest is payable quarterly and the interest rate used in discounting the cash flow is
8% per year compounded quarterly, what is the bond coupon rate b per year?
Problem 4.25
Jamal bought a 5% $1000 20-year bond for $925. He received a semiannual dividend for
8 years, then sold it immediately after the 16th dividend for $800. Did Jamal make the
return of 5% per year compounded semiannually that he wanted? Solve using (a) factors,
and (b) a spreadsheet.
Problem 4.26
An investor thought that market interest rates were going to decline. He paid $19,000 for
a corporate bond with a face value of $20,000. The bond has an interest rate of 10% per
year payable annually. If the investor plans to sell the bond immediately after receiving
the 4th interest payment, how much will he have to receive in order to make a return of
14% per year? Solve using (a) tabulated factors, and (b) the GOAL SEEK tool on a
spreadsheet.
Problem 4.27
An investor pays $30,000 for a convertible bond (one that can be converted into shares of
corporate common stock). The bond conversion rate is 100 shares of stock anytime
within the next five years. What will the stock price have to be in year 3 in order for the
investor to make 10% per year on the investment? Assume the bond interest rate is 4%
per year payable annually.
Problem 4.28
Atari needs $4.5 million in new investment capital to develop and market downloadable
game software for its new GPS2-ZX system. The plan is to sell $10,000 face-value
corporate bonds at a discount of $9000 now. A bond pays a dividend each 6 months
based on a bond interest rate of 5% per year with the $10,000 face value returned after
20 years. Will a purchase make at least 6% per year compounded semiannually?
Problem 4.29
Heidleman Industries is considering two types of materials for roofing its warehouses.
EPDM is an elastomeric polymer synthesized from ethylene, propylene, and a small
amount of diene monomer, compounded with carbon black processing oils and various
cross-linking and stabilizing agents. The 75 mil thickness will cost $4.10 per square foot
and will last for 25 years. A thin sheet aluminum roof will cost $6.00 per square foot, but
it will last for 50 years. Using an interest rate of 10% per year and a present worth
comparison, determine whether the company should install the polymer or the aluminum
roof.
Problem 4.30
Benjamin is an engineer with the Lego Group in Bellund, Denmark, manufacturers of
Lego toy construction blocks. He is responsible for the economic analysis of a new
production method of special-purpose Lego parts. Method 1 will have an initial cost of
$400,000, an annual operating cost of $140,000, and a life of three years. Method 2 will
have an initial cost of $600,000, an operating cost of $100,000 per year, and a six-year
life. Assume 10% salvage values for both methods. If Lego Industries uses a MARR of
15% per year, which method should it select on the basis of a present worth analysis?
Problem 4.31
A mechanical engineer is considering two types of pressure sensors for the low-pressure
steam lines in several of the company plants. Piezoresistive sensors use the change in
conductivity of semiconductors to measure the pressure. Fiber optic sensors use the
properties of fiber optic interferometers to sense nanometer scale displacement of
membranes. The costs for each system are shown below. Which should be selected based
on a present worth comparison at an interest rate of 1% per month?
Problem 4.32
Two mutually
exclusive projects
have the estimated
cash flows shown.
Use a present worth
analysis to determine
which should be
selected at an interest
rate of 10% per year.
Project P
Project Q
First cost, $
-55,000
-95,000
Annual cost,
$/year
-9,000
-5,000 year 1,
increasing by
$1000 per year
Salvage value, $
nil
4,000
Life, years
2
4
Problem 4.33
An industrial engineer is considering two robots for improving efficiency in a fiber-optic
manufacturing company. Robot X will have a first cost of $85,000, an annual
maintenance and operation (M&O) cost of $30,000, and a $35,000 salvage value after its
useful life of two years. A more sophisticated model, Robot Y will cost $157,000, have
an annual M&O cost of $28,000, and a $60,000 salvage value after its four-year life.
Select the better robot on the basis of a future worth comparison at an interest rate of 10%
per year. Solve by (a) tabulated factors, and (b) calculator.
Problem 4.34
Piezoresistive
Fiber Optic
Purchase cost, $
-13,650
-22,900
Maintenance cost,
$/month
-200
-50
Salvage value, $
0
2,000
Life, years
2
4
Virgin Galactic is considering two materials for certain parts in a re-useable space
vehicle: carbon fiber reinforced plastic (CFRP) and fiber reinforced ceramic (FRC). 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? Solve using (a) tabulated factors, and
(b) single-cell spreadsheet functions.
CFRP
FRC
First cost, $
-205,000
-235,000
Maintenance cost, $/year
-29,000
-27,000
Salvage value, $
2,000
20,000
Life, years
2
4
Problem 4.35
A metallurgical engineer is considering the two ceramics estimated below for use in a
high-temperature annealing furnace. (a) Which should be selected on the basis of a
present worth comparison at an interest rate of 12% per year? (b) If the life of material
XX is increased from 3 to 4 years, determine the number of re-purchases for both
alternatives necessary for a present worth analysis based on the equal-service
requirement.
Material XX
Material ZZ
First cost, $
-230,000
-380,000
Maintenance cost, $/year
-9,000
-12,000
Salvage value, $
12,000
140,000
Life, years
3
6
Problem 4.36
An environmental engineer must recommend one of two methods for monitoring high
colony counts of E. coli and other bacteria in watershed area “hot spots.” Estimates are
tabulated and the MARR is 10% per year. Use tabulated factors or a spreadsheet for your
analysis.
a. Use present worth analysis to select the better method.
b. For a study period of 3 years, use PW analysis to select the better method.
Method A
Method B
Initial cost, $
-100,000
-250,000
Annual operating cost,
$ per year
-30,000 in year 1,
-20,000
increasing by $5000
each year
Life, years
3
6
Problem 4.37
Allen Auto Group owns corner property that can be a parking lot for customers or sold
for retail sales space. The parking lot option can use concrete or asphalt. Concrete will
cost $375,000 initially, last for 20 years, and have an estimated annual maintenance cost
of $200 starting at the end of the eighth year. Asphalt is cheaper to install at $250,000,
but it will last 10 years and cost $2500 per year to maintain starting at the end of the
second year. If asphalt is replaced after 10 years, the $2500 maintenance cost will be
expended in its last year. There are no salvage values to be considered. Use i = 8% per
year and PW analysis to select the more economic surface, provided the property is
(a) used as a parking lot for 20 years, and (b) sold after 5 years and the parking lot is
completely removed.
Problem 4.38
The manager of engineering at the 900-megawatt Hamilton Nuclear Power Plant has
three options to supply personal safety equipment to employees. Two are vendors who
sell the items, and the third alternative is to rent the equipment for $50,000 per year, but
for no more than 3 years per contract. These items have relatively short lives due to
constant use. The MARR is 10% per year.
a. Select from the two vendors using the LCM and PW analysis.
b. Determine which of the three options is cheaper over a study period of 3 years.
Problem 4.39
Akash Uni-Safe in Chennai, India, makes Terminator fire extinguishers. It needs
replacement equipment to form the neck at the top of each extinguisher during
production. Select between two metal-constricting systems. Use the corporate MARR of
15% per year with (a) present worth analysis, and (b) future worth analysis.
Machine D
Machine E
First cost, $
-62,000
-77,000
Annual operating cost,
$ per year
-15,000
-21,000
Salvage value, $
8,000
10,000
Life, years
4
6
Problem 4.40
Vendor R
Vendor T
Rental
Initial cost, $
-75,000
-125,000
0
Annual upkeep,
$ per year
-27,000
-12,000
0
Annual rental,
$ per year
0
0
-50,000
Salvage value, $
0
30,000
0
Estimated life, years
2
3
Maximum of 3
HJ Heinz Corporation is constructing a distribution facility in Italy for products such as
Heinz Ketchup, Jack Daniel’s sauces, HP steak sauce, and Lea & Perrins Worcestershire
sauce. A 15- year life is expected for the structure. The exterior of the building has not
yet been selected. One alternative is to use concrete walls as the facade. This will require
painting now and every 5 years at a cost of $80,000 each time. Another alternative is an
anodized metal exterior attached to the concrete wall. This will cost $200,000 now and
require only minimal maintenance of $500 every 3 years. A metal exterior is more
attractive and will have a resale value of an estimated $25,000 more than concrete
15 years from now. Assume painting (for concrete) or maintenance (for metal) will be
performed in the last year of ownership to promote selling the property. Use future worth
analysis and i = 12% per year to select the exterior finish.
Problem 4.41
Three types of bits can be used in an automated drilling operation. A bright high-speed
steel (HSS) bit is the least expensive 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 of use. 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 bits will cost $7000 to buy and will last
6 months with an operating cost of $1200 per month. At an interest rate of 12% per year
compounded monthly, which type of drill bit should be selected? Use a future worth
analysis.
Problem 4.42
Three different plans were presented to the GAO by a high-tech facilities manager for
operating an identity-theft scanning facility. Plan A involves renewable 1-year contracts
with payments of $1 million at the beginning of each year. Plan B is a 2-year contract that
requires four payments of $600,000 each, with the first one made now and the other three
at 6-month intervals. Plan C is a 3-year contract that entails a payment of $1.5 million
now and a second payment of $0.5 million 2 years from now. Assuming that the GAO
could renew any of the plans under the same payment conditions, which plan is best on
the basis of a present worth analysis at an interest rate of 6% per year compounded
semiannually?
Problem 4.43
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 “bare–bones”
design and standard grade construction of walls, windows, doors, and other features.
With this option, heating and cooling costs will be greater, maintenance costs will be
higher, and replacement will occur earlier than proposal B. The initial cost for A will be
$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 for 20 years. They
will have no salvage value. Proposal B will include tailored design and construction costs
of $1.1 million initially with estimated heating and cooling costs of $3000 per month and
maintenance costs of $1000 per month. There will be no salvage value at the end of the
20-year life. Which proposal should be accepted on the basis of a life-cycle cost analysis
at an interest rate of 0.5% per month?
Problem 4.44
A medium-size municipality plans to develop a software system to assist in project
selection during the next 10 years. A life-cycle cost approach has been used to categorize
costs into development, programming, operating, and support costs for each alternative.
There are three alternatives under consideration, identified as A (tailored system),
B (adapted system), and C (current system). The costs are summarized on the next page.
Perform a life-cycle cost analysis to identify the best alternative at 8% per year using
(a) tabulated factors first, then (b) a spreadsheet to verify your selection.
Cost
Alternative Component Cost Estimates
A Development $250,000 now, $150,000 years 1–4
Programming $45,000 now, $35,000 years 1,2
Operation $50,000 years 1 through 10
Support $30,000 years 1 through 5
B Development $10,000 now
Programming $45,000 year 0, $30,000 years 1–3
Operation $80,000 years 1 through 10
Support $40,000 years 1 through 10
C Operation $175,000 years 1 through 10
Problem 4.45
Recently introduced Gatorade Endurance Formula contains more electrolytes (such as
calcium and magnesium) than the original sports drink formula, thus causing Endurance
to taste saltier to some. It is important than the amount of electrolytes be precisely
balanced in the manufacturing process. The currently installed system (called EMOST)
can be upgraded to monitor the amount more precisely. It costs $12,000 per year for
equipment maintenance, $45,000 a year for labor, and the upgrade will cost $25,000 now.
This can serve for 10 more years, the expected remaining time the product will be
financially successful. A new system (UPMOST) will also serve for the 10 years and
have the following estimated costs. All costs are per year for the indicated time periods.
Equipment: $150,000 years 0 and 1
Development: $120,000 years 1 and 2
Maintain and phase-out EMOST: $20,000 years 1, 2, and 3
Maintain hardware and software: $10,000 years 3 through 10
Personnel costs: $90,000 years 3 through 10
Scrapped formula: $30,000 years 3 through 10
Sales of Gatorade Endurance with the UPMOST system installed are expected to go up
by $150,000 per year beginning in year 3 and increase by $50,000 per year through year
10. Use LCC analysis at an MARR of 20% per year to select the better electrolyte
monitoring system. (Choose from tabulated factors, calculator, or spreadsheet to make
your evaluation.)
Problem 4.46
The Golden Gate bridge is maintained by 17 ironworkers, who replace corroding steel
and rivets, and 38 painters. If the painters have an average wage of $120,000 per year
with benefits and the ironworkers get $150,000, what is the capitalized cost today of all
the future wages for bridge maintenance at an interest rate of 8% per year?
Problem 4.47
Determine the capitalized cost of $100,000 now and $50,000 per year in years one
through infinity at an interest rate of 10% per year compounded continuously.
Problem 4.48
Determine the capitalized cost of $1,000,000 at time 0, $125,000 in years 1 through 10,
and $200,000 per year from year 11 on. Use an interest rate of 10% per year.
Problem 4.49
The cost of extending Park Road PR2 in Yellowstone National Park is $1.7 million.
Resurfacing and other maintenance is expected to cost $350,000 every 3 years. What is
the capitalized cost of the road extension at an interest rate of 6% per year?
Problem 4.50
John wants to have the financial ability to withdraw $80,000 per year forever beginning
30 years from now. If his retirement account earns 8% per year interest and dividends,
what is the required balance in (a) year 29, and (b) year 0?
Problem 4.51
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 requires an expenditure of $50,000
every five years forever beginning in year 5. Plan B requires an expenditure of $100,000
every 10 years forever beginning in year 10.
Problem 4.52
An alumna of Ohio State University wants to set up an endowment fund that can award
scholarships to female engineering students totaling $100,000 per year forever. The first
scholarships are to be granted now and continue each year from now on. How much must
the alumna donate now, if the endowment fund is expected to earn interest at a rate of 8%
per year?
Problem 4.53
Field chlorination of reclaimed water can be accomplished via a low-cost system that
uses calcium hypochlorite tables. System components include a 4 foot diameter pipe that
is 4 inches long ($70), a solenoid valve ($50), a float switch ($30), a chlorine analyzer
($1500), and a programmable VFD solution pump ($1900). Assume the following
component lives: pipe – 10 years; solenoid valve – 2 years; float switch – 2 years;
chlorine analyzer – 5 years; solution pump – 5 years. Calculate the capitalized cost of the
system at an interest rate of 8% per year.
Problem 4.54
Compare the cost of the two types of composite materials on the basis of their capitalized
costs. Use an interest rate of 10% per year.
Material J1
Material K2
First cost, $
-55,000
-325,000
Maintenance cost, $/year
-6,000
-1,000
Salvage value
2,000
200,000
Life, years
3
∞
Problem 4.55
The president of Biomed Products is considering a year compounded quarterly. long-term
contract to outsource maintenance and operations that will significantly improve the
energy efficiency of their imaging systems. The payment schedule has two large
payments in the first years with continuing payments thereafter. The proposed schedule is
$200,000 now, $300,000 four years from now, $50,000 every 5 years, and an annual
amount of $8000 beginning 15 years from now and continuing indefinitely. Determine
the capitalized cost at 8% per year.
Problem 4.56
UPS Freight plans to spend $100 million on new long-haul tractor-trailers. Some of these
vehicles will include a new shelving design with adjustable shelves to transport
irregularly sized freight that requires special handling during loading and unloading.
Though the life is relatively short, the director wants a capitalized cost analysis
performed on the two final designs. Compare the alternatives at the MARR of 10% per
year using (a) tabulated factors, and (b) a spreadsheet.
Design A: Design B: movable
shelves adaptable frames
First cost, $ -2,500,000 -1,100,000
AOC, $ per year -130,000 -65,000
Annual revenue,
$ per year 800,000 625,000
Salvage value, $ 50,000 20,000
Life, years 6 4
Problem 4.57
A water supply cooperative plans to increase its water supply by 8.5 million gallons per
day to meet increasing demand. One alternative is to spend $10 million to increase the
size of an existing reservoir in an environmentally acceptable way. Added annual upkeep
will be $25,000 for this option. A second option is to drill new wells and provide added
pipelines for transportation to treatment facilities at an initial cost of $1.5 million and
annual cost of $120,000. The reservoir is expected to last indefinitely, but the productive
well life is only 10 years. Compare the alternatives at 5% per year.
Problem 4.58
Three alternatives to incorporate improved techniques to manufacture computer drives to
play HD DVD optical disc formats have been developed and costed. Compare the
alternatives below using capitalized cost and an interest rate of 12% per year
compounded quarterly.
Alternative Alternative Alternative
E F G
First cost, $ -2,000,000 -3,000,000 -10,000,000
Net income,
$ per quarter 300,000 100,000 400,000
Salvage value, $ 50,000 70,000 —
Life, years 4 8 ∞
Problem 4.59
A small manufacturing company is considering the addition of one or more of four new
product lines. If the total amount of investment capital available for new ventures is
$800,000, which one(s) should the company undertake on the basis of a present worth
analysis? Assume the company uses a 5-year project recovery period and a MARR of
20% per year. All cash flows are in $1000 units.
Product Lines
R1 S2 T3 U4
First cost, $ -200 -400 -500 -700
M&O cost, $/year -50 -200 -300 -400
Revenue, $/year 150 450 520 770
Problem 4.60
Determine which of the following independent projects should be selected for investment
if $240,000 is available and the MARR is 10% per year. Use the PW method to evaluate
mutually exclusive bundles to make the selection.
Project
Initial
Investment, $
Net Cash Flow,
$/year
Life, Years
A
-100,000
50,000
8
B
-125,000
24,000
8
C
-120,000
75,000
8
D
-220,000
39,000
8
E
-200,000
82,000
8
Problem 4.61
Feng Seawater Desalination Systems has established a capital investment limit of
$800,000 for next year for projects that target improved recovery of highly brackish
groundwater. Select any or all of the projects using a MARR of 10% per year. All
projects have a 4-year life.
Initial
Net Cash
Salvage
Project
Investment, $
Flow, $/year
Value, $
X
-250,000
50,000
45,000
Y
-300,000
90,000
-10,000
Z
-550,000
150,000
100.000
Problem 4.62
Dwayne has four independent vendor proposals to contract the nationwide oil recycling
services for the Ford Corporation manufacturing plants. All combinations are acceptable,
except that vendors B and C cannot both be chosen. Revenue sharing of recycled oil sales
with Ford is a part of the requirement. Develop all possible mutually exclusive bundles
under the additional following restrictions and select the best projects. The corporate
MARR is 10% per year.
a. A maximum of $4 million can be spent.
b. A larger budget of $5.5 million is allowed, but no more than two vendors can be
selected.
c. There is no limit on spending.
Initial Life, Annual Net
Vendor Investment, $ Years Revenue, $ per Year
A -1.5 million 8 360,000
B -3.0 million 10 600,000
C -1.8 million 5 620,000
D -2.0 million 4 630,000
Problem 4.63
In the PW method of alternative evaluation, equal service means that:
a. all projects must start at the same time.
b. all projects are evaluated over the same time period.
c. all projects must have the same operating cost.
d. all projects have equal salvage values.
Problem 4.64
The cost of money is 10% per year.
Machine P
Machine Q
Initial cost, $
−
35,000
−
66,000
Annual cost,
$/year
−
20,000
−
15,000
Salvage value, $
10,000
23,000
Life, years
2
4
In comparing the machines on a present worth basis, the present worth of machine P is
closest to:
a. $−82,130
b. $−87,840
c. $−91,568
d. $−112,230
Problem 4.65
The cost of money is 10% per year.
Machine P
Machine Q
Initial cost, $
−
35,000
−
66,000
Annual cost,
$/year
−
20,000
−
15,000
Salvage value, $
10,000
23,000
Life, years
2
4
In comparing the machines on a present worth basis, the present worth of machine Q is
closest to:
a. $−68,445
b. $−97,840
c. $−125,015
d. $−223,120
Problem 4.66
The cost of money is 10% per year.
Machine P
Machine Q
Initial cost, $
−
35,000
−
66,000
Annual cost,
$/year
−
20,000
−
15,000
Salvage value, $
10,000
23,000
Life, years
2
4
The capitalized cost of machine P is closest to:
a. $−35,405
b. $−97,840
c. $−354,050
d. $−708,095
Problem 4.67
The cost of maintaining a public monument in Washington, D.C. occurs as periodic
outlays of $10,000 every 5 years. If the first outlay is 5 years from now, the capitalized
cost of the maintenance at an interest rate of 10% per year is closest to:
a. $−1638
b. $−16,380
c. $−26,380
d. $−29,360
Problem 4.68
A grateful donor wishes to start an endowment at her alma mater that will provide
scholarship money of $40,000 per year beginning now (time 0) and continue indefinitely.
If the funds earn 10% per year, the amount she must donate now is closest to:
a. $340,000
b. $400,000
c. $440,000
d. $493,800
Problem 4.69
A corporate bond has a face value of $10,000, a bond interest rate of 8% 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 market place is 8% per year compounded
semiannually, the size and frequency of the interest payments the person will receive are:
a. $270 every six months
b. $300 every six months
c. $360 every six months
d. $400 every six months
Problem 4.70
The MARR is 12% per year.
Alternative 1 Alternative 2
First cost, $ −40,000 −65,000
Annual cost, $ per year −20,000 −15,000
Salvage value, $ 10,000 25,000
Life, years 3 4
The relation that correctly calculates the present worth of alternative 2 when comparing it
to alternative 1 is:
a. −65,000 − 15,000(P/A,12%,12) + 25,000(P/F,12%,8) + 25,000(P/F,12%,12)
b. −65,000 − 15,000(P/A,12%,4) + 25,000(P/F,12%,4)
c. −65,000 − 15,000(P/A,12%,12) + 25,000(P/F,12%,12)
d. −65,000 −40,000[(P/F,12%,4) +(P/F,12%,8)] −15,000(P/A,12%,12)
+25,000(P/F,12%,12)
Problem 4.71
The MARR is 12% per year.
Alternative 1 Alternative 2
First cost, $ −40,000 −65,000
Annual cost, $ per year −20,000 −15,000
Salvage value, $ 10,000 25,000
Life, years 3 4
The number of life cycles for each alternative when performing a present worth
evaluation based on the LCM for equal service is:
a. 2 for each alternative
b. 1 for each alternative
c. 3 for alternative 1; 4 for alternative 2
d. 4 for alternative 1; 3 for alternative 2
Solution 4.1
The do-nothing alternative is not an option (1) when it is absolutely required that one of
Solution 4.2
Solution 4.3
(a) Do-nothing, which is to leave in place the existing equipment. Annual costs for
Solution 4.4
Solution 4.5
Of the 24 = 16 bundles possible, there are 12 acceptable bundles.
Solution 4.6
Equal service means that the alternatives provide the same services over the same time
Solution 4.7
Solution 4.8
Units are $ million
Solution 4.9
Alternative lives are equal at 10 years. Units are $ billion
Solution 4.10
= $-693,119
Select the robotic system
Solution 4.11
Solution 4.12
System is not justified since PW < 0
Solution 4.13
Interest rate of 12% per year compounded monthly is 1% per month.
Solution 4.14
(a) Determine if the deposit’s F value in year 10 equals the $20 million
(b) A spreadsheet solution follows.
Solution 4.15
Subscripts are C for contract service and B for Burling Coop installed.
Solution 4.16
(a) PWA = -70,000 – 20,000(P/A,12%,3) + 15,000(P/F,12%,3)
Solution 4.17
Solution 4.18
(a) Monetary units are in $1000. Calculate PW values to select Siemens.
(b) By spreadsheet, enter the following into single cells to display the PW values.
Solution 4.19
(a) Monetary units are in $1000. Calculate PW values to select the pull system.
(b) By spreadsheet, enter the following into single cells to display the PW values.
Solution 4.20
(a) Monetary units are in $ million. Calculate PW values to select alternative B.
(b) By spreadsheet, enter the following to display the PW values in $ million
units.
Solution 4.21
(b) A spreadsheet solution follows; select the Professional plan.
Solution 4.22
(a) For Allison (A), use i = 1% per month and n = 60 months to calculate PW.
(b) A spreadsheet solution follows; select Joshua’s plan
Solution 4.23
Solution 4.24
Determine quarterly dividend, then solve for b.
Solution 4.25
(b) A spreadsheet solution follows to obtain PW = $-59.72
Solution 4.26
(a) Let R = amount received
Solution 4.27
Let S = price per share
S = $359.59 per share
Solution 4.28
A spreadsheet solution follows.
Solution 4.29
Evaluate over a 50 year LCM.
Solution 4.30
Solution 4.31
LCM is 48 months; repurchase P after 24 months
Solution 4.32
Solution 4.33
(a) FWX = -85,000(F/P,10%,4)-30,000(F/A,10%,4)-50,000(F/P,10%,2)+35,000
Solution 4.34
(a) LCM is 4 years; repurchased CFRP after 2 years
Solution 4.35
(a) PWXX = -230,000 – 9000(P/A,12%,6) – 218,000(P/F,12%,3)
Solution 4.36
Factors:
(b) Use n = 3 in all calculations and do not repurchase A. Still select method A,
now by a larger margin.
Spreadsheet: Solution for parts (a) and (b) follows.
Solution 4.37
For asphalt, repave after 10 years and re-start maintenance charge in year 12.
Select the concrete option with a marginal advantage.
A spreadsheet solution for parts (a) and (b) follows.
Solution 4.38
(a) LCM is 6 years for R and T evaluation. Select vendor T.
(b) Re-purchase R after 2 years. Rental is paid at the end of each year.
A spreadsheet solution for (a) and (b) follows.
Solution 4.39
(a) PW analysis requires an LCM of 12 years. Select machine D.
A spreadsheet solution for parts (a) and (b) follows.
Solution 4.40
Solution 4.41
Solution 4.42
Draw cash flow diagrams first.
Solution 4.43
Solution 4.44
(a) LCCA = -250,000 – 150,000(P/A,8%,4) – 45,000 – 35,000(P/A,8%,2)
(b) Spreadsheet follows to verify selection of B
Solution 4.45
LCC is determined by spreadsheet much easier than by hand calculator. Select the
UPMOST system with the lower LCC.
Solution 4.46
Solution 4.47
i per year = e0.10 – 1 = 10.517%
Solution 4.48
Solution 4.49
Solution 4.50
Solution 4.51
Find AW of each plan, then take difference, and divide by i
Solution 4.52
Solution 4.53
First find AW and then divide by i to determine CC
Solution 4.54
For J1, first find AW and then divide by i.
Solution 4.55
Monetary terms are in $1000 units.
Solution 4.56
(a) Monetary terms are $1000 units. Determine CC = AW/i values of revenues
minus costs to select design B.
Solution 4.57
Monetary terms are $1000 units. Determine CC values to select wells.
Select wells alternative.
Solution 4.58
Quarterly interest rate is 12/4 = 3% with 4 quarters per year
Solution 4.59
Of the 24 = 16 possible bundles, there are 6 within the $800,000 budget limit, as follows:
R1, S2, T3, U4, R1 & S2, and R1 & T3. Values are in $1000 units.
Solution 4.60
Develop the bundles with less than $240,000 investment, and select the one with the
largest PW value.
Initial
Bundle Projects investment, $ NCF, $/year PW at 10%, $
1 A -100,000 50,000 166,746
PW3 = -120,000 + 75,000(P/A,10%,8)
All other PW values are obtained by adding the respective PW for bundles 1
through 5.
Solution 4.61
Budget = $800,000 i = 10% 6 viable bundles
Bundle Projects NCFj0 NCFjt S PW at 10%
1 X $-250,000 $ 50,000 $ 45,000 $-60,770
Solution 4.62
Determine the PW for each project.
By spreadsheet, enter the following to display the project PW values.
Formulate acceptable bundles from the 24 = 16 possibilities, without both B and C
and select projects with largest total PW of a bundle.
Bundle
Investment,
$ million
PW, $
DN
0
0
A
-1.5
420,564
Solution 4.63
Solution 4.64
Solution 4.65
Solution 4.66
-3.0
686,760
-1.8
550,296
AC
970,860
Bundle
Investment,
$ million
DN
0
420,564
-3.0
686,760
-1.8
550,296
AB
1,107,313
AC
970,860
CCP = A/i
Solution 4.67
CC = A/i
Solution 4.68
Solution 4.69
I = Vb/c
Solution 4.70
Solution 4.71