Engineering Economics Flashcards
If a one-time amount of $500 is invested at an interest rate of 8% per year, what is its future worth at the end of 30 years?
P = $500
i = 8% or .08
n = 30
Use formula F = P(F/P,i,n)
find “F/P” from chart
If you need to have $800 in savings at the end of 4 years, how much do you need to deposit today, assuming 5% annual interest?
F = $800
i = 5% or 0.05
n = 4
Use formula P = F (P/F,i,n) , there is no table for 5%, must use formula
F (1+i)-n = ($800)(1+0.05)-4
A company borrows $100,000 today on a 5-year loan at 12% nominal annual interest, compounded monthly. What would the monthly payment be?
P = $100,000
A = ?
n = 60
i = 1%
A = P(A/P,i,n) = P(A/P,1%,60) = ($100,000)(0.0222)
- The Powerball was won by a single individual. The individual was given to choices: receive 26 payments of $7 million each year, with the first payment to be made now; or receive a single equivalent lump-sum payment. If the state uses an interest rate of 4% per year, the amount of the lump sum payment is closest to.
- A. $72 million
- B. $109 million
- C. $116 million
- D. $135 million
Draw it out on a line.
Trying to determine the present worth “P”
we were given an Annual quantity “A”
P = A(P/A, 4%,25) + Po
= ($7million) (15.6221) + (7million)
- If $10,000 is borrowed now at 10% per year interest, the balance at the end of year 2 after payments of $3000 in years 1 and 2 will be closest to:
- A. $4100
- B. $5800
- C. $6100
- D. $7300
Draw table out.
use P/F table for each payment
P/F,10%,1 for first payment = -$2727.3
P/F,10%,2 for second payment = -$2479.2
add the initial $10,000
= $4793.5
P=4793.5
F = ?
i = 10%
n = 2
F = P(F/P, i, n) = $5800
If a company wants to have $100,000 in a contingency fund 10 years from now, the amount the company must deposit each year in years 6 through 9, at an interest rate of 10% per year, is closest to:
A. $19,591
B. $20,614
C.$21,547
D. 22,389
Draw cash flow diagram
We want $100,000 in year 10 which is the “F”
the company wants to put in some money “A”
Figure out how much 100,000 is in year 9
= $100,000(F/P,10%,1) = $90,910
A = F (A/F, i, n)
=($90,910)(A/F, 10%, 4)
= $19,591.11
Get everything into present worth
P = (-$66,000) + (-$10,000)(P/A,10%,6) + ($10,000)(P/F, 10%, 6)
Then solve
An interest rate of 12% per year, compounded monthly, is closest to an effective rate per year of:
- A. 12.08%
- B. 12.28%
- C. 12.48%
- D. 12.68%
Use formula in handbook to solve this problem
Ie = (1+ r/m)m - 1
Depreciation
Dj = (C-Sn/n) = ($50000 - $10000) / 5 = $8,000
BV in year 3 = Initial cost (C) - 3 years of depreciation (Dj)
= $26,000
Use table in Handbook for MACRS constants
2 years of depreciation
Year 1 20%
year 2 32%
so take 20% of $50,000 and then take 32% of $50,000
$10,000 and $16,000 weve already lost $26,000
$50,000 - $26,000 = $24,000
Capitalized costs are present worth values when the analysis period is infinite.
capitalized costs = A/i
$120,000 + A1(P/A, i, n) + A2/i
This is equal to the operating costs plus the loss salvage value.
Find the “F” of the salvaged cost
F = $9000(F/P, 12%, 1) = $10,080
subtract $9000 from $10,080 to get $1080
The cost of one year of ownership and operation is
C = Operating cost + loss salvage value + opportunity cost
= $2500 + $3000 + $1080
Use straight line formula in handbook
Dj = (C - Sn) / n
($140,000 - $20,000) / 7 = $17,143
$17,143(4) = $68,571
= $70,000
Uniform series capital recovery factor.
A = P(A/P, i, n)
use table to find figure
Use single payment present worth formula,
P = F(P/F, i, n)
use table to find out number
solve
We want to know the future value, “F”
we are given “A”
F = A(F/A, i, n)
Find the effective annual interest rate using formula in handbook
ie = ( 1 + r/m)r - 1
we are given the present value we are looking for the future value “F”
F = P(F/P, i, n)
Interest available is
iavailable = F - P = $6417 - $5000
