1
An annuity is a stream of equal payments.
Same amount, same gap, same interest rate throughout.
R
the regular payment
₱2,000 a year here
i
rate per payment period
0.06
n
number of payments
5
A simple annuity is one where the payment interval matches the compounding
period — yearly payments with yearly compounding, monthly with monthly. When they
differ it is a general annuity, and that is the next lesson.
2
Each payment earns for a different length of time.
deposit 1
2,000(1.06)⁴ = 2,524.95
earns for 4 years
deposit 3
2,000(1.06)² = 2,247.20
earns for 2 years
deposit 5
2,000
earns nothing yet
the shortcut
F = R[(1 + i)ⁿ − 1i]
future value of the whole stream
Adding all five gives ₱11,274.19. The formula is a shortcut for exactly that sum, not
a different idea.
3
Present value asks what the stream is worth today.
P = R[1 − (1 + i)⁻ⁿi]
note the minus n
= 3000[1 − 1.08⁻⁴0.08]
1.08⁻⁴ ≈ 0.735030
= 3000 × 3.312127
the annuity factor
= ₱9,936.38
the fair price today
Present value is always smaller than the total payments, and future value is always
larger. If your answer sits on the wrong side of ₱12,000, the total of the four payments, you have used the wrong
formula.
4
Paying at the start changes the answer.
ordinary annuity
paid at the END of each period
the usual case
annuity due
paid at the START
rent, insurance premiums
the adjustment
multiply by (1 + i)
every payment earns one period more
Read when the payment is made
"At the end of each year" and "at the beginning of each year" are one multiplication
apart: ₱11,274.19 becomes ₱11,950.64. The wording is easy to skim past and it
changes the answer by more than half a payment.
5
Variations: monthly payments, an annuity due in numbers, the payment from a present value.
Monthly payments: ₱500 at the end of each month for 2 years at 6% compounded monthly
i = 0.06 ÷ 12 = 0.005, n = 24
per month, and 24 months
F = 500[1.005²⁴ − 10.005]
the same formula
= 500 × 25.431955 ≈ ₱12,715.98
deposits total ₱12,000
i and n are per PAYMENT period. Using 0.06 and n = 2 would treat 24 payments as two.
An annuity due in numbers: ₱1,500 at the start of each year for 3 years at 4%
ordinary: 1500[1.04³ − 10.04] = 1500 × 3.1216
as if paid at the end
= ₱4,682.40
the ordinary future value
× 1.04 = ₱4,869.70
each payment earns one year more
Work the ordinary annuity first, then multiply by (1 + i) once — not once per payment.
The payment from a present value: what yearly payment for 3 years at 10% is worth ₱24,868.52 today?
24868.52 = R[1 − 1.1⁻³0.1]
P is known, R is not
1.1⁻³ ≈ 0.751315, factor ≈ 2.486852
the annuity factor
R = 24868.52 ÷ 2.486852 ≈ ₱10,000
divide
Three payments of ₱10,000 are worth only ₱24,868.52 today: the later ones are discounted more.
6
Rearrange to find the payment.
F = R[(1 + i)ⁿ − 1i]
F is known, R is not
150000 = R[1.04⁶ − 10.04]
1.04⁶ ≈ 1.265319
150000 = R × 6.632975
the factor
R ≈ ₱22,614.29 a year
less than 150,000 ÷ 6
₱11,274.19 for the opening problem
Saving ₱22,614 a year reaches ₱150,000 in six years, while ₱25,000 a year would be
needed with no interest at all. The gap is what the interest did, and it is the
whole argument for starting early.
On the exam
Periodical exams ask for the future value or the present value of an ordinary annuity, the regular payment for a target, or the difference between ordinary and due; entrance tests mostly test the vocabulary — equal payments at equal intervals, end versus start — and whether F beats the total paid in. Name R, i and n first, choose F for "in the future" and P for "today". For a 10-second check, F must be more than R × n and P must be less.
Remember the flow
Name R, i and n · Forward is F, back is P · Check the payment timing · Rearrange for R when the target is given
Future value beats the total paid in; present value falls short of it.
