Nominal Rate of Interest

A nominal rate \(i^{(m)}\) is convertible m times per year: each sub-period earns \(i^{(m)}/m\), and compounding all m of those sub-periods gives \((1+i^{(m)}/m)^m = 1+i\), the annual effective rate.

By the end you'll be able to convert a nominal rate \(i^{(m)}\) to its annual effective rate \(i\) (and back), for any compounding frequency m.

Predict: holding the nominal rate fixed, does compounding more often per year raise or lower the effective annual rate? Drag m and check.

The nominal rate here is fixed at \(i^{(m)} = 9\%\). Drag m — the number of sub-periods per year — and watch the growth curve reshape: more, smaller sub-periods means interest starts compounding sooner, so the step curve climbs closer to the dashed continuous-compounding limit. Hover the end point or the limit point for exact readouts.

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i(12) = 9.00% (fixed) · per-period rate i(12)/12 = 0.75% · effective annual i = (1 + 0.0075)12 − 1 = 9.38%

Balance over one year: step growth at frequency m vs. the continuous limit
balance under m-thly compounding continuous limit (m→∞)
Formal

Over one year there are m sub-periods, each multiplying the balance by \(1+i^{(m)}/m\). All m of those must compound to one year of growth at the effective rate: \((1+i^{(m)}/m)^m = 1+i\). Solve either direction: \(i = (1+i^{(m)}/m)^m - 1\) (nominal → effective), or \(i^{(m)} = m\big[(1+i)^{1/m} - 1\big]\) (effective → nominal). Because interest compounds within the year, i always exceeds \(i^{(m)}\) for m > 1, and the gap widens as m grows — as m → ∞, \(i^{(m)}\) approaches the force of interest δ.

Applied

This is exactly why loan and savings ads quote a rate "convertible monthly" or "convertible quarterly" instead of just a plain annual number — the compounding frequency changes what you actually earn (or owe), even when the quoted nominal rate looks identical. Before comparing two offers, always convert each to its effective annual rate first; comparing nominal rates directly silently ignores how often each one compounds.

Worked example

A bank quotes a nominal annual rate of 6% convertible monthly: \(i^{(12)} = 0.06\). Monthly rate: \(i^{(12)}/12 = 0.005\). Effective annual rate: \(i = (1.005)^{12} - 1 = 1.061678 - 1 = \) 6.1678%.

Your turn

Same idea, a different frequency: a nominal rate of 8% convertible quarterly, \(i^{(4)} = 0.08\). Quarterly rate: \(i^{(4)}/4 = 0.02\). Raising that to the 4th power: \((1.02)^4 = 1.0824\). Now finish it: \(i = (1.02)^4 - 1 = 1.0824 - 1 = \) ____

Reveal the answer

\(i = 1.0824 - 1 = \) 8.24%. Notice it's above the nominal 8% — as it must be for m > 1 — but by less than the 6%-convertible-monthly example above, since quarterly compounding (m = 4) has fewer, larger sub-periods than monthly (m = 12). Set m = 4 on the slider above to see this same gap on the diagram.

More info — why the gap keeps shrinking (and never disappears)

Each time you double m, the sub-period rate \(i^{(m)}/m\) is cut in half, but there are twice as many compounding events — and compounding always beats simple addition of the same total rate, so i keeps rising with m. That rise has a ceiling, though: as \(m \to \infty\), \((1+i^{(m)}/m)^m \to e^{i^{(m)}}\), the continuous-compounding limit shown as the dashed curve above. Beyond a few thousand sub-periods a year, the effective rate barely moves — daily (m = 365) is already extremely close to that limit. See the Wikibooks formula sheet in Dive deeper for the full derivation.

Check your understanding

Question 1 of 4

A rate is quoted as \(i^{(2)} = 10\%\) convertible semiannually. What is the annual effective rate \(i\)?

Question 2 of 4

Holding a nominal rate fixed, which compounding frequency below produces the highest effective annual rate?

Question 3 of 4

A rate is quoted as \(i^{(52)} = 10.4\%\) convertible weekly. What is the rate credited each week?

Question 4 of 4

A loan quotes a nominal rate of 12% convertible monthly. What is the discount factor \(v\) for one year, and the present value of a 1000 payment due in one year?

Recap

  • A nominal rate \(i^{(m)}\) is compounded m times per year; each sub-period earns \(i^{(m)}/m\). The (m) is a frequency label, not an exponent.
  • Nominal → effective: \(i = (1+i^{(m)}/m)^m - 1\). Effective → nominal: \(i^{(m)} = m\big[(1+i)^{1/m} - 1\big]\).
  • For m > 1, the effective rate i always exceeds the nominal rate \(i^{(m)}\); the gap widens as m grows.
  • As \(m \to \infty\), \(i^{(m)}\) approaches the force of interest δ — the continuous-compounding limit.
  • Always convert to a common effective rate before comparing rates quoted at different frequencies.

Dive deeper

Sources

  • Nominal Rates of Interest Convertible m-thly