04 - Solving Cashflow Problems

Sequence 4_ Solving Cash Flow
Problems_Engineering
bb Economics
Dr. Hassan Ashraf

Engineering Economics _ CU Islamabad _ Wah Campus _ Civil
Engineering Department _ 2018 Session 1
Problem # 01
The ABC Company deposited $100,000 in a bank account on June 15 and withdrew a total
of $115,000 exactly one year later. Compute: (a) the interest which the ABC Company
received from the $100,000 investment, and (b) the annual interest rate which the ABC
Company was paid.
So, the first part is very easy to handle:
The future worth of the Principal amount $100,000 is given as $115,000; we can compute
the interest amount using the formula as under:
F=P+I
115,000 = 100,000 + I
I = $ 15,000
The second part is also very easy to handle. We can solve by employing either of the two
assumptions:
2
Problem # 01
The first assumption is that the accumulated interest that has accumulated is of simple
nature. So, with this assumption we can use the formula:
F = P ( 1 + ni)
115,000 = 100,000 ( 1 + (1)i)
115,000 = 100,000 ( 1 + i )
115,000 = 100,000 + 100,000i
115,000 – 100,000 = 100,000i
15,000/100,000 = i
i = 0.15 or in other words i = 15%
Similarly, if we assume the interest that has accumulated through compounding, we can
use the following formula:
3
Problem # 01
F = P ( 1 + i )n
115,000 = 100,000 ( 1 + i ) 1
115,000/100,000 = 1 + i
1.15 – 1 = I
i = 0.15 or in other words it is 15%
What we have learned from this example ?
We have learned from this example that if the interest period is ‘1’, we can use either of
the two equations to compute the answer. The answer will not be different as you have
seen in this example.
4
Problem # 02
What is the annual rate of simple interest if $265 is earned in four months on an
investment of $15,000?
As the problem mentions about simple interest, and the unknown is interest rate, we can
easily use the formula:
F = P ( 1 + ni )
15,265 = 15,000 ( 1 + (4/12)i)
15265/15000 = 1+ 0.33i
1.018 = 1 + 0.33i
1.018 – 1 = 0.33i
0.018/0.33 = i
0.054 = i or I = 5.4%
5
Problem # 03
Determine the principal that would have to be invested to provide $200 of simple interest
income at the end of two years if the annual rate is 9%?
As it mentions about the simple interest income we can simply use the simple interest
formula as :
F = P ( 1 + ni)
F=P+I
I = 200
So, F = P + 200
P + 200 = P (1 + 2 (0.09))
P + 200 = 1.18 P
200 = 0.18P
6
Problem # 04
Compare the interest earned from an investment of $1000 for 15 years at 10% per annum
simple interest, with the amount of interest that could be earned if these funds were
invested for 15 years at 10% per year, compounded annually?
This a question of comparing the interest earned when it compounded VS when it earned
as a simple interest.
F = P ( 1 + ni)
F = 1000 ( 1 + 15 ( 0.1))
F = 1000 ( 2.5)
F = 2500
F=P+I ; I = F – P ; I = 2500 – 1000 ; I = 1500
7
Problem # 04
Similarly, if the interest is earned through compounding, we can use the formula
F=P(1+i)n
F = 1000 ( 1 + 0.1 ) 15
F = 1000 ( 4.18)
F = 4180
I=F–P
I = 4180 – 1000
I = 3180
So, its 3180 VS 1500

8
Problem # 05
At what annual interest rate is $500 a year ago equivalent to $600 today?
So i = ?
It is not mentioned whether interest is earned through simple interest or compound

interest, we assume the interest is earned through compounding:
F=P(1+i)n
600 = 500 ( 1 + i ) 1
600/500 = 1 + i
1.2 = 1 + i
i = 0.2 or 20%
9
Problem # 06
Suppose that the interest rate is 10% per year, compounded annually. What is the
minimum amount of money that would have to be invested for a two-year period in order
to earn $300 in interest?
F = 300 + P
F=P(1+i)n
300 + P = P ( 1 + 0.1) 2
300 + P = 1.21 P
300 = 0.21 P
P = 300/0.21
P = 142.57
10
Problem # 07
How long would it take for an investor to double his money at 10% interest per year,
compounded annually?
F = 2P
N=?
F = P ( 1 + i )n
2P = P ( 1 + i )n
2 = (1.1)n
Taking Log on both sides of the equation
n = log 2 / log 1.1 ; n = 0.301 / 0.041
n = 7.34 years
11
bb
Thank You
12

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Sequence 4_ Solving Cash Flow

Dr. Hassan Ashraf

So, the first part is very easy to handle:

115,000 = 100,000 ( 1 + (1)i)

115,000 = 100,000 + 100,000i

115,000 – 100,000 = 100,000i

i = 0.15 or in other words i = 15%

i = 0.15 or in other words it is 15%

What we have learned from this example ?

15,265 = 15,000 ( 1 + (4/12)i)

F=P+I ; I = F – P ; I = 2500 – 1000 ; I = 1500

So, its 3180 VS 1500

It is not mentioned whether interest is earned through simple interest or compound

Taking Log on both sides of the equation

n = log 2 / log 1.1 ; n = 0.301 / 0.041

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