Future Value of Ordinary Annuities

ENGINEERING ECONOMICS



1. Find the accumulated amount of the ordinary annuity paying an amortization of 1000P per month at a
rate of 12% compounded monthly for 5 years

GIVEN:

A = 1000P n = 5 years or 60 months
0.12
j = 12% or 0.12 i= 12
i= 0.01

n1 = monthly or 12

FIND: F

FORMULA:

(1+𝑖)𝑛−1
F=A( 𝑖
)

SOLUTION:

(1+0.01)60 −1
F = 1000P ( )
0.01

F = 81670P

FINAL ANSWER: 81,670P will be the accumulated amount of the annuity for five years.




2. What present sum is equivalent to a series of 1000P annual end-of-year payments, if a total of 20
payments are made and interest is 12%?

GIVEN:

A = 1000P

i = 12% = 0.12

n = 20

FIND: P

FORMULA:

1 − (1 + i)−𝑛
P = A( )
i

, ENGINEERING ECONOMICS



SOLUTION:

1−(1+0.12)−20 1−(1.12)−20
P = 1000P ( 0.12
) → 1000P ( 0.12
) → 1000P(7.469443624)

P = 7,469.44P

FINAL ANSWER: 7,469.44P is the present sum equivalent to a series of 1000P annual end-of-year
payments.




3. A man made ten annual-end-of year purchases of 1000P common stock. At the end of 10th year, he sold
all the stock for 12000P. What interest rate did he obtain on his investment?

GIVEN:

F = 12,000

A = 1,000

n = 10

FIND: i

FORMULA:

(1+𝑖)𝑛 −1
F=A( )
𝑖

SOLUTION:

(1+𝑖)𝑛 −1
F=A( 𝑖
)

(1+𝑖)10 −1
12000 = 1000 { 𝑖
}

i = 0.039 ≈ 4%

FINAL ANSWER: He obtained 4% of interest rate from his investment.