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ECLT 5930/SEEM 5740: Engineering Economics
2014–15 Second Term
Master of Science in ECLT & SEEM
Instructors: Dr. Anthony Man–Cho So
Dr. Man Hong Keith Wong
Department of Systems Engineering & Engineering Management
The Chinese University of Hong Kong
January 22, 2015
Recap: Cost Concepts and Analysis
1. Various cost terminologies and their characteristics
2. Cost–volume relationships; breakeven points
3. Cost–driven design optimization
ECLT 5930/SEEM 5740 (2014–15 Second Term)
January 22, 2015
1
Application: To Produce or Not to Produce?
• Department A of a manufacturing plant occupies 100 square meters and produce,
among other things, 576 pieces of product X per day.
• The average daily production costs for product X are summarized as follows:
Direct labor
1 operator working 4 hours per day
at $22.50 per hour;
part–time manager at $30 per day
Direct material
Overhead
$120.00
$86.40
at $0.82 per square meter
Total cost per day
$82.00
$288.40
• One can also outsource the production of X to another company at a cost of
$0.35 per piece. This results in a total purchase cost of 576 × $0.35 = $201.60.
Question: Should the plant shut down the production line for X and purchase it
from the other company?
ECLT 5930/SEEM 5740 (2014–15 Second Term)
January 22, 2015
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This Lecture: Cost Estimation Techniques
• Cost is an integral element in evaluating and analyzing feasible alternatives.
–
–
–
–
provide information for setting prices
determine whether a proposed product can be made and distributed at a profit
evaluate how much capital can be justified for improvements
establish benchmarks for productivity improvement programs
• Previous discussions assume that they are exactly known.
• In reality, various cost parameters must be estimated or forecasted.
• A decision based on these cost estimated are economically sound only to the
extent that those estimates are representative of what would occur.
– Garbage in, garbage out!
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January 22, 2015
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Fundamental Cost Estimation Approaches
• Top–Down
– use historical data from similar projects and adjust them to current levels
– get a rough estimate, best used early in the estimation process
• Bottom–Up
– break down a project into small, manageable units and estimate their economic
consequences
– get a detailed estimate, best used when details concerning the desired output
have been defined
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January 22, 2015
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Integrated Approach for Cost Estimation
• Work Breakdown Structure (WBS): explicitly defining the elements of a project
and their interrelationships at successive levels of detail
• Cost and Revenue Structure: delineation of cost and revenue categories for the
purpose of cash flow estimation at each level of WBS
• Estimating Techniques and Models: use selected mathematical models to
estimate the cost and revenue elements
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January 22, 2015
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Work Breakdown Structure (WBS)
• WBS is a framework for
–
–
–
–
defining all project work elements and their interrelationships,
collecting and organizing information,
developing relevant cost and revenue data, and
integrating project management activities.
• It is developed from top–down in successive levels of detail.
• Both functional (e.g., logistical support, marketing, etc.) and physical (e.g.,
labor, material, etc.) elements are included in the WBS.
• A WBS can include both recurring (e.g., maintenance) and non–recurring (e.g.,
initial construction) work elements.
ECLT 5930/SEEM 5740 (2014–15 Second Term)
January 22, 2015
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Work Breakdown Structure: Example
Commercial Building Project
Physical
• Site Work and Foundation
– Site Gardening
– Excavation
– ...
• Exterior
– Framing
– Siding
– ...
• Interior
– Doors
– Flooring/Stairways
– ...
ECLT 5930/SEEM 5740 (2014–15 Second Term)
Functional
• Sales
– Leasing/Asset Sales
– Adminstrative Support
– ...
• Project Management
– Technical Management
– Legal
– ...
• Engineering Services
– Design
– Consulting
– ...
January 22, 2015
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Cost and Revenue Structure
• This structure is used to identify and categorize the costs and revenues that
need to be included in the analysis.
• Some categories include (but are not limited to)
capital investment
material costs
property taxes/insurance
disposal costs
quality costs
ECLT 5930/SEEM 5740 (2014–15 Second Term)
labor costs
maintenance costs
overhead costs
sales revenues
market/salvage values
January 22, 2015
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Cost Estimation
• The purpose of estimation is not to produce exact data about future cash flows,
but rather a reasonably accurate projection.
• Different levels of accuracy
– order–of–magnitude estimates: used in the planning or initial evaluation stage
– budget estimates: used in the preliminary or conceptual design stage
– definitive estimates: used in the detailed engineering stage
• Sources of Data
– accounting records: good source of historical data, but have their limitations
– sources within a firm: e.g., people/data from engineering, sales, production,
quality, purchasing, personnel, etc.
– sources outside a firm: e.g., government statistics, marketing research,
personal contact, etc.
– research and development (R&D): could be a large undertaking and need
rigorous techniques
ECLT 5930/SEEM 5740 (2014–15 Second Term)
January 22, 2015
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Cost Estimation (Cont’d)
• How estimates are accomplished?
– conference: people with good information are gathered to estimate the
quantities in question
– comparison: use similar situations or designs to extrapolate relevant estimates
– quantitative: use of mathematical techniques to derive estimates
• Question: What are the limitations of each of the above approaches?
We shall focus on quantitative techniques in this lecture.
ECLT 5930/SEEM 5740 (2014–15 Second Term)
January 22, 2015
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Use of Indices
• An index is a dimensionless number that indicates how cost or price has changed
with time with respect to a base year.
• Mathematically, an estimate of the cost of an item in year n, Cn (typically
current), with respect to that of year k, Ck (which is in the past), can be
obtained by
In
Cn = Ck × ,
Ik
where
–
–
–
–
–
–
k = reference year (e.g., 2000) for which cost or price is known;
n = year for which cost or price is to be estimated (note that n > k);
Ck = cost or price of item in year k;
Cn = estimated cost or price of item in year n;
Ik = index value in year k;
In = index value in year n.
• Indices capture the ratio of price changes.
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January 22, 2015
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Use of Indices: Example
• A company installed a boiler for $525,000 in 2000, when the index was 468.
• The company wants to install another boiler of the same type now in 2012,
when the index is 542.
• Question: What is the approximate cost of the new boiler?
– Using the index equation, the approximate cost is given by
C2012 = C2000 ×
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I2012
542
= $608, 013.
= 525, 000 ×
I2000
468
January 22, 2015
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Use of Indices (Cont’d)
• Question: What are the advantages and disadvantages regarding the use of
indices?
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January 22, 2015
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Computation of Indices
• An index can be compiled for a single or multiple items.
Examples
– Consumer Price Index (CPI): tracks the prices of consumer goods and
services (such as food and beverage, transport, entertainment) purchased
by households
– Commodity Price Index: tracks the prices of agricultural and industrial
commodities (such as oil, metal, corn) consumed by industries
• Suppose that we want to incorporate M items in an index. Let
– Ckm = cost or price of item m in year k;
– Cnm = estimated cost or price of item m in year n;
– Wm = weight (relative importance) assigned to item m.
• Then, we can form the following weighted (or composite) index:
In =
W1(Cn1/Ck1) + W2(Cn2/Ck2) + · · · + WM (CnM /CkM )
× Ik .
W1 + W2 + · · · + WM
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January 22, 2015
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Composite Index: Example
• Consider the composite price for gasoline (in cents per gallon) in year 1992,
2006 and 2011:
Premium
Unleaded plus
Regular unleaded
1992
114
103
93
2006
240
230
221
2011
320
303
291
• The weight on regular unleaded is three times that of the other two, based on
the amount sold.
• Suppose that the index in 1992 was 100. Then, the index value in 2011 is
I2011 =
(320/114) + (303/103) + 3(291/93)
× 100 ≈ 302.7172.
1+1+3
• Now, suppose that the index in 2012 is estimated to be I2012 = 327. Then, the
price of gasoline in 2012 can be estimated using the composite index:
cost(premium)2012 = cost(premium)2011 ×
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I2012
≈ 345.67 cents per gallon.
I2011
January 22, 2015
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Unit and Factor Technique
• The unit technique involves using a per unit factor that can be estimated
effectively.
Examples
capital cost of plant per kilowatt capacity
revenue per mile
material cost per unit square
revenue per customer served
• The factor technique extends the unit technique by taking into account multiple
cost components and forming a single cost estimate. Mathematically, the total
cost C estimated by this technique is given by
X
X
fmUm,
C=
Cd +
d
m
where
– Cd = cost of component d that is estimated directly;
– fm = cost per unit of component m;
– Um = number of units of component m.
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January 22, 2015
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Unit and Factor Technique: Example
• Suppose that your company has a commerical building for lease. The annual
revenue can be estimated as
3
X
where
–
–
–
–
–
–
–
–
Sj × uj × dj ,
R = 12 × P × rp + 12 × B × rb +
|
{z
}
direct estimation
{z
}
|j=1
factors
P = number of parking spaces;
B = number of billboards;
rp = monthly rate of parking spaces;
rb = monthly rate of billboards;
j = type of building space;
Sj = available type–j space (in square feet);
uj = utilization rate of space type j;
dj = rate (per square foot utilized per year) of space type j.
ECLT 5930/SEEM 5740 (2014–15 Second Term)
January 22, 2015
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Parametric Cost Estimating
• Basic Idea: use both historical data and statistical techniques to predict future
costs.
• Statistical techniques are used to produce Cost Estimating Relationships (CER),
which take the form
cost (or price) of an item = f (cost driver1, cost driver2, . . .).
Here, f is a function that we specify, which may include parameters that need
to be estimated from historical data.
• Two immediate questions:
– What are the cost drivers?
– What kind of f should we use?
ECLT 5930/SEEM 5740 (2014–15 Second Term)
January 22, 2015
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Cost Drivers
• Cost drivers are design variables that account for large portion of total cost.
• They depend on the product/service being considered.
• Examples
Product
software
turbine engine
housing
cellphone plan
set menu
Cost Drivers
number of lines of codes, target audience
maximum thrust, fuel consumption
size, neighborhood, interest rate
number of features, coverage
quality of food, ambience, rent
• Caution: One can always come up with a large number of cost drivers, but not
all of them are necessarily significant (more on this later).
ECLT 5930/SEEM 5740 (2014–15 Second Term)
January 22, 2015
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CER I: Power–Sizing Technique
• Also known as the exponential model.
• Commonly used to estimate capital investment on industrial plant or equipment.
• Idea: cost varies as some power of the change in capacity or size. Mathematically,
we have
X
X
SA
SA
CA
=
⇐⇒ CA = CB
,
CB
SB
SB
where
–
–
–
–
–
CA = cost of plant A;
CB = cost of plant B;
SA = size of plant A;
SB = size of plant B;
X = cost–capacity factor to reflect economies of scale.
• Note: The costs must be determined at the same point in time for which the
estimate is desired.
ECLT 5930/SEEM 5740 (2014–15 Second Term)
January 22, 2015
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Power–Sizing Technique: Example
• A power plant with capacity 200–MW cost $100M 20 years ago. The cost index
was 400.
• Now in 2012, consider a power plant with capacity of 600 MW, and the cost
index is 1200.
• Suppose that the cost–capacity factor is 0.79.
• Question: Estimate the cost of building such a plant.
– To apply the power–sizing technique, we first need to find the current cost of
a 200–MW plant:
1200
= $300 million.
cost(200 MW)2012 = cost(200 MW)1992 ×
400
– Then, by the power–sizing technique, we have
cost(600 MW)2012 = cost(200 MW)2012 ×
ECLT 5930/SEEM 5740 (2014–15 Second Term)
600
200
0.79
January 22, 2015
≈ $714 million.
21
CER II: Learning Curve
• In practice, it is observed that workers’ efficiency and organizational performance
improve with repetitive production of a good or service.
• To formalize this observation, we can use the learning curve, which measures
the percentage reduction occurs in, say, labor hours, as the number of units
produced is doubled.
Example
– Suppose that there is a 10% reduction in labor hours when the number of
bicycles produced is doubled.
– Suppose that it takes 10 hours to assemble the first bicycle.
– Then, it takes 10 × 0.9 = 9 hours to assemble the second bicycle, 10 × 0.9 ×
0.9 = 8.1 hours to assemble the fourth bicycle, etc.
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January 22, 2015
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CER II: Learning Curve (Cont’d)
• Let
– Zu = number of input resource units needed to produce the u–th unit;
– K = number of input resource units needed to produce the first unit;
– s = learning curve slope parameter (e.g., s = 0.9 means a 90% learning
curve).
• Then, the definition of learning curve says
Z 2n = K × sn .
• Letting n = log2 u, we have the following equivalent form:
Zu = K × slog2 u.
ECLT 5930/SEEM 5740 (2014–15 Second Term)
January 22, 2015
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Learning Curve: Example
• Suppose that the time needed to assemble the first car is 100 hours, and the
learning curve is 80%.
• Questions:
– What is the time needed to assemble the 10th car?
– The total time required to assemble all 10 cars?
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January 22, 2015
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Learning Curve: Example
• Suppose that the time needed to assemble the first car is 100 hours, and the
learning curve is 80%.
• Questions:
– What is the time needed to assemble the 10th car?
Z10 = 100 × (0.8)log2 10 ≈ 47.651 hours.
– What is the total time required to assemble all 10 cars?
total time = Z1 + Z2 + · · · + Z10 ≈ 631 hours.
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January 22, 2015
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CER III: Functional Regression
• Sometimes we may suspect that the cost and the cost drivers follow certain
functional relationships.
Examples
Type
Linear
Power
Logarithmic
Exponential
Functional Form
cost = b0 + b1x1 + b2x2 + · · · + bk xk
cost = b0 + b1xb111 xb212 + · · ·
cost = b0 + b1 log(x1) + b2 log(x2) + · · · + bk log(xk )
cost = b0 + b1 exp(b11x1 ) + b2 exp(b22x2 ) + · · · + bk exp(bkk xk )
• To use a particular functional form to produce cost estimates, we need to
estimate the coefficients (i.e., b0, b1, . . .) from historical data.
– Problem: The historical data may not fit the chosen functional form exactly.
(Why?)
ECLT 5930/SEEM 5740 (2014–15 Second Term)
January 22, 2015
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Functional Regression: Schematic View
ǫi
xi
f
f (xi )
+
(unknown)
yi = f (xi ) + ǫi
Figure 1: Schematic view of functional regression
ECLT 5930/SEEM 5740 (2014–15 Second Term)
January 22, 2015
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Coefficient Estimation in Linear Regression
• Suppose we decide that there is one cost driver (x) for our product, and the
cost (y) is related to this cost driver via
y = b0 + b1x.
• To determine b0 and b1, we collect past data. We observe that when the cost
driver is set to xi, the cost is given by yi, where i = 1, . . . , n, and n is the total
number of observations.
• As mentioned before, these points may not fall exactly on a line.
• To find the “best” line, we may try to find b0, b1 that minimize the following
penalty function:
n
n
X
X
e2i .
(yi − (b0 + b1xi))2 =
i=1
i=1
This is the well–known least squares approach.
ECLT 5930/SEEM 5740 (2014–15 Second Term)
January 22, 2015
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Coefficient Estimation in Linear Regression (Cont’d)
y
(x6 , y6 )
e6
(x1 , y1 )
e1
0
(x3 , y3 )
(x5 , y5 )
e3
e4
e2
y = b0 + b1 x
(x4 , y4 )
(x2 , y2 )
x
Figure 2: Linear regression and the relevant parameters
ECLT 5930/SEEM 5740 (2014–15 Second Term)
January 22, 2015
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Coefficient Estimation in Linear Regression (Cont’d)
• The solution is given by
b1 =
b0 =
n
1
n
Pn
Pn
Pn
i=1 xi yi − (
i=1 xi ) (
i=1 yi )
,
Pn
Pn
2
2
n i=1 xi − ( i=1 xi)
n
X
i=1
yi − b1
n
X
i=1
!
xi .
• Questions:
– How do we know if this solution is good?
– Are there other approaches besides least squares?
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January 22, 2015
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Model Validation
y
(x3 , y3 )
(x2 , y2 )
(x4 , y4 )
(x5 , y5 )
(x1 , y1 )
(x6 , y6 )
x
0
Figure 3: Mis–specification of the model
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January 22, 2015
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Model Validation (Cont’d)
• To decide whether we have a good fit, we need some statistical tools.
• Two widely used measures include standard error and correlation coefficient.
– Standard error measures the deviation of the actual costs from the predicted
costs.
– Correlation coeffcient measures whether there is in fact a linear relationship
between the cost and the cost driver.
ECLT 5930/SEEM 5740 (2014–15 Second Term)
January 22, 2015
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Presence of Outliers
y
(x3 , y3 )
(x5 , y5 )
(x2 , y2 )
(x4 , y4 )
(x1 , y1 )
x
0
Figure 4: Outliers in the data
ECLT 5930/SEEM 5740 (2014–15 Second Term)
January 22, 2015
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Presence of Outliers (Cont’d)
• In general, it is hard to tell which data points are outliers.
• To mitigate the effect of outliers, one can minimize other penalty functions. For
example, consider minimizing the following:
n
X
i=1
|yi − (b0 + b1xi)| =
n
X
|ei|.
i=1
This is known as the least absolute error approach.
• This approach does not admit a closed–form solution. Nevertheless, the solution
can be found using linear programming.
ECLT 5930/SEEM 5740 (2014–15 Second Term)
January 22, 2015
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What’s Next?
• Assignment: Read Chapter 3 of the course textbook.
• Next: The time value of money (Chapter 4 of the course textbook)
ECLT 5930/SEEM 5740 (2014–15 Second Term)
January 22, 2015
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