Blank L., Tarquin A. Engineering Economy (McGraw-Hill Series in Industrial Engineering and Management)
Подождите немного. Документ загружается.
'"
o
U
SECTION
13.1
Breakeven Ana
ly
sis for a Single Project
Nonlinear
Q units per year
(a) Revenue relations
-(l
) increas
in
g and
(2) decrea
in
g revenue per unit
TC
= FC + VC
Q units per y
ear
(b) Linear
co
st relations
'"
o
U
TC = FC + VC
Q units per
ye
ar
(e) Nonlinear
co
st relations
When FC and Y C are added, they form the total cost relation TC. Figure 13-1 b
illustrates the
TC
relation for linear fixed and variable costs. Figure
13-lc
shows
a general
TC
curve for a nonlinear YC in which unit variable costs decrease as the
quantity level rise
s.
At a specific but unknown value Q
of
the decision variable, the revenue and total
cost relations will intersect to identify the breakeven point
Q
SE
(Figure 13-2).
If
Q >
QSE'
there
is
a predictable profit; but
if
Q <
QSE
' there is a loss. For linear mod-
els
of
Rand
YC, the greater the quantity, the larger the profit. Profit
is
calculated
as
Profit = revenue - total cost
=R-TC
[13.1]
A relation for the breakeven point may be derived when revenue and total cost
are linear functions
of
quantity Q by setting the relations for
Rand
TC
equal to
each other, indicating a profit
of
zero.
R
=TC
rQ
=
FC
+
VC
= FC + vQ
where r = revenue per unit
v = variable cost
per
unit
Figure 13-1
Linear and nonlinear
revenue and cost
relations.
445
446
Figure
13-2
Err ect on
th
e brea k
eve
n
point when
th
e variable
cost per unit is reduced.
Figure 1
3-3
Breakeven points a
nd
maximum profit point ror
a nonlinear analysis.
CHAPTE
R 13
$
Loss
$
Q
BE
Breakeven
Analysis
I
I
I
I
I
I
I
I
Br
ea
keven :
point moves :
I .
I I
QU
E
Q units per year
I
Pro
fi
t range
I
I
I
I
I
I
I
I
Qp
Q units
per
year
Profit
maximized
Profit
TC
- I
I
I
\
I
I
R
R
I
Loss
I
Q
BE
TCwith
lowered
VC
SECTION
13.1
Breakeven Analysis for a Single Project
Solve for the breakeven quantity Q
BE
to obtain
Fe
Q
BE
=--
r - v
[13.2]
The breakeven graph is an important management tool because it is easy to under-
stand and may be used in decision making and analysis in a variety
of
ways. For
example,
if
the variable cost per unit
is
reduced, the TC line has a smaller slope
(Figure
13-2)
, and the breakeven point will decrease. This is an advantage because
the smaller the value
of
QB
E' the greater the profit for a given amount
of
revenue.
If
nonlinear R or TC models are used, there may be more than one breakeven
point. Figure L3-3 presents this situation for two breakeven points. The maxi-
mum profit occurs at
Qp
between the two breakeven points where the distance
between the
Rand
TC
relations is greatest.
Of
course, no static
Rand
TC
relations-linear
or nonlinear- are able to esti-
mate exactly the revenue and cost amounts over an extended period
of
time. But
the breakeven point is an excellent target for planning purposes.
EXAMPLE
13.1 .
Lufkin Trailer Corporation assembles
up
to 30 trailers per month for 18-wheel trucks
in
its
east coast facility. Production has dropped to 25 units per month over the last 5 months
due to a worldwide economic slow down
in
transportation services. The following
in
-
formation
is
available.
Fixed costs FC
= $750,000 per month
Variable cost per unit v
= $35,000
Revenue per unit r = $75,000
(a) How does the reduced production level
of
25
units per month compare with the
CUJTent
breakeven point?
(b) What is the current profit level per month for the facility?
(c) What
is
the difference between the revenue and variable cost per trailer that is
necessary to break even at a monthly production level
of
15
units,
if
fixed costs
remain constant?
Solution
(a) Use Equation [13.2] to determine the breakeven number
of
units. All dollar
amounts are in
$1000 units.
Q
=
~
BE
r - v
750
75
_
35
= 18.75 units per month
Figure 13-4 is a plot
of
Rand
TC lines. The breakeven value is 18.75, or
19
in
integer trailer units. The reduced production level
of
25
units
is
above the
breakeven value.
(b)
To
estimate profit
in
$1000 at Q =
25
units per month, nse Equation [13.1].
Profit = R - TC = rQ - (FC + vQ)
= (r - v)Q - FC
= (75 - 35)25 - 750
= $250
There is a profit
of
$250,000 per month currently.
[13.3]
447
448
Payback period
CHAPTER 13 Breakeven Analysis
2000
1500
o
8
~
1000
750
500
Q
BE
=
18.75
o
10
15
20
25
30
Q,
units/month
Figure
13-4
Breakeven graph, Example 13.1.
(c) To determine the required difference r -
v,
use Equation [13.3) with profit = 0,
Q = 15, and FC = $750,000. In $1000 units,
o =
(r
- v)(15) - 750
750
$ 0 .
r - v =
--
= 5 per umt
15
The spread between r and v must be $50,000.
If
v stays at $35,000, the revenue per
trailer must increase from $75,000 to $85,000 just to break even at a production level
of
Q =
15
per month.
In some circumstances, breakeven analysis performed on a per unit basis is
more meanjngful. The value
of
Q
BE
is still calculated using Equation [13.2], but
the TC relation is divided by
Q to obtrun an expression for cost per unit, also
termed
average cost
per
unit
C"~O
C = TC =
FC
+
vQ
= FC + v [13.4]
/I
Q Q Q
At the breakeven quantity Q =
QBE'
the revenue per unit
is
exactly equal to the
cost per unit.
If
graphed, the FC per unit term in Equation [13.4] takes on the
shape
of
a hyperbola.
In
Chapter 5, payback period analysis was discussed. Payback
is
the number
of
years np necessary to recover an imtial investment. Payback analysis at a zero
SECTION 1
3.
1 Breakeven Anal
ys
is
fo
r a Single Project
interest rate is performed o
nl
y when
th
ere is no requirement to earn a rate
of
return
greater than zero
in
addition to recovering the initial investmen
t.
(As discuss
ed
earlier, the technique is best used as a supplement to PW analysis at the MARR.)
If
payback analysis is coupled
wi
th
breakeven, the quantity
of
the decision variable
for different payback pe
ri
ods can be determined, as illustrated in the next
exa
mpl
e.
EXAMPLE
13.2 .
The pres
id
ent
of
a local company, Online Ontime, Inc., expects a product
to
have a prof-
ita
bl
e life
of
between 1 and 5 years. He wa
nt
s
to
know the breakeven number
of
units that
mu
st be so
ld
annua
ll
y to rea
li
ze payback within each
of
the time pedods
of
1 year, 2 years,
and so on up to 5 years. Find
th
e
an
swers using hand and computer solutions. The cost and
revenue estimates are
as
follow
s:
Fi
xed
costs: initial investment of $80,000 with $1000 annual operating cost.
Va
riable cost: $8 per unit.
Revenue: Twice
th
e varia
bl
e cost for the first 5 years and 50%
of
the variable cost
th
ereafter.
Solution by Hand
D
efi
ne X
as
th
e
uni
ts
sold per y
ea
r to break even and np as the payback period, where
np
=
],
2, 3, 4, a
nd
5 years. There are two unknowns and one relation, so it is necessary to
esta
bli
sh
values
of
one varia
bl
e a
nd
solve for
th
e other. The following approach is used:
Esta
bli
sh annual cost and reve
nu
e rela
ti
ons with no time value
of
money c
on
sidered, then
use
IIp
va
lu
es
to
fi
nd the breakeven value
of
X.
80,000
Fixed costs
---
+ 1000
IIp
Varia
bl
e cost
8X
Reve
nu
e
{
16X
4X
Set reve
nu
e
eq
ual to total cost, and solve for
X.
Reve
nu
e = total cost
years
1
th
rough 5
year 6 and
th
ere
af
ter
16X = 80,000 + 1000 + 8X
to~OO
+ 125
II
p
[13
.
5]
Inse
rt
th
e va
lu
es 1 through 5 for
II
p a
nd
solve for X (Figure
13
- 5). For exampl
e,
payback in
2 years requires sales
of
5125 units per year
in
order to break even. There
is
no considera-
ti
on
of
interest
in
this solution;
th
at i
s,
i = 0%.
Solution by Computer
Sol
ve
Equa
ti
on [13.
5]
for
X,
retaining the symbols r and v for use in the spreadsheet.
80,000/
II
p + 1000
X = [13.6]
r - v
The spreadsheet in Figure
13
-6
includes the breakeven Equation [13.6]
in
cells C9 through
C13 as deta
il
ed
in
the cell tag. Column C and the
xy
scatter chart display the results. For
exam
pl
e, payback
in
1 year requires sales
of
X =
10
,125, while a 5-year payback requires
o
nl
y 2J
25
uni
ts
per year.
449
~
E-Solve
450
CHAPTER
13
Breakeven Analysis
Breakeven
12
,000
@
Q)
>,
....
9,000
Q)
0-
Zl
·2
::l
6,000
><
5,125
3,000
Payback
2
3
4
5
n
p
,
years
Figure
13-5
Breakeven sales volumes for different payback periods,
Example
J 3.2.
X M,crosoft Excel l!Il'iI
13
"
"
.,.
~
:J
C
6
00
0
"
>
i
400
0
1;
x·
2000
2 3 ,
n,
payba
ck years
~!.F~!!!£Ll..!!l:!:..!l!:~~Em
analysis:
(SO
,
OOO
ln +
100..Q)
..£.(l6-S)
Ready
Figure
13-6
Spreadsheet solution
of
breakeven values for different payback years, Example 13.2.
SECTION 13.2
Breakeven Analysis between Two Alternatives
13.2
BREAKEVEN ANALYSIS BETWEEN
TWO
ALTERNATIVES
Breakeven analysis involves the determination
of
a common variable or eco-
nomic parameter between two alternatives. The parameter can be the interest rate
i,
first cost
P,
annual operating cost (AOC), or any parameter.
We
have already
performed breakeven analysis between alternatives on several parameters. For
example, the incremental
ROR value
(~i*)
is
the breakeven rate between alter-
natives.
If
the MARR is lower than
~i*,
the extra investment
of
the larger-
investment alternative is justified. In Section 11.3, the replacement value (RV)
of
a defender was determined.
If
the market value is larger than
RV,
the decision
should favor the challenger.
Often breakeven analysis involves revenue or cost variables common to
both alternatives, such
as
price per unit, operating cost, cost
of
materials, and
labor cost. Figure
13-7 illustrates this concept for two alternatives with linear
cost relations. The fixed cost
of
alternative 2 is greater than that
of
alternative
1.
However, alternati ve 2 has a smaller variable cost,
as
indicated by its lower
slope. The intersection
of
the total cost lines locates the breakeven point.
Thus, if the number
of
units
of
the common variable is greater than the
breakeven amount, alternative 2 is selected, since the total cost will be lower.
Conversely,
an
anticipated level
of
operation below the breakeven point favors
alternati ve
l.
Instead
of
plotting the total costs
of
each alternative and estimating the
breakeven point graphically, it may be easier
to
calculate the breakeven point
numerically using engineering economy expressions for the
PW or
AW
at the
MARR. The
AW
is preferred when the variable units are expressed on a yearly
basis, and
AW
calculations are simpler for alternatives with unequal lives. The
following steps may be used to determine the breakeven point
of
the common
variable and
to
select
an
alternative:
1.
Define the common variable and its dimensional units.
2.
Use
AW
or PW analysis to express the total cost
of
each alternative
as
a
function of the common variable.
3. Equate the two relations and solve for the breakeven value
of
the
variable.
'"
o
u
S
o
E-<
'AI!.
1 FC point
Common variable, units
451
Incremental ROR
Figure
13-7
Breakeven between two
alternatives with linear
cost relations.
452
CHAPTER 13
Br
eakeven Analysis
4.
If
the anticipated level is below the breakeven value, select the alternative
with the hjgher variable cost (larger slope
).
If
the level is above
th
e
breakeven point, select the alternative with the lower variable cost. Refer
to
Figure 13-7.
A small aerospace company is evaluating two alternatives: the purchase
of
an automatic
feed machine and a manual feed machine for a finishing process.
The
auto-feed
machine has an initial cost
of
$23,000, an estimated salvage value
of
$4000, and a pre-
dicted life
of
10 years.
One
person will operate the machine
at
a rate
of
$12 per hour.
The expected output is 8 tons per hour. Annual maintenance and operating cost is
expected to be
$3500.
The
alternative manual feed machine has a first cost
of
$8000, no expected salvage
value, a 5-year life, and an output
of
6 tons
per
hour. However, three workers will be re-
quired at $8
per
hour each. The machine will have an annual maintenance and operation
cost
of
$1500. All projects are expected to generate a return
of
10%
per
year.
How
many
tons per year must be finished in order to justify the higher purchase cost
of
the auto-
feed machine?
Solution
Use the steps above to calculate the breakeven point between the two alternatives.
1.
Let
x represent the number
of
tons per year.
2.
For
the auto-feed machine the annual variable cost is
Annual VC
= $12 I hour x tons
hour 8 tons year
= 1.5x
The
VC is developed in dollars per year. The AW expression for the auto-feed
machine is
AW
au1
0
= -
23
,000(A/
P,1O
%
,1O)
+ 4000(A/
F,
I 0%,10) - 3500 - 1.5x
= $- 6992 - 1.5x
Similarly, the annual variable cost and AW for the manual feed machine are
$8
I hour x tons
Annual
VC
=
--
(3 operators) - - - -
--
hour 6 tons year
= 4x
AWm
a
nu
al
= - 8000(A/P,
10
%,
5)
- 1500 - 4x
=
$-3610
- 4x
SECTION
13.2
Breakeven Analysis between Two Alternatives
3.
Equate the two cost relations and solve for
x.
A W auto = A W manual
-6992
1.5x =
-3610
- 4x
x = 1353 tons per year
4.
If
the output
is
expected
to
exceed 1353 tons per year, purchase the auto-feed
machine, since its
VC
slope
of
1.5
is
smaller than the manual feed
VC
slope
of
4.
The breakeven analysis approach is commonly used for make-or-buy deci-
sions. The alternative to buy usually has no fixed cost and a larger variable cost
than the option
to
make. Where the two cost relations cross is the make-buy de-
cision quantity. Amounts above this indicate that the item should be made, not
purchased outside.
EXAMPLE 13.4 '
Guardian is a national manufacturing company
of
home health care appliances.
It
is
faced with a make-or-buy decision. A newly engineered lift can be installed
in
a
cm:
trunk
to
raise and lower a wheelchair. The steel arm
of
the lift can be purchased for
$0.60 per unit or made inhouse.
If
manufactured on site, two machines will be required.
Machine A
is
estimated to cost $18,000, have a life
of
6 years, and have a $2000 salvage
value; machine B will cost
$12,000, have a life
of
4 years, and have a
$-500
salvage
value (carry-away cost). Machine A will require an overhaul after 3 years costing
$3000. The annual operating cost for machine A
is
expected
to
be $6000 per year and
for machine B
$5000 per year. A total of four operators will be required for the two ma-
chines at a rate
of
$12.50 per hour per operator. In a normal 8-hour period, the opera-
tors and two machines can produce parts sufficient
to
manufacture 1000 units. Use a
MARR
of
]
5%
per year
to
determine the following.
(a) Number of units
to
manufacture each year
to
justify the inhouse (make) option.
(b) The maximum capital expense justifiable
to
purchase machine
A,
assuming all
other estimates for machines A and B are as stated. The company expects
to
pro-
duce
125,000 units per year.
Solution
(a) Use steps I
to
3 stated previously
to
determine the breakeven point.
1. Define
x as the number
of
lifts produced per year.
2.
There are variable costs for the operators and fixed costs for the two ma-
chines for the make option.
Annual VC
= (cost per unit)(units per year)
__
4 operators $] 2.50
(8
h )
---
ours x
1000 units hour
= OAx
453
454
Figure
13-8
Breakeven points for
three a
lt
ernatives.
CHAPTER
13
Breakeven Analysis
::;;
o
u
'"
'0
f-
The
annual fixed costs for machines A and B are the
AW
amounts.
AW
A
=
-18,000(A/P,15%,6)
+
2000(A/F,15%,6)
- 6000 -
3000(P/F,15
%,3)(A/
P,15%,6)
AWg
=
-12,000(A/P,15%,4)
-
500(A/F,15
%,4) - 5000
Total cost is the sum
of
AW
A'
AW
B'
and
Vc.
3. Equating the annual costs
of
the buy option (0.60x) and the make option
yields
-0.6
0x =
AW
A
+
AW
B
-
VC
=
-18,000(A/P,15%,6)
+
2000(A/F,15%,6)
- 6000
- 3000(P / F,
15
%,3)(A/
P,
15
%,
6) - 12,000(A/ P,
15
%,4)
- 500
(A/F,
15
%,
4)
- 5000 - 0.4x [13.7]
- 0.2x
= - 20,352.43
x =
101
,762 units per year
A minimum
of
101
,762 lifts must be produced each year to justify the make
option, which has the lower variable cost
of
0.40x.
(b) Substitute 125,000 for x and P A for the to-be-determined
fir
st cost
of
machine A
(currently
$
18
,000) in Equation [13.7]. Solution yields P A = $35,588. This
is
ap-
proximately twice the estimated first cost
of
$
18
,000, because the production
of
125
,000 per year is larger than the breakeven amount
of
101
,762.
,
I
I
I
I
I
I
I
I
I
I
I
I
Alternative I
Alternati
ve
2
I I
Br
eakeven
i~point
s
40
60
Output, units/hour