Time value of money and interest formulas

Why it matters

Every engineering investment decision — buying a CNC machine, leasing versus owning, choosing between two plant layouts — involves money spent and received at different times. A rupee today can be invested and earn interest, so it is worth more than a rupee five years from now. The interest formulas in this topic let you move any cash flow to any point in time so that alternatives can be compared fairly.

Key ideas

Time value of money. Money has earning power. If the going rate is 10% per year, ₹1,000 today is equivalent to ₹1,100 one year from now. Two cash flows are equivalent when they have the same value at a common point in time at the given interest rate.

Interest rate. The interest rate i is the ratio of interest earned in one period to the amount at the start of that period. In economic studies it is often called the minimum attractive rate of return (MARR) or discount rate.

Simple interest is charged only on the original principal; the interest does not itself earn interest. It is rarely used in engineering economy except for short loans.

Compound interest is charged on the principal plus all interest accumulated so far. All standard economic-analysis formulas assume compound interest.

Cash-flow diagram. A horizontal time line divided into periods 0, 1, 2, … n. Receipts are drawn as upward arrows, payments as downward arrows. Conventions used by all the factors below:

• P (present worth) occurs at time 0.
• F (future worth) occurs at the end of period n.
• A (uniform series) occurs at the end of every period from 1 to n — the first A is one period after P, and the last A coincides with F.
• G (arithmetic gradient) is zero at end of period 1, G at period 2, 2G at period 3, … (n−1)G at period n.

If the cash flows do not match these positions (for example payments at the beginning of each year, an annuity due), shift them using single-payment factors.

Nominal and effective rates. A nominal rate r per year compounded m times a year means r/m per sub-period. The effective annual rate is what you really earn in a year and is always ≥ the nominal rate. As m → ∞ (continuous compounding) the effective rate tends to e^r − 1.

Discrete compound interest factors. Each factor is written (X/Y, i, n) and read "find X given Y". Factors come in reciprocal pairs, and the series factors are built from the single-payment ones. They connect to every later topic: present worth, annual worth, rate of return, depreciation (sinking fund) and replacement analysis all use them.

Formulas

• Simple interest: I = P·i·n, amount F = P(1 + i·n)
• Single-payment compound amount: F = P(1 + i)ⁿ — factor (F/P, i, n)
• Single-payment present worth: P = F / (1 + i)ⁿ — factor (P/F, i, n)
• Uniform series compound amount: F = A[(1 + i)ⁿ − 1] / i — (F/A, i, n)
• Sinking fund: A = F·i / [(1 + i)ⁿ − 1] — (A/F, i, n)
• Uniform series present worth: P = A[(1 + i)ⁿ − 1] / [i(1 + i)ⁿ] — (P/A, i, n)
• Capital recovery: A = P·i(1 + i)ⁿ / [(1 + i)ⁿ − 1] — (A/P, i, n)
• Useful identity: (A/P, i, n) = (A/F, i, n) + i
• Gradient present worth: P = G[(1 + i)ⁿ − 1 − n·i] / [i²(1 + i)ⁿ] — (P/G, i, n)
• Gradient to uniform series: A = G[1/i − n / ((1 + i)ⁿ − 1)] — (A/G, i, n)
• Effective annual rate: iₑ = (1 + r/m)ᵐ − 1
• Continuous compounding: iₑ = eʳ − 1, F = P·e^(r·n)
• Perpetuity (n → ∞): P = A / i

Symbols: P, F, A, G in ₹ (or any currency); i = interest rate per period (decimal); n = number of periods; r = nominal annual rate (decimal); m = compounding periods per year. Use i and n in the same period unit — if payments are monthly, use the monthly rate and the number of months.

Worked examples

Example 1 (standard) — equal loan instalments. A firm borrows ₹5,00,000 to buy a lathe, repayable in 5 equal end-of-year instalments at 10% per year. Find the instalment and the total interest paid.

1. Capital recovery formula: A = P·i(1 + i)ⁿ / [(1 + i)ⁿ − 1].
2. (1.10)⁵ = 1.61051, so (A/P, 10%, 5) = 0.10 × 1.61051 / 0.61051 = 0.263797.
3. A = 5,00,000 × 0.263797 = ₹1,31,898.7 per year.
4. Total paid = 5 × 1,31,898.7 = ₹6,59,493.7; interest = 6,59,493.7 − 5,00,000 = ₹1,59,493.7.

Answer: instalment ₹1,31,899 per year; total interest ≈ ₹1,59,494.

Example 2 (GATE level) — gradient series. Maintenance on a press costs ₹20,000 at the end of year 1 and rises by ₹5,000 every year for 6 years. At i = 10% per year, find the present worth of maintenance.

1. Split the series: a uniform series A₁ = ₹20,000 plus a gradient G = ₹5,000 (zero at year 1).
2. (P/A, 10%, 6) = [(1.1)⁶ − 1] / [0.1 × (1.1)⁶] = 0.771561 / 0.177156 = 4.35526.
3. (P/G, 10%, 6) = [(1.1)⁶ − 1 − 6 × 0.1] / [0.01 × (1.1)⁶] = 0.171561 / 0.0177156 = 9.68417.
4. P = 20,000 × 4.35526 + 5,000 × 9.68417 = 87,105.2 + 48,420.9.

Answer: P ≈ ₹1,35,526. (Discounting each year's cost individually gives the same total — a good check.)

Example 3 — nominal vs effective. A loan is quoted at 12% per year compounded monthly. Effective annual rate = (1 + 0.12/12)¹² − 1 = (1.01)¹² − 1 = 0.12683, i.e. 12.68%. With continuous compounding it would be e^0.12 − 1 = 12.75%.

Common mistakes

• Putting the first A at time 0. In (P/A) the first payment is at the end of period 1; for payments at the start of each period multiply by (1 + i) or treat the first one separately.
• Forgetting that the gradient starts at period 2: (P/G) gives the present worth of 0, G, 2G … only; the base amount needs a separate (P/A) term.
• Mixing periods — using an annual rate with monthly payments, or a nominal rate where the effective rate is needed.
• Using simple interest in a multi-year economic comparison.
• Rounding factors to two decimals early; keep at least four or five significant figures until the final answer.
• Treating P/F and F/P as interchangeable; they are reciprocals.

For GATE PI

Expect direct numericals: future or present worth of a single sum, an equal-instalment loan, a sinking fund deposit, or an effective-rate conversion. Harder questions combine factors — a series that starts late (deferred annuity), a gradient, or an annuity due — and ask for a single equivalent value. Practise drawing the cash-flow diagram first, then writing the factor chain before substituting numbers. Know the reciprocal pairs and the identity A/P = A/F + i by heart; tables are not provided, so compute factors from the closed forms.

Quick check

1. What is the future value of ₹5,000 after 2 years at 6% per year compounded annually?
2. Which factor converts a single future sum into an equal end-of-year deposit series?
3. A bank quotes 8% per year compounded quarterly. What is the effective annual rate?
4. In the (P/A, i, n) factor, when does the first payment occur?
5. What is the present worth of ₹10,000 per year forever at 8%?

Answers: 1. ₹5,618; 2. the sinking fund factor (A/F, i, n); 3. (1.02)⁴ − 1 = 8.24%; 4. at the end of period 1; 5. 10,000 / 0.08 = ₹1,25,000.