Chapter 3 - Cost and Revenue Estimation
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Cost and revenue estimation
Why do we need to estimate costs and revenues?
Required in the development of cash flows (often into the
future) for feasible alternatives, projects, etc.
Why is it difficult?
Projecting the future is always difficult due to uncertainties.
Past data is not directly applicable for unique projects
(estimation by analogy may be useful).
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Cost and revenue estimation
How can we do it?
Integrated approach - to develop the net cash flows for all
feasible project alternatives (Step 3 of engineering
economic analysis in Chapter 1). There are three basic
components:
Work Breakdown Structure (WBS): Define the work
elements of a project and their interrelationships at
successive levels of detail.
Cost and revenue structure (classification): Categorize the
costs and revenues.
Estimating techniques (models): Apply mathematical
methods to the available data to estimate the future costs
and revenues during the analysis period.
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Integrated approach - WBS
WBS is developed from the top (project level) down in
successive levels of detail. The following diagram exhibits a
three level WBS to a commercial building project.
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Integrated approach - Cost and revenue structure
Some typical categories of costs and revenues to be used
in an engineering economy study include:
Capital investment
Labor costs
Material costs
Maintenance and equipment costs
Property taxes and insurance
Quality and scrap costs
Overhead costs
Disposal costs
Revenues
Market (salvage) values
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Integrated approach - Estimation
According to detail, accuracy and the intended use, cost
and revenue estimates can be classified as follow:
Order-of-magnitude (rough) estimates: Used for the
high-level planning phase of a project; involve little detail
and accuracy.
Semidetailed (budget) estimates: Used mostly for the
budgeting phase.
Definitive (detailed) estimates: Used for the detailed design
phase; involve high accuracy and detail.
Sources of data:
Accounting records
Lack of the data on incremental costs and opportunity costs.
Both of them are important for engineering economic
analysis.
Other sources within the firm
The data can be obtained from various departments such as
sales, purchasing and quality control.
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Integrated approach - Estimation
Sources of data: (contd.)
Sources outside the firm
Published information - government publications and
journals.
Personal contracts - vendors, salespeople and government
agencies.
Research and development
To conduct R&D to generate it such as developing a pilot
plant.
How to estimate?
Conferences
Comparisons
Quantitative techniques
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Quantitative techniques for estimation
Cost indexes
Unit technique
Factor technique
Parametric cost estimating (Cost Estimating Relationship
(CER) development through mathematical techniques)
Power-sizing technique
Learning curves
Statistical techniques for developing CER (e.g., linear
regression)
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Cost indexes
Indexes are dimensionless numbers that are useful for
estimating the future cost (or price) of items based on the
historical data.
Given:
The cost of an item in year k, (Ck )
The cost index in year k (reference year), (Ik )
We are now in year n (n > k) and we are also given the cost
index in year n, (In ). We want to estimate the cost of this item in
year n, (Cn ).
Cn
Ck
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=
In
Ik
⇒ Cn =
In
Ik Ck
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Example 3.1
A waste water treatment equipment cost $250,000 in 2001,
which is the reference year having a cost index value of 100.
Estimate the cost of this equipment for year 2007 which has a
cost index of 125.
C2007 =
125
100 ($250, 000)
= $312, 500
The cost of this equipment is now $350,000. Calculate the cost
index for the current year.
In =
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Cn
Ck Ik
⇒ I2008 =
$350,000
$250,000 (100)
= 140
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Cost indexes
A composite (general weighted) index is created by averaging
the ratios of selected item costs in a particular year to the cost
of the same items in a reference year. Mathematically, it is
given by
P
m
In =  i=1

C
(Wi ) Cni
ki
m
P
Wi
 Ik
i=1
where
m =
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total number of items in the index;
Cni =
unit cost (or price) of the ith item in year n;
Cki =
unit cost (or price) of the ith item in year k;
Wi =
weight assigned to the ith item;
Ik =
composite index value in year k.
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Example 3.2
Annual cost per apartment in year 2000 and 2005 is as follows:
Electricity
Gas
Water
2000
180,000
85,000
35,000
2005
220,000
98,000
42,000
weight (Wi )
3
2
1
The reference year is 2000 having an index of 98. This index
was created for multiple cost items with different weights:
composite index. Composite index for year 2005:
!
(3)
220,000
180,000
+(2)
98,000
85,000
3+2+1
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+(1)
42,000
35,000
× 98 = 117.15
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Unit technique
We use cost per unit to estimate the total cost.
For example, if constructing one square foot of a house costs
$1,000, the total cost of a house with a 1,500 square feet is
$1,500,000.
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Factor technique
C=
P
Cd +
d
P
fm Um
m
where
C =
cost being estimated;
Cd =
cost of the selected component d that is estimated directly;
fm =
cost per unit of component m;
Um =
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number of units of component m.
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Example 3.3
Consider a house consisting of 1,000 square feet, a garage and
a swimming pool.
Directly estimated components:
Garage: $9,000
Swimming pool: $21,000
Unit cost: $100 per square foot.
By factor technique, we have the cost of the house C as given
by
C = $100 × 1, 000 + $9, 000 + $21, 000 = $130, 000
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Parametric cost estimation
Idea: Estimate the functions to relate cost to the main cost
drivers.
Power-sizing technique
X
CA = CB SSAB
Ci : cost for item i,
Si : size of item i,
X : cost-capacity factor.
X = 1: linear cost relationship with size;
X < 1: decreasing economies of scale (Favourable: a
larger size is expected to be less costly than that of a
purely linear relation.)
X > 1: increasing economies of scale. (Unfavourable: a
larger size is expected to be more costly than that of a
purely linear relation.)
(See the spreadsheet "Economies of scale.xls" for an example.)
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Example 3.4
The cost of building a wafer fabrication facility with a capacity of
500,000 units per year was $2,500,000 in year 2000. We want
to estimate the cost of a similar wafer fabrication facility with a
capacity of 1,500,000 units per year for year 2008. The
technological changes dictate that an additional piece of
equipment must be integrated with the facility and the cost of
this new equipment has been estimated to be $50,000 for year
2008. Suppose that the cost-capacity factor (X) is 0.65 and the
cost index for wafer fabrication facility has increased an
average rate of 12% per year for the past 8 years. How much is
the estimated total cost of the wafer fabrication facility?
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Solution
First, estimate the cost of the old facility in year 2008 using cost
indexes technique:
C2008 = II2008
C2000
2000
I2001 = I2000 (1.12)
I2002 = I2000 (1.12)2
I2003 = I2000 (1.12)3
...
I2008 = I2000 (1.12)8
⇒ C2008 = (1.12)8 (2, 500, 000) = $6, 189, 908
Next, use power sizing technique:
0.65
C(cap@1, 500, 000) = C(cap@500, 000) 1,500,000
=
500,000
6, 189, 908(2.04) = $12, 641, 919.32
Finally, add the estimated cost of the additional equipment
$50,000; total cost is $12,691,919.32.
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Learning curve
As we perform the same task on a repetitive basis, we learn the
task and become more efficient. As a result, the number of
input resources required for producing each additional unit
decreases as the number of units produced increases.
Model assumption:
As the output level doubles, the number of input resources
decreases by a constant percentage.
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Example 3.5
Your records show that the cost of producing the first unit on
your production line was $100, the second unit was $90, the
fourth unit was $81, and the eighth unit was $72.9. Each time
the output doubled the percentage reduction in cost was 10%:
100 − 100(0.1) = 100(0.9) = 90;
90 − 90(0.1) = 100(0.9)2 = 81;
81 − 81(0.1) = 100(0.9)3 = 72.9.
90% learning curve
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Model
Zu = K(u)n
Zu : the number of input resources to produce the uth unit,
K : the number of input resources to produce the first unit,
log s
n = log
2,
s : the learning curve parameter or learning rate (s = 0.9 for a
90% learning curve)
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Example 3.6
The assembly line team in your factory has a learning rate of
75%. If the first unit took 80 hours to be assembled,
a) how much time will it take the team to assemble the 20th
item,
b) how many total number of units does the team have to
produce before reducing the assembly time to 15 hours?
Solution:
log 0.75
a) Z20 = 80(20) log 2 = 80(20)−0.415 = 23.1 hours
log 0.75
−0.415 ⇒ u = 56.47.
b) 15 = 80(u) log 2 ⇒ 15
80 = u
Therefore, the team has to produce 56 units before reducing
the assembly time to 15 hours.
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Example 3.7
In a chip producing facility, you observe that 50th
microprocessor required 3.3 hours to produce, while the 100th
required 2.2 hours to produce. What is the learning curve
model that fits these data?
Solution:
log s
Z50 = K(50) log 2 = 3.3
Z100 = K(100)
log s
log 2
= 2.2
(1);
(2);
Solve equations (1) and (2) simultaneously.
Z50
Z100
log s
= (0.5) log 2 =
3.3
2.2
= 1.5
log s
log 2
⇒
log 0.5 = log 1.5 ⇒ s = 0.67 (67% learning curve. We can
easily find it from the definition of a learning curve, too. Think
aboutthe doubling
issue!) Find K from equation (1) or (2).
K=
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3.3
log 0.67
50 log 2
= 32.5. Then, Zu = 32.5(u)−0.585 .
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Cost estimating relationship (CER)
A CER is a mathematical model that describes the cost of
an engineering project as a function of one or more design
variables.
Steps in CER development:
1
2
3
4
Define the problem
Collect data
Develop CER equation
Validate model
CER equation development involves initial guess of the
shape of the relationship and estimation of the model
parameters.
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Cost estimating relationship (CER)
Typical examples of equation form:
Table 1
Type of Relationship
Linear
Power
Logarithmic
Exponential
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Generalized Equation
C = b0 + b1 x1 + b2 x2 + b3 x3 + · · ·
C = b0 + b1 x1b11 x2b12 + · · ·
C = b0 + b1 log(x1 ) + b2 log(x2 ) + b3 log(x3 ) + · · ·
C = b0 + b1 eb11 x1 + b2 eb22 x2 + · · ·
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Cost estimating relationship (CER)
Use the historical data to build a model that relates the
cost of an engineering design or project to one or more
design variables and cost drivers. Statistical tools such as
linear regression, multiple linear regression models are
useful to develop the models.
Example:
Find a model for the cost per mile of driving a car which is
dependent on the car’s weight, current mileage, speed, and
outside temperature. (Multiple linear regression)
Example:
Find a model for the cost of operating a machine which is
related to the weight of the machine. (Simple linear regression)
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Linear regression
We have collected some observations related with the cost of
operating a machine and the weight of the machine. We have
the following data:
Machine i
1
2
3
4
5
6
Weight xi (kg)
200
250
380
450
560
600
Cost yi ($ thousands)
139
200
250
325
375
425
Homework: Plot these points on a graph with weight in the
x-axis and cost in the y-axis.
Fit a simple linear equation to the available data:
Equation is of the form: y = b0 + b1 x
Here, Cost = b0 + b1 (Weight), so y = Cost, x = Weight.
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How to estimate the parameters b0 and b1 ? Linear Regression.
The linear regression refers to the way on estimating b0 and
b1 in order to make the estimated line as close to the available
data as possible.
Use the available data points. By the linear regression, b0 and
b1 are given by
n
b1 =
n
P
xi yi −
i=1
n
b0 =
xi
i=1
n
P
i=1
n
P
n
P
yi −b1
xi2 −
n
P
i=1
i=1
n
n
P
i=1
n
P
2
yi
,
xi
i=1
xi
.
where n is the number of observations or data points.
Cost = 14.97 + 0.6666x, where x represents the weight of the
machine in kgs.
See the spreadsheet file "Regression.xls".
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Application of linear regression technique
In addition to the linear model, we can also apply the linear
regression technique to estimate some other models in Table 1.
Power Form
Suppose b0 = 0 and there is only one independent variable v
that affects C. The power form can then be expressed in the
form of
C = a0 va1 ,
a0 > 0.
By taking logarithm on both sides, we have
log C = log a0 + a1 log v.
Set y = log C and x = log v, we then have
y = log a0 + a1 x.
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Application of linear regression technique
By applying the linear regression technique,
n n n
P
P
P
yi
xi
xi yi −
n
i=1
i=1
i=1
;
a1 =
n 2
n
P
P
2
xi
xi −
n
i=1
i=1
n
P
log a0 =
yi − a1
i=1
n
P
xi
i=1
n
where yi = log Ci and xi = log vi , and Ci and vi are the data of C
and v respectively.
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Application of linear regression technique
Exponential Form
Suppose b0 = 0 and there is only one independent variable v
that affects C. The exponential form can then be expressed in
the form of
C = a0 ea1 v ,
a0 > 0.
By taking natural logarithm (ln) on both sides, we have
ln C = ln a0 + a1 v.
Set y = ln C and x = v, we then have
y = ln a0 + a1 x.
Similar to the power form, ln a0 and a1 can be estimated
through the linear regression technique.
Homework: Write down the expressions for ln a0 and a1 .
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Determining the selling price and estimating cost
We need to estimate cost in the design process because we
want to make a product that can be sold at a competitive price
and can bring a reasonable profit.
Bottom-up approach:
The cost elements at the lower levels of the cost structure are
estimated and are added together to obtain the total cost of the
product.
Then,
Selling price per unit= total cost per unit (1+profit margin)
Profit margin (in percentages)= profit per unit/cost per unit
Example:
If the unit cost is $100 and you want to obtain a profit margin of
10%, you must sell the item at 100(1+0.1)=$110.
See the spreadsheet file "Bottom-up cost estimation.xls" for an
example.
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Top-down approach:
Determine the target cost such that the desired profit is
obtained.
Target cost=competitor’s price-desired profit
Target cost calculation with profit margin:
Target cost =
Competitor’s price
1+profit margin .
Example:
If the competitor’s price for a product is $33, and you are
planning to sell your product at this price, what would be your
target cost per product if you want a profit margin of 10%?
Target cost=
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$33
1+0.1
= $30.
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Example 3.8
Estimate the per-unit selling price of the following product:
Direct labor rate: $15 per hour
Production material: $375 per 100 items
Factory overhead: 125% of direct labor
Packing costs: 75% of direct labor
Desired profit: 20% of total manufacturing cost
Assume that 80% learning curve applies to the labor. The time
to complete the first item is 1.76 hours. Use the estimated time
to complete the 50th item as the standard time for the purpose
of estimating the unit selling price.
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Solution
By using learning curve method, the estimated direct labor
hours for the 50th unit, Z50 , can be obtained as follow.
K = 1.76; s = 0.8; n =
log 0.8
log 2
= −0.32193.
Z50 = 1.76(50)−0.32193
= 0.5 hr
Direct Labor
Production Material
Factory overhead
Packing costs
Total manufacturing cost
Desired profit
Unit selling price
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= $15 /hr × 0.5 hr
= $375 / 100 units
= 1.25 × $7.5 /unit
= 0.75 × $7.5 /unit
= 0.2 × $26.26 /unit
= $7.5 /unit
= $3.75 /unit
= $9.38 /unit
= $5.63 /unit
= $26.26 /unit
= $5.25 /unit
= $31.51 /unit
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