Valuing One Deposit Across Changing Rates

Why Changing Rates Require Period-by-Period Valuation

A single deposit does not need to earn the same rate throughout its term. A savings account, investment fund, or hypothetical corporate reserve might be credited at one effective rate during the first year and different rates in later years. To find its value at a future date, apply the rate for each period to the balance carried into that period. The result is a product of accumulation factors, not generally the original deposit multiplied by one rate for the entire term.

The key is to match each rate to the period it describes. A rate stated as effective per year applies to an annual period; a rate stated as effective per quarter applies to a quarterly period. Do not convert a rate unless the problem asks for a different rate basis and supplies enough information to do so.

Definition: Suppose a deposit $D$ is invested for $n$ successive periods. Let $i_k$ be the effective interest rate for period $k$, for $k=1,\ldots,n$. The accumulation factor for period $k$ is $1+i_k$. If interest is credited at the end of each period and there are no further deposits or withdrawals, the value at the end of period $n$ is $$ D\prod_{k=1}^{n}(1+i_k) $$ The rates and periods must use matching units. For example, if $i_k$ is a rate per quarter, period $k$ must be one quarter.

This model assumes that the deposit is made at time zero, each listed rate applies to its stated period, interest credited in one period remains invested in later periods, and no other cash flows occur. It does not directly value a stream of deposits or withdrawals. If rates are supplied as annual effective rates but the periods are shorter than a year, first establish how the problem specifies the conversion to those shorter periods.

Why the Accumulation Factors Multiply

Let $B_k$ denote the account balance at the end of period $k$, with $B_0=D$. During period $k$, the starting balance $B_{k-1}$ earns interest at rate $i_k$. The interest credited for that period is $B_{k-1}i_k$, so the new balance is

B_k=B_{k-1}+B_{k-1}i_k=B_{k-1}(1+i_k).

Applying this recurrence successively gives $B_1=D(1+i_1)$, then $B_2=D(1+i_1)(1+i_2)$, and so on. Repeating the same substitution through period $n$ yields

B_n=D(1+i_1)(1+i_2)\cdots(1+i_n) =D\prod_{k=1}^{n}(1+i_k).

This derivation also explains why a rate change does not erase previously earned interest: later interest is calculated on the balance already accumulated. If every period has the same rate $i$, the product reduces to $D(1+i)^n$, the usual compound future-value formula. With changing rates, each factor must instead reflect its own period.

Applying the Rate Schedule

A reliable calculation keeps the rate schedule and the balance timeline together. At each period end, multiply the current balance by that period’s factor. The following process makes both the arithmetic and the timing visible.

Set the valuation timeline.
Identify the deposit date, the target date, and the periods between them.

Match each rate to a period.
Record whether each rate is effective per year, per quarter, or for another specified interval.

Apply each factor once.
Multiply the balance entering a period by $1+i_k$ to obtain the balance leaving it.

Check the result.
Confirm that all intervals through the target date are included exactly once and that the final units are dollars.

Worked Example: Three Annual Rates on One Deposit

A hypothetical reserve receives a single deposit of $8,000 at time zero. The effective annual rate is 4% in year 1, 6% in year 2, and 5% in year 3. Find the value at the end of year 3. Use the period-by-period accumulation method and identify the mistake that would result from applying just one of the rates to all three years.

Method. The deposit remains invested for three annual periods, so the factor for each year is one plus that year’s effective annual rate. The successive factors are $1.04$, $1.06$, and $1.05$.

End of year 1. The first-year interest is $8{,}000(0.04)=\$320$. Therefore, the balance is $8{,}000+\$320=\$8{,}320$.

End of year 2. The second-year interest is $8{,}320(0.06)=\$499.20$. The balance becomes $8{,}320+\$499.20=\$8{,}819.20$.

End of year 3. The third-year interest is $8{,}819.20(0.05)=\$440.96$. Thus, the balance is $8{,}819.20+\$440.96=\$9{,}260.16$.

8{,}000(1.04)(1.06)(1.05) =8{,}320(1.06)(1.05) =8{,}819.20(1.05) =\$9{,}260.16

Check and trap. Each annual factor was applied once, and the final balance exceeds the original deposit, as expected with positive rates. Applying 4% to all three years would give $8{,}000(1.04)^3=\$8{,}998.91$, which is not the stated rate schedule. The correct value is $9,260.16.

Worked Example: Rates Quoted Per Half-Year

A hypothetical fund receives a deposit of $15,000 and remains invested for three consecutive half-year periods. The effective rate per half-year is 1.2% in the first period, 1.8% in the second, and 2.0% in the third. Find the value after 18 months. Explain why treating these percentages as annual rates would be a unit error.

Method. The problem gives an effective rate for each half-year period, and 18 months contains three such periods. Use the rates directly as the period rates; no annual-rate conversion is needed.

After the first half-year. Interest is $15{,}000(0.012)=\$180$, so the balance is $15{,}000+\$180=\$15{,}180$.

After the second half-year. Interest is $15{,}180(0.018)=\$273.24$, giving a balance of $15{,}180+\$273.24=\$15{,}453.24$.

After the third half-year. Interest is $15{,}453.24(0.020)=\$309.0648$, giving an unrounded balance of $15{,}453.24+\$309.0648=\$15{,}762.3048$. Rounding to the nearest cent only at the final step, the value is $15,762.30.

15{,}000(1.012)(1.018)(1.020) =15{,}180(1.018)(1.020) =15{,}453.24(1.020) =\$15{,}762.3048

Check and trap. There are three half-year periods, so exactly three half-year factors appear. Treating 1.2%, 1.8%, and 2.0% as annual rates would mismatch the rates and the intervals and would not follow the given terms. The stated half-year rates apply directly; the final answer is rounded to cents.

Worked Example: Finding an Intermediate and Final Balance

A hypothetical corporate account receives a single deposit of $12,000. Its effective annual rates are 4% in year 1, 5% in year 2, 8% in year 3, and 3% in year 4. Find the balance at the end of year 2 and the future value at the end of year 4. Keep full precision until reporting the final dollar amounts.

Method. First apply the first two annual factors to obtain the requested intermediate balance. Then apply the year 3 and year 4 factors to that balance.

End of year 1. Interest is $12{,}000(0.04)=\$480$, so the balance is $\$12{,}480$.

End of year 2. Interest is $12{,}480(0.05)=\$624$, so the balance is $\$13{,}104$. This is the value at the intermediate valuation date.

End of year 3. Interest is $13{,}104(0.08)=\$1{,}048.32$, so the balance is $\$14{,}152.32$.

End of year 4. Interest is $14{,}152.32(0.03)=\$424.5696$, giving an unrounded balance of $\$14{,}576.8896$. Rounded to the nearest cent, the future value is $14,576.89.

B_2=12{,}000(1.04)(1.05)=\$13{,}104

B_4=13{,}104(1.08)(1.03)=\$14{,}576.8896 \approx \$14{,}576.89

Check and trap. The end-of-year-2 value includes only the first two years’ factors; the end-of-year-4 value includes all four. Stopping the calculation at year 2 would answer the intermediate-value question, not the final-value question. The final value is greater than the year-2 balance, consistent with the two later positive rates.

Why an Arithmetic Average Usually Does Not Work

A tempting shortcut is to average the period rates and use the average rate for every period. This generally gives a different result because the actual growth depends on the product of the factors, whereas an arithmetic average preserves only the sum of the rates. In general,

\prod_{k=1}^{n}(1+i_k) \ne \left(1+\frac{1}{n}\sum_{k=1}^{n}i_k\right)^n.

The two expressions can happen to agree for particular rate schedules, but there is no general rule that makes them equal. A single equivalent constant rate can be defined from the total accumulation factor, but it is not ordinarily the arithmetic average. Unless a problem explicitly asks for such an equivalent rate, use the period-specific factors.

A related timing trap is applying a rate for the wrong number of periods. If an account is invested for four quarters and the problem supplies four effective quarterly rates, there are four factors. If the problem instead supplies annual rates, the quarterly factors cannot be inferred without a stated convention or other information. Always read the unit attached to the rate before calculating.

Key takeaway: For one deposit with changing rates, carry the balance forward one period at a time. The future value is the initial deposit multiplied by every effective accumulation factor between the deposit date and the valuation date, with each rate matched to its own period.

Check Your Understanding

For each problem, identify the applicable periods and apply each stated rate to the balance entering that period. Round only when reporting a final dollar amount.

A hypothetical account receives $5,000 and earns effective annual rates of 3% in year 1, 4% in year 2, and 2% in year 3. Find its value at the end of year 3. (Answer: $5,463.12)
A deposit of $20,000 is invested for two half-year periods at effective rates of 1.5% and 2.5% per half-year, respectively. Find the value after one year. (Answer: $20,807.50)
A single deposit of $9,000 earns 5% in year 1, 2% in year 2, and 6% in year 3, with all rates effective annually. Find both the end-of-year-2 balance and the end-of-year-3 value. (Answer: $9,639.00 at the end of year 2; $10,217.34 at the end of year 3)
A $16,000 deposit earns effective annual rates of 2%, 7%, and 3% over three successive years. Calculate its actual end-of-year-3 value. Then calculate the value obtained by incorrectly using the arithmetic average rate for all three years, and state the dollar difference. (Answer: Actual value $17,986.27; average-rate value $17,997.82; the incorrect result overstates the value by $11.55)
A hypothetical deposit of $8,000 earns 4% effective annually for one year, followed by four quarterly periods at an effective rate of 1.5% per quarter. Find the value at the end of the fifth quarter. (Answer: $8,830.54)

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