Time value of money: interest and annuities

Why it matters

A plant is paid for today but earns money over 10–20 years, so costs and revenues at different dates must be brought to a common basis before they can be compared. Interest formulas do exactly that. They underlie loan repayments, sinking funds for replacement, capitalised cost of equipment, net present value and the comparison of alternatives in every later topic.

Key ideas

Interest is the charge for using money. The rate at which a company can earn on its money elsewhere (its cost of capital, or minimum acceptable rate of return) is the rate used to move sums through time.

Simple and compound interest. With simple interest, interest is earned only on the original principal. With compound interest, interest earned in each period is added to the principal and itself earns interest. Economic evaluation of plants always uses compound interest, because money received is in practice reinvested.

Nominal and effective rates. A nominal annual rate r compounded m times a year means a rate r/m is applied each period. The effective annual rate is the single annual rate that gives the same result; it exceeds r whenever m > 1. In the limit of continuous compounding the effective rate is eʳ − 1.

Cash-flow diagrams. Draw time on a horizontal axis in years (0 = now), receipts as upward arrows and payments as downward arrows. The standard convention is that a payment in a given year occurs at the end of that year, and that an annuity begins one period after time zero.

Annuities. An annuity is a series of equal payments A at equal intervals. Common uses:

• Capital recovery — the equal annual amount that repays a present sum P with interest over n years (loan instalment, equivalent annual cost of equipment).
• Sinking fund — the equal annual deposit that accumulates to a future sum F (for example, a fund to replace equipment at the end of its life).
• Present worth of an annuity — the lump sum today equivalent to a series of future payments.

Capitalised cost. The capitalised cost of an item is its first cost plus the present sum that, invested at rate i, would pay for its replacement every n years for ever. It allows equipment with different lives to be compared on one number. A perpetual annuity A has present worth A/i.

Inflation. If prices rise at rate f, the real interest rate i_r is related to the market (nominal) rate i by (1 + i) = (1 + i_r)(1 + f). Use the market rate with cash flows in inflated (money-of-the-day) rupees, or the real rate with cash flows in constant rupees — never mix the two.

Formulas

Simple interest: I = P·i·n, F = P·(1 + i·n) Compound amount: F = P·(1 + i)ⁿ and present worth P = F / (1 + i)ⁿ Nominal r compounded m times a year for n years: F = P·(1 + r/m)^(m·n) Effective annual rate: i_eff = (1 + r/m)^m − 1; continuous: i_eff = eʳ − 1, F = P·e^(r·n) Future worth of an annuity: F = A·[(1 + i)ⁿ − 1] / i Sinking-fund deposit: A = F·i / [(1 + i)ⁿ − 1] Present worth of an annuity: P = A·[(1 + i)ⁿ − 1] / [i·(1 + i)ⁿ] Capital recovery: A = P·i·(1 + i)ⁿ / [(1 + i)ⁿ − 1] Capitalised cost: K = C_V + (C_R) / [(1 + i)ⁿ − 1], with replacement cost C_R = C_V − S Perpetuity: P = A / i Real rate: (1 + i) = (1 + i_r)·(1 + f)

Symbols: P present sum (₹), F future sum (₹), A end-of-period annuity (₹ per period), i interest rate per period (decimal), n number of periods, r nominal annual rate (decimal), m compounding periods per year, C_V first cost (₹), S salvage value (₹), f inflation rate (decimal). Rate and period must match (monthly rate with months).

Worked examples

Example 1 (standard). ₹5 lakh is invested at 12% per year compounded monthly for 3 years. Find the final amount and the effective annual rate.

1. Rate per month = 0.12/12 = 0.01; periods = 36.
2. F = P·(1 + r/m)^(m·n) = 5 × 1.01³⁶ = 5 × 1.4308 = ₹7.154 lakh.
3. i_eff = (1.01)¹² − 1 = 1.1268 − 1 = 0.1268.
4. F = ₹7.15 lakh; effective rate = 12.68% per year (compared with 12.75% for continuous compounding).

Example 2 (GATE level). A reactor costs ₹40 lakh installed, lasts 8 years and has a salvage value of ₹4 lakh. Money is worth 10% per year. Find (a) the equivalent uniform annual cost of owning it and (b) its capitalised cost.

1. (1 + i)ⁿ = 1.1⁸ = 2.1436.
2. Capital recovery factor: i(1 + i)ⁿ/[(1 + i)ⁿ − 1] = 0.1 × 2.1436/1.1436 = 0.18744.
3. (a) Annual cost = (C_V − S)·(A/P) + S·i = 36 × 0.18744 + 4 × 0.10 = 6.748 + 0.400 = ₹7.148 lakh/yr.
4. (b) K = C_V + (C_V − S)/[(1 + i)ⁿ − 1] = 40 + 36/1.1436 = 40 + 31.48 = ₹71.48 lakh.
5. Check: a perpetual annual cost of 7.148 at 10% has present worth 7.148/0.10 = 71.48 — the two answers are consistent.
6. Annual cost ≈ ₹7.15 lakh/yr; capitalised cost ≈ ₹71.5 lakh.

Common mistakes

• Using simple interest where compounding is implied.
• Using the nominal rate with monthly periods (r instead of r/m) or the annual number of periods with a monthly rate.
• Confusing sinking-fund and capital-recovery factors: they differ by exactly i (A/P = A/F + i).
• Putting the first annuity payment at time zero when the formula assumes end-of-period payments.
• In capitalised cost, using the full first cost instead of (first cost − salvage) as the replacement amount.
• Mixing real rates with inflated cash flows.

For GATE CH

Expect direct application of compound-interest, annuity, sinking-fund and capitalised-cost formulas, effective-rate conversions and equivalent annual cost. Many numericals are one or two steps but need care with which factor applies; practise drawing the cash-flow diagram first.

Quick check

1. ₹1 lakh at 10% compounded annually for 2 years?
2. Effective rate for 8% nominal compounded quarterly?
3. Annual deposit at 10% to accumulate ₹10 lakh in 5 years?
4. Present worth of ₹2 lakh per year for ever at 8%?

Answers: 1. 1 × 1.1² = ₹1.21 lakh. 2. 1.02⁴ − 1 = 8.24%. 3. A = 10 × 0.1/(1.1⁵ − 1) = 10 × 0.1/0.6105 = ₹1.638 lakh/yr. 4. 2/0.08 = ₹25 lakh.