WHY MONEY HAS A TIME VALUE — three reasons, and they are independent of one another: 1. EARNING POWER. A rupee today can be invested and become more than a rupee later. This is the dominant reason and the one the formulae model. 2. INFLATION. Purchasing power falls over time. 3. RISK AND PREFERENCE. A promised future rupee may not arrive, and people prefer certain consumption now. THE NOTATION, used throughout engineering economics: P — PRESENT worth, a single amount at t = 0 F — FUTURE worth, a single amount at t = n A — ANNUAL amount, a uniform series at the end of periods 1 through n G — arithmetic GRADIENT, the constant increase per period i — interest rate per period n — number of periods SIMPLE VERSUS COMPOUND INTEREST, the necessary preliminary: SIMPLE: F = P(1 + i·n) interest only on the principal COMPOUND: F = P(1 + i)ⁿ interest on interest ON Rs 10,000 AT 10% FOR 10 YEARS: simple: 10,000(1 + 1.0) = Rs 20,000 compound: 10,000(1.10)¹⁰ = Rs 25,937 THE DIFFERENCE OF Rs 5,937 IS THE INTEREST EARNED ON INTEREST, and it grows without limit as n increases. ALL ENGINEERING ECONOMICS USES COMPOUND INTEREST. ── THE SIX FACTORS ───────────────────────────────────────── Each converts one cash flow pattern into another. The notation (X/Y, i, n) reads "find X given Y". 1. SINGLE PAYMENT COMPOUND AMOUNT — find F given P F = P(1 + i)ⁿ (F/P, i, n) 2. SINGLE PAYMENT PRESENT WORTH — find P given F P = F / (1 + i)ⁿ (P/F, i, n) THE MOST IMPORTANT FORMULA IN THE SUBJECT: DISCOUNTING, the reverse of compounding. 3. UNIFORM SERIES COMPOUND AMOUNT — find F given A F = A [ ((1+i)ⁿ − 1) / i ] (F/A, i, n) A savings plan: what a regular deposit accumulates to. 4. SINKING FUND — find A given F A = F [ i / ((1+i)ⁿ − 1) ] (A/F, i, n) What must be set aside each year to accumulate a known future sum — how a replacement fund for a machine is sized. 5. UNIFORM SERIES PRESENT WORTH — find P given A P = A [ (1 − (1+i)⁻ⁿ) / i ] (P/A, i, n) THE WORKHORSE OF PROJECT APPRAISAL: the present value of a stream of equal annual benefits. 6. CAPITAL RECOVERY — find A given P A = P [ i / (1 − (1+i)⁻ⁿ) ] (A/P, i, n) THE LOAN INSTALMENT FORMULA, and also the annual cost of owning a capital asset. THE RECIPROCAL RELATIONSHIPS, worth knowing as a check: (F/P) = 1/(P/F) (F/A) = 1/(A/F) (P/A) = 1/(A/P) (A/P) = (A/F) + i ← capital recovery = sinking fund plus interest on the principal FACTOR VALUES AT i = 10%, computed: n F/P P/F F/A P/A A/P ─────────────────────────────────────────────────── 1 1.1000 0.9091 1.0000 0.9091 1.1000 5 1.6105 0.6209 6.1051 3.7908 0.2638 10 2.5937 0.3855 15.9374 6.1446 0.1627 20 6.7275 0.1486 57.2750 8.5136 0.1175 ─────────────────────────────────────────────────── READ THE P/F COLUMN DOWNWARD AND THE WHOLE SUBJECT IS THERE: at 10%, a rupee 20 years away is worth 14.9 paisa today. THIS IS WHY LONG-TERM PROJECTS ARE HARD TO JUSTIFY FINANCIALLY, and why the choice of discount rate is politically contentious for infrastructure, climate and public health investments — a high rate makes the distant future almost worthless. NOTE ALSO THAT P/A APPROACHES A LIMIT: as n → ∞, P/A → 1/i = 10. A perpetual annuity of Rs 1 at 10% is worth only Rs 10 today, so YEARS BEYOND ABOUT 20 ADD ALMOST NOTHING.

PROBLEM 1 — SINGLE SUM. Rs 50,000 invested at 12% compounded annually for 8 years. F = 50,000(1.12)⁸ = 50,000 × 2.4760 = Rs 123,800 PROBLEM 2 — PRESENT WORTH. What is Rs 200,000 receivable in 6 years worth now at 9%? P = 200,000 / (1.09)⁶ = 200,000 / 1.6771 = Rs 119,254 PROBLEM 3 — LOAN INSTALMENT (capital recovery). Rs 1,000,000 borrowed at 12% for 10 years, repaid annually. A = 1,000,000 × [0.12 / (1 − 1.12⁻¹⁰)] = 1,000,000 × 0.176984 = Rs 176,984 per year TOTAL PAID = Rs 1,769,842 on a Rs 1,000,000 loan — the extra Rs 769,842 is the interest, and seeing it stated as a total is a useful corrective to the instalment looking small. PROBLEM 4 — SINKING FUND. A machine must be replaced in 8 years at a cost of Rs 800,000. How much must be set aside each year at 8%? A = 800,000 × [0.08 / (1.08⁸ − 1)] = 800,000 × 0.094015 = Rs 75,212 per year PROBLEM 5 — PRESENT WORTH OF A SERIES. Annual savings of Rs 30,000 for 5 years at 10%: P = 30,000 × 3.7908 = Rs 113,724 SO AN INVESTMENT OF UP TO Rs 113,724 WOULD BE JUSTIFIED, and this single figure is what the NPV topic builds on. ── COMPOUNDING FREQUENCY ─────────────────────────────────── THE DISTINCTION EXAMINERS ALWAYS TEST: NOMINAL RATE (r) — the quoted annual rate, which IGNORES compounding within the year. "12% per annum compounded monthly" means 1% per month, quoted as 12%. EFFECTIVE RATE (i) — the rate that actually applies over a year, INCLUDING intra-year compounding: i = (1 + r/m)^m − 1 m = compounding periods per year AT r = 12% NOMINAL: annually (m = 1) i = 12.0000% semiannually (m = 2) i = 12.3600% quarterly (m = 4) i = 12.5509% monthly (m = 12) i = 12.6825% daily (m = 365) i = 12.7475% CONTINUOUS (m → ∞) i = e^r − 1 = 12.7497% TWO OBSERVATIONS WORTH MAKING: · MORE FREQUENT COMPOUNDING ALWAYS INCREASES THE EFFECTIVE RATE, so a borrower should compare effective rates, not quoted ones. This is why consumer credit regulation in most countries requires the EFFECTIVE rate (APR) to be displayed. · THE INCREASE CONVERGES. Going from annual to monthly gains 0.68 percentage points; going from monthly all the way to continuous gains only 0.067. DAILY COMPOUNDING IS PRACTICALLY CONTINUOUS, so "compounded continuously" is a mathematical convenience rather than a materially different product. THE GRADIENT SERIES, for completeness: When the cash flow increases by a constant G each period, the present worth of the gradient part is P = G [ ((1+i)ⁿ − i·n − 1) / (i²(1+i)ⁿ) ] and it is ADDED to the present worth of the base uniform series. Used for maintenance costs that rise with age.