The Time Value of Money
Press the EQN key, then follow this:
A x 100 x ( 1 - ( 1 + I ÷ 100 )^ -N) ÷ I + F x ( 1 + I ÷ 100 ) ^ -N + P
A= uniform amount per interest period
I=interest rate
N= # of periods
F=Future Value FV
P=Present Value PV (Benefit)
For HP 35S users, when input -N use the "+/-" key instead of the subtraction key "-"
Question: A city engineer is deciding which of two alternatives to select.
Alternative Cost Annual Benefit Salvage Useful Life
A $48,000 $13,000 $0 6 yrs
B $40,000 $12,000 $0 6 yrs
If the interest rate is 10% per year, the Benefit-Cost analysis ratio of the alternative to be selected is most nearly:
a. 1.125
b. 1.179
c. 1.307
d. 1.387
Alternative A:
Press the key SOLVE and select the variable you are looking for, in this case the Present Value "P" , then select the equation pressing EQN and selecting the pv equation
A? 13,000; press "R/S"
I? 10; press "R/S"
N? 6; press "R/S"
F? 0; press "R/S"
P? press "R/S"
The screen will show "RUNNING"
then the answer is 56,618.4 which is the present value of the investment.
the initial cost is $48,000
the Benefit-Cost analysis ratio of alternative A is 56,618.4/48,000 = 1.18%
Same methodology for Alternative B => present value is $52,263
Question: A residential community will need to add an additional unit to the existing water treatment plant when the population reaches 40,000. An average of 6,000 new residents are moving into the community each year. The estimated birthrate for the community is 8 per 1,000 and the death rate is 14 per 1,000 resulting in a negative annual increase of -0.6%. The number of years before the second unit is needed for the new water treatment plant is most nearly:
a. 6.57
b. 6.78
c. 6.91
d. 6.87
Solution: The population must reach 40,000 before the additional unit is needed, so that F = 40,000. The number of new residents expected each year is 6,000, therefore, A = 6,000.
The residents should have a natural rate of negative annual increase of -0.6%, i = -0.006.
Use the F/A equation as follows and solve for n:
F = (A (1+i)^n – 1)/i
n = (ln( F i / A + 1))/(ln (1+i))
n = (ln (40,000 x (-0.006)) / 6,000] + 1))/(ln (1 - 0.006))
n = 6.78
Using the equation it gives you the same result: N=6.78323
Press the EQN key, then follow this:
A x 100 x ( 1 - ( 1 + I ÷ 100 )^ -N) ÷ I + F x ( 1 + I ÷ 100 ) ^ -N + P
A= uniform amount per interest period
I=interest rate
N= # of periods
F=Future Value FV
P=Present Value PV (Benefit)
For HP 35S users, when input -N use the "+/-" key instead of the subtraction key "-"
Question: A city engineer is deciding which of two alternatives to select.
Alternative Cost Annual Benefit Salvage Useful Life
A $48,000 $13,000 $0 6 yrs
B $40,000 $12,000 $0 6 yrs
If the interest rate is 10% per year, the Benefit-Cost analysis ratio of the alternative to be selected is most nearly:
a. 1.125
b. 1.179
c. 1.307
d. 1.387
Alternative A:
Press the key SOLVE and select the variable you are looking for, in this case the Present Value "P" , then select the equation pressing EQN and selecting the pv equation
A? 13,000; press "R/S"
I? 10; press "R/S"
N? 6; press "R/S"
F? 0; press "R/S"
P? press "R/S"
The screen will show "RUNNING"
then the answer is 56,618.4 which is the present value of the investment.
the initial cost is $48,000
the Benefit-Cost analysis ratio of alternative A is 56,618.4/48,000 = 1.18%
Same methodology for Alternative B => present value is $52,263
the Benefit-Cost analysis ratio of alternative B is 52,263/40,000 = 1.31%
Answer= C
Question: A residential community will need to add an additional unit to the existing water treatment plant when the population reaches 40,000. An average of 6,000 new residents are moving into the community each year. The estimated birthrate for the community is 8 per 1,000 and the death rate is 14 per 1,000 resulting in a negative annual increase of -0.6%. The number of years before the second unit is needed for the new water treatment plant is most nearly:
a. 6.57
b. 6.78
c. 6.91
d. 6.87
Solution: The population must reach 40,000 before the additional unit is needed, so that F = 40,000. The number of new residents expected each year is 6,000, therefore, A = 6,000.
The residents should have a natural rate of negative annual increase of -0.6%, i = -0.006.
Use the F/A equation as follows and solve for n:
F = (A (1+i)^n – 1)/i
n = (ln( F i / A + 1))/(ln (1+i))
n = (ln (40,000 x (-0.006)) / 6,000] + 1))/(ln (1 - 0.006))
n = 6.78
Using the equation it gives you the same result: N=6.78323
This problem is from Practice Problems for the Civil Engineering PE Exam book.
ReplyDeleteSection 85 Engineering Economic Analysis, problem # 13: A firm expects to receive $32,000 each year for 15 years from sales of a product. An initial investment of $150,000 will be required to manufacture the product. Expenses will run $7,530 per year. Salvage value is zero, and Straight-depreciation is used. The income tax rate is 48%. What is the after-tax rate?
Solution:
Annual depreciation = 150k/15 yrs = 10k
Income reported to IRS after taxes: (32k-10k-7,530)x0.52= 7,524.4
The depreciation of 10k is used only for taxes purposes, it is actually accounted for in the company's income, then the actual annual income after taxes is: $10,000 + 7,524.4 = $17,524.4
Now we have all the info we need to plug in the time value formula in the HP calculator:
A=17524.2
PV=150k
FV=0
N=15
Press EQN (be sure that the equation Ax100X(1-... is in the first line of the screen, if not, look for it)
press (blue arrow) SOLVE
press I
A? 17524, then R/S (positive because it is an income)
N? 15, then R/S
F? 0, then R/S
P? -150000, then R/S (negative because that initial cost comes out from the company's account)
SOLVING........
I=7.99997 = 8%
Much easier than looking at tables and wasting precious time during the exam.