Workshop On Unknown Rates And Focal Dates

Choosing a Focal Date Before Solving

When a transaction has an unknown interest rate or payment date, the algebra is only part of the work. The equation must reflect the stated interest model, and every cash flow must be valued at the same focal date. A solution is not complete until its units, admissibility, and fit with the timeline have been checked.

The method is familiar from value equations: list the dated cash flows, choose a date, value each cash flow at that date under the stated rate convention, and equate the two sides of the transaction. The choice of focal date can make the equation simpler, but it does not change the cash-flow dates or the interest model.

Definition: A focal date is the date at which all cash flows in a value equation are expressed. Under annual effective compound interest $i$, a cash flow $C$ at time $t$ has value $C(1+i)^{T-t}$ at focal time $T$, when the relevant dates are measured in years. Under simple interest, a cash flow $C$ valued from time $t$ to time $T$ uses the simple-interest factor $1+i(T-t)$. State which model and rate convention apply before writing the equation.

Use a consistent sign convention: one side of the equation represents one set of payments and the other side represents the equivalent set. For a positive effective annual rate, $1+i$ is positive. A rate found by algebra must also satisfy the interest model’s domain and the problem’s stated restrictions. If the problem specifies an annual effective rate, do not silently substitute a nominal rate convertible monthly or semiannually.

Why the Focal Date Can Be Changed Under Compound Interest

For annual effective compound interest, changing the focal date preserves an equivalence. This gives a useful way to check a result, or to choose a focal date that leads to more convenient arithmetic.

Write the equation at time zero.
For cash flows $C_k$ at times $t_k$, an equation with signed cash flows is $\sum_k C_k(1+i)^{-t_k}=0$.

Move the equation to time $T$.
Multiply every term by the same factor $(1+i)^T$, which is positive when $1+i>0$.

Read the accumulated values.
The result is $\sum_k C_k(1+i)^{T-t_k}=0$, the same transaction valued at time $T$.

This argument depends on the compound-interest factors being multiplicative: $(1+i)^a(1+i)^b=(1+i)^{a+b}$. It does not justify moving a simple-interest equation to a new focal date as if simple accumulation factors were multiplicative. Simple interest and compound interest therefore require different checks.

Solving for an Unknown Rate

With an unknown rate, first keep the periods and rate convention visible in the equation. Then solve, consider every algebraic root, and apply the model’s admissibility conditions. A convenient focal date may reduce the number of terms, but it does not guarantee that the resulting equation has only one algebraic root.

Worked Example: Finding an Effective Annual Rate

A payment of $1,000 at time 0 is equivalent under annual effective compound interest to payments of $550 at the end of year 1 and $605 at the end of year 2. Find the annual effective rate. Check the result at a second focal date. Assume the accumulation factor per year is positive.

Method and focal date. Use time 0, so both future payments are discounted to the present. This avoids mixing accumulated and discounted values. Let $i$ be the unknown annual effective rate and set $x=1+i$.

1000=\frac{550}{1+i}+\frac{605}{(1+i)^2}

Multiply by $x^2$, which is positive under the stated assumption:

1000x^2=550x+605,\qquad 1000x^2-550x-605=0.

The discriminant is $550^2+4(1000)(605)=302{,}500+2{,}420{,}000=2{,}722{,}500$, whose square root is 1,650. Thus

x=\frac{550\pm1650}{2000}, \qquad x=1.1\ \text{or}\ x=-0.55.

Since $x=1+i$ must be positive, reject $-0.55$. The admissible root is $x=1.1$, so $i=0.10$, or 10% per year effective.

Check at time 0. The present values are $550/1.1=500$ and $605/1.21=500$. Their sum is $1,000, as required.

Check at time 2. The value of the time-0 payment is $1000(1.1)^2=1{,}210$. The year-1 payment accumulates one year to $550(1.1)=605$, and the year-2 payment is already $605. The total is $605+605=1{,}210$. Both focal dates agree. The common mistake this method avoids is accepting the negative root without checking whether it can represent the accumulation factor $1+i$.

Solving for an Unknown Cash-Flow Date

For an unknown date, use the known interest rate and the stated time-unit convention. Under annual effective compound interest, fractional-year periods are valued with fractional powers: a cash flow at time $t$ has present value $C(1+i)^{-t}$. This assumes the annual effective rate is applied through the compound accumulation function for fractional as well as whole years. If a problem specifies a different convention for fractional periods, use that convention instead.

Worked Example: Finding a Fractional Payment Date

Under annual effective compound interest of 10%, $1,000 at time 0 is equivalent to $550 at time 1 and $600 at an unknown time $t$ years after time 0. Find $t$, assuming $t$ is after year 1. Report the date to three decimal places.

Method and focal date. Use time 0 and discount all future amounts. The first payment has present value $550/1.1=500$, leaving $500 of present value to be supplied by the $600 payment:

1000=\frac{550}{1.1}+\frac{600}{(1.1)^t}, \qquad 500=\frac{600}{(1.1)^t}.

Rearrange to isolate the accumulation factor:

(1.1)^t=\frac{600}{500}=1.2.

Take natural logarithms of both sides:

t=\frac{\ln(1.2)}{\ln(1.1)} =\frac{0.1823216}{0.0953102} \approx 1.91293\text{ years}.

Rounded to three decimal places, the payment date is 1.913 years after time 0, which is indeed after year 1. To check the rounded answer, $1.1^{1.913}$ is approximately 1.200008, so the $600 payment has present value about $600/1.200008=\$499.997$. Together with the first payment’s $500 present value, this is approximately $1,000.00 to the nearest cent.

The common mistake this method avoids is treating $t$ as a number of whole years or rounding before taking logarithms. The exponent is measured in years under the stated annual-effective fractional-period convention.

Rate Conventions and Simple-Interest Limits

A nominal rate is not an effective annual rate. For a nominal annual rate $i^{(m)}$ convertible $m$ times per year, the periodic rate is $i^{(m)}/m$, and the accumulation over $n$ conversion periods is $(1+i^{(m)}/m)^n$. In contrast, annual effective interest uses the factor $(1+i)^t$ over $t$ years under the fractional-period compound convention.

Simple interest uses a different model: accumulation over $t$ years from time 0 is $1+it$, not $(1+i)^t$. Its factors do not generally multiply across separate intervals. Therefore, a value equation under simple interest should be evaluated at the chosen focal date rather than moved to another date using compound-interest reasoning.

Worked Example: Solving a Rate Under Simple Interest

Under simple interest, $1,000 at time 0 is equivalent to $550 at the end of year 1 and $600 at the end of year 2. Find the annual simple-interest rate $i$. Assume the accumulation factors from time 0 to years 1 and 2 are positive.

Method and focal date. Choose time 0. Under simple interest, the factors from time 0 to years 1 and 2 are $1+i$ and $1+2i$, respectively. Thus

1000=\frac{550}{1+i}+\frac{600}{1+2i}.

The assumed positive factors require $1+i>0$ and $1+2i>0$, so the combined domain is $i>-0.5$. Multiply by $(1+i)(1+2i)$:

$$ 1000(1+3i+2i^2)=550(1+2i)+600(1+i). $$

Expanding both sides gives $1000+3000i+2000i^2=1150+1700i$, hence

2000i^2+1300i-150=0, \qquad 40i^2+26i-3=0.

The roots are $(-26\pm\sqrt{26^2+4(40)(3)})/80=(-26\pm34)/80$. They are $i=0.10$ and $i=-0.75$. The second root violates $i>-0.5$, so reject it. The admissible simple-interest rate is 10% per year.

Check at the selected focal date. At time 0, the present values are $550/1.1=\$500$ and $600/1.2=\$500$, totaling $1,000. Do not claim that the equation must also balance at year 2 by accumulating each earlier cash flow with simple interest over its separate interval: simple-interest factors are not multiplicative. The common mistake this example avoids is using $(1+i)^2$ for the year-2 payment or treating a simple-interest equation as freely movable between focal dates.

Common Checks Before Accepting an Answer

An equation can be solved correctly and still produce an unusable answer. The final check should follow the actual timeline and the model, not just the algebra. In particular, distinguish a rate per conversion period from an annual rate, and distinguish an exact date from a rounded date.

Check	Question to ask
Focal date	Are all amounts valued at the same date?
Rate convention	Is the rate effective, nominal with a stated conversion frequency, or simple?
Time unit	Do the exponent and rate period use compatible units?
Admissibility	Does the root satisfy the model’s domain and the problem’s conditions?
Timeline check	Does the resulting equation reproduce the transaction values at the chosen focal date?

For compound-interest transactions, a second focal date is a useful independent check because multiplying an equation by a common accumulation factor preserves equality. For simple-interest transactions, check at the focal date used to construct the equation unless the problem explicitly supplies a consistent alternative valuation rule.

Key takeaway: State the interest model and rate convention, choose one focal date, and write every cash flow’s value using the correct time interval. Solve the resulting equation, reject inadmissible roots, and substitute the answer back into the transaction before reporting it.

Check Your Understanding

For each problem, identify the rate convention and focal date before solving. Answers are included for checking.

Under annual effective compound interest, $900 at time 0 is equivalent to $495 at the end of year 1 and $544.50 at the end of year 2. Find the effective annual rate. (Answer: 10%)
At an annual effective rate of 8%, $1,000 at time 0 is equivalent to $400 at time 1 and $734.40 at an unknown time $t$. Use fractional-year compound interest to find $t$. (Answer: 2 years)
Under simple interest, $1,200 at time 0 is equivalent to $660 at the end of year 1 and $720 at the end of year 2. Find the admissible annual simple-interest rate, assuming positive accumulation factors over both stated terms. (Answer: 10%)
Under a nominal annual rate $i^{(2)}$ convertible semiannually, $1,000 at time 0 is equivalent to $550 at time 0.5 and $605 at time 1. Find $i^{(2)}$. (Answer: 20% nominal, convertible semiannually)
At an unknown annual effective compound rate, $1,000 at time 0 is equivalent to a single payment of $1,331 at time 3. Find the rate, then check the equivalence at time 2. (Answer: 10%; both values at time 2 are $1,210)

What You'll Learn