As discussed in Section 3.3, the standard assumption for annuities and gradients is that the payment period and compounding period are the same. If they are not, the formulas given in this chapter cannot be applied directly. There are three methods for dealing with this situation:
1. Treat each cash flow in the annuity or gradient individually. This is most useful when the annuity or gradient series is not large.
2. Convert the non-standard annuity or gradient to standard form by changing the compounding period.
3. Convert the non-standard annuity to standard form by finding an equivalent standard annuity for the compounding period. This method cannot be used for gradients.
How much is accumulated over 20 years in a fund that pays 4% interest, compounded yearly, if Si000 is deposited at the end of even' fourth year?
The cash flow diagram for this set of payments is shown in Figure 3.9.
Figure 3.9 Non-Standard Annuity for Example 3.9
1000 1000 1000 1000 1000
Method 1: Consider the annuities as separate future payments.
Formula: Known values: Year 4 8 12 16 20
P = Si000, i = 0.04, At = 16, 12, 8, 4, and 0 Future Value
1000(F/P,4%,16) 1000(F/P,4%,12) 1000(F/P,4%,8) 1000(F/P,4%,4) 1000
Total future value
1000(1.8729) 1000(1.6010) 1000(1.3685) 1000(1.1698)
1873 1601 1369 1170 1000
About S7013 is accumulated over the 20 vears.
Method 2: Convert the compounding period from yearly to every four years. This can be done with the effective interest rate formula.
The future value is then
Method 3: Convert the annuity to an equivalent yearly annuity. This can be done by considering the first payment as a future value over the first four-year period, and finding the equivalent annuity over that period, using the sinking fund factor:
In other words, a Si000 deposit at the end of the four years is equivalent to four equal deposits of S235.49 at the end of each of the four years. This yearly annuity is accumulated over the 20 years.
Note that each method produces the same amount, allowing for rounding. Y\Tien you have a choice in methods as in this example, your choice will depend on what you find convenient, or what is the most efficient computationally.B
This year's electrical engineering class has decided to save up for a class party. Each of the 90 people in the class is to contribute SO.2 5 per day which will be placed in a daily interest (7 days a week, 365 days a year) savings account that pays a nominal 8% interest. Contributions will be made five days a week, Monday through Friday, beginning on Monday. The money is put into the account at the beginning of each day, and thus earns interest for the day. The class party is in 14 weeks (a full 14 weeks of payments will be made), and the money will be withdrawn on the Monday morning of the 15th week. How much will be saved, assuming everybody makes payments on time?
There are several good ways to solve this problem. One way is to convert each day's contribution to a weekly amount on Sunday evening/Monday morning, and then accumulate the weekly amounts over the 14 weeks:
Total contribution per day is 0.25 x 90 = 22.50
The interest rate per day is V?f = 0.000219
The effective interest rate for a 1 -week period is i = (1 + 0.08/365)' - 1 = 0.00154 Value of one week's contribution on Friday evening (annuity due formula): 22.50 x (F/P,0.08/365,l) x (F/4,0.08/365,5)
On Sunday evening this is worth
[22.50(F/P,0.08/365,l)(F/4,0.08/365,5)] x (F/P,0.08/365,2) = 22.50(P/P,0.08/365,3) (P7A,0.08/365,5)
Then the total amount accumulated by Monday morning of the 15th week is given by: [22.50(p7p,0.08/365,3)(f/a,0.08/365,5)](f/a, (1 + 0.08/365)7- 1,14) = [22.50(1.000 658) (5.002 19)] (14.1406) = 1592.56
The total amount saved would be S1592.56.^
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