CH.8

1
CH.8 Forecasting
Learning objectives After completing this
chapter, you should be able to 1.Explain the
importance of forecasting in organization. 2.Desc
ribe the three major approaches to
forecasting. 3.Measure the accuracy of a
forecast over time.
2
Summary Decision maker rely on forecasting to
reduce uncertainty about the future, for this
reason forecasts play a vital role in
planning. Decision makers have a wide variety of
forecasting models to choose from, these can be
classified as qualitative, projection of
historical pattern, and explanatory models.
3

Glossary Time series A time- ordered sequence of
observation. Trend A long- term movement in
time series data, often linear, but not
necessarily.
4
CH.8 Forecasting If there is considerable
uncertainty , it is much more difficult to
formulate plans that will produce the desired
results than when there is little or no
uncertainty involved. That is Forecasts provide
decision makers with an improved picture of
probable future events. So, forecasts are
important because they, can help to reduce some
of the uncertainty?
5
In fact, forecasts are vital inputs for almost
all planning processes. Forecasts are used
planning the system Product and service
design, process design, capacity planning, and
equipment investment decision. As well as for
planning the use of system Advertising
production and inventory planning and scheduling,
purchasing, planning the size of work force,
budgeting and cost estimation.
6
Good forecasts are more likely to result using
this process 1. Determine the purpose of the
forecast. 2. Determine the time horizon. 3.
Select an appropriate technique. 4. Identify the
necessary date. 5. Make the forecast. 6. Monitor
forecast errors. For forecasting we use
Qualitative forecasts
Quantitative forecasts

We will use this approach
7
Quantitative models
Associative models 1. Simple regression.
Time series models 1. Naïve 2. Moving
averages. 3. Exponential smoothing. 4. Trend.
Lets go to example to see how we could apply
each type
8
Forecasts that use Time series DATA Time
series data are historical values of a variable
that have been intervals. Daily demand, weekly
sales, quarterly revenues, annual
demand. Forecasting techniques that use time
series data typically involve the assumption that
past experience reflects probable future
experience. So, what are the observed patterns
in historical data? These are Trends,
seasonal variation, cyclical variations, and
random variations.
9
Look at figures
Cyclical variation
Random variation around an average
average
Demand
Demand
0
0
1
2
3
1
2
3
4
month
year
Trend with random
Seasonal variation
Demand
Demand
0
0
1
2
3
4
1
2
3
Year
Year
10
Techniques for averaging 1. Naïve forecasts. 2.
Moving averages. 3. Exponential smoothing.
Naïve forecasts Its is
simple , for any period the previous periods
actual value . How? If demand last week was 50
units the forecasts for the upcoming week is 50
units. If the demand in the upcoming week turns
to be 54 units, the forecast for the following
week would be 54 units.
11
Moving averages Uses a number of the most
recent actual data values in generating
forecast. The moving average forecast can be
computed using the following equation MAn
Ai
n
M
i 1
n
Where i the age of the data I
1,2,3,.. n the number of periods in the
moving average. Ai actual value with age i. Go
to example
12
Example 1. Compute a three-period moving
average forecast given demand for shopping
product for the last 5 years
Period
Age
Demand
1
5
40
2
4
44
3
3
36
MA3 36 42 40 38.33
4
2
42
3
5
1
40
If actual demand in period 6 turns out to be 41,
the moving average forecast for period 7 would be

MA3 42 40 41 41.00
3
If we want to make a weight for each data used in
computing moving average ? Go to the example
13
Example 1. Weighted MA. WMA
Ai (w)
M
Weight determined base number of period MA.
w
M
Period
Weight
Demand
If we use WMA3 36(1) 42(2) 40(3) 40.00
1
40
2
44
6
3
3
36
If actual demand in period 6 turns to be 40,the
weighted MA forecast for period 7 would be
4
2
42
5
1
40
Weight
WMA3 42(1) 40(2) 40(3) 40.33
For 7
6
Note - as we move forward we change the weighted
( its arrangement ).
14
Exponential smoothing model Is an averaging
technique that reduces difficulties associated
the previous model . The exponentially smoothed
forecast can be computed using this equation Ft
F X ( A F ) Where
Ft the forecast for period (t) . F t-1 the
forecast for period ( t-1 ) . X
smoothing constraint . A t-1 actual demand or
sales for period ( t-1 ) . Go to understanding
smoothing constant, and how to use it ?
t-1
t-1
t-1
15
The smoothing constant , represent a
percentage of the forecast error . Each new
forecast is equal to the previous forecast plus a
percentage of the previous error. For example
suppose the previous forecast was 100 units,
actual demand was 90 units, and .10 .
The new forecast would be computed as follows Ft
100 .10 ( 90- 100 ) 99 Then, if the next
actual demand turns out to be 102, the next
forecast would be Ft 99 .10 ( 102- 99 )
99.3 For forecasting which a percentage to use
low or high?
16
The sensitivity of forecast adjustment to error
is determined by the smoothing constant, alpha
. The closer its value ( ) is to zero , the
slower the forecast will be to adjust to forecast
errors ( the greater the smoothing ) conversely ,
the closer the value of is to 1.00 the
greater the sensitivity and the less the
smoothing . Selecting a smoothing constant is,
basically, a matter of judgment and trial and
error. If you want to be sure about that use
.05 and .50
Go to the example
17
.05
.50
Period
A
Ft
Error
A
Ft
Error
1
42
-
-
-
-
-
40
2
3
43
4
40
5
41
6
39
7
46
44
8
9
45
38
10
40
11
12
Ft F - ( A t-1 F t-1 )
t-1
Error A Ft
18
For accuracy use mean absolute deviation ( MAD )
equation MAD e ( A
Ft )
M
M
n
n
value with MAD less is the best for
forecasting? Suppose you have these info .
Alternative
MAD
1
26.3
Best accuracy
2
18.2
3
37.0
Complete the previous example to see that works.
19
Techniques for trend Trend is a persistent
upward or downward movement in a time series. The
trend may be linear or nonlinear. We will focus
on linear trends. There are two important
techniques that can be used to develop forecasts
when trend present. 1. Trend equation. 2. Trend-
adjusted exponential smoothing. Go to examples
20
Trend equation a linear trend equation
Yt a bt Where t a specified
number of time periods from t0 Yt the
forecast for period t . a value of Yt
at t0. b slope of the line . The
coefficient of the line, a and b , can be
computed from historical data using these two
equation b n Ty - t y
M
M
M
M
M
n t - ( t)
2
2
M
M
a Y - b t
n
Where n the number of periods
21
Note - Consider the trend equation Yt 45 5t
. The value of Yt when t 0 is 45 , and the
slope of the line is 5 , which means that the
value of Yt will increase by 5 units for each one
unit increase in t. If t 10, the
forecast, Yt, is 45 5(10) 95 units. Example
Month
Actual demand
Mar
112
Apr
125
May
120
Jun
133
Jul
136
Auq
146
Sept
140
Oct
155
Nov
152
1. Plot the data to determine if linear trend
equation is appropriate. 2. Obtain a trend
equation . 3. Forecast demand for the next two
months.
22
Solution 1. The data seem to show an upward,
roughly linear trend .
160
150
Demand
140
130
120
110
Period
1
2
3
4
5
6
7
8
9
Month
y
t
ty
Mar
112
1
112
Apr
125
2
250
May
120
3
360
Jun
133
4
532
Jul
136
5
660
Auq
146
6
876
Sept
140
7
980
Oct
155
8
1240
Nov
152
9
1360
M
M
M
y
t 45
1218
Ty
6398
23
From the table n 9 , t 45 , t
285
2
M
M
b 9 ( 6398 ) 45 ( 1219 ) 2727 t
5.05
540
9 (285) 45 ( 45 )
a 1219 5,05 ( 45 ) 110.19
9
Yt 110.19 5.05 t
3. The next two months, t 10 , t 1 . Y dec
110.19 5.05 ( 10 ) 160.69 . Y jan 110.19
5.05 ( 11 ) 165.74 .
24
Trend- adjusted exponential smoothing Note we
apply this technique when time series includes
trend. The equation for trend adjusted forecast
is TAFt1 St Tt Where St smoothed error
Tt current trend estimate. And St TAF t
( A t TAF t ) Tt T t-1 ( TAF t
TAF t-1 T t-1 )
2
25
Suppose a manager estimates a trend of 10
units, based on observed changes in the first
four data, and uses a starting forecast of 250,
and .5 and .4 if the next actual
value is 255, TAF for the following period would
be computed in this manager St 250 .5 ( 255
250 ) 252.5 Tt 10 .4 (0) Since this is
the initial forecast, no previous error is
available, so the value zero is used. TAF St
Tt 252.5 10 262.5 If the next actual
value is 265 the next TAF would be St 262.5
05 ( 265 262.5 ) 263.75 Tt 10 .4 ( 262.5
250 10 ) 11.00 TAF 263.75 11.00
274.75 We can use another equation for TAF
2
26
FITL F t T t Where FIT t forecast
including trend F t new forecast simple
smoothing . T t trend for period For F t F
t-1 ( A t-1 F t-1 ) For T t ( 1 B )
T t-1 B ( F t F t-1 ) To compute FIT t
follow these steps 1. Find F t . 2. Compute T t
3. Find FIT t 1 2 . Go to example
27
Example You have the following info
F 2 11 .2 ( 12 11 ) 11.2 T 2 ( 1 - .4)
0 .4 ( 11.2 11.0 ) .08 FIT 2 11.2 .08
11.28 And so on for the other periods.
Month
demand
1
12
2
17
3
20
4
19
5
24
6
26
Assume the initial forecast for month first was
11 unit
7
31
8
32
9
36
.2 B .4
M
d
Ft
Tt
FITt
1
12
11
0 .08 .51 .92 .96 1.30
1.52 1.91 2.02
- 11.28 12.87 14.81 15.87 18.03
20.10 22.98 25.27
2
17
11.2
3
20
12.56 13.89 14.91 16.75 18.58 21.07 23.25
4
19
5
24
6
26
7
31
8
32
9
36
28
Techniques for seasonality Seasonal variation in
time series data are regularly upward or downward
movements. Weather variation and its impact on
travel sports product. To compute seasonal
relative, use this model Centered moving
average, which based on Identical to a moving
average forecast as are the resulting values.
However the values are not projected as in a
forecast, but, instead, they are positioned in
the middle of the periods to compute
average. Go to example
29
Period
Demand
Three-period centered average
1
40
40 46 42
2
46
Average
42.67
3
3
42
It would be positioned of period 2, the average
representative of the series at that point. The
ratio of demand at period 2 to this centered
average at period 2 is an estimate of the
seasonal relative at that point. Because the
ratio is 46 1.08 the series is about 8
above average at this point. What if both trend
and seasonality appear in a series?
42.67
30
Trend seasonality Suppose a furniture
manufacturer wants to predict quarterly demand
for a certain seat for period 15 and 16, which
happen to be the second and third quarters of a
particular year. The series consists of both
trend and seasonality. The trend portion of
demand can be projected the equation Yt 124
7.5t . Quarter relatives are Q1 1.20 Q2
1.10 Q3 .75 Q4 .95 Use the info.
To predict demand for period 15 and 16. Solution
The trend values at t 15 and t 16 Y15
124 7.5 ( 15 ) 235.5 Y 16 124 7.5 ( 16 )
244.0 Period 15 235.5 ( 1.10 )
259.05 Period 16 244.0 ( .75 )
183.00 Quarter relative
31
Forecasts that use explanatory model These
model incorporate one or more variables that are
related to the variable of interest and,
therefore, they can be used to predict future
values of that variable. Are called regression
models We are going to use the simple linear
regression. For multiple regression go to the
computer?
32
Simple linear regression Consists two variable
The variable to be forecasted is dependent
variable ? The other value used to explain or
predict the dependent variable is called
independent variable ? To make forecast , the two
variables must be expressed in a linear equation
of the form Y a bx Where Y the
dependent variable . X the independent
variable. B slope of the line . b n x y
- x y
M
M
M
M
M
A y - b x y bx
2
2
M
M
N x - ( x )
n
33
Note In order to successfully use this approach
, it is necessary to be able to 1. Identify an
appropriate independent variable . 2. Obtain a
sample of at least 10 observations. 3. Develop an
equation . 4. Identify any restriction on
prediction . 5. Measure accuracy in a given
forecast . Go to the example ??
34
Conceder, for example , 10 paired observations
about sales and ads. ( millions of )
observation
Ads expenditure, x
Sales , y
7
1
1.1
8
2
1.4
3
1.4
10
4
2
10
5
0.9
7
6
1.6
10
7
2
11
8
1.7
11
9
1.2
9
10
0.8
6
You can plot the pairs to see if a linear
relationship is appropriate,?? It appear / linear
relationship
35
Solution Calculations for regression
coefficients
2
2
observation
Ads expenditure, x
Sales , y
x y x y
1
1.1
7
7.7
1.21
49
2
1.4
8
11.2
1.96
64
3
1.4
10
14.0
1.69
100
4
2
10
20.0
4.00
100
5
0.9
7
6.3
0.81
49
6
1.6
10
16.0
2.56
100
7
2
11
22.0
4.00
121
8
1.7
11
18.7
2.89
121
9
1.2
9
10.8
1.44
81
10
0.8
6
4.8
0.64
36
14.1 89 131.5
21.47 821
b 10 ( 131.5 ) 14.1 ( 89 ) 60.1
3.78
15.89
10 ( 21.47 ) 14.1 ( 14.1 )
a 89 3.78 ( 14.1 ) 3.57
10
y 3.57 3.87
36
Before we go in prediction process ,we must
measure the strength of the relationship between
the two variable . For that, we use correlation
coefficient ?
M
M
M
r n x y - x y
2
2
M
M
M
M
2
n ( x ) - ( x ) X n ( y )
- ( y )
2
r 10 ( 131.5 ) - ( 14.1 )
( 89 )
10 (21.47 ) - ( 14.1 ) X 10 ( 821 ) - ( 89
)
.887
The same as the sign of b . Relationship is good
to use for prediction .
37
2
Square of the correlation coefficient, r , can be
used as a measure of goodness of fit the closer
the value is to 1.00 the better the fit, for Ads
data r 0.787 this means that .79 of the sum
of the squared deviation of Y value around the
mean of the Y . We say that the independent
variable explain .79 of the variation . There
are a number of important questions that must be
answered if a regression equation is to be used
for forecasting ??
2
38
1. Is the relationship significant ? 2. How
accurate will a particular forecast probably be
? These questions and other depend on the extend
to which the points scatter around the regression
line . This is measured by standard error of
estimate, se, Se n 2 / because we have
two coefficient a and b
M
M
M
y - a y - b x y
2
n 2
39
For our example Se
821 3.57 (89) 3.78 ( 131.5 )
.880
10 2
2
Note if the value of r is less than , say
.40 the computed relationship will not be strong
enough to yield acceptable forecast accuracy. To
test for the existence of relationship ( than the
slope is non zero ) Compute the ration
T test b
S b
Where b slope of regression line . S b Se
1
2
2
M
M
X - ( x )
n
40
For our example S b .880
1
.629
.698
.880
2
21.47 ( 14.1 )
10
Compare this value with value of t .025 for n 2
/ from the table of T test
T test 3.78 5.42
.698
n 10 11 12 13 14 15 16 17 18 19 20 21 22
t 2.31 2.26 2.23 2.20 2.18 2.16 2.15 2.13 2.12
2.11 2.10 2.09 2.08
If the test is greater than the table value, as
it is in our case ( 5.42 ), it can be concluded
that the relationship is significant. Not due to
the chance ??
41
How accurate will a given predication be ? Here
we should use the regression equation to compute
expected value of the dependent variable for a
selected or given value of X the construct a 95
confidence interval around that value . Suppose
expected value of X is 1.3 ? Y t S reg
Where Y a b x t from the T
test table
S reg
2
( x g x )
1 1
n
2
2
x - ( x )
M
M
n
X g value of x to be used
42
X 1.3 , X 14.1 1.41
10
S reg .880
2
( 1.3 1.41 )
1 1
.926
2
10
21.47 - (14.1)
10
For our case X g 1.3 , 95 confidence
interval For the predicted value of y is Y a
b x 3.57 3.78 (1.3) 8.48 95
interval 8.48 2.31 ( .926 )
8.48 2.14 or
6.34 to 10.62
for test table .