# Calculating monthly payments of a savings plan

A savings plan requires you to make payments of £250 each at the end of every month for a year. The bank will then make six equal monthly payments to you, with its first payment due one month after the last payment you make to the bank. Compute the size of each monthly payment made by the bank, assuming a nominal interest rate of 4% p.a. payable monthly.

So I've been pondering on this question for a while now, and I'm not making any progress. I've tried various methods without any confidence in what I'm doing. Every method I try brings me a figure not mentioned in the answer. The only parts I'm confident in is the AER = 4.074%. What is the line of logic I should take with this question? Help is much appreciated as always.

• The yield is the interest rate that makes the present value of the cash inflows equal to the present value of the cash outflows. You are given the yield and the outflows. For the outflows, the only unknown is the amount of the constant monthly payment. Apr 6, 2019 at 14:59
• @saulspatz Thanks for the response, and yes I understand this conceptually. Although my main problem(s) comes in actually acquiring the answer given, through calculation/ method. Apr 6, 2019 at 15:10

Firstly we need a reference date for your payments and the payments of the bank. I´ve made a time line. The converted montly interest rate is $$i_{12}=\frac{0.04}{12}$$

Your payments start at the last day of January. We calculate the future value of your 12 payments at 31.12. We pretend that you start at 01.01, but make the payments at the end of each month. Then the future value is at $$\color{red}{t=0} \ (31.12)$$ This is the left hand side of the equation. Then we calculate the present value of the payments made by the bank. Since the payments ($$+x$$) start one month after $$t=0$$ we just discount the future value 6 times. The equation is

$$250\cdot\frac{(1+\frac{0.04}{12})^{12}-1}{\frac{0.04}{12}}=x\cdot\frac{(1+\frac{0.04}{12})^{6}-1}{\frac{0.04}{12}}\cdot\frac1{\left(1+\frac{0.04}{12}\right)^6}$$

$$3055.6157=x\cdot 5.930618 \Rightarrow x=\large{\color{grey}{£}} \ \normalsize 515.23 \ \$$

• Brilliant, it makes perfect sense now. Thank you so much. Apr 8, 2019 at 13:27
• I was asked a second part of the question just recently. I was asked that if the bank instead made equal payments in perpetuity three years after the last payment you paid, what would be the size of the annual payments. I took the LHS of the equation and times through by 0.04, but I was told this was wrong. Do you know why? Cheers. Apr 8, 2019 at 14:00
• @user657675 First of all you have to compound the LHS three years. Then set it equal with the present value of a perpetuity. $$250\cdot\frac{(1+\frac{0.04}{12})^{12}-1}{\frac{0.04}{12}}\cdot \left(1+\frac{0.04}{12}\right)^{36}=...$$ Apr 8, 2019 at 14:15
• Thanks for this. I did exactly that but still am off from the answer. I got an answer of 137.78, however apparently the actual answer is 134.84. Can't see why. Apr 8, 2019 at 14:23
• @user657675 At the moment I have no idea. I recommend to ask a new question in a new post. You just have to copy the exercise (including the two questions ) and add your thoughts. Apr 8, 2019 at 14:51