**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

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