Annuities Payable m-thly

Real payments rarely land once a year — mortgages are monthly, bonds are semiannual. An annuity payable m-thly splits a total annual payment of 1 into m equal installments, and its present value \(a_n^{(m)} = (1-v^n)/i^{(m)}\) is always a little larger than the ordinary \(a_n\) — because paying sooner, even in small pieces, is worth more.

By the end you'll be able to compute \(a_n^{(m)}\) and \(\ddot a_n^{(m)}\) directly, and convert between an m-thly annuity and its annual counterpart using the factor \(i/i^{(m)}\).

Predict: if you split the same annual total of 1 into MORE frequent payments — say monthly instead of yearly — does the present value rise or fall? Drag m below and check.

Fix i = 7% and n = 6 years. The top timeline shows a total annual payment of 1 split into m equal installments of 1/m (first two years shown; the pattern repeats for all n). The chart below plots \(a_6^{(m)}\) for m = 1, 2, 4, 12 — watch it climb toward the dashed limit as payments get more frequent.

i(1) = 7.000% · a₆(1) = 4.767 · conversion factor i/i(m) = 1.000

Splitting 1 into m installments per year — payment size shrinks as m grows, but the present value rises
a₆(m) by m continuous-payment limit

Because only the denominator changes from the annual case, an m-thly annuity is always a scalar multiple of the ordinary one — which is why one conversion factor, \(i/i^{(m)}\), connects every m-thly formula back to the annuity you already know.

Formal

Normalize the total annual payment to 1, so each of the m installments inside a period is 1/m. Interest still runs on the annual effective rate i, but discounting mn small payments requires the nominal rate i(m), defined by \((1+i^{(m)}/m)^m = 1+i\), and its discount counterpart \(d^{(m)}\), defined by \((1-d^{(m)}/m)^m = v\). Summing the geometric series of mn payments collapses to the SAME numerator as the annual annuity:

\(a_n^{(m)} = \dfrac{1-v^n}{i^{(m)}}\)  (end of each m-th)    \(\ddot a_n^{(m)} = \dfrac{1-v^n}{d^{(m)}}\)  (start of each m-th)

Since only the denominator changed, \(a_n^{(m)} = (i/i^{(m)})\cdot a_n\) and \(\ddot a_n^{(m)} = (i/d^{(m)})\cdot a_n\). The factor \(i/i^{(m)}\) is slightly above 1 for m > 1 (since \(i^{(m)} < i\)): paying in installments through the period, rather than lumping everything at year-end, pulls each payment's value up a little. The bridge \(\ddot a_n^{(m)} = (1+i^{(m)}/m)\cdot a_n^{(m)}\) mirrors the familiar annual \((1+i)\) relationship between annuity-due and annuity-immediate.

Applied

Think of a mortgage: quoting "1,000 per month" is really an m-thly annuity with m = 12 and total annual payment 12,000. To value it, first convert the annual rate i to the nominal \(i^{(12)}\) — never discount monthly cash flows with i directly — then plug into \(a_n^{(12)}=(1-v^n)/i^{(12)}\) and multiply by the annual total.

Worked example

A loan is repaid by 1,000 per month for 3 years; annual effective i = 8%. Total annual payment = 12,000, n = 3, m = 12.

\(i^{(12)} = 12\big((1.08)^{1/12}-1\big) = 12\times0.006434 = 0.077208\).
\(v^3 = 1.08^{-3} = 0.793832\), so \(1-v^3 = 0.206168\).
\(a_3^{(12)} = 0.206168/0.077208 = 2.670283\).
\(PV = 12{,}000\times2.670283 = \) 32,043.40.

Your turn

Rent of 500 is paid at the end of each quarter (m = 4) for 4 years; annual effective i = 5%. Total annual payment = 2,000.

\(i^{(4)} = 4\big((1.05)^{1/4}-1\big) = 4\times0.012273 = 0.049092\).
\(v^4 = 1.05^{-4} = 0.822702\), so \(1-v^4 = 0.177298\).
Now finish it: \(a_4^{(4)} = 0.177298/0.049092 = \) ____, then \(PV = 2{,}000\times a_4^{(4)} = \) ____

Reveal the answer

\(a_4^{(4)} = 0.177298/0.049092 = \) 3.6115, so \(PV = 2{,}000\times3.6115 = \) 7,223.09. Set m = 4 on the slider above (with the demo's own i and n) and compare how the conversion factor moved the value up from the annual case.

More info — where the conversion factor comes from

Both \(a_n^{(m)}\) and \(a_n\) share the numerator \(1-v^n\) — it only depends on the discount factor over the full n periods, not on how payments are sliced within each period. Dividing one by the other cancels that shared numerator entirely: \(a_n^{(m)}/a_n = i/i^{(m)}\). That's why you never need to re-derive the m-thly formula from scratch — compute the ordinary \(a_n\) you already know, then scale by one ratio of rates. See the Finan guide in Dive deeper for the full derivation from the geometric payment series.

Check your understanding

Question 1 of 4

An annuity pays a total of 800 per year, split into 2 equal installments per year (m = 2), for 10 years at effective annual i = 6%. Its present value uses \(a_{10}^{(2)} = (1-v^{10})/i^{(2)}\) with \(i^{(2)} = 2\big((1.06)^{1/2}-1\big) \approx 5.9126\%\). What is the present value?

Question 2 of 4

Why is the conversion factor \(i/i^{(m)}\) greater than 1 for any m > 1?

Question 3 of 4

What does \(a_n^{(m)}\) equal when m = 1?

Question 4 of 4

You know the annuity-immediate value \(a_{10}=7.36009\) at i = 6% for n = 10 years, and you're told the nominal rate \(i^{(4)}\approx5.870\%\) (m = 4). Using the conversion factor \(i/i^{(m)}\), find \(a_{10}^{(4)}\).

Recap

  • \(a_n^{(m)} = (1-v^n)/i^{(m)}\) (end of each m-th); \(\ddot a_n^{(m)} = (1-v^n)/d^{(m)}\) (start of each m-th) — same numerator as the annual annuity.
  • Total annual payment is normalized to 1, so each of the m installments is 1/m; scale by the total payment amount to price a real cash flow.
  • \(a_n^{(m)} = (i/i^{(m)})\cdot a_n\); the factor \(i/i^{(m)}\) exceeds 1 for m > 1 because \(i^{(m)} < i\) — earlier, smaller payments are discounted less.
  • Immediate m-thly annuities use \(i^{(m)}\) in the denominator; due m-thly annuities use \(d^{(m)}\) — never mix them up.

Dive deeper

Sources

  • Annuities Payable m-thly