Unit 7: Financial Mathematics
Perpetuity and Sinking Funds
Perpetuity
A Perpetuity is an annuity in which the periodic payments begin on a fixed date and continue indefinitely.
- The Present Value (PV) of a perpetuity of \( R \) per period at an interest rate of \( i \) per period is:
\[ PV = \frac{R}{i} \]
Sinking Fund
A Sinking Fund is a fund created by accumulating predetermined equal periodic payments to pay off a given sum of money at a specified future date. Businesses use sinking funds to retire debts, replace machinery, or fund major future expenses.
Unlike a standard savings account, the purpose of a sinking fund is explicitly defined, ensuring that a specific financial obligation is met on time.
\[ A = R \left[ \frac{(1 + i)^n - 1}{i} \right] \] where \( A \) = Accumulated Amount, \( R \) = Periodic Payment, \( i \) = Interest Rate per period, \( n \) = Number of periods.
Valuation of Bonds
A bond is a debt security issued by a company or government. Valuation of a bond means determining its present fair market value. The bond’s value equals the present value of future interest payments (Coupons) plus the present value of the face value at maturity.
\[ V = C \left[ \frac{1 - (1+i)^{-n}}{i} \right] + F(1+i)^{-n} \] where \( V \) = Valuation, \( C \) = Periodic Coupon Payment, \( i \) = Required Yield to Maturity, \( n \) = Number of periods to maturity, \( F \) = Face Value.
Calculation of EMI (Equated Monthly Installment)
EMI is the fixed payment made by a borrower to a lender at a specified date each calendar month. EMIs are used to pay off both interest and principal each month so that over a specified number of years, the loan is paid off in full.
There are two primary methods:
- Flat-Rate Method: Interest is computed on the entire original principal amount for the entire duration of the loan.
- Reducing-Balance Method: Interest is calculated strictly on the outstanding balance.
For Reducing Balance: \[ E = P \cdot \frac{i(1+i)^n}{(1+i)^n - 1} \] where \( P \) is the loan amount, \( i \) is the monthly interest rate, and \( n \) is the number of months.
Compound Annual Growth Rate (CAGR) and Depreciation
CAGR represents the smoothed annualized growth rate of an investment over a specified time period. \[ CAGR = \left( \frac{EV}{BV} \right)^{\frac{1}{n}} - 1 \] where \( EV \) = Ending Value, \( BV \) = Beginning Value, \( n \) = Number of years.
Linear Method of Depreciation (Straight-Line Method) \[ D = \frac{Cost - Residual\ Value}{Useful\ Life} \]
Competency-Based Questions
Q1. A scholarship fund has been established to pay ₹ 50,000 annually. If the fund earns 5% interest compounded annually, what must be the initial endowment (Present Value of the Perpetuity) for the scholarship to last indefinitely?
Q2. For a loan of ₹ 10 Lakhs at 9% p.a. interest, to be repaid in 5 years, calculate the Equated Monthly Installment (EMI) using the Reducing-Balance Method. Why might a borrower prefer this method over the Flat-Rate Method?
Q3. A bond with a Face Value of ₹ 1,000 pays a semi-annual coupon of 4% (annual rate of 8%). If the current required yield to maturity is 6% per annum, find the present fair market value of the bond if it matures in 5 years.
Q4. A company invests in advanced machinery costing ₹ 50 Lakhs. They estimate its useful life to be 10 years, and its scrap value thereafter to be ₹ 5 Lakhs. Calculate the annual depreciation charge using the Linear Method. What will be the book value of the machinery at the end of Year 6?
Answers
Ans 1. \( PV = \frac{R}{i} = \frac{50000}{0.05} = 10,00,000 \). The initial endowment must be ₹ 10 Lakhs.
Ans 2. \( P = 10,00,000 \), \( i = \frac{0.09}{12} = 0.0075 \), \( n = 5 \times 12 = 60 \). \( E = 1000000 \times \frac{0.0075(1+0.0075)^{60}}{(1+0.0075)^{60} - 1} pprox 20,758 \). Borrowers prefer it as interest is charged only on the remaining balance, meaning lower total interest paid overall.
Ans 3. Semi-annual yield \( i_y = 3% = 0.03 \). \( n = 10 \) periods. Coupon \( C = 4% \times 1000 = 40 \). \( V = 40 \left[ \frac{1 - (1.03)^{-10}}{0.03} \right] + 1000(1.03)^{-10} = 40(8.53) + 1000(0.744) = 341.2 + 744.1 = 1085.3 \). Market value is ₹ 1085.3.
Ans 4. Annual Depreciation \( = \frac{50 - 5}{10} = 4.5 \) Lakhs per year. Book value after Year 6 = \( 50 - (4.5 \times 6) = 50 - 27 = 23 \) Lakhs.