522  
523 1+m - n
524 m
525
526T = £177.60 =  Dn 1 + 
527 n=1  100 
528    
529 2
530  
531 = £100(= D1 ) 1 +  + £50(= D2 ) 1 + 
532  100   100 
533
534Trying  = 10.5 gives £177.353; 10.6 gives £177.624; 10.59 gives £177.596.
535This process of iterative computation yields the AER of 10.59% to two decimal places.
536
537
538
5394. If in the above example there were not an additional deposit contracted in the
540 second year, the calculation is simpler and formula (b) can be used:
541
542
543    10(= i )   11(= i )   
544 =   2 1+ 1
545  1+ 2
546   - 1  100 = 10.498...
547   100  100   
548 
549
550giving an AER of 10.50% to two decimal places.
551
552
553
5545. If an account pays interest more often than once a year, then the AER is
555 calculated by adding each interest payment to the deposit and calculating the
556 next interest payment on the total – compounding the interest.
557
558The treatment of monthly income accounts shows the basic use of formula (d).
559
560Suppose a fixed deposit offers two options:
561 A. interest paid annually of 6% per annum
562 B. interest paid monthly at a rate of 5.8% per annum
563
564Option A will have an AER of 6.00% (see example 1 above).
565
566For option B, the AER is calculated using formula (d) as:
567
568
569  5.8(= i) 
570 12
571
572
573 = 1 +  - 1 100 = 5.956...
574   12(months per year)100  
575  
576
577giving an AER of 5.96% to two decimal places.
578
579This demonstrates the value to the depositor of the monthly interest but also shows that the
580two options are not quite identical in terms of return.
581
582
583
5846. A short-term bond, for example an 8-month bond paying 5.5% per annum, has to
585 be treated using a combination of formulae (c) and (d).
586
587 11