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

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}

• Please show your work, so we can see where you got stuck and why. You won't find anybody solving your problems here if you don't show interest. – vonbrand Feb 4 '14 at 16:04
• I added what I think it is, but I'm very rusty on this and think I'm wrong somewhere – Josh_at_Savings_Champion Feb 4 '14 at 16:12
• That's not correct, and I don't know latex to fix it. – Josh_at_Savings_Champion Feb 4 '14 at 16:57

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%

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%

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 $\frac 14[(2.25+I)+(2.25+2I)+(2.25+3I)+(2.25+4I)]=2.25+\frac {10I}4=R$

• 5root(A/P) - 1 = R rearranged for R, then use (4/10)R - (9/10) = I? – Josh_at_Savings_Champion Feb 4 '14 at 16:53
• You don't say what that is applying to. If it is Bob's account (the only thing with a 5 in it), it is not correct: it should be $1+R$ on the right. We know the interest Bob is getting. What do you get for the value of Bob's account at the end? – Ross Millikan Feb 4 '14 at 16:56
• edited, and I'm after I – Josh_at_Savings_Champion Feb 4 '14 at 16:57
• Note that I gave instructions for calculating $R$ in the second paragraph. It will be (one less than) the fourth root of (Bob's account value at end of year 5)/(Sue's account value at the end of year 1) Do you see why? – Ross Millikan Feb 4 '14 at 17:01
• I see where we've gotten confused, The first equation above in comments is the Compound interest equation rearranged for R, I personally ended up doing as you've just said, then plugged that value of R into the second equation to retrieve I. But my answer is negative which is not expected. – Josh_at_Savings_Champion Feb 4 '14 at 17:07