IENG 217
Cost Estimating for Engineers
Break Even &
Sensitivity
Motivation
Suppose that by investing in a new information
system, management believes they can reduce
inventory costs. Your boss asks you to figure
out if it should be done.
Motivation
Suppose that by investing in a new information system,
management believes they can reduce inventory costs.
After talking with software vendors and company
accountants you arrive at the following cash flow
diagram.
25,000
i = 15%
1
2
100,000
3
4
5
Motivation
Suppose that by investing in a new information system,
management believes they can reduce inventory costs.
After talking with software vendors and company
accountants you arrive at the following cash flow
diagram.
25,000
i = 15%
1
2
3
4
5
100,000
NPW = -100 + 25(P/A,15,5) = -16,196
Motivation
Suppose that by investing in a new information system,
management believes they can reduce inventory costs.
After talking with software vendors and company
accountants you arrive at the following cash flow
diagram.
25,000
i = 15%
1
2
3
4
5
100,000
NPW = -100 + 25(P/A,15,5) = -16,196
Motivation
Boss indicates $25,000 per year savings is too
low & is based on current depressed market.
Suggests that perhaps $40,000 is more
appropriate based on a more aggressive market.
40,000
1
2
100,000
3
4
5
Motivation
Boss indicates $25,000 per year savings is too low & is
based on current depressed market. Suggests that
perhaps $40,000 is more appropriate based on a more
aggressive market.
40,000
1
2
3
4
5
100,000
NPW = -100 + 40(P/A,15,5) = 34,086
Motivation
Boss indicates $25,000 per year savings is too low & is
based on current depressed market. Suggests that
perhaps $40,000 is more appropriate based on a more
aggressive market.
40,000
1
2
3
4
5
100,000
NPW = -100 + 40(P/A,15,5) = 34,086
Motivation
Tell your boss, new numbers indicate a go.
Boss indicates that perhaps he was a bit hasty.
Sales have fallen a bit below marketing forecast,
perhaps a 32,000 savings would be more
appropriate
32,000
1
2
100,000
3
4
5
Motivation
Tell your boss, new numbers indicate a go. Boss
indicates that perhaps he was a bit hasty. Sales have
fallen a bit below marketing forecast, perhaps a 32,000
savings would be more appropriate
32,000
1
2
3
4
5
100,000
NPW = -100 + 32(P/A,15,5) = 7,269
Motivation
Tell your boss, new numbers indicate a go. Boss
indicates that perhaps he was a bit hasty. Sales have
fallen a bit below marketing forecast, perhaps a 32,000
savings would be more appropriate
32,000
1
2
3
4
5
100,000
NPW = -100 + 32(P/A,15,5) = 7,269
Motivation
Tell your boss, new numbers indicate a go.
Boss leans back in his chair and says, you know
....
Motivation
Tell your boss, new numbers indicate a go.
Boss leans back in his chair and says, you know
....
I’ll do anything, just
tell me what numbers
you want to use!
Motivation
A
1
2
3
4
5
100,000
NPW = -100 + A(P/A,15,5) > 0
Motivation
A
1
2
3
4
5
100,000
NPW = -100 + A(P/A,15,5) > 0
A > 100(A/P,15,5)
> 29,830
Motivation
A
1
2
3
100,000
A > 29,830
A < 29,830
4
5
Fixed vs Variable
 Fixed
- do not vary with production
general admin., taxes, rent, depreciation
 Variable - costs vary in proportion to the
quantity of output
material, direct labor, material handling
Fixed vs Variable
 Fixed
- do not vary with production
general admin., taxes, rent, depreciation
 Variable - costs vary in proportion to the
quantity of output
material, direct labor, material handling
TC(x) = FC + VC(x)
Fixed vs Variable
TC
VC
FC
TC(x) = FC + VC(x)
Break Even
R
TC
FC
Profit = R(x) - FC - VC(x)
Break-Even Analysis
Site
Fixed Cost/Yr
A=Austin
$ 20,000
S= Sioux Falls 60,000
D=Denver
80,000
TC = FC + VC * X
Variable Cost
$ 50
40
30
Break-Even (cont)
Break-Even Analysis
250,000
200,000
Austin
Total Cost
150,000
S. Falls
Denver
100,000
50,000
0
0
500
1,000
1,500
2,000
Volume
2,500
3,000
3,500
4,000
Example
 Company
produces crude oil from a field
where the basis of decision is the number of
barrels produced. Two methods for
production are:
 automated
tank battery
 manually operated tank battery
Example
 Automated
 annual
tank battery
depreciation = $3,200
 annual maintenance = $5,200
 Other fixed & variable costs
Automated Tank Battery
Automatic Tank Battery Operations
Fixed Cost / day
Control panel power
Circulating pump
Maintenance
Meter calibration
Chemical pump power
Total
2.69/day or $982/yr
Variable Cost / day
Pipeline pump (5 hsp @ 50% util)
Chemical additives (7.5 qts/day)
Inhibitor (2 qts/day)
Gas (10.8 MCF/day x 0.0275/MCF)
Total
5.68/day / 500 barrels
= $0.01136 / barrel
$0.15
0.82
1.00
0.40
0.32
$2.69
$0.63
3.75
1.00
0.30
$5.68
TC(x) = (982 + 3,200 + 5,200) + 0.01136 X
Example
 Manual
Tank Battery
 annual
depreciation = $2,000
 annual maintenance = $7,500
 other costs
Manual Tank Battery
Automatic Tank Battery Operations
Fixed Cost / day
Chemical pump power
Circulating pump
Total
0.98/day or $358/yr
Variable Cost / day
Chemical additives (7.5 qts/day)
Gas (10.8 MCF/day x 0.0275/MCF)
Total
4.05/day / 500 barrels
= $0.00810 / barrel
$0.16
0.82
$0.98
3.75
0.30
$4.05
TC(x) = (2,000 + 7,500 + 358) + 0.00810 X
BreakEven
TCA(x) = TCM(x)
BreakEven
TCA(x) = TCM(x)
9,382 + 0.01136 x = 9,858 + 0.0081 x
BreakEven
TCA(x) = TCM(x)
9,382 + 0.01136 x = 9,858 + 0.0081 x
0.0033 x = 476
BreakEven
TCA(x) = TCM(x)
9,382 + 0.01136 x = 9,858 + 0.0081 x
0.0033 x = 476
x* = 145,000
Example
Cost of Production
Automatic vs Manual
12,000
11,000
Automatic
10,000
Manual
9,000
8,000
-
50,000
100,000 150,000 200,000
Barrels per Year
Average vs Marginal Cost
TC ( x )
AC ( x ) =
x
MC ( x ) =
TC ( x )
x
Example
 Cost
of running an automobile is
TC(x) = $950 + 0.20 x
where $950 covers annual depreciation and
maintenance and x is the number of miles driven
per year
Example
TC ( x) 950
=
+ 0.20
AC ( x) =
x
x
TC ( x)  (950 + 0.20 x)
=
= 0.20
MC ( x) =
x
x
Example
Average vs Marginal Cost
(Automobile)
cost
1.5
1.0
Average
Marginal
0.5
0.0
0
10,000
20,000
Miles per year
30,000
Marginal Returns
Example
 Small
firm sells garden chemicals.
x = number of tons sold per year
SP(x) = selling price per ton (to sell x tons)
= $(800 - 0.8x)
TR(x) = total revenue at x tons
= $(800 - 0.8x) x
TC(x) = total production cost for x tons
= $(8,000 + 400x)
Example
TP(x) = total profit at x tons
= TR(x) - TX(x)
= (800x - 0.8x2) - (8,000 + 400x)
= -0.8x2 + 400x - 8,000
Compute
a. x at which revenue is maximized
b. marginal revenue at max revenue
c. x at which profit is maximized
d. average profit at max profit
Example
TR(x) = -0.8x2 + 800x
a. max R
TR( x)
(-0.8 x 2 + 800 x)
=0=
x
x
= - 1.6 x + 800
x = 500 tons
Example
TR(x) = -0.8x2 + 800x
b. Marginal Revenue
MR(500) = -1.6(500) + 800
= $0
Example
TP(x) = -0.8x2 + 400x - 8,000
c. max profit
TP( x)
(-.8 x 2 + 400 x + 8,000)
=0=
x
x
= - 1.6 x + 400
x = 250
Example
TP(x) = -0.8x2 + 400x - 8,000
c. average profit
AP ( x ) =
- 0.8 x 2 + 400 x - 8,000
x
= - 0.8 x + 400 - 8,000 / x
AP ( 250) = $168 / ton
Break-Even Analysis
Site
Fixed Cost/Yr
A=Austin
$ 20,000
S= Sioux Falls 60,000
D=Denver
80,000
TC = FC + VC * X
Variable Cost
$ 50
40
30
Break-Even (cont)
Break-Even Analysis
250,000
Total Cost
200,000
Austin
150,000
S. Falls
Denver
100,000
50,000
0
0
500
1,000
1,500
2,000
Volume
2,500
3,000
3,500
4,000
Class Problem
A firm is considering a new product line and the
following data have been recorded:
Sales price
Cost of Capital
Overhead
Oper/maint.
Material Cost
Production
Planning Horizon
MARR
$ 15 / unit
$300,000
$ 50,000 / yr.
$ 50 / hr.
$ 5 / unit
50 hrs / 1,000 units
5 yrs.
15%
Compute the break even point.
Class Problem
Class Problem
Profit Margin = Sale Price - Material - Labor/Oper.
50 hrs
= $15 - 5 $50 / hr
1000 units
= $ 7.50 / unit
Class Problem
Profit Margin = Sale Price - Material - Labor/Oper.
50 hrs
= $15 - 5 $25 / hr
1000 units
= $ 7.50 / unit
7.5X
1
50,000
300,000
2
3
4
5
Class Problem
Profit Margin = Sale Price - Material - Labor/Oper.
50 hrs
= $15 - 5 $25 / hr
1000 units
= $ 7.50 / unit
7.5X
300,000(A/P,15,5) + 50,000 = 7.5X
1
50,000
300,000
2
3
4
5
139,495 = 7.5X
X = 18,600
Sensitivity
Suppose we consider the following cash flow
diagram:
35,000
i = 15%
1
2
3
4
5
100,000
NPW = -100 + 35(P/A,15,5) = $ 17,325
Sensitivity
Suppose we don’t know A=35,000 exactly but
believe we can estimate it within some
percentage error of + X.
35,000(1+X)
1
2
100,000
3
i = 15%
4
5
Sensitivity
Then,
35,000(1+X)
i = 15%
1
2
3
4
5
100,000
EUAW = -100(A/P,15,5) + 35(1+X) > 0
35(1+X) > 100(.2983)
X > -0.148
Sensitivity (cont.)
NPV vs. Errors in A
50,000
40,000
30,000
NPV
20,000
10,000
0
-0.30
-0.20
-0.10
0.00
(10,000)
(20,000)
Error X
0.10
0.20
Sensitivity (Ao)
Now suppose we believe that the initial
investment might be off by some amount X.
35,000
i = 15%
1
2
3
100,000(1+X)
4
5
Sensitivity (Ao)
NPV vs Initial Cost Errors
50,000
40,000
30,000
NPV
20,000
10,000
0
-0.30
-0.20
-0.10
0.00
(10,000)
(20,000)
Error X
0.10
0.20
Sensitivity (A & Ao)
NPV vs Errors
50,000
40,000
30,000
NPV
20,000
10,000
0
-0.30
-0.20
-0.10
0.00
(10,000)
(20,000)
Error X
Errors in initial cost
Errors in Annual receipts
0.10
0.20
Sensitivity (PH)
Now suppose we believe that the planning
horizon might be shorter or longer than we
expected.
35,000
i = 15%
1
2
100,000
3
4
5
6
7
Sensitivity (PH)
NPV vs Planning Horizon
50,000
40,000
30,000
PH
20,000
10,000
0
(10,000) 0
1
2
3
4
(20,000)
(30,000)
NPV
5
6
7
Sensitivity (Ind. Changes)
NPV vs Errors
50,000
n=7
40,000
30,000
NPV
20,000
10,000
0
-0.30
-0.20
-0.10
n=3
0.00
(10,000)
0.10
0.20
(20,000)
Error X
Errors in initial cost
Planning Horizon
Errors in Annual receipts
MARR
Multivariable Sensitivity
Suppose our net revenue is composed of $50,000
in annual revenues which have an error of X and
$20,000 in annual maint. costs which might have
an error of Y (i=15%).
50,000(1+X)
1
2
3
20,000(1+Y)
100,000
4
5
Multivariable Sensitivity
Suppose our net revenue is compose of $50,000
in annual revenues which have an error of X
and $20,000 in annual maint. costs which might
have an error of Y (i=15%).
50,000(1+X)
1
2
3
4
5
Multivariable Sensitivity
Suppose our net revenue is compose of $50,000 in annual
revenues which have an error of X and $20,000 in annual
maint. costs which might have an error of Y.
50,000(1+X)
1
2
3
4
5
You Solve It! ! !
20,000(1+Y)
100,000
20,000(1+Y)
100,000
You Solve It!!!
Multivariable Sensitivity
Multivariable Sensitivity
50,000(1+X)
1
2
3
4
5
20,000(1+Y)
100,000
EUAW = -100(A/P,15,5) + 50(1+X) - 20(1+Y) > 0
50(1+X) - 20(1+Y) > 29.83
Multivariable Sensitivity
50,000(1+X)
1
2
3
4
5
20,000(1+Y)
100,000
EUAW = -100(A/P,15,5) + 50(1+X) - 20(1+Y) > 0
50(1+X) - 20(1+Y) > 29.83
50X - 20Y > -0.17
X > 0.4Y - 0.003
Multivariable Sensitivity
Simultaneous Errors (Rev. vs. Cost)
0.4
0.3
Error Y
Unfavorable
0.2
+ 10%
0.1
0
-0.15
-0.1
-0.05
-0.1 0
-0.2
-0.3
-0.4
Error X
0.05
Favorable
0.1
0.15
Mutually Exclusive Alt.
Suppose we work for an entity in which the MARR is
not specifically stated and there is some uncertainty as
to which value to use. Suppose also we have the
following cash flows for 3 mutually exclusive
alternatives.
t
A
A
A
0
1
2
3
4
5
1t
(50,000)
18,000
18,000
18,000
18,000
18,000
2t
(75,000)
25,000
25,000
25,000
25,000
25,000
3t
(100,000)
32,000
32,000
32,000
32,000
32,000
Mutually Exclusive Alt.
t
0
1
2
3
4
5
A1t
(50,000)
18,000
18,000
18,000
18,000
18,000
A2t
(75,000)
25,000
25,000
25,000
25,000
25,000
A3t
(100,000)
32,000
32,000
32,000
32,000
32,000
4.0%
6.0%
8.0%
10.0%
12.0%
14.0%
16.0%
18.0%
20.0%
30,133
25,823
21,869
18,234
14,886
11,795
8,937
6,289
3,831
36,296
30,309
24,818
19,770
15,119
10,827
6,857
3,179
(235)
42,458
34,796
27,767
21,305
15,353
9,859
4,777
69
(4,300)
MARR =
NPV1
NPV2
NPV3
Mutually Exclusive Alt.
NPV vs. MARR
50,000
40,000
NPV
30,000
NPV1
20,000
NPV2
NPV3
10,000
0
0.0%
(10,000)
5.0%
10.0%
MARR
15.0%
20.0%