ࡱ> ^`]5@ 6bjbj22 ŨXX&}.vvv1118J162lZz24444>I>I>I1Z3Z3Z3Z3Z3Z3Z$,\R~^FWZv>IGNH@>I>IWZ44lZpNpNpN>I2@4:R41ZpN>I1ZpNzpNNVU@"T"U42 pRӽ m1pKEU Y|Z0ZQUT^vL^U^vU>I>IpN>I>I>I>I>IWZWZ(/BN./ You should already know how to calculate the APR in the simplest case where a sum of money is borrowed at a particular time and paid back, with interest, in a single payment at a later date using the following formula C =  EMBED Equation.3  which gives the APR, i, as a decimal when a loan of C is paid back by a single repayment A after n years. In more difficult cases when the repayment will be made in more than one instalment, the formula involves a summation. For example, when there are 4 annual instalments, A1, A2,, A3, A4, and the formula becomes  EMBED Equation.3   Checking a given APR value This simply requires substitution of the given value of the APR into the formula to check whether it gives the correct value of the loan, C. Example A loan of 5000 is repaid in three equal annual instalments of 2000. The APR is quoted as 9.7%. Is this correct? For 3 annual instalments the formula is C =  EMBED Equation.3  Substituting i = 0.097 and A1 = A2 = A3 = 2000 gives:  C =  EMBED Equation.3  = 1823.15 + 1661.95 + 1514.99 = 5000.09 As this is very near 5000, the statement that the APR is 9.7% is likely to be correct. If you wish to be certain that 9.7% is correct to 1 decimal place, then you can do this by finding the value of C when i = 0.0965 and when 0.0975 as shown below: When i = 0.0965 C =  EMBED Equation.3  = 5005 When i = 0.0975 C =  EMBED Equation.3  = 4996 As the first of these gives a loan value above 5000 and the second gives a loan value below 5000, the value of i must lie between 0.0965 and 0.0975. So the APR must lie between 9.65% and 9.75% i.e. the APR is 9.7% correct to 1 decimal place. Some to try: 1 Check the value of the APR given in each of the following advertisements:   2 A borrower repays a debt of 4000 in 3 equal annual instalments of 1500. Show that the APR is 6.1% correct to 1 decimal place. 3 A loan of 12 000 is repaid in 4 equal annual instalments of 4000. Show that the APR is 12.6% correct to 1 decimal place. 4 A debt of 11 000 is repaid in 4 annual instalments of 2000, 3000, 4000 and 5000. Show that the APR is 9.0% correct to 1 decimal place. Finding a value for the APR When there is only one repayment, the formula for C can be rearranged to find a value for i relatively easily you should already have used this idea previously. However, when there is more than one instalment, it is not possible to do this and finding a value for the APR becomes much more difficult. One method that can be used is called the interval bisection method. Essentially this involves starting with a range of possible values for i and then repeatedly halving this range until the range becomes so small that it is possible to give an accurate value for i and hence the APR. If you require a very accurate value for i the method is time-consuming and tedious to carry out by hand and the more instalments there are, the worse it gets. The use of a computer makes this much less onerous. The next example explains how the method works using the relatively simple case where a loan is repaid by 4 instalments. Note that this is the most difficult case that the specification suggests you will need, but in real life things can become far more complex.  Example A loan of 24 000 is repaid in 4 annual instalments of 7500, 7500, 8000 and 8000. Use interval bisection to find the APR correct to 1 decimal place. Solution Suppose we start by trying the interval 6% to 12%. When i = 0.06, C =  EMBED Equation.3  = 26 804 When i = 0.12, C =  EMBED Equation.3  = 23 454 As the true value of C, 24 000 lies between these values, then the true value of i must lie between 0.06 and 0.12. 1st bisection The mid-point of the interval  EMBED Equation.3  is 0.09 When i = 0.09, C =  EMBED Equation.3  = 25 038 This is too high, indicating that 0.09 is too low, so i must lie between 0.09 and 0.12. This sketch illustrates the results so far:  2nd bisection The mid-point of  EMBED Equation.3  is 0.105 When i = 0.105, C =  EMBED Equation.3  = 24 225 This is still too high, so 0.105 is too low and i must lie between 0.105 and 0.12. 3rd bisection The mid-point of  EMBED Equation.3  is 0.1125 When i = 0.1125, C =  EMBED Equation.3  = 23 834 This is too low, so 0.1125 is too high and i must lie between 0.105 and 0.1125. This means that the APR must lie between 10.5% and 11.25%. Hence the APR = 11% to the nearest %. For a more accurate value of the APR, the process can be continued: 4th bisection The mid-point of  EMBED Equation.3  is 0.10875 When i = 0.10875, C =  EMBED Equation.3  = 24 028 This is too high, so 0.10875 is too low, so i must lie between 0.10875 and 0.1125. 5th bisection The mid-point of  EMBED Equation.3  is 0.110625 When i = 0.110625, C =  EMBED Equation.3  = 23 931 This is too low, so 0.110625 is too high and i must lie between 0.10875 and 0.110625. 6th bisection The mid-point of  EMBED Equation.3  is 0.1096875 When i = 0.1096875, C =  EMBED Equation.3  = 23 980 This is too low, so 0.1096875 is too high and i must lie between 0.10875 and 0.1096875. 7th bisection The mid-point of  EMBED Equation.3  is 0.10921875 When i = 0.10921875, C =  EMBED Equation.3  = 24 004 This is slightly too high, so 0.10921875 is too low, so i must lie between 0.10921875 and 0.1096875. 8th bisection The mid-point of the range  EMBED Equation.3  is 0.109453125 When i = 0.109453125, C =  EMBED Equation.3  = 23992 This is slightly too low, so 0.109453125 is too high, so i must lie between 0.10921875 and 0.109453125. This means that the APR must lie between 10.92% and 10.945%. Hence the APR = 10.9% to 1 decimal place. Some to try: 1 A borrower repays a debt of 4000 in 2 annual instalments of 2300. (a) Show that the APR lies between 8% and 12% (b) Use the interval bisection method to find the APR to the nearest %. 2 A loan of 8000 is repaid in 3 equal annual instalments of 3000. (a) Show that the APR lies between 8% and 10%. (b) Use the interval bisection method to find the APR correct to 1 decimal place. 3 A debt of 12 500 will be repaid in 4 annual instalments of 4000, 4250, 4500 and 4750. (a) Show that the APR lies between 10% and 20%. (b) Use the interval bisection method to find the APR correct to 1 decimal place. 4 A loan of 6000 is repaid in 3 annual instalments of 2000, 2500 and 3000. (a) Use the interval bisection method to find the APR to the nearest %. (b) Continue to use the interval bisection method until you find the value of the APR correct to 1 decimal place. 5 A debt of 3000 is repaid in 4 equal annual instalments of 1000. (a) Use the interval bisection method to find the APR to the nearest %. (b) Continue to use the interval bisection method until you find the value of the APR correct to 1 decimal place. 6 A borrower repays a loan of 7500 by paying 5000 after one year and 4000 after another year. Find the APR correct to 1 decimal place. 7 To repay a loan of 5500 a borrower pays annual instalments of 1000, 1500, 2000 and 2500. Find the APR correct to 1 decimal place. 8 A lender offers a choice of two ways of repaying a loan of 10 000. Repayment Method A: Repay in 3 annual instalments of 4000, 4500 and 5000 Repayment Method B: Repay in 4 equal annual instalments of 3500. (a) Find the total amounts repaid by each method. (b) Calculate the APR in each case correct to 1 decimal place. (c) (i) Give a reason why a borrower may prefer to use Method A to repay the loan of 10 000. (ii) Give a reason why a borrower may prefer to use Method B to repay the loan of 10 000.  Unit Advanced Level, Mathematical Principles for Personal Finance Notes on Activity An earlier Nuffield resource for this FSMQ called 'APR Annual Percentage Rate' shows learners how to find the APR of a loan in the simplest case where the loan is repaid in a single instalment. This resource involves checking and finding the APR in more difficult cases when a loan or debt is repaid in 2, 3 or 4 annual instalments. Pages 1 and 2 show how checks can be made by substituting given values of i into an equation of the form:  EMBED Equation.3  Pages 3 to 5 then show how the APR can be found using the interval bisection method to solve the above equation for i. There are also examples for students to try on pages 2 and 6 the answers to those on page 6 are given below. The accompanying spreadsheet gives two worksheets that show how spreadsheet formulae can be used to find the APR in the example that starts on page 3. This can be used to demonstrate to learners how a spreadsheet can be used to find an APR or learners can use it themselves to check the answers to the questions on page 6. Examples of APRs in more complex cases can be found in 'Credit Charges and APR' which is available from the Office of Fair Trading ( HYPERLINK "http://www.oft.gov.uk" www.oft.gov.uk). If you have time you could ask students to check values of APR that are given in real advertisements for credit cards and mortgages as well as loans. Answers Page 6 1 (b) 10% 2 (b) 6.1% 3 (b) 14.5% 4 (a) 11% (b) 11.2% 5 (a) 13% (b) 12.6% 6 13.6% 7 9.0% 8 (a) Method A 13 500 Method B 14 000 (b) Method A 16.0% Method B 15.0% (c) (i) A borrower may prefer Method A because it keeps repayments costs to a minimum. (ii) A borrower may prefer Method B because it spreads the repayments over a longer time and the size of each instalment is lower. A Resource for Free-standing Mathematics Qualifications APR in more difficult cases  The Nuffield Foundation  PAGE 7 General APR formula for cases involving more than one instalment C =  EMBED Equation.3  where i is the APR expressed as a decimal, k is the number identifying a particular instalment, Ak is the amount of the instalment k, tk is the interval in years between the payment of the instalment and the start of the loan. Note these are the present values of the 3 future instalments. Can you use your calculator in an efficient way to check these values? Teacher Notes APR 20% Borrow 10 000. Repay in 2 annual instalments of 6000 and 7200 Borrow 3600. Repay in 2 annual instalments of 2250 and 2025 APR 12.5% 26 804 too high Finding the APR using the interval bisection method This 'trial and improvement' method is summarised below: Choose an interval of values within which you believe the APR lies. Substitute the i value at each end of your interval into C =  EMBED Equation.3  to check that the range you have chosen does include the value of i. Next use the mid-point of your range as i and find C using C =  EMBED Equation.3  If the calculated value of C is too low, this implies that the value used for i was too high, so you now know that the correct value of i lies in the lower half of the interval you used. If the calculated value of C is too high, this implies that the value used for i was too low, so you now know that the correct value of i lies in the upper half of the interval you used. Repeat the last two steps using the new interval within which you know the correct value of i lies. The fact that this is half of the previous interval gives this method its name of 'interval bisection'. Repeat the process again and again, until the interval is narrow enough to give you an accurate value of i. 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