Definition: A sinking fund is an annuity created for accumulating money over a period of time, which can be used for paying off a loan or for replacement of some asset or other financial obligations.
Formula
S = R × [(1+i)^n − 1] / iWhere: S = Total amount accumulated (Future Value) R = Periodic payment (money set aside each period) i = Interest rate per period (annual rate ÷ number of compoundings per year) n = Total number of payment periodsTo find R (periodic payment): R = iS / [(1+i)^n − 1]
📝 NOTE: If compounded half-yearly: i = rate/(2×100), n = years × 2. If quarterly: i = rate/(4×100), n = years × 4.
SOLVED EXAMPLE: Ex.1 — Company Sinking Fund
Problem: A company sets aside ₹5000 annually for 10 years in a sinking fund at 5% p.a. compound interest. Find the surplus after paying debenture of ₹60,000.
Given: R = ₹5000, n = 10, i = 5/100 = 0.05
Using: S = R × [(1+i)^n − 1] / i
S = 5000 × [(1.05)^10 − 1] / 0.05
Let x = (1.05)^10 → log x = 10 × log(1.05) = 10 × 0.0212 = 0.2120
Problem: A sinking fund is created to accumulate ₹10,00,000 for child’s education at end of 25 years. Interest = 4% p.a. Find annual payment. [Given (1.04)^25 = 2.6658]
S = ₹10,00,000, n = 25, i = 4/100 = 0.04
R = iS / [(1+i)^n − 1] = (0.04 × 1000000) / [(1.04)^25 − 1]
SOLVED EXAMPLE: Ex.3 — Half Yearly Sinking Fund [CBSE 2023]
Problem: Accumulate ₹1,00,000 for daughter’s education after 10 years. Invest at 5% p.a. compounded half-yearly. [Given (1.025)^20 = 1.6386]
S = ₹1,00,000, rate = 5% half-yearly → i = 5/(2×100) = 0.025
n = 10 × 2 = 20 periods
R = iS / [(1+i)^n − 1] = (0.025 × 100000) / [1.6386 − 1]
= 2500 / 0.6386 = ₹3,914.81
Answer: Each half-yearly payment = ₹3,914.81
THINGS TO REMEMBER
a) Sinking Fund = money set aside periodically to accumulate a future amount
b) Use S = R[(1+i)^n − 1]/i when you know R and want S
c) Use R = iS/[(1+i)^n − 1] when you know target amount S and want to find R
d) Always convert annual rate to per-period rate: i = annual rate / (100 × compoundings per year)
e) Always convert years to number of periods: n = years × compoundings per year
f) Surplus = Amount accumulated − Amount to be paid
2. PERPETUITY
Definition: An annuity in which payments continue forever (infinite term). Example: Property rent received forever.
Two Types of Perpetuity
(1) Perpetuity payable at END of each period
(2) Perpetuity payable at BEGINNING of each period
Type 1 (End): P = R / iType 2 (Beginning): P = R + R/i = R(1 + 1/i)Where: P = Present value of perpetuity R = Periodic payment amount i = Interest rate per period
NOTE: When interest is calculated only at the end of a year, nominal rate = effective rate.
Solved Examples — Perpetuity
SOLVED EXAMPLE: Perpetuity at End (Annual)
Problem: Find present value of perpetuity of ₹10,000 payable at end of each year if money is worth 5% p.a.
R = ₹10,000, i = 5/100 = 0.05
P = R/i = 10000/0.05 = ₹2,00,000
Answer: Present value = ₹2,00,000
SOLVED EXAMPLE: Perpetuity at End (Semi-Annual) [CBSE 2023]
Problem: Present value of perpetuity of ₹2400 at end of each 6 months = ₹1,20,000. Find rate of interest (convertible half-yearly).
P = ₹1,20,000, R = ₹2400
P = R/i → 120000 = 2400/i
i = 2400/120000 = r/200 → r = 2400 × 200/120000 = 4% p.a.
Answer: Rate = 4% p.a. convertible half-yearly
SOLVED EXAMPLE: Perpetuity at Beginning
Problem: Find present value of perpetuity of ₹8400 payable at beginning of each year if money is worth 6% compounded annually.
R = ₹8400, i = 6/100 = 0.06
P = R + R/i = 8400 + 8400/0.06 = 8400 + 1,40,000 = ₹1,48,400
Answer: ₹1,48,400
SOLVED EXAMPLE: Father investing for son’s perpetual income
Problem: A father wants his son to receive ₹60,000 at end of each year indefinitely. Money is worth 6% p.a. How much should father invest?
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