465It follows that if an account pays interest once a year on say the 31 March then the AER, no
466matter when the account is advertised, will always equal the gross rate.
467
468
469
4702. If an account pays interest at intervals greater than 1 year, the AER is the rate
471 which will give the same answer if applied and compounded each year.
472
473In the simple interest example of 7% for two years quoted above, using £100:
474
475
476  
477 1 + 2 -1(= 2)
478   0(= i1 )   14(= i2 ) 
479 £100 1 +  = £100 1 +  1 +  ( = £114 )
480  100   100   100 
481
482
483
484  114 
485 so  =  2 - 1  100 = 6.771...
486  100 
487
488
489
490(which can also be reached using formula (c) directly) and the resultant AER is 6.77% to two
491decimal places.
492
493
494
4953. Now suppose a deposit of £100 is to be made at the beginning of the first year
496 and £50 to be added at the start of Year 2 (as part of the product requirement, not
497 an option), and interest is to be 10% for Year 1 and 11% for Year 2 (e.g. an
498 escalating interest rate, not a prediction of interest rate movements).
499
500The calculation should be approached as follows:
501
502First work out what the eventual return at the end of year 2 (T) is going to be, using the right
503hand side of the formula:
504
505 2  2  i j 
506T =  Dn   1 +  
507   100  
508 n=1  j = n
509
510
511   10(= i1 )   11(= i 2 )    11(= i 2 ) 
512 = £100(= D1 )  1 +  1 +   + £50(= D2 ) 1 + 
513  100   100    100 
514 = £177.60
515
516Then find a value for  that satisfies the left hand side:
517
518
519
520
521 10