Fractional Periods and Irregular Reinvestment Dates

Why Fractional Dates Need an Explicit Convention

Cash flows do not always occur on annual anniversaries. A distribution might arrive after nine months, or an investment might be valued eighteen months after its start. To find the accumulated value at a target date, each cash flow must be accumulated only over the time from its own payment date to that target. The fractional part of that time must be handled according to the stated interest convention.

The phrase “an annual effective rate of 8%” does not by itself tell you whether a partial year is accumulated using a fractional power or simple interest for the fraction. Those are different conventions. In a problem, follow the convention given; if one is not specified, state the assumption you use. Do not silently replace a fractional interval with a whole number of years.

Definition: Let $t_j$ be the time, in years from a common origin, at which a signed cash flow $C_j$ occurs. Let $T$ be the target valuation time, with $T\ge t_j$. Under accumulation function $a(s)$, the accumulated value at $T$ of all the cash flows is $$ A(T)=\sum_j C_j\,a(T-t_j). $$ A positive $C_j$ is an amount added to the investment; a negative $C_j$ is an amount withdrawn. Each amount is carried forward from its own date, not from the common origin.

Fractional Time Under Compound Accumulation

Under the fractional-exponent compound convention, an effective annual rate $i$ applies over an interval of $s$ years through the accumulation factor $a(s)=(1+i)^s$. This convention treats the force of interest as constant over time, with $\delta=\ln(1+i)$, so that $a(s)=e^{\delta s}$. Use it when the problem specifies fractional powers or otherwise establishes this convention. The rate must satisfy $i>-1$, and the time interval $s$ is measured in years.

a(s)=(1+i)^s,\qquad A(T)=\sum_j C_j(1+i)^{T-t_j}.

This is not the same as applying simple interest to the fractional part of a year. For example, with $i=10\%$, one and a half years under this convention has factor $1.1^{1.5}$, not $1.1(1+0.5(0.10))$. A problem may specify either method; the method is part of the model.

Why the fractional-power formula works: Under the stated convention, the force $\delta$ is constant, so accumulation over $s$ years is $e^{\delta s}$. Since $e^\delta=1+i$ for one year, $e^{\delta s}=(e^\delta)^s=(1+i)^s$. The exponential form also gives consistent accumulation across split intervals: $a(u)a(v)=e^{\delta u}e^{\delta v}=e^{\delta(u+v)}=a(u+v)$. Thus carrying an amount through two consecutive intervals gives the same result as carrying it directly through their combined length.

Worked Example: Several Cash Flows on Fractional-Year Dates

A fund receives \$800 at time $0.25$, \$1,200 at time $1.00$, and \$500 at time $1.50$, where times are in years from the start. Find the accumulated value at time $2.00$. The effective annual rate is $21.550625\%$, and the stated convention is compound accumulation with fractional exponents.

Method. Calculate the remaining time from each payment to time $2.00$, then accumulate each amount separately. The annual accumulation factor is $1+i=1.21550625=1.05^4$, so one quarter-year of accumulation has factor $1.05$.

The \$800 payment accumulates for $2.00-0.25=1.75$ years, or seven quarter-years: $$ 800(1.21550625)^{1.75}=800(1.05)^7 =800(1.40710042265625)=\$1,125.680338125. $$

The \$1,200 payment accumulates for $2.00-1.00=1.00$ year: $$ 1,200(1.21550625)^1=1,200(1.21550625)=\$1,458.6075. $$

The \$500 payment accumulates for $2.00-1.50=0.50$ years, or two quarter-years: $$ 500(1.21550625)^{0.5}=500(1.05)^2 =500(1.1025)=\$551.25. $$

Adding the amounts at the common target date gives $$ 1,125.680338125+1,458.6075+551.25=\$3,135.537838125. $$ Rounded to the nearest cent, the accumulated value is \$3,135.54. The check is that each payment is carried forward a nonnegative interval to time $2.00$, and each of the three accumulated values is included once.

Common mistake avoided: using the same accumulation period for every payment. The earliest payment has 1.75 years to grow, while the latest has only 0.50 years.

Simple Interest for a Fractional Stub

Some problems specify annual compounding for each complete year and simple interest for the remaining fraction. If an elapsed interval is $s=n+f$, where $n$ is a nonnegative integer and $0\le f<1$, the corresponding factor under that convention is $(1+i)^n(1+fi)$. This convention is applied to the full interval from the cash-flow date to the target date. Do not substitute it for fractional-exponent accumulation unless it is the stated rule.

a(n+f)=(1+i)^n(1+fi),\qquad n\in\{0,1,2,\ldots\},\quad 0\le f<1.

Worked Example: Annual Compounding With Simple Interest in the Stub

A hypothetical account receives \$1,000 at time $0.40$ and \$500 at time $1.20$. Find its value at time $2.30$. The rate is $12\%$ effective per complete year; the stated convention is compound interest for complete years and simple interest for the remaining fraction of a year.

Method. Find each elapsed interval and express it as a whole number of years plus a fractional stub. Apply the specified convention separately to each payment.

For the \$1,000 payment, the interval is $2.30-0.40=1.90$ years. Thus $n=1$, $f=0.90$, and $$ a(1.90)=(1.12)^1(1+0.90(0.12)) =1.12(1.108)=1.24096. $$ The accumulated payment is $1,000(1.24096)=\$1,240.96$.

For the \$500 payment, the interval is $2.30-1.20=1.10$ years. Thus $n=1$, $f=0.10$, and $$ a(1.10)=(1.12)^1(1+0.10(0.12)) =1.12(1.012)=1.13344. $$ The accumulated payment is $500(1.13344)=\$566.72$.

At time $2.30$, the total is $$ \$1,240.96+\$566.72=\$1,807.68. $$ The calculation checks because both amounts have been accumulated using their own elapsed intervals and the same stated convention.

Common mistake avoided: using $1.12^{1.90}$ and $1.12^{1.10}$. Those fractional powers belong to the fractional-exponent compound convention, not the simple-stub convention specified here.

Valuing at an Intermediate Date

A useful way to manage a timeline with several irregular dates is to accumulate all cash flows to an intermediate valuation date, add them there, and then carry the resulting value forward. Under fractional-exponent compound accumulation, this gives the same result as carrying each cash flow directly to the final target, because accumulation factors multiply over consecutive intervals. Be careful that a cash flow occurring exactly on the intermediate date receives no accumulation before it is included.

Choose one time origin.
Express every payment date, intermediate valuation date, and final target date on the same scale.

Apply the stated convention.
For each cash flow, calculate the interval from its date to the valuation date where it will be included.

Combine and carry forward.
Add amounts only after they are at the same date; then accumulate that combined value to the final target if needed.

Worked Example: An Intermediate Valuation Date

A fund receives \$900 at time $0.50$, \$700 at time $1.25$, and \$300 at time $1.75$. Find the value at time $2.25$. The effective annual rate is $21.550625\%$, and the convention is fractional-exponent compound accumulation. As above, the quarter-year factor is $1.05$.

Method. First value the cash flows at time $1.75$, where the last payment occurs, then accumulate that total for $0.50$ years to time $2.25$.

The \$900 payment grows for $1.75-0.50=1.25$ years, or five quarter-years: $$ 900(1.05)^5=900(1.2762815625)=\$1,148.65340625. $$ The \$700 payment grows for $1.75-1.25=0.50$ years, or two quarter-years: $$ 700(1.05)^2=700(1.1025)=\$771.75. $$ The \$300 payment occurs at time $1.75$, so its value on that date is \$300. The combined value at time $1.75$ is $$ 1,148.65340625+771.75+300=\$2,220.40340625. $$

From time $1.75$ to $2.25$ is another $0.50$ years, or two quarter-years. Therefore, $$ 2,220.40340625(1.05)^2 =2,220.40340625(1.1025) =\$2,447.994755390625. $$ Rounded to the nearest cent, the value at time $2.25$ is \$2,447.99.

Check. Carrying each payment directly to time $2.25$ gives $900(1.05)^7+700(1.05)^4+300(1.05)^2$, or $\$1,266.390380390625+\$850.854375+\$330.75=\$2,447.994755390625$. The direct calculation agrees.

Common mistake avoided: accumulating the \$300 payment from time $0$, or allowing it to earn interest before it exists. It is included at time $1.75$, not earlier.

Common Errors and Exam Checks

Fractional-date questions are usually manageable once the timeline and convention are explicit. The most consequential errors are not complicated algebra; they are mismatches between dates, units, and the stated accumulation rule.

Rounding a duration to a whole year. A cash flow paid 0.25 years before the target does not have zero time to accumulate. Keep the fractional duration.
Using the wrong fractional convention. $(1+i)^f$ and $1+fi$ generally differ. Use fractional powers only when that compound convention applies, and use simple-stub accumulation only when specified.
Mixing valuation dates. Do not add a payment at its original date to an amount already accumulated to a later date. Move every amount to the same date first.
Starting accumulation at the wrong date. Use $T-t_j$, not $T$, as the time interval for a cash flow paid at $t_j$.
Rounding too early. Retain unrounded factors and amounts through the sum, then round the final monetary value to cents.

For example, in the first worked example, a tempting but incorrect calculation is to accumulate all \$2,500 from time $0$ to time $2$: $2,500(1.21550625)^2=\$3,693.64$ to the nearest cent. That treats every payment as if it were present from the start. The correct date-by-date calculation is $800(1.21550625)^{1.75}+1,200(1.21550625)+500(1.21550625)^{0.50}=\$3,135.54$. The difference is caused by ignoring the payment dates, not by rounding.

Key takeaway: Record each cash-flow date and the target date on one timeline. Apply the stated fractional-period convention to the elapsed time from each payment to the valuation date, and combine amounts only after they are expressed at the same date.

Check Your Understanding

For each problem, identify the applicable convention and keep the cash-flow dates aligned with the target date.

A fund receives \$400 at time $0.50$ and \$600 at time $1.50$. Find the value at time $2.00$ under fractional-exponent compound accumulation, using an effective annual rate $i=1.04^4-1$. (Answer: \$1,155.09)
A single \$1,000 payment is made at time $0.25$. Find its accumulated value at time $1.75$ at an effective annual rate of $10\%$, using fractional-exponent compound accumulation. Round to cents. (Answer: \$1,153.69)
A \$2,000 payment is made at time $0.60$. Find its value at time $2.20$ at a rate of $5\%$ per complete year, with simple interest for the fractional stub after annual compounding. (Answer: \$2,163.00)
A fund receives \$500 at time $0.25$, receives \$300 at time $1.25$, and withdraws \$100 at time $1.75$. Find its value at time $2.25$ using fractional-exponent compound accumulation with a quarter-year factor of $1.02$. (Answer: \$806.52)
A \$1,000 payment is made at time $0.80$. Find its accumulated value at time $2.30$ at an effective annual rate of $8\%$, using fractional-exponent compound accumulation. Round to cents. (Answer: \$1,122.37)

What You'll Learn