Build financial mathematics fluency through 20 original five-choice questions and 60 worked concepts covering cash-flow valuation, annuities, loans, bonds, yield curves and immunization. Each question includes a solution and explanations of realistic alternative answers. This representative preview follows the supplied October 2026 SOA FM syllabus allocation, applicable as of October 16, 2026; it is not a full-length examination or the complete ASA qualification. The published examination has 30 questions over 2.5 hours. Bond valuation stays on coupon dates. Coverage includes the duration and convexity competencies associated with required study note FM-24-17, sections 1–4. Citations establish syllabus relevance; calculations use established financial mathematics.
What this guide covers
Original SOA FM practice with 20 solved questions and 60 worked lessons for the October 2026 syllabus.
Time Value of Money
1. Simple and compound accumulation functions
An accumulation function a(t) gives the value at time t of one unit invested at time zero, with a(0) = 1. Simple interest uses a(t) = 1 + it; compound interest uses a(t) = (1 + i)ᵗ for a constant effective rate. Compound accumulation is multiplicative across equal intervals, while simple-interest accumulation generally depends on the original investment date. For an investment already outstanding at time s, its accumulation to time t is a(t)/a(s), rather than automatically a(t − s). The distinction matters when comparing dated offers.
Worked example: A $2,000 deposit earning 6% simple interest reaches $2,360 after three years. With 6% annual compound interest it reaches $2,382.03. Under the simple-interest contract, accumulation from year one to year three is 1.18/1.06, not 1.12.
Common mistake: Applying the original simple-interest rate independently to each later interval incorrectly grants a stationary compounding property to simple interest.
Official syllabus reference: [1]
2. Converting interest and discount quotations
Equivalent rates produce the same accumulation over the same interval. For annual effective interest i and discount d, 1 − d = 1/(1 + i), so d = i/(1 + i). A nominal interest rate j convertible m times yearly satisfies 1 + i = (1 + j/m)ᵐ. A nominal discount rate k convertible m times yearly satisfies 1 + i = (1 − k/m)⁻ᵐ. Conversion requires identifying whether the quoted subperiod rate is interest charged on initial value or discount deducted from terminal value. Nominal rates themselves are not annual accumulation rates.
Worked example: A nominal interest rate of 12% convertible monthly gives a monthly rate of 1%. Its annual effective interest rate is 1.01¹² − 1, approximately 12.6825%. Its equivalent annual effective discount rate is 1 − 1.01⁻¹², approximately 11.2551%.
Common mistake: Dividing an annual effective interest rate by twelve generally does not produce the equivalent monthly effective interest rate.
Official syllabus reference: [1]
3. Integrating a variable force of interest
The force of interest δ(t) is the instantaneous proportional growth rate a'(t)/a(t). Accumulation between times s and t is exp(∫ₛᵗδ(u)du), and discounting uses the negative of that integral. For constant force δ, the annual effective rate is eᵟ − 1 and δ = ln(1 + i). A variable force must be integrated over the actual interval; its endpoint values are not interchangeable with its average. Piecewise forces are handled by adding interval integrals before exponentiating, which is equivalent to multiplying their accumulation factors.
Worked example: Suppose δ(t) = 0.03 + 0.002t. Between years two and five, the integrated force is 0.03(3) + 0.001(25 − 4) = 0.111. A $4,000 balance at year two therefore becomes 4,000e⁰·¹¹¹ at year five.
Common mistake: Multiplying the terminal force by elapsed time ignores the path of a changing force and generally gives the wrong accumulation.
Official syllabus reference: [1]
4. Equations of value at a focal date
An equation of value compares cash flows after moving every amount to one common focal date. With constant annual effective interest i, a cash flow C at time t has value C(1 + i)^(T − t) at focal date T. Earlier payments accumulate and later payments discount. Under a coherent accumulation function, changing the focal date preserves the economic equivalence. Signed cash flows can instead be summed to zero. The rate must describe the same valuation basis for both sides; mixing accumulation assumptions can destroy focal-date consistency.
Worked example: A debtor replaces $1,000 due now and $1,100 due in two years with one payment in one year. At 10% annual effective interest, the replacement is 1,000(1.10) + 1,100/1.10 = $2,100, with both obligations valued at year one.
Common mistake: Adding payments from different dates before applying interest treats dated dollars as interchangeable and can misprice an otherwise equivalent settlement.
Official syllabus reference: [1]
5. Inflation and real investment returns
A nominal return measures growth in money, while a real return measures growth in purchasing power. If the nominal effective return is i and inflation over the identical interval is π, the real effective return r satisfies 1 + r = (1 + i)/(1 + π). Subtracting inflation from interest is only an approximation because the exact relationship includes a cross-product. Cash flows may be valued consistently in nominal money with nominal rates or in purchasing-power units with real rates. Mixing real cash flows and nominal discount rates changes the valuation basis.
Worked example: An account grows from $10,000 to $10,800 while a representative consumption basket rises in price by 3%. Its real return is 1.08/1.03 − 1 = 4.8544%. The investor gains purchasing power, but less than the 8% monetary return suggests.
Common mistake: Calling the nominal return minus inflation an exact real return overlooks the multiplicative relationship between money growth and price growth.
Official syllabus reference: [1]
6. Discount functions and interval accumulation
A discount function v(t) is the present value of one unit payable at time t, so v(t) = 1/a(t). The time-s value of an amount C payable at time t is Cv(t)/v(s), provided both values use the same underlying accumulation function. This ratio distinguishes a discount factor to a date from a factor over an interval. When rates vary, the interval factor usually depends on both endpoints. The approach supports current-value calculations without assuming a constant rate, and it makes the direction of discounting explicit.
Worked example: Suppose a discount function gives v(2) = 0.92 and v(5) = 0.78. The value at year two of $5,000 payable at year five is 5,000(0.78/0.92) = $4,239.13. The ratio discounts only the interval still remaining.
Common mistake: Using v(5) alone to obtain a year-two value discounts to time zero instead of to the requested valuation date.
Official syllabus reference: [1]
Annuities/cash flows with non-contingent payments
7. Annuity-immediate present-value factors
An n-payment annuity-immediate pays one unit at the ends of periods 1 through n. Its present-value factor is aₙ = ∑ₖ₌₁ⁿvᵏ = (1 − vⁿ)/i, where v = 1/(1 + i). Multiplying by a level payment gives the stream's value one payment period before the first payment. The interest rate must be effective for that payment period. At zero interest, the limiting factor is n. The formula is useful for funding withdrawals, valuing contracts and determining equivalent level payments from a known present value.
Worked example: A fund will pay $3,000 annually for four years, first payment in one year. At 5%, a₄ = 3.54595, so the required initial fund is $10,637.85. The calculation includes exactly four discounted withdrawals and no payment today.
Common mistake: Using an annuity-immediate factor for a payment made today discounts that first payment unnecessarily and understates the required fund.
Official syllabus reference: [1]
8. Annuity-due timing and endpoint replacement
An n-payment annuity-due pays at times 0 through n − 1. Each payment is one period earlier than its annuity-immediate counterpart, so äₙ = (1 + i)aₙ. An equivalent identity is äₙ = 1 + aₙ₋₁: one payment now plus n − 1 future payments. Shifting a finite annuity earlier preserves the number of payments. If converting by adding a time-zero payment, the original final payment must also be removed. Due factors appear naturally in rent, prepaid contributions and withdrawals beginning immediately.
Worked example: Five annual payments of $800 begin today. At 6%, their value is 800(1.06)a₅. Alternatively, value the immediate $800 separately and add the value of four later payments, obtaining 800 + 800a₄ with the same result.
Common mistake: Adding one immediate payment to an n-payment annuity-immediate creates n + 1 payments unless the original final payment is removed.
Official syllabus reference: [1]
9. Accumulated annuity values
The accumulated-value factor sₙ = [(1 + i)ⁿ − 1]/i values n unit payments made at period ends at the time of the final payment. It also equals aₙ(1 + i)ⁿ. The last payment earns no interest before that valuation instant, while the first earns n − 1 periods. For payments at period beginnings accumulated to time n, the factor is s̈ₙ = (1 + i)sₙ. The valuation date must be specified separately from the last payment date, especially for annuities-due.
Worked example: An investor deposits $500 at each year end for three years at 4%. Immediately after the third deposit, the balance is 500(1.04² + 1.04 + 1) = $1,560.80. The final deposit contributes its face amount without additional interest.
Common mistake: Giving the last end-of-period deposit a full period of interest moves the accumulated-value date one period too far forward.
Official syllabus reference: [1]
10. Deferred annuities and first-payment placement
A deferred annuity is valued by locating its standard annuity factor first and then moving that value to the requested date. If n level payments begin at time m + 1, their present value is Rvᵐaₙ. The immediate-annuity factor belongs at time m, one period before the first payment. Alternatively, if payments begin at time m, use Rvᵐäₙ. These expressions agree when they describe the same dates. Drawing the first and last payments prevents confusing a deferral period with the first payment's absolute time.
Worked example: Six payments of $2,000 occur at years four through nine. At annual rate i, their time-zero value is 2,000v³a₆. The a₆ factor places value at year three, and the extra v³ then carries that value back to today.
Common mistake: Discounting an immediate-annuity factor by the first payment's full time delays every payment by one additional period.
Official syllabus reference: [1]
11. Payment frequency and equivalent periodic rates
When payments occur m times yearly, convert the annual effective rate i to the effective payment-period rate q = (1 + i)^(1/m) − 1. An n-year stream then contains mn payments. If the stated annual payment total is R, each installment is R/m when installments are equal. Present value can be computed as (R/m)aₘₙ at rate q. This avoids ambiguity in actuarial m-thly notation, whose unit benefit is often a total annual amount of one rather than one at every installment.
Worked example: A contract pays $1,200 per year in twelve equal monthly installments for two years. If annual effective interest is 1.01¹² − 1, each payment is $100 and the monthly rate is 1%. Present value is therefore 100a₂₄ at 1%.
Common mistake: Combining monthly payments with an annual effective rate in the same annuity factor mismatches the rate and payment periods.
Official syllabus reference: [1]
12. Continuous payment streams
A continuously paid annuity is described by a payment rate c(t), measured in money per unit time. Its present value is ∫c(t)v(t)dt over the payment interval. With constant annual payment rate R and constant force δ over n years, PV = R(1 − e⁻ᵟⁿ)/δ. Accumulated value at time n is R(eᵟⁿ − 1)/δ. The rate R is not an immediate lump sum, and isolated endpoints contribute no separate payment unless the contract explicitly includes one. Continuous valuation naturally accommodates changing income rates.
Worked example: A project earns revenue continuously at $6,000 per year for three years. With force 4%, present value is 6,000(1 − e⁻⁰·¹²)/0.04. This discounts each small receipt for its own waiting time rather than assigning all revenue to year ends.
Common mistake: Treating a continuous annual payment rate as a lump sum paid at each year end changes the cash-flow timing and value.
Official syllabus reference: [1]
13. Finite arithmetic payment progressions
An arithmetic payment stream has payment R + (k − 1)G at time k, where G is a fixed monetary increment. Its present value is Raₙ + G[(Ia)ₙ − aₙ], with (Ia)ₙ = ∑ₖ₌₁ⁿkvᵏ = (äₙ − nvⁿ)/i. Negative G handles decreasing payments while scheduled amounts remain as specified. The distinction between k and k − 1 determines the first payment. Decomposition into a level portion and an increasing portion is usually clearer than memorizing a separate formula for every starting amount.
Worked example: A maintenance contract pays $200, $250 and $300 at the next three year ends. At 10%, value it directly as 200/1.10 + 250/1.10² + 300/1.10³. Equivalently, use 200a₃ + 50[(Ia)₃ − a₃].
Common mistake: Using R + kG when the stated first payment is R increases every scheduled payment by one extra increment.
Official syllabus reference: [1]
14. Arithmetic increasing perpetuities
For positive interest i, a perpetuity paying R + (k − 1)G at time k has present value R/i + G/i². The level component contributes R/i, while the increments contribute G∑(k − 1)vᵏ = G/i². A stream paying Gk instead has value G(1 + i)/i² because its first payment already contains one increment. Positive interest makes polynomially increasing payments summable under compound discounting. A perpetually decreasing arithmetic schedule may eventually become negative, so its mathematical continuation must match the contract's intended meaning.
Worked example: Annual payments begin at $400 and rise by $30 each year forever. At 6%, present value is 400/0.06 + 30/0.06² = $15,000. The $30 increment starts with the second payment, which is $430.
Common mistake: Using the increasing-perpetuity formula for payments Gk without adjusting the starting payment adds an increment in the first year.
Official syllabus reference: [1]
15. Finite geometric annuities
A geometric annuity pays R(1 + g)^(k − 1) at time k. Its present value for n payments is R[1 − ((1 + g)/(1 + i))ⁿ]/(i − g), provided i differs from g. This comes from summing a geometric series with ratio (1 + g)/(1 + i). Unlike a perpetuity, a finite stream remains valuable when growth equals or exceeds interest because it contains only finitely many payments. A due version shifts every payment earlier by multiplying the immediate value by 1 + i.
Worked example: A contract pays $1,000 in one year and increases payments by 3% annually for five payments. At 7%, present value is 1,000[1 − (1.03/1.07)⁵]/0.04. The fifth payment is $1,000(1.03)⁴, not $1,000(1.03)⁵.
Common mistake: Applying five growth increases to a five-payment annuity overstates the final payment because the first payment is the growth baseline.
Official syllabus reference: [1]
16. Growing perpetuities and convergence
A perpetuity with first payment R one period from its valuation date and subsequent growth g has value R/(i − g), assuming ordinary positive payment factors and i > g. The condition makes the discounted-payment ratio (1 + g)/(1 + i) less than one. The formula values the stream one period before its first payment, regardless of when that segment begins. If positive payments grow as fast as or faster than the discount rate forever, the present-value series does not converge to a finite amount.
Worked example: A stream first pays $900 at year six and then grows by 2% annually forever. With interest 5%, its value at year five is 900/(0.05 − 0.02) = $30,000. Its value today is 30,000/1.05⁵.
Common mistake: Using a growing-perpetuity formula when growth exceeds interest produces a negative expression that is not a valid finite present value.
Official syllabus reference: [1]
17. The equal-growth-and-interest boundary
When geometric payment growth g equals interest i, the usual finite-annuity formula appears to divide zero by zero. Evaluate the cash flows directly instead. Every discounted payment R(1 + i)^(k − 1)/(1 + i)ᵏ equals R/(1 + i), so n payments have present value nR/(1 + i). Their value at the last payment date is nR(1 + i)^(n − 1). This is a finite-stream boundary result, not a convergent perpetuity formula. Recognizing it avoids unnecessary numerical instability near equal rates.
Worked example: Six annual deposits start at $200 and increase by 5% each year, while the account earns 5%. At year six, each deposit contributes $200(1.05)⁵. The total is therefore six times that amount, approximately $1,531.54.
Common mistake: Concluding that equal growth and interest makes a finite annuity undefined confuses a removable formula singularity with the underlying cash flows.
Official syllabus reference: [1]
18. Decomposing irregular and piecewise cash flows
Present value is linear: the value of a combined stream equals the sum of its component values at a common date. Irregular schedules can therefore be separated into level annuities, growth segments and isolated payments. Each component requires its own first-payment date, last-payment date and payment count before values are combined. Direct summation remains valid when no standard pattern fits. Decomposition is particularly useful when a contract changes payment rules partway through, but boundary payments must belong to exactly one segment to prevent omission or double counting.
Worked example: A contract pays $700 at years one and two, then $1,000 at years three through five, plus a $2,000 bonus at year five. Its present value is 700a₂ + 1,000v²a₃ + 2,000v⁵ at the chosen annual rate.
Common mistake: Applying one annuity factor to a piecewise schedule can silently move the changeover date or count a boundary payment twice.
Official syllabus reference: [1]
19. Solving an annuity's payment count
For a level annuity-immediate with present value P, payment R and positive periodic rate i, P = R(1 − vⁿ)/i implies n = −ln(1 − iP/R)/ln(1 + i). A finite positive solution requires R > iP. The resulting real number describes an exact mathematical annuity term, while actual contracts may require an integer payment count and an adjusted final payment. If payments occur only on specified dates, do not automatically interpret a fractional n as permission for an unscheduled fractional-period payment.
Worked example: A $5,000 fund supports annual withdrawals of $1,000 at 5%. Solving gives n = −ln(0.75)/ln(1.05), approximately 5.90. If withdrawals must occur at year ends, determine the feasible full withdrawals and calculate the remaining balance separately.
Common mistake: Rounding a calculated term without recomputing the terminal payment generally leaves either unpaid value or an unnecessary excess withdrawal.
Official syllabus reference: [1]
20. Solving an annuity's implied interest rate
An implied annuity rate solves P = ∑Rₖ(1 + i)⁻ᵗᵏ for the supplied payment schedule. For strictly positive future receipts, present value decreases as i increases over i > −1, so a feasible target price has at most one rate on that domain. A financial calculator or numerical root finder can solve the equation. Payment timing, frequency and signs must be entered consistently. Trial values provide a useful bracket: a calculated present value above the price means the trial discount rate is too low.
Worked example: A stream pays $1,000 at each of the next three year ends and costs $2,723.25. At 5%, 1,000a₃ is approximately $2,723.25, identifying the implied annual effective rate. A 4% trial would produce a higher present value.
Common mistake: Changing payment timing to obtain a convenient calculator result solves a different cash-flow equation, even if the numerical rate looks plausible.
Official syllabus reference: [1]
21. Current value within a payment stream
A stream's current value at an intermediate date can be computed by moving each included cash flow to that date. Past payments accumulate, while future payments discount. This total-stream value differs from the remaining obligation, which excludes payments already made. Immediately before and after a scheduled payment, a remaining-obligation value differs by that payment amount. The question's wording determines which cash flows belong in the valuation. Separating total economic value from an outstanding reserve prevents a common error when a focal date falls inside an annuity.
Worked example: Three $100 payments occur at years one, two and three. At 10%, their total-stream value at year two is 110 + 100 + 100/1.10 = $300.91. Immediately after year two, the remaining obligation alone is $90.91.
Common mistake: Including accumulated past receipts in a remaining-obligation value confuses the value of the entire stream with what is still payable.
Official syllabus reference: [1]
Loans
22. Level-payment amortization
A fully amortizing loan equates its principal L to the present value of repayments. For n equal end-of-period payments at effective periodic interest i, L = Raₙ and R = Li/[1 − (1 + i)⁻ⁿ]. Each payment covers interest on the opening balance and then reduces principal. The formula assumes the first payment is one period after advance and no separate terminal balance remains. A contractual payment rounded to cents may require a small final adjustment, even when the unrounded theoretical payment amortizes the loan exactly.
Worked example: A $12,000 loan is repaid by three annual payments at 5%. Since a₃ = 2.723248, the unrounded payment is 12,000/2.723248, approximately $4,406.50. Using the exact payment makes the balance zero immediately after the third payment.
Common mistake: Dividing principal by the number of payments ignores interest and generally produces a payment too small to fully amortize the loan.
Official syllabus reference: [1]
23. Retrospective outstanding loan balances
The retrospective balance reconstructs outstanding principal from the original advance and payments already made. Immediately after payment k, a level-payment loan has Bₖ = L(1 + i)ᵏ − Rsₖ. More generally, accumulate each actual payment from its payment date to the balance date and subtract from the accumulated advance. This method works with extra payments and irregular schedules without assuming future contractual payments remain unchanged. The phrase immediately after means the current payment has already been deducted; immediately before requires adding that payment back at the same instant.
Worked example: A $10,000 loan earns 6% annually and receives $2,000 at each of the first two year ends. Immediately after the second payment, its balance is 10,000(1.06)² − 2,000(1.06 + 1) = $7,116.
Common mistake: Subtracting past payments without accumulating them omits the interest avoided after those payments reduced the outstanding principal.
Official syllabus reference: [1]
24. Prospective outstanding loan balances
The prospective balance is the value at the loan's contractual interest rate of payments still required. Immediately after payment k on an n-payment level loan, Bₖ = Raₙ₋ₖ. This equals the retrospective balance when both methods use a consistent contract and rate. Prospective valuation is often quickest near maturity because few payments remain. It is a contractual balance, not automatically the loan's market value under a new discount rate. An extra payment may change the future schedule, so that revised schedule must be identified before applying this method.
Worked example: A loan has four remaining annual payments of $900 immediately after today's payment. At the contractual 5% rate, the balance is 900a₄ = $3,191.36. Discounting those same repayments at a market rate of 7% would instead estimate market value.
Common mistake: Using the original total payment count in a prospective balance includes payments already made and overstates the outstanding obligation.
Official syllabus reference: [1]
25. Splitting payments into interest and principal
For a payment R at the end of a period, interest is iBₖ₋₁, where Bₖ₋₁ is the balance immediately after the preceding payment. Principal repaid is R − iBₖ₋₁, and the new balance is Bₖ = Bₖ₋₁(1 + i) − R. This recurrence is the foundation of an amortization schedule. With constant payments and a positive rate, declining balances produce declining interest and increasing principal components. Fees, separate charges or changing interest conventions require explicit treatment rather than being silently included in the principal reduction.
Worked example: A loan opens a year with $8,000 outstanding, charges 7%, and receives a $2,500 year-end payment. Interest is $560, principal repayment is $1,940, and the closing balance is $6,060. These components add back to the payment.
Common mistake: Calculating interest on the closing balance reverses the period's sequence: interest accrues before the end-of-period payment reduces principal.
Official syllabus reference: [1]
26. Geometric growth of principal repayments
Under a fixed rate and equal end-of-period payments, successive principal repayments form a geometric progression. If Kₖ is principal repaid in payment k, then Kₖ₊₁ = (1 + i)Kₖ. This follows because reducing principal by Kₖ lowers the next period's interest by iKₖ, freeing that amount for additional principal repayment. For an n-payment fully amortizing loan, Kₖ = Rvⁿ⁻ᵏ⁺¹. The relationship fails when payments or rates change, but it provides a fast route to selected schedule entries without constructing every balance.
Worked example: The principal component of a loan's fourth annual payment is $600, and the annual rate is 8%. With unchanged level payments, the fifth principal component is $648 and the sixth is $699.84, reflecting the increasing share available after interest.
Common mistake: Assuming equal total payments imply equal principal repayments overlooks the declining interest charge as the outstanding balance falls.
Official syllabus reference: [1]
27. Drop payments after rounded regular payments
A drop payment is a final payment smaller than the regular installment. It commonly arises when a chosen regular payment does not produce an integer amortization term. Determine the number of full regular payments that can occur without overpaying, compute the balance after them, and accumulate that balance to the final payment date. The last payment equals the balance then due, including the final period's interest. Payment dates remain those specified by the contract; a fractional mathematical term does not automatically establish a permissible midperiod settlement date.
Worked example: A loan balance immediately after its last full installment is $420. The next scheduled payment date is one month later, with monthly interest 1%. The final drop payment is $424.20, rather than another full $500 installment or the unchanged $420 balance.
Common mistake: Using the balance after the preceding payment as the final payment omits interest accruing before the final scheduled settlement.
Official syllabus reference: [1]
28. Balloon payments and residual balances
A balloon structure leaves a substantial balance to be paid at a specified terminal date. If regular payment R is made at each of n period ends and an additional balloon F is paid with the nth payment, then L = Raₙ + Fvⁿ. Thus F = L(1 + i)ⁿ − Rsₙ. If the quoted final payment replaces rather than supplements the nth regular installment, the equation must instead include only n − 1 regular payments. Distinguishing an additional balloon from a total final payment prevents double counting.
Worked example: A $10,000 loan charges 5% annually and receives three $2,000 year-end installments plus an additional balloon with the third. The additional balloon is 10,000(1.05)³ − 2,000s₃ = $5,271.25; the total third-date payment is $7,271.25.
Common mistake: Calling the residual balloon the entire final payment can omit a regular installment when the contract specifies both at maturity.
Official syllabus reference: [1]
29. Payment recalculation after a rate reset
When a loan rate resets, first determine the outstanding balance at the reset instant under the old terms. That balance becomes the principal for the remaining schedule. If n payments remain and the new effective payment-period rate is j, the revised level payment is B/aₙ at j. The old rate governs earlier accruals and the new rate governs later ones. A reset does not retroactively reprice the original advance unless explicitly stated. The reset's position relative to a scheduled payment also determines whether that payment remains outstanding.
Worked example: Immediately after a scheduled payment, a loan has $30,000 outstanding and five annual payments remaining. Its new rate is 6%, so the revised payment is 30,000/a₅ at 6%. Past interest is already reflected in the $30,000 balance.
Common mistake: Recalculating from the original principal at the new rate disregards principal already repaid and applies the reset retroactively.
Official syllabus reference: [1]
30. Refinancing as a dated cost comparison
A refinancing decision compares the present value of remaining old costs with replacement repayments and transaction costs at a stated comparison rate. The new loan advance and old settlement offset when they are equal, leaving payment differences and fees. If a fee is financed, it increases new principal; if paid immediately, it is a time-zero outflow. A lower quoted loan rate need not produce savings after fees or a longer term. Keep contractual balance calculations separate from the discount rate used to compare alternatives economically.
Worked example: Old repayments have present value $25,400 at the chosen comparison rate. New repayments have present value $24,900, but refinancing costs $700 immediately. The saving is 25,400 − 24,900 − 700 = −$200, so refinancing increases valued costs under these assumptions.
Common mistake: Comparing only installment sizes can hide both upfront fees and additional payments caused by extending the replacement loan's term.
Official syllabus reference: [1]
31. Loan quotations with mismatched frequencies
A loan's quotation frequency and payment frequency need not coincide. Convert the quoted rate to an effective rate for one payment interval before using amortization formulas. If nominal annual interest j is convertible m times yearly and payments occur p times yearly, the payment-period rate is (1 + j/m)^(m/p) − 1 under the stated compound model. The number of payments over T years is pT. Using j/p is valid only when that division matches the quotation's actual conversion convention, not merely because payments occur p times yearly.
Worked example: A loan quotes 12% nominal interest convertible monthly but requires quarterly payments. Its quarterly effective rate is 1.01³ − 1 = 3.0301%, and a two-year loan has eight payments. Use those inputs together rather than a 3% quarterly rate.
Common mistake: Dividing a nominal quotation by the payment frequency ignores its stated conversion frequency when the two frequencies differ.
Official syllabus reference: [1]
32. Negative amortization and interest-only boundaries
A payment amortizes principal only when it exceeds interest accrued for the period. With opening balance B and rate i, payment R produces principal reduction R − iB. If R = iB, the payment is interest-only and the balance remains unchanged. If R < iB and unpaid interest is capitalized under the stated model, the balance grows: this is negative amortization. A level payment below initial interest cannot fully amortize a positive-rate loan without later payment changes or an outside payoff. The conclusion follows from the balance recurrence.
Worked example: A $20,000 loan charges 1% monthly and receives $150 monthly. First-month interest is $200, so $50 is added to principal and the new balance is $20,050. A $200 payment would instead leave principal unchanged for that first month.
Common mistake: Assuming any positive payment reduces principal overlooks the requirement to cover the period's accrued interest before principal can fall.
Official syllabus reference: [1]
33. Solving loan term and integer maturity
For a fixed-rate loan with principal L and regular payment R, the exact amortization term satisfies n = −ln(1 − iL/R)/ln(1 + i), provided R > iL. This is derived from the loan's annuity equation. A noninteger answer indicates that equal scheduled payments do not end exactly at an integer payment date. The contract may use a smaller final installment, a larger earlier final installment, or another specified convention. Determine the actual final payment from the balance recurrence rather than treating a rounded term as exact.
Worked example: A $1,000 loan charges 10% annually and permits $400 year-end payments. After two payments the balance is $370; one year later $407 is due. Thus two $400 payments followed by $407 settle it at year three under that specified final-payment arrangement.
Common mistake: Interpreting a fractional calculated term as an exact number of unchanged scheduled payments leaves the maturity convention unresolved.
Official syllabus reference: [1]
Bonds
34. Face amount, coupon amount and redemption
A bond's face amount is the reference principal used to calculate coupons, while redemption value is the amount repaid when the bond terminates. They need not be equal. If annual coupon rate r applies to face F with m equal coupons yearly, each coupon is Fr/m. Yield calculations use actual coupon amounts and actual redemption, not merely the quoted coupon rate. A call price can replace scheduled redemption under a permitted call outcome. Keeping these quantities separate is essential when bonds redeem above par or have call premiums.
Worked example: A bond has face $2,000, annual coupon rate 6%, semiannual coupons and redemption $2,100. Each coupon is $60, calculated from face rather than redemption. Its final scheduled cash flow is $2,160, combining the last coupon with redemption.
Common mistake: Calculating coupons from redemption value instead of the stated face amount changes the contractual cash flows unless the two amounts coincide.
Official syllabus reference: [1]
35. Bond price from coupon-period yield
Immediately after a coupon date, a bond with n remaining level coupons C and redemption R has price P = Caₙ + Rvⁿ at effective yield i per coupon period. The annuity includes the final coupon; redemption is added separately at the same final date. For nominal yield convertible at the coupon frequency, divide by that frequency to obtain i. If an annual effective yield is given, use an equivalent-period conversion instead. The supplied October 2026 syllabus excludes valuation between coupon payment dates, so this framework stays on coupon dates.
Worked example: A bond has four remaining semiannual coupons of $30 and redemption $1,000. At 4% nominal yield convertible semiannually, use i = 2% and calculate P = 30a₄ + 1,000v⁴. The fourth coupon and redemption occur together.
Common mistake: Using the annual yield directly with semiannual cash-flow periods discounts every payment at an inconsistent rate and understates price.
Official syllabus reference: [1]
36. Premium or discount relative to redemption
The identity P = R + (C − iR)aₙ reveals whether a bond trades above or below redemption. If coupon C exceeds iR, price exceeds R; if coupon is smaller, price is below R. The familiar comparison of coupon rate and yield assumes face equals redemption and rate periods match. When redemption differs from face, compare coupon dollars with interest on redemption instead. This identity also isolates the present value of the excess or shortfall coupons and helps explain how book value converges toward redemption.
Worked example: A bond pays $55 annually and redeems for $1,100. At 5% annual yield, C = iR = $55, so its price equals $1,100 regardless of its positive finite term. Its coupon rate on a $1,000 face amount would be 5.5%.
Common mistake: Comparing coupon rate with yield without checking redemption can misclassify a bond when its face and redemption values differ.
Official syllabus reference: [1]
37. Book value under the acquisition yield
A bond's book value immediately after a coupon is the present value of remaining contractual cash flows at the original book yield. It also follows Bₖ = Bₖ₋₁(1 + i) − C for constant coupon C and book yield i. The interest portion of each coupon is iBₖ₋₁. Book value reaches redemption immediately before redemption is paid, after the final coupon has been accounted for. It does not automatically change when market yields change, because market valuation and amortized book valuation use different rate bases.
Worked example: A bond's book value after a coupon is $980, its annual book yield is 6%, and its next coupon is $50. The next post-coupon book value is 980(1.06) − 50 = $988.80, moving upward toward redemption.
Common mistake: Replacing the acquisition yield with today's market yield calculates market value rather than continuing the bond's original book-value schedule.
Official syllabus reference: [1]
38. Premium amortization by effective interest
For a premium bond, coupon C exceeds effective interest income iB on the opening book value. Premium amortized during the period is C − iB, reducing book value toward redemption. With level coupons and constant yield, successive amortization amounts increase by factor 1 + i. Total premium amortized across the remaining schedule equals current book value minus redemption. Effective-interest amortization differs from dividing premium equally among periods; it allocates a constant proportional yield to the changing book balance. The coupon is cash received, not entirely interest income.
Worked example: A premium bond opens a coupon period at book value $1,040, yields 5%, and pays a $60 annual coupon. Interest income is $52, premium amortization is $8, and closing book value is $1,032. The next amortization is $8.40.
Common mistake: Treating the full coupon as interest income ignores the portion that returns the premium originally paid above redemption.
Official syllabus reference: [1]
39. Discount accumulation toward redemption
For a discount bond, effective interest income iB exceeds coupon C. Their difference, iB − C, is discount accumulated and is added to book value. This is the counterpart of premium amortization: the investor's yield includes both cash coupons and growth of book value toward redemption. Under fixed coupons and yield, successive discount-accumulation amounts increase geometrically by 1 + i. The total remaining accumulation equals redemption minus current book value. Even a zero-coupon bond earns effective interest through this accretion despite paying no interim cash.
Worked example: A bond opens the year at book value $950, has 6% book yield, and pays a $40 coupon. Interest income is $57, so discount accumulation is $17 and closing book value is $967. The coupon alone understates the year's effective income.
Common mistake: Calling a discount bond's coupon its entire investment income omits the book-value accretion that forms part of its yield.
Official syllabus reference: [1]
40. Solving yield from a bond price
Yield to redemption is the coupon-period rate that equates purchase price with discounted coupons and redemption. For positive contractual receipts, price is strictly decreasing in yield over i > −1, giving at most one solution for a feasible positive price. Numerical calculation may be needed because yield appears in several powers. After solving, report the requested basis: effective annual yield is (1 + i)ᵐ − 1, while nominal yield convertible m times yearly is mi. A callable bond requires a separately identified redemption scenario.
Worked example: A bond costs $1,000 and pays $30 semiannually for two years, redeeming for $1,000. Its semiannual yield is 3% because coupon equals yield times redemption. The nominal semiannual-convertible yield is 6%, while annual effective yield is 6.09%.
Common mistake: Reporting twice a semiannual yield as an annual effective yield overlooks compounding and instead reports a nominal convertible quotation.
Official syllabus reference: [1]
41. Finding remaining term from book value
At a coupon date, book value with n periods remaining satisfies B − R = (C − iR)aₙ. If C differs from iR, solve for the annuity factor and then obtain vⁿ = 1 − iaₙ. Taking logarithms gives n. The result must be compatible with the scheduled coupon dates and the supplied rounding. If C = iR, book value equals redemption for every term, so book value alone cannot determine n. This identifiability boundary is as important as performing the algebra.
Worked example: A bond pays $60 annually, redeems for $1,000 and has 5% book yield. If its post-coupon book value is $1,027.23248, then aₙ = (1,027.23248 − 1,000)/10 = 2.723248, identifying three remaining annual coupon periods.
Common mistake: Trying to infer term from book value on a par-priced bond ignores that identical book value can persist for every remaining term.
Official syllabus reference: [1]
42. Callable bonds and the binding yield scenario
To guarantee a specified minimum yield on a callable bond, value every permitted redemption scenario at that yield. The maximum admissible purchase price is the minimum of those scenario present values. Each scenario includes coupons through its redemption date and the applicable call or maturity amount. The binding scenario need not be the earliest call when call prices vary, so comparing dates alone is insufficient. This calculation protects yield under contractual redemption possibilities; it does not predict which action the issuer will actually choose.
Worked example: At the target yield, a bond's year-three call scenario is worth $1,040 and its year-five maturity scenario $1,025. Paying no more than $1,025 protects the yield in both cases. The later date binds despite the available earlier call.
Common mistake: Automatically selecting the earliest call can miss a lower scenario value created by a later date or different redemption amount.
Official syllabus reference: [1]
43. Coupon reinvestment and realized return
Terminal wealth from holding a bond equals redemption plus each coupon accumulated from receipt to the terminal date at its actual reinvestment rate. With n annual coupons C and constant reinvestment rate j, wealth is R + Csₙ at j. A realized annual compound return over n years is then (wealth/P)^(1/n) − 1. It equals acquisition yield only under suitable reinvestment and holding assumptions. Yield to maturity is a cash-flow equation; it does not ensure every received coupon can actually be reinvested at that yield.
Worked example: A two-year bond costs $1,000, pays $80 annually and redeems for $1,000. Reinvesting the first coupon at 3% produces terminal wealth 1,000 + 80(1.03) + 80 = $1,162.40. Realized annual return is √1.1624 − 1, approximately 7.81%.
Common mistake: Accumulating purchase price at acquisition yield assumes a reinvestment outcome that may differ from the actual coupon reinvestment opportunities.
Official syllabus reference: [1]
44. Market value versus amortized book value
At a coupon date, market value discounts remaining bond cash flows at current required yields, while amortized book value uses the acquisition yield. Both values refer to the same remaining contract but answer different questions. A market-yield increase generally lowers market value for fixed positive cash flows without changing the predetermined effective-interest book schedule. The difference can represent an unrealized gain or loss relative to that book basis. Neither measure includes coupons already received, and the October 2026 scope keeps these valuations on coupon dates.
Worked example: A bond has two $50 coupons and $1,000 redemption remaining. Its book yield is 5%, giving book value $1,000. If current market yield is 6%, market value is 50a₂ + 1,000v² at 6%, approximately $981.67.
Common mistake: Updating book value whenever market yield changes confuses amortized-cost progression with a new market valuation of the remaining payments.
Official syllabus reference: [1]
45. Zero-coupon pricing and yield extraction
A zero-coupon bond has one contractual payment, its redemption amount R at maturity n. At a flat annual effective yield i, price is P = R(1 + i)⁻ⁿ and i = (R/P)^(1/n) − 1. There are no interim coupons to reinvest, so holding to redemption removes coupon-reinvestment uncertainty under the assumed payment certainty. Its price can still vary before maturity when required yields change. With a term structure, its maturity-specific yield is the corresponding spot rate rather than an average coupon-bond yield.
Worked example: A zero-coupon bond costs $800 and redeems for $1,000 after five years. Its annual effective yield is 1.25^(1/5) − 1, approximately 4.564%. The entire dollar return comes from the increase from purchase price to redemption.
Common mistake: Dividing total percentage gain by years gives a simple annual average, not the compound annual yield implied by zero-coupon pricing.
Official syllabus reference: [1]
General Cash Flows, Portfolios, and Asset Liability Management
46. Net present value with signed cash flows
Net present value is the sum of signed cash flows discounted to one valuation date. With annual effective rate i, NPV = ∑Cₜ(1 + i)⁻ᵗ, including any time-zero investment without discounting. Positive NPV means receipts exceed costs on the specified valuation basis. It does not mean that every interim cash balance is positive or that financing is unnecessary. A term structure can replace the flat rate with maturity-specific discount factors. Consistent signs and explicit treatment of terminal costs are essential for projects with later outflows.
Worked example: A project costs $1,000 today and returns $600 at each of the next two year ends. At 10%, NPV = −1,000 + 600/1.10 + 600/1.10² = $41.32. The positive result measures value relative to the chosen 10% discount basis.
Common mistake: Comparing undiscounted total receipts with initial cost ignores both timing and the return required by the chosen valuation rate.
Official syllabus reference: [1]
47. Internal rates of return and multiple roots
An internal rate of return is a rate that makes the signed cash-flow net present value zero. A conventional initial outflow followed only by positive receipts has a strictly decreasing present-value function and, when feasible, a unique IRR. Later negative cash flows can create multiple roots or no economically relevant root. Multiple sign changes permit, but do not guarantee, multiple positive IRRs. A numerical solver may find only one root depending on its starting value. Project comparison should therefore retain an explicit valuation rate rather than assuming every IRR is unique.
Worked example: Cash flows of −$100 now, $230 in one year and −$132 in two years produce zero NPV at both 10% and 20%. Multiplying the equation by (1 + i)² reveals two roots, so neither rate alone completely describes the investment.
Common mistake: Accepting the first calculator root as the only possible yield overlooks the nonmonotonic value function of nonconventional cash flows.
Official syllabus reference: [1]
48. Spot curves and maturity-specific discounting
An annual effective t-year spot rate sₜ prices a single payment at time t through discount factor (1 + sₜ)⁻ᵗ. A stream's value is the sum of its payments discounted at their respective spot rates. The spot curve is not a schedule of successive one-year rates: each quoted rate covers the whole interval from today to its maturity. A coupon bond's single yield generally summarizes its own price and cash-flow pattern, whereas the spot curve supplies discount factors usable across different payment patterns.
Worked example: A liability pays $100 in one year and $1,000 in two years. With one-year spot 3% and two-year spot 4%, value is 100/1.03 + 1,000/1.04². The two-year payment uses the 4% rate for both years together.
Common mistake: Applying the one-year spot rate in year one and the two-year spot rate only in year two misinterprets their maturity coverage.
Official syllabus reference: [1]
49. Forward rates from spot accumulation factors
A forward rate links accumulation over adjacent dated intervals. For annual effective spot rates sₘ and sₙ, the annual effective forward rate f for years m through n satisfies (1 + sₙ)ⁿ = (1 + sₘ)ᵐ(1 + f)ⁿ⁻ᵐ. Thus f = [(1 + sₙ)ⁿ/(1 + sₘ)ᵐ]^(1/(n − m)) − 1. The relationship follows from consistent discount factors, not a claim that future realized short rates are known. Forward rates can value later payments when the relevant earlier spot factors are supplied.
Worked example: If the one-year spot rate is 4% and the two-year spot rate is 5%, the one-year forward rate starting in one year is 1.05²/1.04 − 1 = 6.0096%. This makes the linked accumulation equal the two-year spot accumulation.
Common mistake: Subtracting spot rates directly ignores their different compounding horizons and does not produce the forward rate implied by consistent prices.
Official syllabus reference: [1]
50. Bootstrapping discount factors from bond prices
Bootstrapping solves discount factors sequentially from instruments whose earlier cash flows can already be valued. If a two-year bond pays coupon C at year one and C + R at year two, its price satisfies P = Cd₁ + (C + R)d₂. Once d₁ is known, d₂ = (P − Cd₁)/(C + R). Convert to the two-year spot rate using s₂ = d₂⁻¹ᐟ² − 1. The method requires enough independent pricing information and consistent cash-flow dates; a coupon bond's yield alone is not its final-maturity spot rate.
Worked example: Suppose d₁ = 0.95 and a two-year bond paying $50 then $1,050 costs $1,000. The second discount factor is (1,000 − 50 × 0.95)/1,050 = 0.9071429. Its corresponding two-year spot rate is approximately 4.994%.
Common mistake: Treating a coupon bond's yield as a spot rate overlooks that its price combines payments at several different maturities.
Official syllabus reference: [1]
51. Macaulay duration as a present-value weighted time
For positive cash flows valued at flat annual effective yield i, Macaulay duration is Dmac = ∑tCₜvᵗ/P. It is the average payment time using present-value shares as weights, not an undiscounted average of dates. A zero-coupon payment has duration equal to its maturity, while interim coupons usually shorten duration relative to final maturity. With signed cash flows, weights may be negative and the intuitive average-time interpretation can fail. Duration summarizes first-order sensitivity under a specified common-yield model rather than describing every possible yield-curve movement.
Worked example: Two payments have present values $400 at year two and $600 at year seven. Total value is $1,000 and Macaulay duration is (2 × 400 + 7 × 600)/1,000 = 5 years. The larger later present-value weight pulls duration upward.
Common mistake: Averaging payment dates without present-value weights ignores both payment amounts and discounting, so it generally does not calculate duration.
Official syllabus reference: [1]
52. Modified duration and the yield variable
Modified duration measures proportional sensitivity to the particular yield variable being changed: Dmod = −P'(i)/P(i). For annual effective yield i, Dmod = Dmac/(1 + i). Thus a small change Δi produces approximate relative price change −DmodΔi. If the variable is instead a nominal rate convertible m times yearly, the conversion denominator becomes 1 + j/m when Macaulay duration is measured in years. Always identify the yield convention before applying a sensitivity coefficient. Duration values tied to different yield variables cannot be interchanged without conversion.
Worked example: A portfolio has Macaulay duration 6.3 years at annual effective yield 5%. Its modified duration is 6.3/1.05 = 6. A yield increase of 0.002 therefore implies an approximate 1.2% price decline under the annual-effective-yield model.
Common mistake: Using Macaulay duration directly as sensitivity to annual effective yield omits the division by one plus the current yield.
Official syllabus reference: [1]
53. Macaulay and modified convexity conventions
For flat annual effective yield i, modified convexity is Cmod = P''(i)/P = ∑t(t + 1)Cₜvᵗ/[P(1 + i)²]. Under the force-of-interest convention, Macaulay convexity is the present-value weighted second moment Cmac = ∑t²Cₜvᵗ/P. Consequently Cmod = (Cmac + Dmac)/(1 + i)². Some materials use different convexity labels, so the defining derivative or formula should govern calculations. Positive fixed cash flows produce positive modified convexity, reflecting an upward-curving price-yield relationship rather than a straight line.
Worked example: A five-year zero-coupon payment at 5% has Macaulay duration 5 and Macaulay convexity 25. Its modified convexity is (25 + 5)/1.05² = 27.2109. The distinction arises because derivatives use annual effective yield rather than force.
Common mistake: Substituting the weighted second moment directly for P''/P ignores the extra duration term and the annual-yield conversion denominator.
Official syllabus reference: [1]
54. First-order modified-duration approximation
A first-order Taylor approximation gives P(i + Δi) ≈ P(i)[1 − DmodΔi]. The yield change is expressed as a decimal change in the same rate variable used to define modified duration. For positive fixed payments, an increased yield lowers price, explaining the minus sign. The approximation is local and omits curvature, so accuracy generally decreases for larger yield moves. It is useful for quick sensitivity estimates, but does not establish exact future prices or protect against nonparallel changes in a multi-rate yield curve.
Worked example: A portfolio worth $80,000 has modified duration 3.5. For an annual effective yield increase of 0.004, estimated value is 80,000[1 − 3.5(0.004)] = $78,880. The estimated dollar loss is $1,120 before a convexity adjustment.
Common mistake: Entering a 0.4-percentage-point yield move as 0.4 rather than 0.004 exaggerates the estimated price change by a factor of one hundred.
Official syllabus reference: [1]
55. Macaulay-duration power approximation
Holding the original present-value weights fixed gives the Macaulay-duration approximation P(i₁) ≈ P(i₀)[(1 + i₀)/(1 + i₁)]ᴰᵐᵃᶜ. Equivalently, approximate log price change by −Dmac times the change in ln(1 + i). This retains a compounded form rather than a linear dollar response. It is exact for a single zero-coupon payment because its duration equals its payment time. For a dispersed positive stream, the weights themselves change with yield, so the approximation remains distinct from exact repricing and from the modified-duration linear estimate.
Worked example: A portfolio has value $10,000 and Macaulay duration four at 5%. For a new yield of 6%, the power approximation gives 10,000(1.05/1.06)⁴, approximately $9,627.93. A four-year zero-coupon payment would follow this expression exactly.
Common mistake: Treating the compounded Macaulay approximation as universally exact ignores the changing present-value weights of payments at different times.
Official syllabus reference: [1]
56. Second-order duration-convexity repricing
Including curvature gives P(i + Δi) ≈ P(i)[1 − DmodΔi + ½Cmod(Δi)²], when Cmod is defined as P''/P. For positive fixed cash flows, the convexity term is positive for either direction of yield movement. It moderates the estimated loss from a yield increase and enlarges the estimated gain from a decrease. This is a second-order local approximation, not an exact valuation formula. The one-half belongs to the Taylor expansion; check whether a supplied convexity definition has already incorporated it before applying the coefficient.
Worked example: A $50,000 portfolio has modified duration 5 and modified convexity 36. For Δi = −0.01, estimated value is 50,000[1 + 0.05 + 0.0018] = $52,590. Convexity adds $90 beyond the first-order estimate of $52,500.
Common mistake: Giving convexity the same sign as the yield change overlooks its squared argument and reverses its effect for a yield decrease.
Official syllabus reference: [1]
57. Aggregating portfolio sensitivity by value weights
When all positions are valued using the same yield variable, portfolio modified duration and modified convexity are market-value weighted averages of the component measures. For values Pⱼ, portfolio duration is ∑PⱼDⱼ/∑Pⱼ, and convexity aggregates similarly. Holdings must first be converted from units to values; equal unit counts rarely imply equal weights. Signed liability or short positions require signed dollar sensitivities. If net surplus value is zero, dividing by net value is undefined, so compare asset and liability dollar derivatives rather than attempting a surplus duration.
Worked example: A portfolio invests $3,000 in duration-two assets and $7,000 in duration-six assets. Its duration is 0.3(2) + 0.7(6) = 4.8. Ten units of each asset would give different weights if their unit prices differed.
Common mistake: Weighting duration by face amounts or unit counts instead of current values misstates the portfolio's dollar sensitivity to yield.
Official syllabus reference: [1]
58. Redington immunization as a local surplus minimum
Let surplus S(i) equal asset value minus liability value under a common flat yield. Strict Redington conditions at i₀ are S(i₀) = 0, S'(i₀) = 0 and S''(i₀) > 0. They match present values and dollar durations while requiring greater asset dollar convexity. These conditions make the base yield a strict local minimum of surplus, protecting against sufficiently small yield changes in either direction. They do not guarantee protection from arbitrary yield movements or nonparallel curve shifts. Passage of time and changed cash flows may require rebalancing.
Worked example: Assets and liabilities each have present value $100,000 and modified duration five. If asset modified convexity is 40 and liability convexity is 32, surplus second derivative is 100,000(40 − 32) = $800,000 per unit yield squared, satisfying the strict curvature condition.
Common mistake: Duration matching alone does not establish Redington immunization because present-value equality and a favorable second derivative are also required.
Official syllabus reference: [1]
59. Full immunization with surrounding asset dates
Full immunization protects surplus for all admissible values of a common flat interest rate under its stated cash-flow assumptions, rather than only locally. A standard construction uses two nonnegative asset payments whose dates surround all nonnegative liability payments, together with matched present value and duration at the base rate. Relative discounting is an exponential function of payment time; the chord joining the two surrounding asset dates dominates that function across the enclosed liability dates. This stronger timing structure supports global protection, while ordinary Redington conditions alone provide only local protection.
Worked example: A single liability is due in five years. Buy zero-coupon assets payable in years two and eight, allocating present values so their weighted duration is five and their total equals liability value. Equal present-value weights work here because five is midway between two and eight.
Common mistake: Calling every duration-and-convexity match fully immunized overstates a local condition unless the stronger cash-flow structure also supports global protection.
Official syllabus reference: [1]
60. Exact cash-flow matching by backward substitution
Exact cash-flow matching selects asset quantities so receipts equal liability payments at each required date. In a triangular coupon schedule, begin with the latest liability, choose the asset quantity needed there, then subtract its earlier coupons before solving the preceding date. Continue backward until every date is matched. Unlike duration matching, this addresses payment amounts and timing directly under the assumed certainty of asset receipts. Negative quantities indicate that a proposed exact solution requires shorting or borrowing; such positions are infeasible when the scenario prohibits them. Asset prices then determine the construction cost.
Worked example: Liabilities are $120 in one year and $1,100 in two years. One unit of a two-year bond pays $100 then $1,100, so buy one unit. Its first coupon leaves only $20 to fund at year one using a one-year instrument.
Common mistake: Matching each liability independently without subtracting coupons from longer assets overfunds earlier dates and does not produce an exact cash-flow match.
Official syllabus reference: [1]