Cash Flow - Inflow & Outflow of Money
Examples of Cash
Inflows - Revenues, cost reductions, salvage receipts, Receipts of loan
principal.
Examples of Cash Outflows - Purchase of assets, operating costs, maintenance
costs,Interest
payments, Repayment of
loan principle.
Cash Flow is not an obligation or accrual
of unpaid receipts.
Point of View
Cash flows are classified dependent upon the point of view of which party is paying and which is receiving payment. Let's look at several examples.
$2000
loan : You =
+$2000 Inflow
Bank =
-$2000 Outflow
$50 Telephone bill: You =
-$50 Outflow
Verizon =
+$50 Inflow
Both happening in
You =
$1950 Net Inflow
one
period
Bank =
-$2000
Outflow
Verizon =
$50
Inflow
Using our first
example from Lesson 1: P =
$1000 i =
10%(Simple) N =
5 yrs
However, let's pay I yearly.
We would have:
Year CashFlow Principal Interest
0
$1000
$1000
--
1
-$100
0 -$100
2 -$100
0
-$100
3
-$100
0 -$100
4
-$100
0 -$100
5
-$1100
-$1000 -$100
The latter two columns would be considered to be an auxiliary table, breaking down the cash flow into a more explanatory fashion. Cash flows can be broken into convenient groupings, but auxiliary totals must equal the total cash flow.
End of Year Convention
For practicality (sacrificing a bit of accuracy)
transactions are assumed to occur at the end of a period, no matter where they
occur during each year.
Note in the example above: Start at year 0 = Begin
year 1 = Start of time. Years 1 - 5 signify the end of year values.
Cash Flow Diagrams
Graphical
representation of Cash Flow
Cash flow diagrams help to visualize the exchange of funds. We will use them in many of our problems. Every Cash flow diagram contains the following components:
Time line -- with discrete periods
Cash flow vectors -- Up (+) = Inflow =
Benefit
or
-- Down (-) = Outflow =
Cost
Interest
rate
From our Simple Interest Example:
Just as with auxiliary tables, Cash flow
diagrams can be split into separate equivalent diagrams. Vectors are
additive.
Determining cash flows and drawing diagrams is part of every
engineering economics problem.
Interest (i) applies to total amount (P + sum of all I) during each period.
Consider the following Cash flow:
A
$1000 deposit for 5 years at 10% /
year would result
in:
Amount
accrued
Year Begin
Year
Interest End year
1 P =
1000 100 F1 =
1100
F_{1}
= P + Pi_{}
2 1100 110 F2
=
1210
F_{2} = P + Pi+ (P + Pi) i
3
1210
121 F3 =
1331 F_{3} = P + Pi+ (P + Pi) i + (P + Pi+ (P + Pi)
i) i
4 1331 133
1464
=
P + Pi+ Pi+Pi^{2} + Pi+Pi^{2} + Pi^{2} + Pi^{3 }
5
1464
146 1610
=
P +
3Pi + 3Pi ^{2} + Pi^{3}
= P(1+3i +3i^{2}
+ i^{3}) = P(1 + i)^{3}
Hence F = P(1 + i)^{N
}where (1 + i)^{N }is called Compound Amount
Factor.
Solving Problems
It is now easy to solve problems regarding the Equivalence of Present (P) and Future (F) values over time (N) with interest (i). The
following steps will help:
1. Identify cash flows.
2. Identify P,
F, i & N.
3. Determine the missing
value.
4. Solve for missing value using
equation.
Example : What annual interest rate must
you get if you need $7000 in 4 years and
have $5000
to
invest
now?
P = $5000, F =
$7000, N = 4 years, i = ?
7000 = 5000(1 +
i)^{4
}7000
/ 5000 = (1 + i)^{4 }(1.4)^{1/4 } = 1 +
i
1.0878
= 1 +
i
0.0878 = i =
8.78% / year
Functional Notation
Standardized notation has been established to avoid writing the equation each time, and to give a logical method by which to find the correct factor to use.
Read the
factor as saying "Find F given P at i% for N periods".
Example: What is F for $1000 deposit for 5 years @
10% /
year?
F = 1000(
F/P,10,5)
( F/P,10,5) = 1.611 ---
See table on Page
198
F
= 1000(1.611) = $1611
Example : Find i for P = $5000, F =
$7000, N = 4
years
Using
F = P(F/P, i,
N)
7000 = 5000(F/P, i,
4)
1.4 = (F/P, i, 4)
Search the Compound Interest Factor
tables (Pages 180 - 208) to find the Interest rate that matches the
factor.
i =
9% (F/P, 9, 4) = 1.412
i =
8% (F/P, 8, 4)
=1.360
Since an even i value can't be found to match (F/P,i,4)
= 1.4, you
must interpolate to find the solution. Use the following as
a guide for interpolation.
Note that the solution
here by interpolation does not match exactly to the one found previously. This is
due to linear interpolation of a Non-linear function. You may
use tabled values or the equation, which ever fits the situation and allows an
easier solution.
More Frequent Compounding
Compounding can take place at intervals more frequently than yearly, say
quarterly, daily, weekly, monthly etc.
One must make adjustments for the
frequency of compounding.
Example:
Using the same $1000 @ 10% now compounded
quarterly.
There are 4 periods in 1 year which yields i =
10% / yr divided by 4
qtrs/yr = 2.5% /
qtr
Year Quarter Beginning Period Interest End Period
1 1
1000
25
1025
1 2 1025
25.62
1050.62
1
3
1050.62
26.26 1076.88
1 4
1076.88
26.92 1103.80
which is>1100
with
yearly
compounding
Hence more frequent compounding increases
F.
What is the actual yearly interest
rate?
Effective annual i = i
_{eff}= (1103.83 - 1000)/1000 =
0.1038
i _{eff}= 10.38% = 10% compounded quarterly >
10%
yearly
One could find the effective interest rate without the aid of
P & F values, by using the following equation :
i _{eff }= (1 + r
/ m)^{m}
-1
where r = nominal rate / yr (Usually quoted
with a Compounding Frequency)
m= # of compounding periods for
nominal period
i = r/m =
Interest rate/Period
In future discussions, we will use i most exclusively
to set up problems. The key is to determine the Frequency of Compounding.
Assume i is a yearly rate with yearly compounding, unless it is stated otherwise.
For the above example: i =
10% per
year compounded
quarterly
Hence
r = 10% = 0.10
m= 4 (qtrs)
i _{eff }= (1 + 0.10/ 4)^{4}
-1
= (1.025)^{4}-1
=
1.1038-1
= 0.1038 = 10.38%
Remember:
Nominal is the annual rate with a Frequency of
Compounding given.
Effective is annual rate with compounding already considered .
Example: A $1000 deposit for
5 years at 10% / yr compounded quarterly yields what future
value?
Quarterly rate = 10% / 4 = 2.5%
# of Periods (qrts) in 5 yrs = (4 qrts/yr)(5 yrs) = 20
Periods
Find an F given P using 2.5%/Period for 20 Periods.
F = P(F/P,2.5,20)
Not (F/P,10,5)
or (F/P,2.5,5) or (F/P,10,20)
F = 1000(1.639) = $1639
Remember to use # of periods that corresponds to frequency of
compounding
i = r / m where i = Interest rate / period
r = nominal rate / year
m
= # of compounding periods/year
Added to the Web: May
20, 2002. Last updated: April 26, 2002. Web page design by Dan Solarek. |
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