Operations
Management
Forecasting
Chapter 4
4-1
Learning Objectives
When you complete this chapter, you should be able
to :
Identify or Define:
Forecasting
 Types of forecasts
 Time horizons
 Approaches to forecasts

4-2
Learning Objectives - continued
When you complete this chapter, you should be
able to :
Describe or Explain:
Moving averages
 Exponential smoothing
 Trend projections
 Regression and correlation analysis
 Measures of forecast accuracy

4-3
What is Forecasting?
 Process of predicting a
future event
 Underlying basis of
all business decisions




Production
Inventory
Personnel
Facilities
4-4
Types of Forecasts by Time
Horizon
Short-range forecast
Up to 1 year; usually less than 3 months
 Job scheduling, worker assignments

Medium-range forecast
3 months to 3 years
 Sales & production planning, budgeting

Long-range forecast
 3+ years

New product planning, facility location
4-5
Short-term vs. Longer-term Forecasting
Medium/long range forecasts deal with more
comprehensive issues and support
management decisions regarding planning and
products, plants and processes.
Short-term forecasting usually employs different
methodologies than longer-term forecasting
Short-term forecasts tend to be more accurate
than longer-term forecasts.
4-6
Influence of Product Life Cycle
Introduction, Growth, Maturity, Decline
 Stages of introduction and growth require longer
forecasts than maturity and decline
 Forecasts useful in projecting



staffing levels,
inventory levels, and
factory capacity
as product passes through life cycle stages
4-7
Types of Forecasts
Economic forecasts

Address business cycle, e.g., inflation rate, money
supply etc.
Technological forecasts
Predict rate of technological progress
 Predict acceptance of new product

Demand forecasts

Predict sales of existing product
PowerPoint presentation to accompany Heizer/Render –
Principles of Operations Management, 5e, and Operations
Management, 7e
4-8
© 2004 by Prentice Hall, Inc., Upper Saddle River, N.J. 07458
Seven Steps in Forecasting
Determine the use of the forecast
Select the items to be forecasted
Determine the time horizon of the forecast
Select the forecasting model(s)
Gather the data
Make the forecast
Validate and implement results
4-9
Product Demand Charted over 4
Years with Trend and Seasonality
Demand for product or service
Seasonal peaks
Trend component
Actual
demand line
Random
variation
Year
1
Year
2
4-10
Average demand
over four years
Year
3
Year
4
Actual Demand, Moving Average,
Weighted Moving Average
35
Sales Demand
30
25
Weighted moving average
Actual sales
20
15
10
5
Moving average
0
Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec
Month
4-11
Realities of Forecasting
Forecasts are seldom perfect
Most forecasting methods assume that there is
some underlying stability in the system
Both product family and aggregated product
forecasts are more accurate than individual
product forecasts
4-12
Forecasting Approaches
Qualitative Methods
Quantitative Methods
 Used when situation is
vague & little data exist
 Used when situation is
‘stable’ & historical data
exist
 New products
 New technology
 Existing products
 Current technology
 Involves intuition,
experience
 Involves mathematical
techniques
 e.g., forecasting sales on
Internet
 e.g., forecasting sales of
color televisions
4-13
Overview of Qualitative Methods
Jury of executive opinion

Pool opinions of high-level executives, sometimes
augment by statistical models
Delphi method

Panel of experts, queried iteratively
Sales force composite

Estimates from individual salespersons are reviewed
for reasonableness, then aggregated
Consumer Market Survey

Ask the customer
4-14
Jury of Executive Opinion
 Involves small group of high-level managers

Group estimates demand by working together
 Combines managerial experience with statistical
models
 Relatively quick
 ‘Group-think’
disadvantage
4-15
Sales Force Composite
 Each salesperson projects
his or her sales
 Combined at district &
national levels
 Sales reps know
customers’ wants
 Tends to be overly
optimistic
4-16
Delphi Method
Iterative group
process
3 types of people
Decision makers
 Staff
 Respondents

Decision Makers
Staff
(What will
(Sales?)
(Sales will be 50!)
sales be?
survey)
Reduces ‘group-think’
Respondents
(Sales will be 45, 50, 55)
4-17
Consumer Market Survey
 Ask customers about
purchasing plans
 What consumers say,
and what they
actually do are often
different
 Sometimes difficult to
answer
How many hours will
you use the Internet
next week?
4-18
Overview of Quantitative Approaches
Naïve approach
Moving averages
Exponential smoothing
Trend projection
Time-series
Models
Linear regression
Associative
models
4-19
Quantitative Forecasting Methods
(Non-Naive)
Quantitative
Forecasting
Associative
Models
Time Series
Models
Moving
Average
Exponential
Smoothing
Trend
Projection
4-20
Linear
Regression
What is a Time Series?

Set of evenly spaced numerical data


Forecast based only on past values


Obtained by observing response variable at regular time periods
Assumes that factors influencing past and present will continue
influence in future
Example
Year:
Sales:
1998
78.7
1999
63.5
4-21
2000
89.7
2001
93.2
2002
92.1
Time Series Components
Trend
Cyclical
Seasonal
Random
4-22
Trend Component
Persistent, overall upward or downward pattern
Due to population, technology etc.
Several years duration
Response
Mo., Qtr., Yr.
4-23
Seasonal Component
Regular pattern of up & down fluctuations
Due to weather, customs etc.
Occurs within 1 year
Summer
Response
Mo., Qtr.
4-24
Common Seasonal Patterns
Period of
Pattern
“Season”
Length
Week
Day
Number of
“Seasons” in
Pattern
7
Month
Week
4–4½
Month
Day
28 – 31
Year
Quarter
4
Year
Month
12
Year
Week
52
4-25
Cyclical Component
Repeating up & down movements
Due to interactions of factors influencing economy
Usually 2-10 years duration
Cycle
Response

Mo., Qtr., Yr.
PowerPoint presentation to accompany Heizer/Render –
Principles of Operations Management, 5e, and Operations
Management, 7e
4-26
© 2004 by Prentice Hall, Inc., Upper Saddle River, N.J. 07458
Random Component
Erratic, unsystematic, ‘residual’ fluctuations
Due to random variation or unforeseen events

Union strike

Tornado
Short duration &
nonrepeating
4-27
General Time Series Models
Any observed value in a time series is the
product (or sum) of time series components
Multiplicative model

Yi = Ti · Si · Ci · Ri (if quarterly or mo. data)
Additive model

Yi = Ti + Si + Ci + Ri (if quarterly or mo. data)
4-28
Naive Approach
 Assumes demand in next
period is the same as demand
in most recent period

e.g., If May sales were 48, then
June sales will be 48
 Sometimes cost effective &
efficient
4-29
Moving Average Method
 MA is a series of arithmetic means
 Used if little or no trend
 Used often for smoothing

Provides overall impression of data over time
 Equation
Demand in Previous n Periods

MA 
n
4-30
Moving Average Example
You’re manager of a museum store that sells
historical replicas. You want to forecast sales
(000) for 2003 using a 3-period moving average.
1998
4
1999
6
2000
5
2001
3
2002
7
4-31
Moving Average Solution
Time
1998
1999
2000
2001
2002
2003
Response
Yi
4
6
5
3
7
Moving
Total
(n=3)
NA
NA
NA
4+6+5=15
NA
4-32
Moving
Average
(n=3)
NA
NA
NA
15/3 = 5
Moving Average Solution
Time
1998
1999
2000
2001
2002
2003
Response
Yi
4
6
5
3
7
Moving
Total
(n=3)
NA
NA
NA
4+6+5=15
6+5+3=14
NA
4-33
Moving
Average
(n=3)
NA
NA
NA
15/3 = 5
14/3=4 2/3
Moving Average Solution
Time
1998
1999
2000
2001
2002
2003
Response
Yi
4
6
5
3
7
NA
Moving
Total
(n=3)
NA
NA
NA
4+6+5=15
6+5+3=14
5+3+7=15
4-34
Moving
Average
(n=3)
NA
NA
NA
15/3=5.0
14/3=4.7
15/3=5.0
Moving Average Graph
Sales
8
Actual
6
Forecast
4
2
95
96
97 98
Year
4-35
99
00
Weighted Moving Average Method
Used when trend is present

Older data usually less important
Weights based on intuition

Often lay between 0 & 1, & sum to 1.0
Equation
WMA =
Σ(Weight for period n) (Demand in period n)
ΣWeights
4-36
Actual Demand, Moving Average,
Weighted Moving Average
35
Sales Demand
30
25
Weighted moving average
Actual sales
20
15
10
5
Moving average
0
Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec
Month
4-37
Disadvantages of
Moving Average Methods
Increasing n makes forecast less
sensitive to changes
Do not forecast trend well
Require much historical
data
4-38
Exponential Smoothing Method
Form of weighted moving average
Weights decline exponentially
 Most recent data weighted most

Requires smoothing constant ()
Ranges from 0 to 1
 Subjectively chosen

Involves little record keeping of past data
4-39
Exponential Smoothing Equations
 Ft = At - 1 + (1-)At - 2 + (1- )2·At - 3
+ (1- )3At - 4 + ... + (1- )t-1·A0
Ft = Forecast value
 At = Actual value
  = Smoothing constant

 Ft = Ft-1 + (At-1 - Ft-1)

Use for computing forecast
4-40
Exponential Smoothing Example
During the past 8 quarters, the Port of Baltimore has unloaded large
quantities of grain. ( = .10). The first quarter forecast was 175..
Quarter
Actual
1
2
3
4
5
6
7
8
9
180
168
159
175
190
205
180
182
?
Find the forecast
for the 9th quarter.
4-41
Exponential Smoothing Solution
Ft = Ft-1 + 0.1(At-1 - Ft-1)
Quarter
Actual
1
180
2
168
3
159
4
175
5
190
6
205
Forecast, F t
(α = .10)
175.00 (Given)
175.00 +
4-42
Exponential Smoothing Solution
Ft = Ft-1 + 0.1(At-1 - Ft-1)
Forecast, F t
(α = .10)
Quarter Actual
1
180
2
168
3
159
4
175
5
190
6
205
175.00 (Given)
175.00 + .10(
4-43
Exponential Smoothing Solution
Ft = Ft-1 + 0.1(At-1 - Ft-1)
Quarter
Actual
1
180
2
168
3
159
4
175
5
190
6
205
Forecast, Ft
(α = .10)
175.00 (Given)
175.00 + .10(180 -
4-44
Exponential Smoothing Solution
Ft = Ft-1 + 0.1(At-1 - Ft-1)
Forecast, Ft
(α = .10)
Quarter Actual
1
180
2
168
3
159
4
175
5
190
6
205
175.00 (Given)
175.00 + .10(180 - 175.00)
4-45
Exponential Smoothing Solution
Ft = Ft-1 + 0.1(At-1 - Ft-1)
Forecast, Ft
(α = .10)
Quarter Actual
1
180
2
168
3
159
4
175
5
190
6
205
175.00 (Given)
175.00 + .10(180 - 175.00) = 175.50
4-46
Exponential Smoothing Solution
Ft = Ft-1 + 0.1(At-1 - Ft-1)
Forecast, F t
(α = .10)
Quarter
Actual
1
180
2
168
175.00 + .10(180 - 175.00) = 175.50
3
159
175.50 + .10(168 - 175.50) = 174.75
4
175
5
190
6
205
175.00 (Given)
4-47
Exponential Smoothing Solution
Ft = Ft-1 + 0.1(At-1 - Ft-1)
Forecast, F t
(α = .10)
Quarter Actual
1995
180
175.00 (Given)
1996
168
175.00 + .10(180 - 175.00) = 175.50
1997
159
175.50 + .10(168 - 175.50) = 174.75
1998
175
174.75 + .10(159 - 174.75)= 173.18
1999
190
2000
205
4-48
Exponential Smoothing Solution
Ft = Ft-1 + 0.1(At-1 - Ft-1)
Forecast, F t
(α = .10)
Quarter Actual
1
180
175.00 (Given)
2
168
175.00 + .10(180 - 175.00) = 175.50
3
4
159
175.50 + .10(168 - 175.50) = 174.75
175
174.75 + .10(159 - 174.75) = 173.18
5
190
173.18 + .10(175 - 173.18) = 173.36
6
205
4-49
Exponential Smoothing Solution
Ft = Ft-1 + 0.1(At-1 - Ft-1)
Forecast, F t
(α = .10)
Quarter Actual
1
180
175.00 (Given)
2
168
175.00 + .10(180 - 175.00) = 175.50
3
159
175.50 + .10(168 - 175.50) = 174.75
4
175
174.75 + .10(159 - 174.75) = 173.18
5
190
173.18 + .10(175 - 173.18) = 173.36
6
205
173.36 + .10(190 - 173.36) = 175.02
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Principles of Operations Management, 5e, and Operations
Management, 7e
4-50
© 2004 by Prentice Hall, Inc., Upper Saddle River, N.J. 07458
Exponential Smoothing Solution
Ft = Ft-1 + 0.1(At-1 - Ft-1)
Actual
Forecast, F t
(α = .10)
4
175
174.75 + .10(159 - 174.75) = 173.18
5
190
173.18 + .10(175 - 173.18) = 173.36
6
7
205
180
173.36 + .10(190 - 173.36) = 175.02
175.02 + .10(205 - 175.02) = 178.02
Time
8
9
4-51
Exponential Smoothing Solution
Ft = Ft-1 + 0.1(At-1 - Ft-1)
Time
Forecast, F t
(α = .10)
Actual
4
175
174.75 + .10(159 - 174.75) = 173.18
5
190
173.18 + .10(175 - 173.18) = 173.36
6
7
205
180
8
9
182
?
173.36 + .10(190 - 173.36) = 175.02
175.02 + .10(205 - 175.02) = 178.02
178.02 + .10(180 - 178.02) = 178.22
178.22 + .10(182 - 178.22) = 178.58
4-52
Forecast Effects of
Smoothing Constant 
Ft =  At - 1 + (1- )At - 2 + (1- )2At - 3 + ...
Weights
=
Prior Period
2 periods ago 3 periods ago

= 0.10
(1 - )
10%
= 0.90
4-53
(1 - )2
Forecast Effects of
Smoothing Constant 
Ft =  At - 1 + (1- ) At - 2 + (1- )2At - 3 + ...
Weights
=
= 0.10
Prior Period
2 periods ago 3 periods ago

(1 - )
10%
9%
= 0.90
4-54
(1 - )2
Forecast Effects of
Smoothing Constant 
Ft =  At - 1 + (1- )At - 2 + (1- )2At - 3 + ...
Weights
=
= 0.10
Prior Period
2 periods ago 3 periods ago

(1 - )
(1 - )2
10%
9%
8.1%
= 0.90
4-55
Forecast Effects of
Smoothing Constant 
Ft =  At - 1 + (1- )At - 2 + (1- )2At - 3 + ...
Weights
=
Prior Period
2 periods ago 3 periods ago

(1 - )
(1 - )2
= 0.10
10%
9%
8.1%
= 0.90
90%
4-56
Forecast Effects of
Smoothing Constant 
Ft =  At - 1 + (1- ) At - 2 + (1- )2At - 3 + ...
Weights
=
Prior Period
2 periods ago 3 periods ago

(1 - )
(1 - )2
= 0.10
10%
9%
8.1%
= 0.90
90%
9%
4-57
Forecast Effects of
Smoothing Constant 
Ft =  At - 1 + (1- ) At - 2 + (1- )2At - 3 + ...
Weights
=
= 0.10
= 0.90
Prior Period
2 periods ago 3 periods ago

(1 - )
(1 - )2
10%
9%
8.1%
90%
9%
0.9%
4-58
Choosing 
Seek to minimize the Mean Absolute Deviation (MAD)
If:
Then:
Forecast error = demand - forecast
MAD 
 forecast errors
n
4-59
Exponential Smoothing with
Trend Adjustment
Forecast including trend (FITt)
= exponentially smoothed forecast (Ft)
+ exponentially smoothed trend (Tt)
4-60
Exponential Smoothing with
Trend Adjustment - continued
or
Ft = Last period’s forecast
+ (Last period’s actual – Last period’s forecast)
Ft = Ft-1 +  (At-1 – Ft-1)
Tt = (Forecast this period - Forecast last period)
+ (1-)(Trend estimate last period
or
Tt = (Ft - Ft-1) + (1- )Tt-1
4-61
Exponential Smoothing with
Trend Adjustment - continued
Ft = exponentially smoothed forecast of the data
series in period t
Tt = exponentially smoothed trend in period t
At = actual demand in period t
 = smoothing constant for the average
 = smoothing constant for the trend
4-62
Comparing Actual and Forecasts
40
35
Actual
Demand
30
Demand
25
20
15
Smoothed
Forecast
Forecast including
trend
10
Smoothed Trend
5
0
1
2
3
4
5
6
Month
4-63
7
8
9
10
Regression
4-64
Least Squares
Values of Dependent Variable
Actual
observation
Deviation
Deviation
Deviation
Deviation
Deviation
Deviation
Deviation
Yˆ  a  bx
Time
4-65
Point on
regression
line
Linear Trend Projection
 Used for forecasting linear trend line
 Assumes relationship between response
variable, Y, and time, X, is a linear function
Yi  a  bX i
 Estimated by least squares method

Minimizes sum of squared errors
4-66
Least Squares Equations
Equation:
Ŷi  a  bx i
n
Slope:
 x i y i  nx y
b  i n
 x i  nx 
i 
Y-Intercept:
a  y  bx
4-68
Computation Table
Xi
X1
Yi
Y1
2
Xi
X1
2
2
X2
Y2
X2
:
:
:
Xn
ΣX i
2
X iY i
Y1
2
X 1Y 1
Y2
2
X 2Y 2
Yi
:
2
Yn
Xn
ΣYi
2
ΣX i
4-69
:
2
X nY n
2
ΣY i
ΣX iY i
Yn
Using a Trend Line
Year
1997
1998
1999
2000
2001
2002
2003
The demand for
electrical power at
N.Y.Edison over the
years 1997 – 2003 is
given at the left. Find
the overall trend.
Demand
74
79
80
90
105
142
122
4-70
Finding a Trend Line
Year
1997
1998
1999
2000
2001
2002
2003
Time
Power
x2
xy
Period Demand
1
74
1
74
2
79
4
158
3
80
9
240
4
90
16
360
5
105
25
525
6
142
36
852
7
122
49
854
x=28 y=692 x2=140 xy=3,063
4-71
The Trend Line Equation
x
Σx 28

4
n
7
b
Σxy - nxy 3,063  (7)(4)(98. 86) 295


 10.54
2
2
2
28
Σx  nx
140  (7)(4)
y
Σy 692

 98.86
n
7
a  y - bx  98.86 - 10.54(4)  56.70
Demand in 2004  56.70  10.54(8)  141.02 megawatts
Demand in 2005  56.70  10.54(9)  151.56 megawatts
4-72
Multiplicative Seasonal Model
 Find average historical demand for each “season” by summing
the demand for that season in each year, and dividing by the
number of years for which you have data.
 Compute the average demand over all seasons by dividing the
total average annual demand by the number of seasons.
 Compute a seasonal index by dividing that season’s historical
demand (from step 1) by the average demand over all seasons.
 Estimate next year’s total demand
 Divide this estimate of total demand by the number of seasons,
then multiply it by the seasonal index for that season. This
provides the seasonal forecast.
4-73
Linear Regression Model
Shows linear relationship between dependent &
explanatory variables

Example: Sales & advertising (not time)
Y-intercept
Slope
^
Yi = a + bX i
Dependent
(response) variable
Independent (explanatory)
variable
4-74
Linear Regression Equations
Equation:
Ŷi  a  bx i
n
Slope:
b 
 x i y i  nx y
i 1
n
 x i2  nx 2
i 1
Y-Intercept:
a  y  bx
4-76
Computation Table
Xi
X1
Yi
2
Xi
2
Yi
X iY i
Y1
X1
2
Y1
2
X 1Y 1
2
Y2
2
X 2Y 2
X2
Y2
X2
:
:
:
Xn
ΣXi
:
2
Yn
Xn
ΣYi
2
ΣXi
4-77
:
2
X nY n
2
ΣYi
Σ X iY i
Yn
Interpretation of Coefficients
Slope (b)

Estimated Y changes by b for each 1 unit increase in X
 If b = 2, then sales (Y) is expected to increase by 2 for each 1
unit increase in advertising (X)
Y-intercept (a)

Average value of Y when X = 0
 If a = 4, then average sales (Y) is expected to be
advertising (X) is 0
4-78
4 when
Random Error Variation
Variation of actual Y from predicted Y
Measured by standard error of estimate
Sample standard deviation of errors
 Denoted SY,X

Affects several factors
Parameter significance
 Prediction accuracy

4-79
Least Squares Assumptions
Relationship is assumed to be linear. Plot the
data first - if curve appears to be present, use
curvilinear analysis.
Relationship is assumed to hold only within or
slightly outside data range. Do not attempt to
predict time periods far beyond the range of the
data base.
Deviations around least squares line are
assumed to be random.
4-80
Standard Error of the Estimate
n
2


y

y
 i c
S y,x 
i 1
n2
n

y
i 1
2
i
n
n
i 1
i 1
 a  y i  b xi y i
n2
4-81
Correlation
Answers: ‘how strong is the linear relationship
between the variables?’
Coefficient of correlation Sample correlation
coefficient denoted r
Values range from -1 to +1
 Measures degree of association

Used mainly for understanding
4-82
Sample Coefficient of Correlation
r
n
n
n
i 
i 
i 
n  x i yi   x i  yi
 n   n   n   n  
n  x i    x i   n  yi    yi  
 i     i 
 i   
 i 
4-83
Coefficient of Correlation and
Regression Model
Y
r=1
Y
Y^i = a + b X i
r = -1
Y^i = a + b X i
X
Y
X
r = .89
Y^i = a + b X i
Y
r=0
Y^i = a + b X i
X
r2 = square of correlation coefficient (r), is the percent of the
variation in y that is explained by the regression equation
4-85
X
Guidelines for Selecting
Forecasting Model
You want to achieve:

No pattern or direction in forecast error
^
 Error = (Y - Y ) = (Actual - Forecast)
i
i
 Seen in plots of errors over time

Smallest forecast error
 Mean square error (MSE)
 Mean absolute deviation (MAD)
4-86
Pattern of Forecast Error
Trend Not Fully
Accounted for
Desired Pattern
Error
Error
0
0
Time (Years)
Time (Years)
4-87
Forecast Error Equations
 Mean Square Error (MSE)
 (y  ŷ )  forecast
n
2
MSE 
i 1
i
i
n

errors
2
n
 Mean Absolute Deviation (MAD)
 | y  yˆ |  | forecast
n
MAD 
i
i 1
n
i

errors |
n
 Mean Absolute Percent Error (MAPE)
n
MAPE  100

i 1
actual i  forecast i
actual i
n
4-88
Selecting Forecasting Model
Example
You’re a marketing analyst for Hasbro Toys. You’ve forecast sales with a linear
model & exponential smoothing. Which model do you use?
Actual
Linear Model
Year
Sales
Forecast
Exponential
Smoothing
Forecast (.9)
1998
1999
2000
2001
2002
1
1
2
2
4
0.6
1.3
2.0
2.7
3.4
1.0
1.0
1.9
2.0
3.8
4-89
Linear Model Evaluation
Year
Yi
Y^ i
1998
1999
2000
2001
2002
1
1
2
2
4
0.6
1.3
2.0
2.7
3.4
Total
Error Error2
|Error|
0.4
-0.3
0.0
-0.7
0.6
0.16
0.09
0.00
0.49
0.36
0.4
0.3
0.0
0.7
0.6
0.0
1.10
2.0
|Error|
Actual
0.40
0.30
0.00
0.35
0.15
1.20
MSE = Σ Error2 / n = 1.10 / 5 = 0.220
MAD = Σ |Error| / n = 2.0 / 5 = 0.400
MAPE = 100 Σ|absolute percent
4-90 errors|/n= 1.20/5 = 0.240
Exponential Smoothing Model
Evaluation
Year
Y
1998
1999
2000
2001
2002
1
1
2
2
4
i
Y^
i
1.0
1.0
1.9
2.0
3.8
Total
Error
Error2
|Error|
0.0
0.0
0.1
0.0
0.2
0.00
0.00
0.01
0.00
0.04
0.0
0.0
0.1
0.0
0.2
0.3
0.05
0.3
|Error|
Actual
0.00
0.00
0.05
0.00
0.05
0.10
MSE = Σ Error2 / n = 0.05 / 5 = 0.01
MAD = Σ |Error| / n = 0.3 / 5 = 0.06
MAPE = 100 Σ |Absolute percent errors|/n = 0.10/5 = 0.02
PowerPoint presentation to accompany Heizer/Render –
Principles of Operations Management, 5e, and Operations
Management, 7e
4-91
© 2004 by Prentice Hall, Inc., Upper Saddle River, N.J. 07458
Exponential Smoothing Model
Evaluation
Linear Model:
MSE = Σ Error2 / n = 1.10 / 5 = .220
MAD = Σ |Error| / n = 2.0 / 5 = .400
MAPE = 100 Σ|absolute percent errors|/n= 1.20/5 = 0.240
Exponential Smoothing Model:
MSE = Σ Error2 / n = 0.05 / 5 = 0.01
MAD = Σ |Error| / n = 0.3 / 5 = 0.06
MAPE = 100 Σ |Absolute percent errors|/n = 0.10/5 = 0.02
4-92
Tracking Signal
Measures how well the forecast is predicting
actual values
Ratio of running sum of forecast errors (RSFE)
to mean absolute deviation (MAD)

Good tracking signal has low values
Should be within upper and lower control limits
4-93
Tracking Signal Equation
RSFE
TS 
MAD
n
 y i
 ŷ i 
 i 
MAD

 forecast
MAD
4-94
error
Tracking Signal Computation
Mo Fcst
Act Error RSFE Abs Cum MAD
Error |Error|
1
100
90
2
100
95
3
100 115
4
100 100
5
100 125
6
100 140
4-95
TS
Tracking Signal Computation
Mo Forc
Act Error RSFE Abs Cum MAD
Error |Error|
1
100
90
2
100
95
3
100 115
4
100 100
5
100 125
6
100 140
-10
Error = Actual - Forecast
= 90 - 100 = -10
4-96
TS
Tracking Signal Computation
Mo Forc
Act Error RSFE Abs Cum MAD
Error |Error|
1
100
90
2
100
95
3
100 115
4
100 100
5
100 125
6
100 140
-10
-10
RSFE =  Errors
= NA + (-10) = -10
4-97
TS
Tracking Signal Computation
Mo Forc
Act Error RSFE Abs Cum MAD
Error |Error|
1
100
90
2
100
95
3
100 115
4
100 100
5
100 125
6
100 140
-10
-10
10
Abs Error = |Error|
= |-10| = 10
4-98
TS
Tracking Signal Computation
Mo Forc
Act Error RSFE Abs Cum MAD
Error |Error|
1
100
90
2
100
95
3
100 115
4
100 100
5
100 125
6
100 140
-10
-10
10
TS
10
Cum |Error| =  |Errors|
= NA + 10 = 10
4-99
Tracking Signal Computation
Mo Forc
Act Error RSFE Abs Cum MAD
Error |Error|
1
100
90
2
100
95
3
100 115
4
100 100
5
100 125
6
100 140
-10
-10
10
10 10.0
MAD =  |Errors|/n
= 10/1 = 10
4-100
TS
Tracking Signal Computation
Mo Forc
Act Error RSFE Abs Cum MAD
Error |Error|
1
100
90
2
100
95
3
100 115
4
100 100
5
100 125
6
100 140
-10
-10
10
10 10.0
TS = RSFE/MAD
= -10/10 = -1
4-101
TS
-1
Tracking Signal Computation
Mo Forc
Act Error RSFE Abs Cum MAD
Error |Error|
1
100
90
-10
2
100
95
-5
3
100 115
4
100 100
5
100 125
6
100 140
-10
10
10 10.0
Error = Actual - Forecast
= 95 - 100 = -5
4-102
TS
-1
Tracking Signal Computation
Mo Forc
Act Error RSFE Abs Cum MAD
Error |Error|
1
100
90
-10
-10
2
100
95
-5
-15
3
100 115
4
100 100
5
100 125
6
100 140
10
10 10.0
RSFE =  Errors
= (-10) + (-5) = -15
4-103
TS
-1
Tracking Signal Computation
Mo Forc
Act Error RSFE Abs Cum MAD
Error |Error|
1
100
90
-10
-10
10
2
100
95
-5
-15
5
3
100 115
4
100 100
5
100 125
6
100 140
10 10.0
Abs Error = |Error|
= |-5| = 5
4-104
TS
-1
Tracking Signal Computation
Mo Forc
Act Error RSFE Abs Cum MAD
Error |Error|
1
100
90
-10
-10
10
2
100
95
-5
-15
5
3
100 115
4
100 100
5
100 125
6
100 140
10 10.0
15
Cum Error =  |Errors|
= 10 + 5 = 15
4-105
TS
-1
Tracking Signal Computation
Mo Forc
Act Error RSFE Abs Cum MAD
Error |Error|
1
100
90
-10
-10
10
2
100
95
-5
-15
5
3
100 115
4
100 100
5
100 125
6
100 140
10 10.0
15
MAD =  |Errors|/n
= 15/2 = 7.5
4-106
7.5
TS
-1
Tracking Signal Computation
Mo Forc
Act Error RSFE Abs Cum MAD
Error |Error|
1
100
90
-10
-10
10
2
100
95
-5
-15
5
3
100 115
4
100 100
5
100 125
6
100 140
10 10.0
-1
15
-2
TS = RSFE/MAD
= -15/7.5 = -2
4-107
TS
7.5
Forecasting in the Service Sector
Presents unusual challenges
special need for short term records
 needs differ greatly as function of industry and product
 issues of holidays and calendar
 unusual events

4-108