Compound interest

1. Compound Interest
Engineering Economics

2. Cash Flow Diagram
0 4
3
2
1 5 6 7
time
Year 1
End of year 1
Beginning
of year 1
Cash Inflow
Cash Outflow
Receipt (Positive Cash Flow or Cash Inflow)
Disbursement (Negative Cash Flow or Cash Outflow)

3. Sample Cash Flow
• A loan of P100.00 at simple interest of 10% will
become P170.00 after 7 years.
0 4
3
2
1 5 6 7
years
P170.00
P100.00 Cash flow diagram on the viewpoint of the lender.

4. Sample Cash Flow
• A loan of P100.00 at simple interest of 10% will
become P170.00 after 7 years.
0 4
3
2
1 5 6 7
years
P100.00
P170.00
Cash flow diagram on the viewpoint of the borrower.

5. Compound Interest
• The interest earned by the principal which is added to the
principal will also earn an interest for the succeeding
periods.
𝐹 = 𝑃 1 +
𝑟
𝑚
𝑚𝑡
𝑃 = 𝐹 1 +
𝑟
𝑚
−𝑚𝑡

6. Compound Interest
where:
P = present worth
F = compound amount the end of “n” periods
m = number of periods per year
(annually m = 1; semi-annually m = 2; quarterly m = 4;
monthly m = 12, bi-monthly m = 6)
t = number of years

7. Rate of Interest
• The nominal interest specifies the rate of interest and a
number of interest period in one year.
𝑖 =
𝑟
𝑚
where:
i = rate of interest per interest period
r = nominal interest rate
m = number of compounding periods per year

8. Example 1
• If the nominal interest is 10% compounded quarterly, then
the rate of interest per interest period is
𝑖 =
𝑟
𝑚
𝑖 =
0.10
4
𝒊 = 𝟎. 𝟎𝟐𝟓 𝒐𝒓 𝟐. 𝟓𝟎%

9. Rate of Interest
• The effective rate interest is the actual or exact rate of
interest on the principal during one year.
𝑖𝑒 = 1 +
𝑟
𝑚
𝑚
− 1

10. Example 2
• If the nominal interest is 10% compounded quarterly, then
the effective interest will be
𝑖𝑒 = 1 +
𝑟
𝑚
𝑚
− 1
𝑖𝑒 = 1 +
0.10
4
4
− 1
𝒊 = 𝟎. 𝟏𝟎𝟑𝟖 𝒐𝒓 𝟏𝟎. 𝟑𝟖%

11. Example 3
• Find the nominal rate which is converted quarterly could be used instead
of 12% compounded monthly. What is the corresponding effective rate?
1 +
𝑟4
4
4
= 1 +
0.12
12
12
𝒓 = 𝟏𝟐. 𝟏𝟐%
1 +
𝑟1
1
1
= 1 +
0.12
12
12
𝒊𝒆 = 𝟎. 𝟏𝟐𝟑𝟖 ≈ 𝟏𝟐. 𝟑𝟖%

12. Example 4
• Find the amount at the end of two years and seven months if P1,000.00
is invested at 8% computed quarterly using simple interest for anytime
less than a year interest period.
𝑡 = 2 𝑦𝑒𝑎𝑟𝑠 & 7 𝑚𝑜𝑛𝑡ℎ𝑠; 𝑃 = 𝑃1,000.00; 𝑟 = 8%; 𝑚 = 4; 𝐹 =?
𝐹 = 𝑃 1 +
𝑟
𝑚
𝑚𝑡
= 1,000 1 +
0.08
4
4(2)
= 𝑃1,171.66
𝐹 = 1,171.66 + 1,171.66 0.08
7
12
= 𝑷𝟏, 𝟐𝟐𝟔. 𝟑𝟒

13. Example 5
• Find the present worth of a future payment of P300,000.00 to be made
in 5 years with an interest rate of 8% per annum.
𝐹 = 𝑃300,000.00; 𝑡 = 5 𝑦𝑒𝑎𝑟𝑠; 𝑟 = 8%; 𝑚 = 1; 𝑃 =?
𝑃 = 𝐹 1 +
𝑟
𝑚
−𝑚𝑡
𝑃 = 300,000 1 +
0.08
1
−1(5)
𝑷 = 𝑷𝟐𝟎𝟒, 𝟏𝟕𝟒. 𝟗𝟔

14. Example 6
• Find the present worth of a future payment of P100,000.00 to be made
in 10 years with an interest of 12% compounded quarterly.
𝑃 =? ; 𝐹 = 𝑃100,000.00; 𝑡 = 10 𝑦𝑒𝑎𝑟𝑠; 𝑟 = 12%; 𝑚 = 4
𝑃 = 𝐹 1 +
𝑟
𝑚
−𝑚𝑡
𝑃 = 100,000 1 +
0.12
4
−4(10)
𝑷 = 𝑷𝟑𝟎, 𝟔𝟓𝟓. 𝟔𝟖

15. Example 7
• In how many years is required for P2,000.00 to increase by P5,000.00 if
interest rate at 12% compounded semi-annually?
𝑡 =? ; 𝑃 = 𝑃2,000.00; 𝐹 = 𝑃5,000.00; 𝑟 = 12%; 𝑚 = 2
𝐹 = 𝑃 1 +
𝑟
𝑚
𝑚𝑡
5,000 = 2,000 1 +
0.12
2
2(𝑡)
𝒕 = 𝟕. 𝟖𝟔 𝒚𝒆𝒂𝒓𝒔

16. Example 8
• John borrowed P50,000.00 from the bank at 25% compounded semi-
annually. What is the equivalent effective rate of interest?
𝑃 = 𝑃50,000.00; 𝑟 = 25%; 𝑚 = 2; 𝑖𝑒 =?
𝑖𝑒 = 1 +
𝑟
𝑚
𝑚
− 1
𝑖𝑒 = 1 +
0.25
2
2
− 1
𝒊𝒆 = 𝟎. 𝟐𝟔𝟓𝟔 ≈ 𝟐𝟔. 𝟓𝟔%

17. Example 9
• A sum of P1,000.00 is invested now and left for four years, at which
time the principal is withdrawn. The interest has accrued is left for
another three years. If the effective annual interest rate is 5%, what will
be the withdrawal amount at the end of the 7th year?
0 4
3
2
1 5 6 7
year
F=P249.48
P1,000.00
P1,000.00
Investor’s viewpoint
5% 5%
P215.51
𝑭 = 𝑷 𝟏 +
𝒓
𝒎
𝒎𝒕
𝑭 = 𝟏, 𝟎𝟎𝟎 𝟏 +
𝟎. 𝟎𝟓
𝟏
𝟏(𝟒)
𝑭 = 𝟐𝟏𝟓. 𝟓𝟏 𝟏 +
𝟎. 𝟎𝟓
𝟏
𝟏(𝟑)
𝑭 = 𝑷𝟏, 𝟐𝟏𝟓. 𝟓𝟏
𝑭 = 𝑷 𝟏 +
𝒓
𝒎
𝒎𝒕