accumulated value

Question

.Find the accumulated value at the end of 10 years of 4 payments of £200 each payable half yearly in advance, if the interest rate is:

(a) 3% per annum effective.

(b) 5% per annum convertible monthly.

Answer 1

Two details before I begin my explanation:

1) I live in the United States. I don't have a GBP symbol on my keyboard, so I'll use the $ symbol instead. I hope that's all right with you.

2) Throughout this explanation, I'll go through a number of less-than-optimally efficient ways to solve it, to motivate the reasoning behind the best way to do it, then present the most efficient solution. If you just want to skip to that part, skip to the part that says "A faster method:"

For part a), we know the interest rate is 3% per year. This means that for every $1 in the account at some time, the amount will accumulate to $1.03 in one year. After n years, even if n is not a whole number, each $1 will grow to $1.03ⁿ.

The phrase "payable in advance" means that the first payment happens right away and each payment happens some time afterward (in this case, six months afterward). So, the first payment happens now, the second one six months from now, the third one year from now, and the fourth a year and a half from now.

Let's think about each of these four payments individually. The first one starts as $500 and grows for 10 years, so it accumulates to $500 * (1.03)¹⁰.

The second one starts as $500 and grows for only 9.5 years because it was deposited half a year in, only giving it 9.5 years to collect interest. It accumulates to $500 * (1.03)^9.5.

The third one starts as $500 and grows for 9 years, so it accumulates to $500 * (1.03)⁹.

Finally, the last one starts as $500 and grows for 8.5 years, so it accumulates to $500 * (1.03)^8.5.

Thus, the total accumulation function would be

$500 * (1.03)¹⁰ + $500 * (1.03)^9.5 + $500 * (1.03)⁹ + $500 * (1.03)^8.5.

You could calculate that number directly by brute force, but there is a shorter way.

A faster method:

Let X be the accumulation amount at the end of the 10 years. We start with the equation

$500 * (1.03)¹⁰ + $500 * (1.03)^9.5 + $500 * (1.03)⁹ + $500 * (1.03)^8.5 = X.

We divide by (1.03)^0.5 to get

$500 * (1.03)^9.5 + $500 * (1.03)⁹ + $500 * (1.03)^8.5 + $500 * (1.03)⁸ = X * (1.03)^-0.5

We subtract this equation from the previous one to get

$500 * ((1.03)¹⁰ - (1.03)⁸) = X *(1 - (1.03)^-0.5)

Which we can use to isolate X and get

X = [$500 * ((1.03)¹⁰ - (1.03)⁸)] / [1 - (1.03)^-0.5] = $2629.26.

Part b):

The only thing that has changed is the interest structure. Now it earns 5% nominal, compounded every month. This means that it earns 5/12 of a percent (roughly 0.42%) every month. For a six month period, this means each $1 grows to $(1.0042)⁶. Analogous to our solution in part a), we get the equation

$500 * (1.0042)¹²⁰ + $500 * (1.0042)¹¹⁴ + $500 * (1.0042)¹⁰⁸+ $500 * (1.0042)¹⁰²= X.

Dividing by (1.0042)⁶, we get

$500 * (1.0042)¹¹⁴ + $500 * (1.0042)¹⁰⁸ + $500 * (1.0042)¹⁰²+ $500 * (1.0042)⁹⁶= X * (1.0042)^-6.

Subtracting the second equation from the first, we get

$500 * ((1.0042)¹²⁰ - (1.0042)⁹⁶) = X * (1 - (1.0042)^-6).

X = [$500 * ((1.0042)¹²⁰ - (1.0042)⁹⁶) ] / (1 - (1.0042)^-6) = 3186.00 (rounded)