Interest Calculation

One Time Payment

Note that:

• $P$ is current value.
• $F$ is future value.

For recursive interests:

Calculate F from P

$$ F = P \cdot (1 + i)^n $$

Which could also be written at:

$$ F = P \cdot (F/P, i , n) $$

Calculate P from F

$$ \begin{aligned} P &= F \cdot \frac{1}{(1+i)^n} \\ &= F (1 + i)^{-n} \end{aligned} $$

Or using the formatted one:

$$ P = F \cdot (P / F, i, n) $$

Multiple Payments

Now let’s consider:

• $A_i$ to be the payment value at end of $i$-th period, then we would have:

Calculate F from At

$$ \begin{aligned} F &= \sum_{t=1}^{n}A_t \cdot (1 + i)^{n-t} \\ &= \sum_{t=1}^{n}A_t (F/P, i, n-t) \\ \end{aligned} $$

Calculate P from At

$$ \begin{aligned} P &= \sum_{t=1}^{n}A_t \cdot \frac{1}{(1 + i)^{t}} \\ &= \sum_{t=1}^{n}A_t (P/F, i, t) \\ \end{aligned} $$

Special $A_t$ Sequence

Here we only discuss the situation when all $A_t$ are the same value.

Calculate $F$ from Identical $A_t$

In this case, $A_i = A_j$ for any $i,j \in [1,n]$. Means each payment in different time period has the same value.

$$ \begin{aligned} F =& A(1+i)^{n-1} + A(1+i)^{n-2} + \cdots+A(1+i)^0 \\ =& A \sum_{t=1}^{n}(1+i)^{n-t} \\ =& A \cdot 1 \frac{1-(1+i)^n}{1-(1+i)} \\ =& A \frac{(1+i)^n - 1}{i} \\ =& A (F/A, i, n) \end{aligned} $$

Notice we used the Summation formula of geometric sequence.

Calculate Identical $A_t$ From $F$

We could just convert the formula above to get something like below:

$$ A = F \frac{i}{(1+i)^{n} -1} $$