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Chp 2 Equipment Economics
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10th ed. Problem Solutions
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A walk-through of problems 2.7 and 2.17 is available for Access Engineering Subscribers via
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www.accessengineeringlibrary.com
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Please check with your respective institution’s librarian if they offer this benefit and/or to request it.
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2.1
If $130,000 is borrowed for 5 years, what total amount must be paid back? Use an
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interest rate equal to 3.5% compounded annually to solve this problem.
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Step 1Provided data:
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• Borrow$130,000
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• Time period5years
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• Interest rate3.5%
compounded annually
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Step 2
Calculate the single payment compound amount factor (SPCAF)
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SPCAF =(1 + 0.035)5=1.187686306
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Step 3
Use SPCAF to calculate the amount
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$130,000X1.187686306$154,399.22
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2.2
A company’s interest rate for acquiring outside capital is 4.5% compounded
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annually. If $1,035,000 must be borrowed for 4 years, what is the total amount of
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interest?
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Step 1Provided data:
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• Borrow$1,035,000
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• Time period4years
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• Interest rate4.5%
compounded annually
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Step 2
Calculate the single payment compound amount factor (SPCAF)
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SPCAF =(1 + 0.045)4=1.192518601
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Step 3
Use SPCAF to calculate the amount paid back at the end of 4 years
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$1,035,000X1.192518601=$1,234,256.75
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Step 4
Subtract the amount borrowed to fine the interest part
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$1,234,256.75
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$1,035,000.00
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$199,256.75total interest
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2.3
To purchase a new truck, you borrow $46,000. The bank offers a 10-year loan at an
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interest rate of 3.5% compounded annually. If you make only one payment at the
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end of the loan period, repaying the principal and interest:
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a. What is the number of time periods (n) you should use in solving this problem?
10years
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b. What rate of interest (i), per period of time, should be used in solving this
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problem?3.5%
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c. Is the present single amount of money (P) known? (Yes or No)
$46,000yes
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d. Which time value factor should be used to solve this problem?
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The single payment compound amount factor
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(1 + 0.035)101.410598761
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e. What is the total amount to be paid back?
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$46,000X1.410598761=$64,887.54
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f. How much of the total amount repaid represents interest?
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$64,887.54
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$46,000.00
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$18,887.54interest
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2.4
What amount must be invested today at an annual interest rate of 4.5% to purchase
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a $950,000 machine 5 years in the future?
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a. What is the number of time periods (n) you should use in solving this problem?
5
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b. What rate of interest (i), per period of time, should be used in solving this
4.5%
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 problem?
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c. Is the present single amount of money (P) known? (Yes or No)
No
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d. What is the desired future amount?
$950,000
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e. Which time value factor should be used to solve this problem?
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The present worth compound amount factor
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1/(1+i)n1/1.24618191.2461819
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f. What is the total amount you must invest today?
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$950,000 =$762,328.49
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P = 1.2461819
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2.5
To purchase a new truck, you borrow $67,000. The bank offers an interest rate of 5.25%
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compounded monthly. If you take a 5-year loan and will be making monthly payments, what
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is the total amount that must be paid back?
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Step 1Provided data:
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• Borrow$67,000
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• Time period5years
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• Interest rate5.3%
compounded monthly
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Step 2
What is the number of time periods (n) you should use
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5years X12months/yr =60
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Step 3
What rate of interest (i), per period of time, should be use in solving this problem?
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Annual interest rate
=0.0043750
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12 months/yr
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Step 4
Is the present single amount of money (P) known?
Yes
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Step 5
Which time value factor should be used to solve this problem?
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The uniform series capital recovery factor
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A =0.0056850.01898598384
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0.2994323
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What is the amount you must pay each month?
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Step 6$67,000X0.01898598384=$1,272.061
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Step 7
What is the total amount that will be paid back over the life of the loan?
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$1,272.061X60=$76,323.66
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$67,000.00
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Step 8
What is the total amount of interest you will pay?
$9,323.66
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2.6
A track dozer cost $1,650,000 to purchase. Fuel, oil, grease, and minor maintenance are
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estimated to cost $35.00 per operating hour. A major engine repair costing $46,000
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will probably be required after 7,200 hours of use. The expected resale price (salvage
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value) is 21% of the original purchase price. The machine is expected to have a useful
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life of 10,800 hours. How much should the owner of the machine charge per hour