Ordinary Annuities and Loan Balances
Coverage and the first decision
The current course syllabus explicitly assigns Annuities and Sinking Funds to Week 8 Quiz 2 (1.5 hours). This unit is core coverage. The exact calendar date was not established from Canvas. Treat the written syllabus scope as the coverage evidence, not a prediction of the instructor's questions.
An annuity is a series of equal cash flows separated by equal time intervals. An ordinary annuity pays at the end of each period. With n monthly payments, the timeline is time 0 (valuation), then payments R at months 1, 2, …, n. An annuity due pays at the beginning; that timing belongs in Unit 4. An annuity certain has a specified payment count; a contingent annuity depends on an uncertain event.
Before calculating, mark: payment R; payment frequency; first and last payment dates; desired valuation date; effective rate i per payment interval; and n payments. Do not use an annuity formula for irregular amounts or irregular intervals without breaking the stream into correctly dated pieces.
Derive the factors so their dates make sense
Immediately after the nth end-of-period deposit, the first deposit has earned n − 1 periods of interest and the last has earned none. Therefore F = R[(1 + i)^(n−1) + … + (1 + i) + 1]. Summing this geometric series gives F = R s(i,n), where s(i,n) = [(1 + i)^n − 1]/i.
At time 0, the first end-of-period payment is discounted once and the last n times. Thus P = R[(1 + i)^−1 + … + (1 + i)^−n] = R a(i,n), where a(i,n) = [1 − (1 + i)^−n]/i.
The useful identity s(i,n) = (1 + i)^n a(i,n) moves the same stream from time 0 to time n. At i = 0 both factors approach n, so P = F = nR. A zero-rate case uses that limit instead of dividing by zero.
| Task | Formula | Where the value belongs |
|---|---|---|
| Accumulate regular savings | F = R s(i,n) | At the last ordinary payment |
| Value installments or withdrawals now | P = R a(i,n) | One full period before first payment |
| Find a loan installment | R = P/a(i,n) | P is financed principal at time 0 |
| Find deposit for a future target | R = F/s(i,n) | Sinking-fund model in Unit 5 |
Convert rates before substituting
At 12% nominal compounded monthly, i = .12/12 = .01; 3 years of monthly payments gives n = 36. At 8% effective annually with annual payments, i = .08. If payment and compounding frequencies differ, use the equivalent payment-period rate (1 + j/m)^(m/p) − 1. That last conversion is an explicitly labeled extension; never divide an effective annual rate by p.
Worked savings case: distinguish deposits from earnings
A student deposits PHP 4,000 at each month end for 18 months into an account earning 6% nominal compounded monthly. Timeline: payments at months 1–18, valued immediately after payment 18. i = .005 and n = 18.
F = 4,000[(1.005)^18 − 1]/.005 = PHP 75,143.15. Deposits total 18 × 4,000 = PHP 72,000; investment earnings = PHP 3,143.15. Multiplying 72,000 by (1.005)^18 wrongly treats every deposit as if it existed at month 0. An extra interest factor would value the stream one month after payment 18 instead of immediately after it.
Worked cash-price case: down payment plus financing
An appliance contract requires PHP 20,000 now and PHP 6,000 at each month end for 24 months. The financing rate is 12% nominal compounded monthly. Financed PV = 6,000 a(.01,24) = PHP 127,460.32. Equivalent cash price = 20,000 + financed PV = PHP 147,460.32. The down payment is already at time 0; it is neither discounted nor included among the 24 installments.
Worked loan case: payment, interest and balance
A PHP 180,000 loan is repaid by 36 equal month-end installments at 12% nominal compounded monthly, with the first installment one month after borrowing. R = 180,000/a(.01,36) = PHP 5,978.58. Use the unrounded R for theoretical balances; actual lenders may round each installment and adjust the last.
For an amortization row, interest = opening balance × i; principal repaid = R − interest; closing balance = opening balance − principal repaid = opening balance(1 + i) − R. First-month interest is PHP 1,800, so first-month principal repaid is PHP 4,178.58. A payment exceeding interest reduces principal. If R is no greater than accrued interest, the usual finite positive amortization model cannot settle the debt under those terms.
Immediately after payment k, only n − k future payments remain, each one or more full periods away. Prospective balance B_k = R a(i,n−k). After payment 12 in this example, B_12 = R a(.01,24) = PHP 127,005.20. Just before payment 12, balance is B_12 + R = PHP 132,983.78. The installment due at that instant has not been removed yet.
The retrospective check is B_k = P(1 + i)^k − R s(i,k). Both methods must agree when evaluated at the same instant. Do not use B_k = P − kR, because each installment includes interest.
Deferred streams and balloon payments
Four PHP 12,000 annual payments are due at years 3, 4, 5 and 6, discounted at 8% annually. The ordinary-annuity PV anchor is year 2, one year before the first payment. Value at year 2 = 12,000 a(.08,4). Discount two more years to time 0: P_0 = 12,000 a(.08,4)/(1.08)^2 = PHP 34,075.38. A common error discounts three years after already using a factor that discounts the first payment once.
If a loan leaves a separate balloon F at time n in addition to n ordinary payments, P = R a(i,n) + F(1 + i)^−n. Rearrange R = [P − F(1 + i)^−n]/a(i,n). Label whether a stated final payment replaces or adds to the regular installment; those are different cash flows.
Unknown time and rate
For savings, F = R[(1 + i)^n − 1]/i gives n = ln(1 + Fi/R)/ln(1 + i). If the task asks the first complete deposit count reaching a target, round n up and check the preceding count.
For a loan, P = R[1 − (1 + i)^−n]/i gives n = −ln(1 − Pi/R)/ln(1 + i), requiring R > Pi for finite repayment at positive i. A fractional result normally means a smaller adjusted final installment, not a fractional cash payment date.
To solve for i, use a financial calculator or bounded numerical trial in P = R a(i,n). There is generally no elementary algebraic isolation. For positive fixed installments and n, PV decreases as i increases. Test the calculated rate in the original equation. If a problem asks for nominal annual j with monthly compounding, return j = 12i rather than the monthly rate or EAR. A trial table can bracket i; a rounded table factor gives an approximation, so report that precision honestly.
Worked unknown-rate case
A PHP 100,000 loan has 12 equal PHP 9,000 month-end installments, first one month from now. Solve 100,000 = 9,000 a(i,12). At i = .005, payment PV is PHP 104,570.39, above the borrowed amount; at i = .02 it is PHP 95,178.07, below it. The rate is therefore between .5% and 2% monthly. Repeatedly test the midpoint and replace the bound on the same side of PHP 100,000. The root is i ≈ 0.012043457, or 1.204346% monthly. Substitution returns PV ≈ PHP 100,000. If compounded monthly, nominal annual j = 12i ≈ 14.452148%; EAR = (1+i)^12 − 1 ≈ 15.448936%. These are different answers to different rate questions.
Exam checklist
- Read end versus beginning, and immediately before versus after, before selecting a factor.
- For a loan use PV; for a savings target use FV.
- The ordinary PV anchor is one period before payment 1; the ordinary FV anchor is at payment n.
- Count actual payments, not merely calendar years.
- Distinguish balance, total remaining installments and remaining interest: remaining interest is (n − k)R − B_k.
- A positive-rate ordinary savings FV exceeds nR, while its PV is below nR.
- Solve with unrounded factors; round money only at the end unless a schedule explicitly rounds each row.
Source basis
Paraphrased from Slater Chapter 20, the course Time Value of Money annuity slides, Ordinary Annuities and the current course syllabus. All cases are independently authored for study; no copied assessment or student submission is included.