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Lesson 26: Compounding More than Once a Year Compounding More than Once a Year
Lesson 26: Compounding More than Once a Year
Learning Outcome(s)
Learning Outcome(s): At the end of the lesson, the learner is able to compute maturity value, interest, and present value, and solve problems involving compound interest when compound interest is computed more than once a year
Lesson Outine:
Lesson Outine:
1. Compounding more than once a year
2. Maturity value, interest, and present value when compound interest is computed more than once a year
Sometimes, interest may be compounded more than once a year. Consider the following example.
Example 1.
Example 1. Given a principal of PhP 10,000, which of the following options will yield greater interest after 5 years:
OPTION A
OPTION A: Earn an annual interest rate of 2% at the end of the year, or OPTION B
OPTION B: Earn an annual interest rate of 2% in two portions —1% after 6 months, and 1% after another 6 months?
Solution.
OPTION A: Interest is compounded OPTION A: Interest is compoundedannually
Time (t) in years
Principal = PhP 10,000
Annual Int. rate = 2%, compounded annually Amount at the end of the year
1 10,000 1.02 = 10,200 2 10,200 1.02 = 10,404 3 10,404 1.02 = 10,612.08 4 10,612.08 1.02 = 10,824.32 5 10,824.32 1.02 = 11,040.81
D E P E D C O P Y
OPTION B: Interest is compounded
OPTION B: Interest is compoundedsemi-annually, or every 6 months.or every 6 months.
Under this option, the interest rate every six months is 1% (2% divided by 2).
Time
Annual Int. rate = 2%, compounded semi-annually2%, compounded semi-annually Amount at the end of the year 2 10,303.0110,303.01 1.01 = 10,406.0410,406.04 2½ 10,406.0410,406.04 1.01 = 10,510.1010,510.10
3 10,510.1010,510.10 1.01 = 10,615.2010,615.20 3½ 10,615.2010,615.20 1.01 = 10,721.3510,721.35
4 10,721.3510,721.35 1.01 = 10,828.5610,828.56 4½ 10,828.5610,828.56 1.01 = 10,936.8510,936.85
5 10,936.8510,936.85 1.01 = 11,046.2211,046.22
Answer:
Answer: Option B will give the higher interest after 5 years.If all else is equal, aa more frequent compounding will result in a higher interest,
more frequent compounding will result in a higher interest, which is why Option B gives a higher interest than Option A.
The investment scheme in Option B introduces new concepts because interest is compounded twice a year, the conversion periodconversion period is 6 months, and the frequencyfrequency of conversion
of conversion is 2. As the investment runs for 5 years, the total number oftotal number of conversion periods
conversion periods is 10. The nominal ratenominal rate is 2% and the rate of interest forrate of interest for each conversion period
each conversion period is 1%. These terms are defined generally below.
Definition of Terms:
Definition of Terms:
Frequency of conversion (m)
Frequency of conversion (m) – number of conversion periods in one year Conversion or interest period
Conversion or interest period – time between successive conversions of interest Total number of conversion periods n
Total number of conversion periods n
n = mt = (frequency of conversion)(time in years) Nominal rate (i
Nominal rate (i(m)(m))) – annual rate of interest Rate (j) of interest for each conversion Rate (j) of interest for each conversion periodperiod
=
D E P E D C O P Y
Note on rate notation: r, i Note on rate notation: r, i(m)(m), j, j
In earlier lessons, r was used to denote the interest rate. Now that an interest rate can refer to two rates (either nominal or rate per conversion period), the symbols i(m) and j will be used instead.
Examples of nominal rates and the
Examples of nominal rates and the corresponding frequencies of conversioncorresponding frequencies of conversion and interest rate for each period
and interest rate for each period:
i(m) = Nominal Rate
From Lesson 25, the formula for the maturity value F when principal P is invested at an annual interest rate j compounded annually isF= P(1+j)F= P(1+j)tt.
Because the rate for each conversion period is j =
, then in t years, interest is compounded mt times. The following formula is obtained.
Maturity Value, Compounding m times a Maturity Value, Compounding m times a year year
where F = maturity (future) value P = principal
i(m) = nominal rate of interest (annual rate) m = frequency of conversion
t = term/ time in years
D E P E D C O P Y
Example 2.
Example 2. Find the maturity value and interest if P10,000 is deposited in a bank at 2% compounded quarterly for 5 years.
Given: P = 10,000 i(4)= 0.02 t = 5 years m = 4 Find: a. F
b. P Solution.
Compute for the interest rate in a conversion period by
j =
=
= 0.005.Compute for the total number of conversion periods given by n = mt = (4)(5) = 20 conversion periods.
Compute for the maturity value using F= P(1+j)n
= (10,000)(1 + 0.005)20 F= 11,048.96.
The compound interest is given by
Ic = F – P = 11,048.96 – 10,000 = P1,048.96.
Example 3.
Example 3. Find the maturity value and interest if P10,000 is deposited in a bank at 2% compounded monthly for 5 years.
Given: P = 10,000 i(12)= 0.02 t = 5 years m= 12 Find: a. F
b. P Solution.
Compute for the interest rate in a conversion period by j =
=
Compute for the total number of conversion periods given by n = mt = (12)(5) = 60 conversion periods.
Compute for the maturity value using F= P(1+j)n
= (10 000)
D E P E D C O P Y
F = P11,050.79
.The compound interest is given by
Ic = F – P =11,050.79 – 10,000 = P1,050.79.
Example 4.
Example 4. Cris borrows P50,000 and promises to pay the principal and interest at 12% compounded monthly. How much must he repay after 6 years?
Given: P = P50,000 i(12) = 0.12 t = 6 m = 12 Find: F
Solution.
You may also use the other formula to compute for the maturity value
F =
F = (50,000)(1.01)72 F = P102,354.97 Thus, Cris must pay P102,354.97 after 6 years.
Example 5.
Example 5. Find the present value of P50,000 due in 4 years if money is invested at 12% compounded semi-annually.
Given: F = 50,000 t = 4 i(2) = 0.12
Find: P
Present Value P at Compound Int Present Value P at Compound Interesterest
where F = maturity (future) value P = principal
i(m) = nominal rate of interest (annual rate) m = frequency of conversion
t = term/ time in years
D E P E D C O P Y
Solution.
First, compute for the interest rate per conversion period given by j =
=
= 0.06.The total number of conversion periods is n = tm = (4)(2) = 8.
The present value can be computed by substituting these values in the formula P =
.Thus,
P =
=
= P31,370.62.Example 6.
Example 6. What is the present value of P25,000 due in 2 years and 6 months if money is worth 10% compounded quarterly?
Given: F = 25,000 t = 2½ years i(4) = 0.10 Find: P
Solution.
The interest rate per conversion period given by j =
=
= 0.025,and the total number of conversion periods is n = tm = (2 ½ )(4) = 10.
The present value can be computed by substituting these values in the formula P =
.Thus,
P =
=
= P19,529.96.Solved Examples Solved Examples
A. Debbie wants to compare the simple interest to compound interest on a P 60,000 investment.
1. Find the simple interest if funds earn 8% simple interest for 1 year.
2. Find the interest if funds earn 8% compounded annually for 1 year 3. Find the interest if funds earn 8% compounded semi-annually for 1 year.
4. Find the interest if funds earn 8% compounded quarterly for 1 year.
D E P E D C O P Y
5. Which is the best investment? Why?
Given: P = 60,000 r = i(m)= 0.08 t = 1 year 1. Is = Prt = (60,000)(0.08)(1) = 4,800
2. F = P(1+r)t = (60,000)(1+0.08)1 = 64,800 Ic = F – P = 64,800 – 60,000 = 4.800
3. F =I
=
= 64,896c = F – P = 64,896 – 60,000 = 4.896
4. F =
=
= 64,945.93 Ic = F – P = 64,945.93 – 60,000 = 4.945.935. The investment that yields the highest interest is the one that earns 8%
compounded quarterly.
B. James aims to accumulate 1 million pesos in 12 years. Which investment will require the smallest present value?
6. 8% simple interest 7. 8% compounded annually 8. 8% compounded semi-annually 9. 8% compounded quarterly 10. 8% compounded monthly
Given: F = 1,000,000 t = 12 years r=i(m) = 0.08 6. P =
P =
= 510,204.08 7. P =
P =
= 397,113.76 8. P
P
= 390,121.479. P
P
= 386,537.61 10. P
P
= 384,114.67 (smallest present value)D E P E D C O P Y
Lesson 26 Supplementary Exercises Lesson 26 Supplementary Exercises
A. Complete the table by computing the interest rate per period and total number of conversion periods.
Nominal Rate
B. Complete the table by computing for the maturity values, compound interests and present values.
C. Solve the following problems on compound interest.
13. Cian lends P45,000 for 3 years at 5% compounded semi-annually. Find the future value and interest of this amount.
14. Tenten deposited P10,000 in bank which gives 1% compounded quarterly and let it stay there for 5 years. Find the maturity value and interest.
15. How much should you set aside and invest in a fund earning 9%
compounded quarterly if you want to accumulate P200,000 in 3 years and 3 months?
16. How much should you deposit in a bank paying 2% compounded quarterly to accumulate an amount of P80,000 in 5 years and 9 months?
17. Miko has P250,000 to invest at 6% compounded monthly. Find the maturity value if he invests for (a) 2 years? (b) 12 years? (c) How much is the additional interest earned due to the longer time.
18. Yohan sold his car and invested P300,000 at 8.5% compounded quarterly.
Find the maturity value if he invests for (a) 3 years? (b) 6 years? (c) How much is the additional interest earned due to the longer time.
19. Maryam is planning to invest P150,000. Bank A is offering 7.5% compounded semi-annually while Bank B is offering 7% compounded monthly. If she plans to invest this amount for 5 years, in which bank should she invest?
20. Yani has a choice to make short term investments for her excess cash P60,000. She can invest at 6% compounded quarterly for 6 months or (b) 5%
compounded semi-annually fo 1 year. Which is larger?