4.7 Compound Interest February 05, 2009
4.7 Compound Interest
Objective: Determine the future value of a lump sum of money.
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4.7 Compound Interest February 05, 2009
Simple Interest Formula:
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I = Prt
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4.7 Compound Interest February 05, 2009
A credit union pays interest of 8% per annum compounded quarterly on a savings account. If $1,000 is deposited and the interest is left to accumulate, how much is in the account after 1 year?
I = Prt
1
I = ($1000)(0.08)( )
4
I = $20
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4.7 Compound Interest February 05, 2009
At the end of one quarter, we have earned $20 in interest.
My principal is now $1020. So at the end of the second quarter, how much do I earn?
1
I = ($1020)(0.08)( )
4
I = $20.40
Adding that interest to the money in my account, I now have $1040.40 ­ 6 months into the year.
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4.7 Compound Interest February 05, 2009
At the end of the third quarter, how much do I have?
1
I = ($1040.40)(0.08)( )
4
I = $20.81
This brings my principal to $1,061.21.
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4.7 Compound Interest February 05, 2009
At the end of the fourth quarter, or at the end of the year what do I have?
1
I = ($1061.21)(0.08)( )
4
I = $21.22
This brings the total in my bank account to $1,082.43 just by collecting interest.
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4.7 Compound Interest February 05, 2009
This process can be very lengthy especially if the interest were being compounded daily. This leads to a more general formula for calculating interest that is compounded.
Compounded interest formula:
r nt
A = P(1 + )
n
n = the number of compounds per year
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4.7 Compound Interest February 05, 2009
Compounded interest is interest that is paid on principal and previously earned interest. Common forms of compounding are:
annually: once per year
semiannually: twice per year
quarterly: four times per year
monthly: 12 times per year
daily: 365 times per year
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4.7 Compound Interest February 05, 2009
Lets compare investments using different compounding periods.
nt
r
A = P(1 + )
n
n = the number of compounds per year
Find the resulting amount of an investment of $3000 when it is compounded annually, semiannually, quarterly, monthly and daily at a rate of 4% for a period of 3 years.
(1 3)
0.04
A = (3000)(1 + )
1
3
A = (3000)(1 + 0.04)
3
A = (3000)(1.04)
A = $3,374.59
Feb 5­3:46 PM
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4.7 Compound Interest February 05, 2009
Lets compare investments using different compounding periods.
nt
r
A = P(1 + )
n
n = the number of compounds per year
Find the resulting amount of an investment of $3000 when it is compounded annually, semiannually, quarterly, monthly and daily at a rate of 4% for a period of 3 years.
0.04 (2 3)
A = (3000)(1 + )
2
A = (3000)(1 + 0.02)
6
6
A = (3000)(1.02)
A = $3,378.49
Feb 5­3:46 PM
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4.7 Compound Interest February 05, 2009
Lets compare investments using different compounding periods.
nt
r
A = P(1 + )
n
n = the number of compounds per year
Find the resulting amount of an investment of $3000 when it is compounded annually, semiannually, quarterly, monthly and daily at a rate of 4% for a period of 3 years.
0.04 (4 3)
A = (3000)(1 + )
4
12
A = (3000)(1 + 0.01)
12
A = (3000)(1.01)
A = $3,380.48
Feb 5­3:46 PM
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4.7 Compound Interest February 05, 2009
Lets compare investments using different compounding periods.
nt
r
A = P(1 + )
n
n = the number of compounds per year
Find the resulting amount of an investment of $3000 when it is compounded annually, semiannually, quarterly, monthly and daily at a rate of 4% for a period of 3 years.
0.04 (12 3)
A = (3000)(1 + )
12
A = (3000)(1 + 0.0033333333)36
A = (3000)(1.0033333333)36
A = $3,381.82
Feb 5­3:46 PM
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4.7 Compound Interest February 05, 2009
Lets compare investments using different compounding periods.
nt
r
A = P(1 + )
n
n = the number of compounds per year
Find the resulting amount of an investment of $3000 when it is compounded annually, semiannually, quarterly, monthly and daily at a rate of 4% for a period of 3 years.
0.04 (365 3)
A = (3000)(1 + )
365
A = (3000)(1 + 0.0001095890411)1095
1095
A = (3000)(1.0001095890411)
A = $3,382.47
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4.7 Compound Interest February 05, 2009
What if the investment is compounded more often than daily?
Compounding Interest Continuously:
rt
A = Pe
Find the resulting amount of an investment of $3000 when it is compounded continuously at a rate of 4% for a period of 3 years.
A = Pe
rt
A = (3000)e
A = (3000)e
(0.04 3)
0.12
A = $3,382.49
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4.7 Compound Interest February 05, 2009
How much principle is needed now to have $4000 after 2 years at 6% compounded quarterly?
nt
r
A = P(1 + )
n
n = the number of compounds per year
0.06 (4 2)
4000 = P(1 + )
4
4000 = P(1 + 0.015) 8
4000 = P(1.015)8
4000 = P(1.126492587)
P = $3,550.84
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4.7 Compound Interest February 05, 2009
How long will it take for an investment $250 to reach $675 in value if it earns 9% compounded monthly?
nt
r
A = P(1 + )
n
n = the number of compounds per year
(12 t)
0.09
675 = (250)(1 + )
12
675 = (250)(1 + 0.0075)
12t
12t
675 = (250)(1.0075)
12t
2.7 = (1.0075)
log 1.00752.7 = 12t
log 2.7
= 12t
log 1.0075
132.9295772 = 12t
t ≈ 11.08 years
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4.7 Compound Interest February 05, 2009
How long will it take for an investment to double in value if it earns 5% compounded continuously?
A = Pe
rt
2P = Pe
0.05t
2 = e 0.05t
ln 2 = ln e
0.05t
ln 2 = 0.05t ln e
ln 2 = 0.05t
t = 13.86 years
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4.7 Compound Interest February 05, 2009
Homework
page 322
(4, 5, 8, 9, 12, 14, 16, 18, 19, 28, 32, 36, 37, 48)
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