THE MEASUREMENT OF INTEREST
CONTENTS • THE ACCUMULATION AND AMOUNT FUNCTIONS • THE EFFECTIVE RATE OF INTEREST • SIMPLE INTEREST
THE ACCUMULATION AND AMOUNT FUNCTIONS The Financial Transaction • An amount of money or capital (Principal) is invested for a period of time. • At the end of the investment period, a total amount (Accumulated Value) is returned. • Difference between the Accumulated Value and the Principal is the Interest Earned, đ??źđ?‘› Accumulation Function : đ?‘Ž(đ?‘Ą) • Let t be the number of investment year where đ?‘Ž(0) = 1 • Assume that đ?‘Ž(đ?‘Ą) is continuously (limit) or increasingly (double differentiation). • đ?‘Ž(đ?‘Ą) defines the pattern of accumulation for an investment of amount 1. Amount Function : đ??´(đ?‘Ą) = đ?‘˜ . đ?‘Ž(đ?‘Ą) • Let đ?‘˜ the initial principal invested (đ?‘˜ > 0) where đ??´(0) = đ?‘˜. • đ??´(đ?‘Ą) is continuously increasing. • đ??´(đ?‘Ą) defines the Accumulated Value that amount đ?‘˜ grows to in đ?‘Ąyears. Interest Earned during the nth period : • đ??źđ?‘› = đ??´ đ?‘› – đ??´ đ?‘› − 1 • Interest earned is the difference between the Accumulated Value at the end of a period and the Accumulated Value at the beginning of the period.
THE ACCUMULATION AND AMOUNT FUNCTIONS Example 1 :
An investment of $20,000 is made into a fund at time t = 0. The fund develops the following balances over the next 4 years : t
A(t)
0
20,000.00
1
20,600.00
2
21,130.00
3
21,575.00
4
22,153.96
If $10,000 is invested at time t = 2, under the same interest environment, find the accumulated value of the $10,000 at time t = 4.
Answer : Let K be the accumulated value of the $10,000. K can be determined by ratio and proportion đ??ž đ??´(4) 22,153.96 = = 10,000 đ??´(2) 21,130.00 And solving K we have đ??ž = $10,484.60
THE ACCUMULATION AND AMOUNT FUNCTIONS
Example 2:
For the $10,000 investment given in Example 1, find the amount of interest earned during the second year of investment, i.e. between t=3 and t=4.
đ??ž đ??´(3) 21,575.20 = = 10,000 đ??´(2) 21,130.00 K = $10,210.70
đ??ž đ??´(4) 22,153.96 = = 10,000 đ??´(2) 21,130.00 K = $10,484.60 đ??źđ?‘› = đ??´(đ?‘›) – đ??´(đ?‘› − 1) = 10,484.60 – 10,210.70 = $273.90
THE EFFECTIVE RATE OF INTEREST : i Definition
•The amount of interest earned over a one-year period when 1 is invested. •Let I be the effective rate of interest earned during the nth period of the investment where interest is paid at the end of the period. •Can also be defined as the ratio of the amount of Interest Earned during the period to the Accumulated Value at the beginning of the period
đ?‘–đ?‘› =
đ??´ đ?‘› − đ??´(đ?‘› − 1) đ??źđ?‘› = đ??´(đ?‘› − 1) đ??´(đ?‘› − 1)
for integral � ≼ 1
THE EFFECTIVE RATE OF INTEREST : i EXAMPLE 1:
For the $20,000 investment given in Example 1 in 1.2, find the effective rate of interest for each of the four years. Answer : From the table of values given in Example 1 in 1.2, we can apply formula đ?‘–đ?‘› = đ??´ đ?‘› − đ??´(đ?‘›âˆ’1) đ??´(đ?‘›âˆ’1)
four times to obtain : đ??´ 1 −đ??´ 0 20,600.00 − 20,000.00 đ??ź1 = = = 0.03, 3% đ??´(0) 20,000.00 đ??´ 2 −đ??´ 1 21,130.00 − 20,600.00 đ??ź2 = = = 0.025,2.5% đ??´(1) 20,600.00 đ??´ 3 −đ??´ 2 21,575.20 − 21,130.00 đ??ź3 = = = 0.021,2.1% đ??´(2) 21,130.00 đ??´ 4 −đ??´ 3 22,153.96 − 21,575.20 đ??ź4 = = = 0.027, 2.7% đ??´(3) 21,575.20
THE EFFECTIVE RATE OF INTEREST : i
EXAMPLE 2:
Assume that 𝐴(𝑡) = 1000 + 50𝑡. Find 𝑖5 Answer :
𝑖5 = = =
𝐴 5 −𝐴(4) 𝐴(4) 1000+50(5) − 1000+50(4) 1000+50(4) 1 24
SIMPLE INTEREST Since đ?‘Ž 0 = 1 and đ?‘Ž(1) = (1 + đ?‘–) , thus consider the investment of one unit such that the amount of interest earned during each period is constant. The accumulated value of 1 at the end of the first period is 1 + đ?‘–, at the end of the second period it is 1 + 2đ?‘–, etc. Thus, in general, we have a linear accumulation function.
� � = 1 + �� for � = 1,2,3,‌..
The accruing of interest according to this pattern is called simple interest.
−đ?‘Ž(đ?‘›âˆ’1) đ?‘–đ?‘› = đ?‘Ž đ?‘›đ?‘Ž(đ?‘›âˆ’1) =
1+đ?‘–đ?‘› −[1+đ?‘–(đ?‘›âˆ’1)] 1+đ?‘–(đ?‘›âˆ’1)
đ?‘– = 1+đ?‘–(đ?‘›âˆ’1)
SIMPLE INTEREST EXAMPLE 1: Find the accumulated value of $4000 invested for 6 years, if the rate of simple interest is 7% per annum. ANSWER: đ?‘Ą = 6đ?‘– = 0.07
đ?‘˜ = 4000
đ??´ đ?‘Ą = đ?‘˜. đ?‘Ž đ?‘Ą đ??´ đ?‘Ą = đ?‘˜. (1 + đ?‘–đ?‘Ą) = 4000. 1 + 0.07 6 = 4000 1 + 0.42 = $5680
SIMPLE INTEREST EXAMPLE 2: At what rate of simple interest will $750 accumulated to $846 in 4 years.
ANSWER: đ?‘Ą = 4đ??´ đ?‘Ą = 846
đ?‘˜ = 750
đ??´ đ?‘Ą = đ?‘˜. đ?‘Ž(đ?‘Ą) đ??´ đ?‘Ą = đ?‘˜. (1 + đ?‘–đ?‘Ą) 846 = 750. 1 + đ?‘– 4 1 + đ?‘– 4 = 1.128 đ?‘–(4) = 0.128 đ?‘– = 0.032 đ?‘– = 3.2%