Management Money Value Lec 1

1. Money Value
Simple interest
Total interest after n years = principal * (interest rate) * n
Total due after n years = principal * [ 1+ n*(interest rate)]
Example
Given a loan of 1000 L.E. for 3 years at 5% (simple interest).
Get the total interest and the total amount due after 3 years.
Solution
Total interest for 3 years= 1000 *5/100*3= 150 L.E.
Total amount due after 3 years = 1000 [ 1 + 3*(5/100)] = 1150 L.E.
Compound Interest
Total amount due after n years= principal (1+interest rate)n
Total interest after n years= principal [(1+interest rate)n
- 1]
Example
Same as above with the consideration of compound interest calculations
Solution
Total amount due after 3 years = 1000 * (1+ .05)3
= 1157.63 L.E.
Total Interest after 3 years=1157.63 -1000 = 157.63

2. Terminology and symbols
- The equations and procedures of engineering economy utilize the following terms and
symbols
P = Value or amount of money at a time designated as the present or time zero.
F = Value or amount of Money at some future time.
A = Series of consecutive, equal ,end_of_period amounts of money.
n = number of interest periods.
i = interest rate
t = time
- Every person or company has cash receipts – revenues and income (inflow) and cash
disbursements – expenses and costs ( outflow).
- A graphical representation of cash flow is called the cash flow diagram
Example from the shown
cash flow diagram the
following notes can be
made:-
- Cash inflow of 600
L.E. at the end of
year 1
- Cash outflow of 300
L.E. at the end of
year 2
- Cash out flow of 400
L.E. at the end of
year 3

3. The mathematical equations and flow diagrams relating P, F, & A
Equation Cash flow diagram
F/P = (1+i)n
A/P=
( )
( )
/
=
(1 + ) − 1

4. Example
A new software is purchased for 5000 L.E. now and annual payments of 500 L.E. for 6 years
starting 3 years from now. What is the present worth of the payments if interest rate is 8% per
year?
notes Cash flow diagram
- Out cash flow
for the given
problem
- We lump the
annual
payments by an
equivalent
amount P’
at the
end of year 2
- I=8%, n=6,
A=500
- P’
=
( )
( )
=
500* 4.6229
- We get an
equivalent of p’
which is P”
attime 0.
- I=8%, n=2,
P’
=500*4.6229
- P”
=
( )
=1981.6
- Software
present worth
value= 5000+
1981.6 = 6981.6
L.E.

5. Effective annual interest rate
Sometimes compounding does not occur every year (i.e. it occurs semiannually, quarterly,
monthly, weekly, and daily). The nominal interest rate/year does not consider compounding and
there is a need for computing the effective interest rate that considers compounding.
Let
r = nominal interest rate/year
m = number of compounding periods/year
i = effective interest rate/ compounding period =r/m
ia = effective interest/year
Then
ia=(1+i) -1
Example
If nominal interest rate/year is 10% and compounding occurs quarterly
Then
r=10%
m = 4
i = r/m=2.5%
ia=(1+i) -1=.104=10.4%
Compounding period can be as follows (e.g. for 1 year):
Compounding occurs Compounding period m
Semiannually 6 month 2
quarterly 3 month 4
monthly 1 month 12
weekly 1 week 52
daily 1 day 365
Example
Find the future value for P=1000 L.E. after 2 years. The nominal interest/year is 5%.
Compounding occurs monthly.
Solution
m=12
Effective interest by compounding period =
%
In a 2 year's time there are 24 periods=m'
F=p(1+i) =1000(1+
%
) = 1104.941