§1Interest and the time value of money
A dollar today outranks a dollar next year, because today’s dollar can be put to work. Interest is the exchange rate between now and later.
$5000 at 8 % compounded annually: F = 5000 × 1.08¹⁰ = $10 794.62 — a doubling in nine years, as the rule of 72 (72/8 = 9) predicted. Simple interest would have paid only $9000; the gap is interest earning interest.
Every method on this page is compound interest wearing different clothes. The rate i is per period, n counts periods, and the two must match — annual with annual, monthly with monthly.
Contents§2Nominal and effective rates
“12 % per annum, compounded monthly” is not 12 %. The nominal rate divides among the periods; the effective rate is what the year actually costs.
r = 12 % monthly-compounded: i_eff = 1.01¹² − 1 = 12.68 %. Comparing finance offers on nominal rates with different compounding is comparing apples with apple-flavoured things; convert both to effective first.
§3Cash-flow diagrams
Before any formula, draw the money: a timeline with receipts up and payments down, each arrow at the period’s end. Half of all economics errors are diagram errors.
The hero drawing above is the standard machine-purchase picture: price P down at year 0, savings A up at each year’s end, salvage S up at the final year. Conventions worth keeping rigid: cash flows land at period ends unless stated; the rate i applies per period of the same length; and “year 0” is now. With the picture right, every question on this page becomes “what single number at one point in time is this set of arrows worth?” — which is what the factors of §4 compute.
Contents§4Equivalence factors
Six factors move money along the timeline. Learn the notation (X/Y, i, n) as “X given Y” and the table does the arithmetic.
| Factor | Converts | Formula |
|---|---|---|
| (F/P, i, n) | present → future | (1 + i)ⁿ |
| (P/F, i, n) | future → present | (1 + i)⁻ⁿ |
| (F/A, i, n) | annual series → future | [(1 + i)ⁿ − 1]/i |
| (A/F, i, n) | future → annual series | i/[(1 + i)ⁿ − 1] |
| (P/A, i, n) | annual series → present | [1 − (1 + i)⁻ⁿ]/i |
| (A/P, i, n) | present → annual series | i/[1 − (1 + i)⁻ⁿ] — the loan-repayment factor |
| n | F/P | P/F | F/A | A/F | P/A | A/P |
|---|---|---|---|---|---|---|
| 1 | 1.08000 | 0.92593 | 1.00000 | 1.00000 | 0.92593 | 1.08000 |
| 2 | 1.16640 | 0.85734 | 2.08000 | 0.48077 | 1.78326 | 0.56077 |
| 3 | 1.25971 | 0.79383 | 3.24640 | 0.30803 | 2.57710 | 0.38803 |
| 4 | 1.36049 | 0.73503 | 4.50611 | 0.22192 | 3.31213 | 0.30192 |
| 5 | 1.46933 | 0.68058 | 5.86660 | 0.17046 | 3.99271 | 0.25046 |
| 6 | 1.58687 | 0.63017 | 7.33593 | 0.13632 | 4.62288 | 0.21632 |
| 7 | 1.71382 | 0.58349 | 8.92280 | 0.11207 | 5.20637 | 0.19207 |
| 8 | 1.85093 | 0.54027 | 10.63663 | 0.09401 | 5.74664 | 0.17401 |
| 9 | 1.99900 | 0.50025 | 12.48756 | 0.08008 | 6.24689 | 0.16008 |
| 10 | 2.15892 | 0.46319 | 14.48656 | 0.06903 | 6.71008 | 0.14903 |
| Any other rate follows from the formulas above; note A/P = A/F + i and F/P × P/F = 1 — quick self-checks. | ||||||
§5Net present value
Drag every arrow back to year 0 at the required rate and add them up. Positive means the project beats the rate; negative means the money works harder elsewhere.
Cost $42 000 now; saves $9500/yr for 6 years; salvage $5000. NPV = −42 000 + 9500 × (P/A, 10 %, 6) + 5000 × (P/F, 10 %, 6) = −42 000 + 9500 × 4.3553 + 5000 × 0.5645 = +$2197 — buy it. At a 12 % requirement the same arrows fall below zero; the verdict belongs to the rate as much as to the machine.
§6Capitalised cost and EUAC
Two reshapings of the same idea: capitalised cost turns a forever-payment into a lump sum; EUAC turns any lumpy project into a level yearly rent — the fairest way to compare machines of unequal life.
Machine A: $30 000, 5-yr life, $4000 salvage, $2600/yr to run. EUAC = 30 000 × 0.25709 − 4000 × 0.16709 + 2600 = $9644/yr.
Machine B: $45 000, 8-yr life, $6000 salvage, $1500/yr. EUAC = 45 000 × 0.18067 − 6000 × 0.09067 + 1500 = $9086/yr — B wins by $558 a year, despite the dearer sticker, and the unequal lives never had to be reconciled: that is what EUAC is for.
§7Rate of return and payback
Instead of judging a project at a chosen rate, ask what rate the project itself earns: the i that zeroes its NPV. Payback asks a blunter question — how fast the cash comes home.
$20 000 buys $6000/yr for 5 years. Solve (P/A, i, 5) = 20 000/6000 = 3.333. No closed form exists — exactly the situation Newton–Raphson and bisection (Algebra and Equations, §6) were built for: the rate brackets between 15 % (P/A = 3.352) and 16 % (3.274), closing on i = 15.24 %. If the firm’s hurdle is 12 %, the project clears it.
Payback = 20 000/6000 = 3.3 years — instantly communicable, and blind to everything after year 3.3 and to interest itself. Use it as a risk screen (“how long is the money exposed?”), never as the deciding test.
Contents§8Benefit–cost ratio
Public-works cousin of NPV: discount benefits and costs separately, then divide.
Benefits of $130 000 present value against costs of $100 000 give B/C = 1.3. When ranking competing schemes, compare the incremental ratio of each step up in cost — a habit that stops a merely-good big scheme displacing an excellent small one.
Contents§9Depreciation
Depreciation spreads an asset’s cost over its working life for the books and the tax return. Three classical schedules cover practice; the running asset costs $24 000, lasts 5 years, salvages at $4000.
| Year | SL dep. | SL book | SOYD dep. | SOYD book | DDB dep. | DDB book |
|---|---|---|---|---|---|---|
| 1 | 4000.00 | 20000.00 | 6666.67 | 17333.33 | 9600.00 | 14400.00 |
| 2 | 4000.00 | 16000.00 | 5333.33 | 12000.00 | 5760.00 | 8640.00 |
| 3 | 4000.00 | 12000.00 | 4000.00 | 8000.00 | 3456.00 | 5184.00 |
| 4 | 4000.00 | 8000.00 | 2666.67 | 5333.33 | 1184.00 | 4000.00 |
| 5 | 4000.00 | 4000.00 | 1333.33 | 4000.00 | 0.00 | 4000.00 |
| All three land on the $4000 salvage; they differ only in timing. DDB year 4 is limited so book value never falls below salvage. | ||||||
Tax law dictates its own methods and lives — in Australia the prime-cost (straight-line) and diminishing-value methods with legislated effective lives, elsewhere systems such as MACRS. The mathematics above is the machinery underneath all of them; the percentages come from the legislation of the day, so take rates from the current rules, not from memory.
§10Break-even analysis
Fixed costs are the flat line you pay regardless; every unit sold climbs away from it by its contribution. Break-even is where the climb crosses the flat.
Selling price $46, variable cost $28.50: contribution $17.50/unit. Q* = 18 000/17.50 = 1028.6 → 1029 units to break even. The same algebra compares two processes: tooling-heavy versus labour-heavy methods break even against each other at the volume where their total-cost lines cross — the quantitative case for hard tooling on long runs only.
§11Overhead and the machine-hour rate
Rent, power, supervision and depreciation attach to no single job — yet every job must carry a share, or the quotes are fiction.
A cell carrying $260 000/yr of overhead runs 6500 productive hours: rate = $40/machine-hour. A 3.2-hour job therefore carries $128 of overhead before material and labour are counted. Choosing the base — machine hours, labour hours, floor area — is a policy decision; consistency matters more than the choice, and under-absorbed overhead at year’s end is the signal the base or the hours were optimistic.
§12Quick reference
The working core of the page on one card rack.
Compound
F = P(1+i)ⁿ
i_eff = (1+r/m)ᵐ − 1
Series ↔ present
(P/A) = [1−(1+i)⁻ⁿ]/i
(A/P) = i/[1−(1+i)⁻ⁿ]
Decision tests
NPV ≥ 0 · B/C ≥ 1
IRR: rate where NPV = 0
EUAC
P(A/P) − S(A/F) + running
compares unequal lives fairly
Depreciation
SL: (P−S)/n
DDB: (2/n) × book value
Break-even
Q* = fixed/(price − variable)
overhead rate = $/machine-hr