#13 Engineering Economics : Problems

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Cost estimates for a propose public facility are being evaluated. Initial construction cost is anticipated to be $120, 000, and annual maintenance expenses are expected to be $6500 for the first 20 years and $2000 for every year thereafter. The facility is to be used and maintained for an indefinite period of time. Using an interest rate of 10% per year, the capitalized cost of this facility is most nearly ?
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NCEES. capitalized costs EQN. P= A/I., P = A(P/A, I%, n) SOLN. Ptotal = P construction + P maintenance. Pmaintance = P/A1 + A2/I Present worth = Annual cost. Pmaintenance of first 20 years: 6500 – 2000 = 4500 ( P/A, I%,n) $180,000.
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A machine has an initial cost of $18,000 and operating costs of $2500 each year. The salvage value decreases y $3000 each year. The machine is now three years old. Assuming an effective annual interest rate of 12%, the cost of owning and operating the machine for one more year is most nearly?
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NCEES. N/A. **memorize! EQN. Costs = C = operating cost + lost salvage value + opportunity cost. SOLN. C = operating cost annually + decrease salvage value annually + interest ( yr * decrease salvage value annually) $6600.
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A Texas baseball team purchased a $140,000 pitching machine that has a useful life of seven years. If the machine has a salvage value of $20,000 at the end of its life, and straight line depreciation is used, the book value at the end of year 4 is most nearly?
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NCEES. book value. depreciation. straight line value. EQN. BV. Dj. SOLN. Dj = C – Sn / 7 = yearly depreciation. BV = initial cost – Dj * # yrs. $70,000.
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Calculate the approximate rate of return for an investment with the following characteristics: initial cost $20,000 project life 10 yr salvage value $5000 annual receipts $7500 annual disbursements $3000
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NCEES. rate-of-return. present worth. EQN. rate of return = i. P =0. SOLN. P = 0 = -initial cost + salvage value – annual receipts – annual disbursements. (P/F, i%, n) single payment = salvage value. (P/A, i%, n) uniform series = annual occurrence = receipts & disbursements. trial & error of guesses until i gets both sides of equation to be nearly equal. 20%
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On January 1, $5000 is deposited into a high-interest savings account that pay 8% interest compounded annually. If all of the money is withdrawn in five equal end-of-year sums beginning, December 31 of the first year, most nearly how much will each withdrawal be?
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NCEES. factor name. (gets you to discount factors table for discrete compounding = first thing in engineering econ) EQN. capital recovery. convert to A given P. A = P(A/P, i%,n) SOLN. 1. need table. 2. given present worth = $5000 = P. 3. need to find annual withdrawals = A. 4. plug into eqn. $1250.
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If you needed to have $800 in savings at the end of four years and your savings account yielded 5% interest paid annually, most nearly how much would you need to deposit today?
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NCEES. factor name. EQN. P/F SOLN. 1. discount factors table. 2. deposit how much today to get 800 in 4 years? 3. deposit = single payment = P. 4. give $800 = future worth. 5. P= F(P/F). $660.00
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At the end of each year for five years, $500 is deposited into a credit union account. The credit union pays 5% interest compounded annually. At the end of five years (immediately following the fifth deposit), most nearly, how much will be in the account?
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NCEES. factor name. EQN. F/A. SOLN. 1. discount factors table. 2. $500 = annual deposits = A. 3. find how much will be in account end of 5th deposit = future worth = F. C. $2800.
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$5000 is put into an empty savings account with a nominal interest rate of 5% No other contributions are made to the account. With monthly compounding, approximately how much interest will have been earned after five years?
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NCEES. nominal interest = “non-annual compounding. monthly compounding = “factor name” EQN. ie. F/P. i available. SOLN. 1. problem asking how much $ = interest. 2. given %5 nominal interest rate, find effective interest = ie. * m = 12. 3. use the calculated ie & given $5000 = present worth = P. 4. find the $ = future worth = F for n = 5 years. 5. F= P(F/P). 6. the $ = interest: i available = F – P. $1420.
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An investment currently costs $28,000, If the current inflation rate is 6% and the effective annual return on investment is 10%, approximately how long will it take for the investment’s future value to reach $40,000.
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NCEES. inflation. factor name. EQN. d = . F/P. SOLN. 1. find the adjusted interest rate d accounting inflation. 2. given present worth & future worth = F/P equation. 3. Solve algebraically until find n. 4. to get rid of exponential, take log of both sides. 5. plug in answer choices until get both sides of equation to equal. 2.32 yr.
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The purchase price of a car is $25,000. Mr. Smith makes a down payment of $5000 and borrows the balance fro a bank at 6% nominal annual interest, compounded monthly for 5 years. Calculate the nearest value of the required monthly payments to pay off loan.
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NCEES. factor name. EQN. i. A/P. SOLN. 1. convert nominal annual interest to monthly interest. 2. total being borrowed from bank = P. 3. monthly payments = A. 4. A/P capital recovery equation. 5. n = convert years to mo. $400.
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45. 1. A 40-year-old consulting engineer wants to set up a retirement fund to be used starting at age 65. $20,000 is invested now at 6% compounded annually. The amount of money that will be in the fund at retirement is most nearly? units $
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NCEES. EQN. SOLN. $86,000.
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$2000 will become available on January 1 in year 8. If interest is 5%, what is the present worth of this some on January 1 of year 1?
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NCEES. EQN. SOLN. year 1 to year 8 = 7 years. $1420.
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$20,000 is deposited at the end of each year into a fund earning 6% interest. At the end of ten years, the amount accumulated is most nearly?
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NCEES. EQN. SOLN. $260,000.
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At the end of each year, an investor deposits some money in a fund earning 7% interest. The same amount is deposited each year, and after six years the account contains $1600. The amount deposited each time is most nearly?
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NCEES. EQN. SOLN. $220.
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A sum of money is deposited into a fund at 5% interest. $400 is withdrawn at the end of each year for nine years, leaving nothing in the fund at the end. The amount originally deposited is most nearly?
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NCEES. EQN. SOLN. $2800.
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The maintenance cost on a house is expected to be $1000 the first year and to increase $500 per year after that, Assuming an interest rate 6% compounded annually, the maintenance cost over 10 years is most nearly equivalent to an annual maintenance cost of?
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NCEES. EQN. SOLN. $3000.
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Money is invested at 5% per annum and compounded quarterly. The effective annual interest rate is most nearly?
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NCEES. EQN. SOLN. 5.1%.
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A computer will be purchased at $3900. The expected salvage value at the end of its service life of 10 years is $1900. Using the straight line method the annual depreciation for this computer is most nearly?
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NCEES. EQN. SOLN. $210.
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A groundwater treatment system costs $2,500,000. It is expected to operate for a total of 130,000 hours over a period of 10 years, and then have a $250,000 salvage value. During the system’s first year in service, it is operated for 6500 hours Using the MACRS method, its depreciation in the third year is most nearly?
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NCEES. EQN. SOLN. $360,000.
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A machine initially costing $25,000 will have a salvage value of $6000 after five years, Using MACRS depreciation, its book value after the third year will be most nearly?
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NCEES. EQN. SOLN. $7200.
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The construction of a volleyball court will cost $1200 and annual maintenance cost is expected to be $300. At an effective annual interest rate of 5%, the project’s capitalized cost is most nearly?
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NCEES. EQN. SOLN. $7000.
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An investment of $20,000 earns an effective annual interest of 10%. The value of the investment in five years, adjusted for an annual inflation rate of 6%, is most nearly?
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NCEES. EQN. SOLN. $43,000.
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A large sewer system will cost $175,000 annually. There will be favorable consequences to the general public equivalent to $500,000 annually, and adverse consequences to a small segment of the public equivalent to %50,000 annually. The benefit-cost ratio is most nearly?
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NCEES. EQN. SOLN. 2.6.
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find unknown given known.
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unknown = known ( unknown/known, i %, n )
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given annual costs, what is present worth
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P = A(P/A, I%, n)
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time
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n
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annual cost changes after n years
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annual cost = annual cost year 1 – annual cost after n years.
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capitalized costs: P total =
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P construction + P maintenance.
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cost of owning & operating a machine for one more year.
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C = operating cost annually + decrease salvage value annually + interest ( yr * decrease salvage value annually)
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depreciation straight line
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Dj = initial cost – salvage value / length of life
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rate of return for investment.
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= interest where Benefits = Costs. Present worth = 0. P = 0 = -initial cost + salvage value – annual receipts – annual disbursements.
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salvage value in investment case in determining rate of return
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(P/F, i%, n) single payment = salvage value.
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annual occurrence and annual disbursements for investment case in determining rate of return
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(P/A, i%, n) uniform series = annual occurrence = receipts & disbursements.
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how to find rate of return for an investment in equation
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trial & error of guesses until i gets both sides of equation to be nearly equal. or plug in answer choices given that gives the closest results where both sides of equation are equal.
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nominal interest.
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nominal interest = “non-annual compounding.
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monthly compounding
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monthly compounding = “factor name”
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to get rid of exponential,
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take log of both sides.
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how to calculate monthly interest given nominal interest rate and period
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i = nominal percentage/ 12 months
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how to calculate monthly payment given nominal interest rate and period of 5 years
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i = nominal percentage/ 12 months. n = 5 yr * 12 mo = 60 mo.
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year 1 to year 8 = ? yr
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7 yrs.
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annual maintenance cost use which discount factor
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find A given G. find annual worth given cash flow increasing by uniform amount. A/G. uniform gradient uniform series
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total effective cost
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A total = A1 + A (annual worth).
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effective annual interest rate
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interest per annum = compounded quarterly = non-annual compounding.
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depreciation MACRS
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1. use table. 2. factor = desired year, period.
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depreciation MACRS factor desired year
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row
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depreciation MACRS factor period
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column = recovery period.
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depreciation MACRS D =
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factor * cost. factor = for value given in table, move 2 decimals to left.

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