Integrated Workshop on Unknown Payments, Rates, and Dates

One Timeline, One Value Equation

Integrated problems often give several cash flows and ask for one missing quantity: a payment, an interest rate, or the date an annuity begins. The main challenge is usually not complicated algebra. It is translating the wording into the correct cash-flow dates before writing the equation. A deposit made at time 0 and a deposit made at time 1 do not earn interest for the same number of periods when both are valued at time 2.

The reliable method is to draw or list every cash flow with its time, choose a focal date, and move each amount to that date using the stated interest model. This approach applies the earlier material on interest and annuities without relying on a memorized formula whose payment timing may not match the problem. The methods here are aligned with the SOA Exam FM (Financial Mathematics) syllabus.

Definition: A focal date is the date at which all cash flows in an equation of value are expressed. Under a constant effective annual rate $i$, a cash flow $C$ at integer time $t$ has value $C(1+i)^{T-t}$ at focal time $T$. If $t<T$, this accumulates forward; if $t>T$, it is discounted back. Use a consistent sign convention, such as deposits received as positive amounts and payments made as negative amounts.

Unless stated otherwise, the examples use a constant effective annual rate and annual cash-flow dates. This means one year of accumulation multiplies a value by $1+i$. The model does not apply unchanged if rates differ by period, if cash flows occur at fractional times under a different convention, or if the stated rate is nominal but is treated as effective.

Why a Common Focal Date Works

The equation of value follows from the definition of equivalent amounts under the assumed interest model. Moving an amount from one date to another changes its numerical value, but not its value after all cash flows have been moved to the same date.

Write the cash flows with dates.
Record each amount $C_j$ at time $t_j$, including whether it is received or paid.

Choose a focal date $T$.
Select a date that makes the unknown and the known amounts convenient to compare. The result is the same whichever common date is chosen, provided every amount is moved correctly.

Move each amount to $T$.
For $t_j\le T$, multiply by $(1+i)^{T-t_j}$. For $t_j>T$, divide by $(1+i)^{t_j-T}$.

Equate the values and solve.
Combine signed values at $T$, solve for the requested unknown, and substitute the result back into the original timeline.

For example, if a deposit $C$ occurs at time $t\le T$, its accumulated value at $T$ is $C(1+i)^{T-t}$, because it earns interest for $T-t$ years. For a cash flow after $T$, discounting reverses that accumulation: multiplying its value at $T$ by $(1+i)^{t-T}$ must recover its amount at time $t$. This is why the exponent counts the actual number of periods between the cash-flow date and the focal date, not the number of dates listed in the problem.

Solving for an Unknown Payment

Worked Example: Finding a Missing Deposit

An account earns 10% effective annual interest. A deposit of $1,000 is made at time 0, another deposit of $500 is made at time 1, and an unknown deposit $X$ is made at time 2. The accumulated value of these deposits at time 3 is $2,596. Find $X$. Use the method of accumulating every deposit to time 3.

Timeline and method. The deposits occur at times 0, 1, and 2, and the focal date is time 3. Thus they earn interest for 3, 2, and 1 years, respectively. All are deposits, so their accumulated values are added.

$$ 1000(1.10)^3+500(1.10)^2+X(1.10)=2596 $$

Calculate the known accumulated values.

1000(1.10)^3=1000(1.331)=1331,\qquad 500(1.10)^2=500(1.21)=605

The known deposits accumulate to $1331+605=1936$. Therefore the accumulated value contributed by $X$ is $2596-1936=660$. Since $X$ earns interest for one year, $1.10X=660$, so $X=660/1.10=600$.

Check. At time 3, the deposits accumulate to $1331+605+600(1.10)=1331+605+660=2596$, as required. The missing deposit is $600.00.

Common mistake avoided. Giving the time-1 deposit three years of interest would use $500(1.10)^3$. It is made at time 1, so it earns interest for $3-1=2$ years, not three.

Solving for an Unknown Rate

When the rate is unknown, the cash-flow dates still determine the exponents. The resulting equation may be nonlinear. For annual periods, it is often useful to set $y=1+i$; under the usual effective-rate model, $y>0$, or equivalently $i>-1$. After solving, retain only a root that satisfies the model and the original equation.

Result: If all deposits are accumulated to a later focal date and the unknown accumulation factor is $y=1+i>0$, the accumulated value is strictly increasing in $y$ whenever at least one positive deposit occurs before the focal date. Consequently, a target accumulated value can have at most one admissible positive value of $y$ in this setting. Algebraic roots outside $y>0$ are not valid accumulation factors.

Worked Example: Finding the Effective Annual Rate

An investor deposits $1,000 at time 0 and $500 at time 1. At time 2, the account is worth $1,760. Assume a constant effective annual rate $i$. Find $i$. Use time 2 as the focal date.

Set up the equation. The first deposit earns interest for two years; the second earns interest for one year. The rate is not given, so let $y=1+i$.

$$ 1000y^2+500y=1760 $$

Divide by 500 to simplify: $2y^2+y-3.52=0$. Its discriminant is $1-4(2)(-3.52)=1+28.16=29.16$, and $\sqrt{29.16}=5.4$. Therefore

y=\frac{-1\pm 5.4}{4}

The roots are $y=1.1$ and $y=-1.6$. Since an accumulation factor must satisfy $y>0$, reject $-1.6$. Thus $1+i=1.1$, so $i=0.10$, or 10% effective annually.

Check. At 10%, the time-2 value is $1000(1.1)^2+500(1.1)=1210+550=1760$, matching the stated value.

Common mistake avoided. Applying two years of interest to both deposits gives $1000y^2+500y^2=1760$, which produces $y=\sqrt{1760/1500}$ and an incorrect rate of about 8.32%. The $500 deposit occurs at time 1 and earns only one year of interest before time 2.

Solving for an Unknown Annuity Date

A date unknown often represents an integer shift in a regular payment pattern. Write the first payment date explicitly, then generate all later payment dates from it. If the question specifies that a value is reported to the nearest cent, compare candidate dates using their computed values at that precision; do not treat a rounded value as an exact, unrounded equality.

Worked Example: Identifying the Start of a Three-Payment Annuity

Three payments of $200 are made at the ends of years $k+1$, $k+2$, and $k+3$, where $k$ is a nonnegative whole number. At 4% effective annually, the time-0 present value is $533.67 to the nearest cent. Find $k$ and the first payment date. Use time 0 as the focal date.

Write the present value for a candidate $k$. Each payment is discounted from its own date. With $v=1/1.04$, the value is

PV(k)=200\left(v^{k+1}+v^{k+2}+v^{k+3}\right) =200v^{k+1}(1+v+v^2)

Evaluate the candidate $k=1$. Then the payment dates are years 2, 3, and 4, and

v^2=0.924556213,\qquad v^3=0.888996359,\qquad v^4=0.854804191

Thus $PV(1)=200(0.924556213+0.888996359+0.854804191)=200(2.668356763)=533.6713526$, which rounds to $533.67.

Check that no other nonnegative integer works. The present value decreases as $k$ increases, since $PV(k+1)=PV(k)/1.04$. The adjacent candidates give $PV(0)=PV(1)(1.04)$, approximately $555.02, and $PV(2)=PV(1)/1.04$, approximately $513.15. Values for $k<1$ are no smaller than $PV(0)$, and values for $k>1$ are no larger than $PV(2)$. Neither can round to $533.67. Therefore $k=1$, and the first payment is at the end of year $k+1=2$.

Common mistake avoided. A positive-time payment must be discounted to time 0. In particular, for $k=1$, the first payment is at year 2, not year 1; using $v$ in place of $v^2$ changes the timeline.

Checks That Catch Timing and Algebra Errors

A computed answer is not complete until it has been checked against the original cash flows. A useful check is to substitute the answer into the equation at the same focal date. For an unknown date, also compare adjacent integer candidates when the value moves monotonically with the date. For an unknown rate, check both the accumulated value and the admissibility of the accumulation factor.

Keep these distinctions clear when setting up an equation:

Payment time versus focal time: the number of interest periods is their difference. A time-1 payment valued at time 4 earns three years of interest.
Interest rate versus accumulation factor: if $y=1+i$, the rate is $y-1$, not $y$.
Unknown annuity start versus first payment: if payments occur at $k+1,k+2,\ldots$, the first payment is one period after $k$.
Rounded target versus exact value: if a problem reports cents, compute candidate values accurately enough to round and distinguish them at that precision.

Changing the focal date can be a useful independent check, but it is not a substitute for correct timing. A cash-flow equation is valid only when every amount is valued at the same date under the same interest assumptions. If the equation fails after substitution, first recheck the dates and exponents before changing the algebra.

Key takeaway: Put every cash flow on one timeline, count periods from each payment date to the chosen focal date, and solve the resulting value equation. Then substitute the answer and verify that it satisfies both the arithmetic and the stated timing conditions.

Check Your Understanding

For each problem, make a timeline before writing the value equation. Rates are effective annually unless stated otherwise.

An account earns 6% annually. Deposits of $800 at time 0 and $300 at time 1, together with an unknown deposit $X$ at time 2, accumulate to $2,000 at time 3. Find $X$, to the nearest cent. (Answer: $669.91)
A deposit of $700 is made at time 0 and $400 at time 1. Their accumulated value at time 2 is $1,287. Find the effective annual rate. (Answer: 10%)
Three payments of $150 occur at times $k+1,k+2,k+3$, where $k$ is a nonnegative integer. At 5%, their time-0 present value is $370.51 to the nearest cent. Find $k$ and the first payment date. (Answer: $k=2$; the first payment is at the end of year 3)
At 10% annually, three deposits of $200 occur at times $k+1,k+2,k+3$, for a nonnegative integer $k$. Their accumulated value at time 5 is $728.20. Find $k$ and list the deposit dates. (Answer: $k=1$; the deposit dates are years 2, 3, and 4)

What You'll Learn