CONTINUOUS CASH FLOWS AND CONTINUOUS COMPOUNDING
Problems
1. Suppose that a savings account pays 6% annual interest, compounded continuously. How
much must be invested now to have P100,000 in the account 30 years from now?
Given:
F= P100,000
n=30 years
r= 6%
Required:
P
Diagram:
Solution: P16,529.89 must be invested
−𝑟𝑛
𝑃 = 𝐹𝑒
𝑃 = (100,000)𝑒 −(0.06)(30)
𝑷 = 𝑷𝟏𝟔, 𝟓𝟐𝟗. 𝟖𝟗
2. P250.00 is left in a savings account at 4.0% and the interest is compounded continuously.
If the balance is now P330.78, then how many years was the money been in the account?
Given:
F= P330.78
P=P250.00
r=4%
Required:
n
Diagram:
Solution:
𝐹 = 𝑃𝑒 𝑟𝑛 n is approximately 7 years.
330.78 = 250𝑒 (0.04)(𝑛)
330.78 = 250𝑒 (0.04)(𝑛)
250
1.32312 = 𝑒 (0.04)(𝑛)
𝑙𝑛(1.32312) = 𝑙𝑛( 𝑒 (0.04)(𝑛) )
0.279992584 = (0.04)𝑛
0.279992584 = (0.04)𝑛
(0.04)
𝟔. 𝟗𝟗𝟗𝟖𝟏𝟒𝟓𝟗𝟗 = 𝒏
3. Determine the interest rate required with continuous compound interest for an investment
to triple in 7.75 years.
Given:
n=7.75 years
Required:
r
Diagram:
Solution:
𝐹 = 𝑃𝑒 𝑟𝑛
The interest rate is 14.18%
(𝑟)(7.75)
3= 𝑒
𝑙𝑛(3) = 𝑙𝑛(𝑒 (𝑟)(7.75) )
1.098612289 = 7.75𝑟
1.098612289=7.75𝑟
7.75
0.1417564243 = 𝑟
𝟏𝟒. 𝟏𝟕𝟓𝟔𝟒𝟐𝟒𝟑% = 𝒓
4. Determine the time necessary for the principal in an account to double if it is invested at
an interest rate of 5.5% compounded continuously.
Given:
r=5.5%
Required:
n
Diagram:
Solution:
𝐹 = 𝑃𝑒 𝑟𝑛 It will take 12.60 years.
2 = 𝑒 (0.055)𝑛
𝑙𝑛(2) = 𝑙𝑛(𝑒 0.055𝑛) )
0.6931471806 = 0.055𝑛
0.6931471806=0.055𝑛
0.055
𝟏𝟐. 𝟔𝟎𝟐𝟔𝟕𝟔𝟎𝟏 = 𝒏
5. Your older sister is about to make you an aunt/uncle. As a gift you deposit P100.00 into
an account that compounds interest quarterly. In 50 years, the account has a balance of
P347.68. What is the interest rate? For educational purposes, would you expect a bigger
balance if as a gift you decided deposit P100.00 into an account that compounds interest
continuously at the same rate? What is the account balance after 50 years? Were you
correct?
Given:
F= P347.68
P= P100.00
n= 50 years
Required:
r
Diagram:
Solution:
𝑟
𝐹 = 𝑃(1 + 𝑀)𝑀 𝒓 ≈ 𝟎. 𝟎𝟐𝟓
𝑟
347.68 = 100(1 + 4)4(50) 𝐹 = 𝑃𝑒 𝑟𝑛
𝑟
347.68=100(1+ )4(50)
4
𝐹 = 100𝑒(0.025)(50)
100
𝑟
3.4768 = (1 + 4)200 𝑭 = 𝟑𝟒𝟗. 𝟎𝟑𝟒𝟐𝟗𝟓𝟕
1 1
𝑟
(3.4768)200 = [(1 + 4)200 ]200
The interest rate is 2.25%, and the
𝑟 new balance is P349.03
1.006250012 = (1 + 4)
𝑟 = 0.025000048
6. P1200 is placed in an account at 4% compounded continuously for 2 years. It is then
withdrawn and placed in another bank at the rate of 5% compounded continuously for 4
years. What is the balance in the second account after the 4 years?
Given:
P= P1200.00
r1= 4%
r2=5%
n1= 2 years
n2=4 years
Required:
F
Diagram:
Solution:
First two years:
The balance in the second account is
𝑟𝑛
𝐹 = 𝑃𝑒 P1587.76
𝐹 = 1200𝑒 (0.04)(2)
𝐹 = 1299.944481
Last four years:
𝐹 = 1299.944481𝑒 (0.05)(4)
𝑭 = 𝟏𝟓𝟖𝟕. 𝟕𝟓𝟓𝟕𝟕𝟓
7. A principal of P120 is deposited in a 7% account and compounded continuously. At the
same time a principal of P150 is deposited in a 5% account and compounded annually.
How long does it take for the amounts in the two accounts to be equal?
Given:
P1=P120.00
r1=7%
P2=P150.00
r2=5%
Required:
n
Diagram:
Solution:
Continuous Compounding Interest:
[𝑙𝑛(150)−𝑙𝑛(120)
𝐹1 = 𝑃𝑒 𝑟𝑛 𝑛 = [0.07−𝑙𝑛(1+0.05)]
𝐹1 = 120𝑒 (0.07)(𝑛) 𝐧 = 𝟏𝟎. 𝟓𝟐𝟎𝟕𝟓𝟖𝟎𝟔
Annual Compounding Interest:
It takes approximately 10.52 years
𝑟 𝑀
𝐹2 = 𝑃(1 + )
𝑀
𝐹2 = 150(1 + 0.05)𝑛
𝐹1 = 𝐹2
120𝑒 (0.07)(𝑛) = 150(1 + 0.05)𝑛
𝑙𝑛(120𝑒 (0.07)(𝑛) ) = 𝑙𝑛(150(1 + 0.05)𝑛 )
𝑙𝑛(120) + 𝑙𝑛(𝑒 (0.07)(𝑛) ) = 𝑙𝑛(150) + 𝑙𝑛((1 + 0.05)𝑛 )
𝑙𝑛(120) + (0.07)𝑛 = 𝑙𝑛(150) + 𝑛𝑙𝑛(1 + 0.05)
𝑛(0.07 − 𝑙𝑛(1 + 0.05)) = 𝑙𝑛(150) − 𝑙𝑛(120)
8. A person deposited P1,000 in a 2% account compounded continuously. In a second
account, he deposited P500 in an 8% account compounded continuously. When will the
total amounts in both accounts be equal? When will the total amount in the second
accounts be 50% more than the total amount in the second account?
Given:
P1=P1000.00
r1=2%
P2=P500.00
r2=8%
Required:
n
Diagram:
Solution:
𝐹1 = 𝑃𝑒 𝑛𝑡 500𝑒 (0.08)(𝑛) = 1.51000𝑒 (0.02)𝑛
𝐹1 = 1000𝑒 (0.02)𝑛 500𝑒(0.08)(𝑛) =(1.5)𝑥1000𝑒(0.02)𝑛
𝐹2 = 𝑃𝑒 𝑛𝑡 500𝑒(0.02)𝑛
𝑒 (0.08𝑛)−(0.02𝑛) = 3
𝐹2 = 500𝑒 (0.08)(𝑛)
𝑙𝑛(𝑒 (0.08𝑛)−(0.02𝑛) ) = 𝑙𝑛(3)
1000𝑒 (0.02)𝑛 = 500𝑒 (0.08)(𝑛)
0.06𝑛 = 𝑙𝑛(3)
1000𝑒 (0.02)𝑛 =500𝑒 (0.08)(𝑛)
500𝑒 (0.02)𝑛 𝒏 = 𝟏𝟖. 𝟑𝟏𝟎𝟐𝟎𝟒𝟖𝟏
2 = 𝑒 (0.06)𝑛
The amount will be the same in
𝑙𝑛(2) = 𝑙𝑛(𝑒 (0.06)𝑛 ) approximately 11.55 years, and the
amount in the second account will be
𝑙𝑛(2) = (0.06)𝑛 50% more in approximately 18.31
𝟏𝟏. 𝟓𝟓𝟐𝟒𝟓𝟑𝟎𝟏 = 𝒏 years.
𝐹2 = 1.5𝐹1
9. A continuously compounded savings account had an initial deposit of P10,000.00 and 10
years later has a balance of P13,125.87. At what interest rate was the savings account?
Given:
P=P10,000.00
n=10 years
F=P13,125.87
Required:
r
Diagram:
Solution:
𝐹 = 𝑃𝑒 𝑟𝑛
13,125.87 = 10,000𝑒 (10)𝑟 The interest rate was approximately
3,125.87=10,000𝑒 (10)𝑟
2.72%
10,000
1.312587 = 𝑒 10𝑟
𝑙𝑛(1.312587) = 𝑙𝑛(𝑒10𝑟 )
𝑙𝑛(1.312587) = 10𝑟
0.02719999 = 𝑟
𝟐. 𝟕𝟐% = 𝒓
10. Lucas wants to make a loan, if he plans to repay a loan of P20,000 in 8 years if the
interest rate is 8% compounded continuously, what is the required quarterly payment?
Given:
P= P20,000
r=8%
n=8 years
Required:
A
Diagram:
Solution:
𝑒 𝑟 −1 The required quarterly payment is
𝐴 = 𝑃 [1−𝑒 −𝑟𝑛 ]
approximately P854.71
𝑒 0.02 −1
𝐴 = 20,000[1−𝑒 −(0.02)(32)]
𝑨 = 𝟖𝟓𝟒. 𝟕𝟎𝟕𝟔𝟗𝟎𝟓
