Chapter 2
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Conestoga College *
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MGMT74115
Subject
Industrial Engineering
Date
Dec 6, 2023
Type
Pages
38
Uploaded by ProfDragonMaster4
Chapter
Factors:
How
Time
and
Interest
Affect
Money
Royalty-Fraa/COREIS
n
Chapter
|
we
leamed
the
basic
concepts
of
engineering
economy
and
their
role
in
decision
making.
The
cash
flow
is
fundamental
to
every
economic
study.
Cash
flows
occur
in
many
configurations
and
amounts—isolated
single
values,
series
that
are
uniform,
and
series
that
increase
or
decrease
by
constant
amounts
or
constant
percentages.
This
chapter
develops
the
commonly
used
engineering
economy
factors
that
consider
the
time
value
of
money
for
cash
flows.
The
application
of
factors
is
illustrated
using
their
mathematical
forms
and
a
standard
notation
format.
Spreadsheet
and
calculator
functions
are
illustrated.
Purpose:
Use
tabulated
factors
or
spreadsheet/calculator
functions
to
account
for
the
time
value
of
money.
LEARNING
QUTCOMES
.
Use
the
compound
amount
factor
and
present
worth
factor
for
F/P
and
P/F
factors
single
payments.
'
.
Use
the
uniform
series
factors.
|
P/A,
A/P,
F/A
and
A/F
factors
3.
Use
the
arithmetic
gradient
factors
and
the
geometric
gradient
Gradients
formula.
-
.
Use
uniform
series
and
gradient
factors
when
cash
flows
are
Shifted
cash
flows
shifted.
.
Use
a
spreadsheet
or
calculator
to
make
equivalency
Spreadsheets/Calculators
calculations.
2.1
Single-Payment
Formulas
(F/P
and
P/F)
2.1
SINGLE-PAYMENT
FORMULAS
(F/P
AND
P/F)
The
most
fundamental
equation
in
engineering
economy
is
the
one
that
determines
the
amount
of
money
F
accumulated
after
n
years
{or
periods)
from
a
single
pres
ent
worth
P,
with
interest
compounded
one
time
per
year
(or
period).
Recall
that
compound
interest
refers
to
interest
paid
on
top
of
interest.
Therefore,
if
an
amount
P
is
invested
at
time
¢
=
(0,
the
amount
F;
accumulated
|
year
hence
at
an
interest
rate
of
i
percent
per
year
will
be
Fi
=P+
Pi
=
P(1
+
i)
where
the
interest
rate
is
expressed
in
decimal
form.
At
the
end
of
the
second
year,
the
amount
accumulated
F,
is
the
amount
after
year
|
plus
the
interest
from
the
end
of
year
|
to
the
end
of
year
2
on
the
entire
F,.
F,=F,
+
Fji
=P+
+
P+
The
amount
£,
can
be
expressed
as
Fo=P(l
+i+i+i)
=P(1
+2i+
%)
=Pl
+i)*
Similarly,
the
amount
of
money
accumulated
at
the
end
of
year
3
will
be
Fy=P(l
+i)
By
mathematical
induction,
the
future
worth
F
can
be
calculated
for
n
years
using
F=PFP1+i"
[2.1]
The
term
(1
+
i)
is
called
a
factor
and
is
known
as
the
single-payment
compound
amount
factor
(SPCAF),
but
it
is
wsually
referred
to
as
the
F/P
factor
This
is
the
conversion
factor
that
yields
the
future
amount
F
of
an
initial
amount
P
after
n
years
at
interest
rate
i.
The
cash
flow
diagram
is
seen
in
Figure
2.la.
Reverse
the
situation
to
determine
the
P
value
for
a
stated
amount
£
Simply
solve
Equation
[2.1]
for
P.
.
1
"
'LL',I
+
:"]"‘
The
expression
in
brackets
is
known
as
the
single-payment
present
worth
factor
(SPPWF),
or
the
P/F
factor.
This
expression
determines
the
present
worth
P
of
a
given
future
amount
F
after
n
years
at
interest
rate
i.
The
cash
flow
diagram
is
shown
in
Figure
2.15.
Note
that
the
two
factors
derived
here
are
for
single
payments;
that
is,
they
are
used
to
find
the
present
or
future
amount
when
only
one
payment
or
receipt
is
involved.
P
=
given
Chapter
2
Factors:
How
Time
and
Interest
Affect
Money
—
Fod
==
F
-
i
=
given
I
i=given
“
}
I
!
1
l
i
:
]
l
|
1
n-2
n-1
‘u
F=17
F
=
given
{a)
(5)
FIGURE
2.1
Cash
flow
diagrams
for
single-payment
factors:
(a)
find
F
and
(b)
find
P.
A
standard
notation
has
been
adopted
for
all
factors.
Tt
is
always
in
the
general
form
(X/Y¥.in).
The
letter
X
represents
what
is
sought,
while
the
letter
¥
represents
what
is
given.
For
example,
F/P
means
find
F
when
given
P
The
i
is
the
interest
rate
in
percent,
and
n
represents
the
number
of
periods
involved.
Thus,
(F/P.6%.20)
represents
the
factor
that
is
used
to
calculate
the
future
amount
F
accumulated
in
20
penods
if
the
interest
rate
is
6%
per
period.
The
P
is
given.
The
standard
nota
tion,
simpler
to
use
than
formulas
and
factor
names,
will
be
used
hereafter.
Table
2.1
summarizes
the
standard
notation
and
equations
for
the
F/P
and
P/F
factors.
To
simplify
routine
engineering
economy
calculations,
tables
of
factor
values
have
been
prepared
for
a
wide
range
of
interest
rates
and
time
periods
from
1
to
large
n
values,
depending
on
the
i
value.
These
tables
are
found
at
the
end
of
this
book.
For
a
given
factor,
interest
rate,
and
time,
the
correct
factor
value
1s
found
at
the
intersection
of
the
factor
name
and
n.
For
example,
the
value
of
the
factor
(P/F.5%.10)
is
found
in
the
P/F
column
of
Table
10
at
period
10
as
0.6139.
When
it
is
necessary
to
locate
a
factor
value
for
an
§
or
n
that
is
not
in
the
interest
tables,
the
desired
value
can
be
obtained
in
one
of
several
ways:
(1)
by
using
the
formulas
derived
in
Sections
2.1
to
2.3,
(2)
by
linearly
interpolating
between
the
tabulated
values.
or
(3)
by
using
a
spreadsheet
or
calculator
function
as
discussed
in
Section
2.5.
For
many
cash
flow
series
or
for the
sake
of
speed,
a
spreadsheet
function
may
be
used
in
lieu
of
the
tabulated
factors
or
their
equations.
For
single-payment
series,
no
annual
series
A
is
present
and
three
of
the
four
values
of
F,
P,
i,
and
n
are
known.
When
solving
for
a
future
worth
by
spreadsheet,
the
F
value
is
calculated
TABLE
2.1
F/P
and
P/F
Factors:
Notation,
Equation
and
Function
Factor
Standard
Notation
Equation
with
Spreadsheet
Calculator
Notation
Name
Equation
Factor
Formula
Function
Function
(F/P.in)
Single-payment
F
=
P(F{P.in)
F=mMl+
F¥V(i%.n,
Py
FV{inAP)
compound
amount
(P/F.i.n)
Single-payment
P
=
Fi(P{F.in)
P
=
F1/1
+
07
PVii%.n
.
F)
PV(inAF)
|JIL.'-_‘iL.',Il1
worth
2.1
Single-Payment
Formiulas
(F/P
and
P/F)
by
the
FV
function,
and
the
present
worth
P
is
determined
using
the
PV
function.
The
formats
are
included
in
Table
2.1.
(Refer
to
Section
2.5
and
Appendix
A
for
more
information
on
the
FV
and
PV
spreadsheet
functions.)
The
calculator
formats
for
FV
and
PV
functions
detailed
in
Table
2.1
include
the
annual
series
A,
which
15
entered
as
()
when
only
single-amounts
are
involved.
In
standard
notation
form,
the
relation
used
by
calculators
to
solve
for
any one
of
the
parameters
P,
F,
A,
i,
or
n
is
A(P/Ai%.n)
+
F(P/Fi%n)
+
P
=10
Using
the
factor
eguations,
this
relation
expresses
the
present
worth
of
uniform
series,
future
values
and
present
worth
values
as:
(1=
(1
+i/100)"
_
o
-’t(
e
)+
F1+
(/100"
+P=10
[2.3]
i/
100
F
'
When
only
single
payments
are
present,
the
first
term
is
(0.
37
An
engineer
received
a
bonus
of
512,000
that
he will
invest
now.
He
wants
to
calculate
the
equivalent
value
after
24
years,
when
he
plans
to
use
all
the
result
ing
money
as
the
down
payment
on
an
island
vacation
home.
Assume
a
rate
of
retum
of
8%
per
year
for
each
of
the
24
years.
Find
the
amount
he
can
pay
down,
using
the
tabulated
factor,
the
factor
formula,
a
spreadsheet
function,
and
a
calculator
function.
Solution
The
symbols
and
their
values
are
P
=
512,000
F=1
i
=
8%
per
year
n
=
24
years
The
cash
flow
diagram
is
the
same
as
that
in
Figure
2.1a.
Tabulated:
Determine
F,
using
the
F/P
factor
for
8%
and
24 years.
Table
13
provides
the
factor
value.
F
=
P(F/P.in)
=
12,000(F
/P
8%
24)
=
12,00006.3412)
=
$76,004.40
Formula:
Apply
Equation
[2.1]
to
calculate
the
future
worth
F.
F
=
P(l
+
0"
=
12,000(1
+
0.08)*
=
120000634
1181)
=
$76.094.17
Spreadsheet:
Use
the
function
=
FV(i%,n,A,P).
The
cell
entry
is
=
FV(8%,
24..12000).
The
F
wvalue
displayed
is
($76.094.17)
in
red
or
—376,094.17
in
black
to
indicate
a
cash
outflow.
EXAMPLE
2.1
Chapter
2
Factors:
How
Time
and
Interest
Affect
Money
Caleulator:
Use
the
TVM
function
FV{in.A.F).
The
numerical
values
are
FVIR.24,0,120000),
which
displays
the
future
worth
value
$—76,094.17,
which
indicates
it
is
a
cash
outflow.
The
slight
difference
in
answers
between
tabulated,
formula,
and
calculator
solutions
15
due
to
round-off
error
and
how
different
methods
perform
equiva
lence
calculations.
An
equivalence
imterpretation
of
this
result
is
that
$12,000
today
15
worth
376,094
after
24
years
of
growth
at
8%
per
year
compounded
annually.
-
EXAMPLE
2.2
Hewlett-Packard
has
completed
a
study
indicating
that
$50,000
in
reduced
main
tenance
this
year
(i.e.,
year
zero)
on
one
of
its
processing
lines
resulted
from
improved
wireless
monitoring
technology.
a.
If
Hewlett-Packard
considers
these
types
of
savings
worth
209
per
year,
find
the
equivalent
value
of
this
result
after
5
years.
b.
If
the
$50,000
maintenance
savings
occurs
now,
find
its
eguivalent
value
3
years
earlier
with
interest
at
20%
per
year.
Solution
a.
The
cash
flow
diagram
appears
as
in
Figure
2.1a.
The
symbols
and
their
vilues
are
P
=
350,000
F=1
i
=
20%
per
year
n
=
5
years
Use
the
F/P
factor
to
determine
F
after
5
years.
F
=
P(F/Pin)
=
$50,000(F/P.20%.5)
=
50,000(2.4883)
—
§124,415.00
The
function
=
FV(209%.5,,50000)
also
provides
the
answer.
See
Figure
2.2.
.
VN
Y
"N
NN
-
SN
NN
-3
-
S
Y
N
N
N
_—
1
|
I
I
——
I
3
Example
2.2a
Example
2.2b
4
5
F=
-$124,416
P
=
-$28,935
6
[
)
i3]
.
[
|
|
|
|
]
=5
Spreadsheet
function
with
}
|
Spreadsheet
function
with
|
|
A
omitted:
|
A
omitted:
1':'
=
FV(20%5,.50000)
=
PV(20%
3,
50000)
12
13
FIGURE
2.2
Use
of
single-cell
spreadsheet
functions
to
find
F
and
P
values,
Example
2.2.
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