Interest for different Compounding Periods

Financial Mathematics → Interest → Simple Interest → Compound Interest → Varying Interest Rate → Interest for different Compounding Periods

Financial Institutions often pay or charge interest more than a single time per year. For example, monthly, quarterly, weekly etc.

If the sum of C is invested for a term h at time t. Then the accumulated sum at t+h will be

Where,
C = Original cash
h = Compounding period
i_h(t) = Nominal interest rate per annum
A = Accumulated amount

The value of h

Term	h	Reason
daily	1/365	There are 365 days in one year
weekly	1/52	There are 52 weeks in one year
monthly	1/12	There are 12 months in one year
quarterly	1/4	There are 12 months in one year
biannually	1/2	Two times in one year

Example 1. The nominal interest rate is 12% per annum on transactions of term a month (interest compounded monthly). Calculate the accumulation of $100 invested at this rate after:
(a) 1 month
(b) 2 months
(c) 3 months
Solution:
(a) after 1 month
$A=C[1+hi_h(t)] \\ = 100[1 + (1/12)(0.12)] \\ = 100(1.01) \\ = \$101$ Answer

(b) after 2 months
$A = C[1 + hi_h(t)]^2 \\ = 100[1 + (1/12)(0.12)]^2 \\ = 100(1.01)2 \\ = 100(1.0201) \\ = \$102.01$ Answer

(c) after 3 months
$A = C[1 + hi_h(t)]^3 \\ = 100[1 + (1/12)(0.12)]^3 \\ = 100(1.01)^3 \\ = 100(1.0303) \\ = \$103.03$ Answer

Example 2. The nominal interest rate of interest per annum quoted in the financial press for local authority deposits on a particular day are as follows:

Term	Nominal rate of interest (%)
1 day	11.75
1 week	11.5
1 month	11.375
1 quarter	11.25

Find the accumulation of an investment at this time of $100 for
(a) for 2 days
(b) for 3 weeks
(c) for 1 month
(d) for 7 days, and compare with 1 week
(e) for 90 days, and compare with 1 quarter
Solution:
(a) for 2 days
$A = C[1 + hi_h(t)]^2 \\ = 100[1 + (1/365)(0.1175)]^2 \\ = 100(1.00064) \\ = \$100.064$ Answer

(b) for 3 weeks
$A = C[1 + hi_h(t)]^3 \\ = 100[1 + (1/52)(0.115)]^3 \\ = 100(1.000664) \\ = \$100.664$ Answer

(c) for 1 month
$A = C[1 + hi_h(t)] \\ = 100[1 + (1/12)(0.11375)] \\ = 100(1.00948) \\ = \$100.948$ Answer

(d) for 7 days $A = C[1 + hi_h(t)]^7 \\ = 100[1 + (1/365)(0.1175)]^7 \\ = 100(1.00225) \\ = \$100.2255$ Answer	for 1 week $A = C[1 + hi_h(t)] \\ = 100[1 + (1/52)(0.115)] \\ = 100(1.002211) \\ = \$100.221$ Answer
(e) for 90 days $A = C[1 + hi_h(t)]^{90} \\ = 100[1 + (1/365)(0.1175)]^{90} \\ = 100(1.02939) \\ = \$102.94$ Answer	for 1 quarter $A = C[1 + hi_h(t)] \\ = 100[1 + (1/4)(0.1125)] \\ = 100(1.028) \\ = \$102.8$ Answer

Next: Continuous Interest