Find percentage increase per period to match a different percentage over the same period

Question

Bob has an account with £1000 that pays 3.5% interest that is fixed for 5 years and he cannot withdraw that money over the 5 years

Sue has an account with £1000 that pays 2.25% for one year, and is also inaccessible for one year.

Sue wants to take advantage of better rates and so moves accounts each year to get the better rates.

How much does the interest rate need to increase per year (on average) for Sue to beat Bob's 5 year account?

Compound interest formula: $A = P(1 + Q)^T$

Where:

$A$ = Amount Earned $P$ = Amount Deposited $R$ = Sues Interest Rate $T$ = Term of Account $Q$ = Bobs Interest rate $I$ = Interest Increase Per Period

My method of working thus far:

\begin{align} \text{First I calculate Bobs money at 5 years}\\ P(1 + Q)^T &= A \\ 1000(1 + 0.035)^5 &= A \\ 1187.686 &= A\\ 1187.68 &= A (2DP)\\ \text{Now work out Sues first years interest}\\ 1000(1 + 0.0225) ^ 1 &= A \\ 1022.5 &= A\\ \text{Then I work out the next 4 years compound interest}\\ ((1187.686/1022.5) ^ {1/4}) - 1 &= R \\ -0.7096122249388753 &= R\\ -0.71 &= R (2DP)\\ \text{Then I use the rearranged formula from Ross Millikan}\\ 4/{10}R - 9/{10} &= I\\ 4/{10}*-0.71 - 9/{10} &= I\\ 0.0 &= I\\ \end{align}

Answer 1

You should just be able to plug the values into your first equation to get the value of Bob's account at the end of 5 years. Note that A is the total value, not the amount earned. Also note that the interest rate needs to be expressed as a decimal.

For Sue, first calculate how much she has at the end of the first year: $1000\cdot (1+0.0225)=1022.50$ When she deposits that into the new account, that becomes P. You should be able to find what her R needs to be for $T=4$ to match Bob's value at the end of 5 years. To a good approximation, that is what here average needs to be over the four, so you have $\frac 14[(2.25+I)+(2.25+2I)+(2.25+3I)+(2.25+4I)]=2.25+\frac {10I}4=R$

Answer 2

A bit of trail and error is needed here as I cant see a closed form solution.

For bob He ends up with $1000\cdot(1+0.035)^5 \approx 1187.686$

For Sue its

$1000 \cdot (1+0.0225)\cdot(1+0.0225+I)\cdot(1+0.0225+2I)\cdot(1+0.0225+3I)\cdot(1+0.0225+4I)$

There are various ways you can solve this. However I just put it into a spreadsheet and played with the values. Its more than 0.6268% and less than 0.6269%

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Note: Its not an average increase as If it were all to come in year 2 a smaller increase would be required

After the first year Sue has $1000\cdot (1+0.0225) = 1022.50$

Now with 4 years compound interest and only one rate

$1187.686 = 1022.50 \cdot (1+0.0225+I)^4 \Rightarrow (1+0.0225+I)^4 = \frac{1187.686}{1022.50}$

$ \Rightarrow 1.0225 + I = \sqrt[4]{\frac{1187.686}{1022.50}}$

So $1.0225 + I = 1.03799 \Rightarrow I = 0.01549 = 1.548\%$

Which averaged over 4 years is 0.3875%