CHAPTER - 3

FORECASTING

4-1
 What is Forecasting?
 A prediction, projection, or
estimate of some future
activity, event, or
occurrence.
??
 Underlying basis of all
business decisions
 Production
 Inventory
 Personnel
 Facilities
4-2
 Types Forecasting
 Economic forecasts
• Predict a variety of economic indicators,
like money supply, inflation rates,
interest rates, etc.
 Technological forecasts
• Predict rates of technological progress
and innovation.
 Demand forecasts
• Predict the future demand for a
company’s products or services. 4-3
 The Realities!
 Forecasts are seldom perfect
 Most techniques assume an underlying
stability in the system
 Product family and aggregated forecasts
are more accurate than individual product
forecasts

4-4
 Forecasting Time Horizons
 Short-range forecast
– Up to 1 year, generally less than 3 months
– Purchasing, job scheduling, workforce
levels, job assignments, production levels
 Medium-range forecast
– 3 months to 3 years
– Sales and production planning, budgeting
 Long-range forecast
– 3+ years
– New product planning, facility location,
research and development
4-5
 Forecasting Techniques
1. Qualitative Methods
 These types of forecasting methods are
based on judgments, opinions, perception,
emotions, or personal experiences.
 They are subjective in nature.
 They do not rely on any rigorous
mathematical computations.

4-6
Forecasting Techniques

4-7
 Forecasting Techniques
2. Quantitative Approaches
 Used when situation is ‘stable’ and historical
data exist
 These types of forecasting methods are based
on mathematical (quantitative) models, and
are objective in nature.
 They rely heavily on mathematical computations.

4-8
Forecasting Techniques

4-9
 Quantitative Methods
1. Naive approach
2. Moving averages
3. Weighted moving Time-series
averages Models

4. Exponential
smoothing
5. Trend projection
Associative
6. Linear regression Model

4 - 10
Time Series Components

Trend Cyclical

Seasonal Random

4 - 11
 Components of Demand
Trend
component
Demand for product or service

Seasonal peaks

Actual demand
line

Average demand
over 4 years

Random variation
| | | |
1 2 3 4
Time (years)
Figure 4.1
4 - 12
 Trend Component
 Persistent, overall upward or
downward pattern
 Changes due to population,
technology, age, culture, etc.
 Typically several years
duration

4 - 13
 Seasonal Component
 Regular pattern of up and
down fluctuations
 Due to weather, customs, etc.
 Occurs within a single year
Number of
Period Length Seasons
Week Day 7
Month Week 4-4.5
Month Day 28-31
Year Quarter 4
Year Month 12
Year Week 52

4 - 14
 Cyclical Component
 Repeating up and down movements
 Affected by business cycle,
political, and economic factors
 Multiple years duration

0 5 10 15 20
4 - 15
 Random Component
 Erratic, unsystematic, ‘residual’
fluctuations
 Due to random variation or unforeseen
events
 Short duration
and nonrepeating

M T W T F
4 - 16
 Naive Approach
 Assumes demand in next period is the
same as demand in most recent period

• e.g., If January sales were 68, then

February sales will be 68

 Sometimes cost effective and efficient

and it Can be good starting point

4 - 17
 Moving Average Method

 MA is a series of arithmetic means and Used

if little or no trend
 Used often for smoothing Provides overall
impression of data over time

∑ demand in previous n periods

Moving average = n

4 - 18
 Moving Average Example
Actual 3-Month
Month Shed Sales Moving Average
January 10
February 12
March 13
April 16 (10 + 12 + 13)/3 = 11 2/3
May 19 (12 + 13 + 16)/3 = 13 2/3
June 23 (13 + 16 + 19)/3 = 16
July 26 (16 + 19 + 23)/3 = 19 1/3
(19 + 23 + 26)/3 = 22 2/3

4 - 19
 Graph of Moving Average
Moving
30 –
Average
28 –
Forecast
26 – Actual
24 – Sales
Shed Sales

22 –
20 –
18 –
16 –
14 –
12 –
10 –
| | | | | | | | | | | |
J F M A M J J A S O N D

4 - 20
Moving Average Example

4 - 21
 Weighted Moving Average
 Used when some trend might be present
– Older data usually less important
 Weights based on experience and intuition

Weighted ∑ [(weight for period n)

moving average = x (demand in period n)]
∑ weights

4 - 22
 Weights Applied Period
Weighted Moving
3 Average
Last month
2 Two months ago
1 Three months ago
6 Sum of weights

Actual 3-Month Weighted

Month Shed Sales Moving Average
January 10
February 12
March 13
April 16 [(3 x 13) + (2 x 12) + (10)]/6 = 121/6
May 19 [(3 x 16) + (2 x 13) + (12)]/6 = 141/3
June 23 [(3 x 19) + (2 x 16) + (13)]/6 = 17
July 26 [(3 x 23) + (2 x 19) + (16)]/6 = 201/2

4 - 23
 Potential Problems With Moving
Average
 Increasing n smoothes the forecast but
makes it less sensitive to changes
 Do not forecast trends well
 Require extensive historical data

4 - 24
 Moving Average And Weighted
Moving Average
Weighted
30 – moving
average
25 –
Sales demand

20 – Actual
sales
15 –
Moving
10 – average

5 –
| | | | | | | | | | | |
J F M A M J J A S O N D
Figure 4.2
4 - 25
 Exponential Smoothing
 A Form of weighted moving average
– Weights decline exponentially
– Most recent data weighted most
 Requires smoothing constant ()
– Ranges from 0 to 1 which is Subjectively chosen
 Involves little record keeping of past data

4 - 26
 Exponential Smoothing
New forecast = Last period’s forecast
+  (Last period’s actual demand
– Last period’s forecast)
Ft = Ft – 1 + (At – 1 - Ft – 1) ……1

Ft =  At – 1 + (1-)Ft – 1) …… 2

Where Ft = new forecast

Ft – 1 = previous forecast
 = smoothing (or weighting)
constant (0 ≤  ≤ 1)
4 - 27
Exponential Smoothing Example
Predicted demand(t-1) = 142 Ford
Mustangs
Actual demand (t-1)= 153
Smoothing constant  = .20
Ft = Ft – 1 + (At – 1 - Ft – 1)

New forecast (t) = 142 + .2(153 – 142)

= 142 + 2.2
= 144.2 ≈ 144 cars

4 - 28
.

4 - 29
Stability Vs. Responsiveness In Forecasting
Forecasting methods that react very strongly
(or quickly) to demand changes are said to be
responsive.
Forecasting methods that do not react quickly
to demand changes are said to be stable.

1. Moving Average Approach

 Using more periods in your moving average
forecasts will result in more stability in the
forecasts.
 Using fewer periods in your moving average
forecasts will result in more responsiveness in
the forecasts.
4 - 30
Stability Vs. Responsiveness In Forecasting
2. Weighted Moving Average Approach

 Using more periods in your weighted moving

average forecasts will result in more stability in
the forecasts.
 Using fewer periods in your weighted moving
average forecasts will result in more
responsiveness in the forecasts.
 placing lower weights on the more recent
demand will result in more stability in the
forecasts.
 Placing higher weights on the more recent
demand will result in more responsiveness in the
forecasts. 4 - 31
Stability Vs. Responsiveness In Forecasting

3. Simple Exponential Smoothing Approach

 Using a lower alpha (α) value will result in more

stability in the forecasts.

 Using a higher alpha (α) value will result in more

responsiveness in the forecasts.

4 - 32
 Forecasting errors

Forecast error = Actual demand - Forecast value

E = At - Ft

4 - 33
Common Measures of Error

Mean Absolute Deviation (MAD)

∑ |Actual - Forecast|
MAD =
n

Mean Squared Error (MSE)

∑ (Forecast Errors)2
MSE =
n

4 - 34
Common Measures of Error

Mean Absolute Percent Error (MAPE)

n
∑100|Actuali - Forecasti|/Actuali
MAPE = i=1
n

4 - 35
 Example 4
• During the past 8 quarters, the Port of
Baltimore has unloaded large quantities of
grain from ships. The Port’s Operations
Manager wants to test the forecasting
method exponential smoothing to see how
well the this method works in predicting
tonnage unloaded.
• He guesses that the forecast of grain
unloaded in the first quarter was 175 tons.

4 - 36
 Comparison of Forecast
Error
Rounded Absolute Rounded Absolute
Actual Forecast Deviation Forecast Deviation
Tonnage with for with for
Quarter Unloaded  = .10  = .10  = .50  = .50
1 180 175 5.00 175 5.00
2 168 175.5 7.50 177.50 9.50
3 159 174.75 15.75 172.75 13.75
4 175 173.18 1.82 165.88 9.12
5 190 173.36 16.64 170.44 19.56
6 205 175.02 29.98 180.22 24.78
7 180 178.02 1.98 192.61 12.61
8 182 178.22 3.78 186.30 4.30
82.45 98.62

4 - 37
 Comparison of Forecast
Error
∑ |deviations|
Rounded Absolute Rounded Absolute
MADActual
= Forecast Deviation Forecast Deviation
Tonnage n
with for with for
Quarter Unloaded  = .10  = .10  = .50  = .50
1
For 180
= .10 175 5.00 175 5.00
2 168 = 82.45/8
175.5 = 10.31
7.50 177.50 9.50
3 159 174.75 15.75 172.75 13.75
4 For 175
= .50 173.18 1.82 165.88 9.12
5 190 173.36 16.64 170.44 19.56
6 205 = 98.62/8
175.02 = 12.33
29.98 180.22 24.78
7 180 178.02 1.98 192.61 12.61
8 182 178.22 3.78 186.30 4.30
82.45 98.62

4 - 38
 Comparison of Forecast
Error2
∑ (forecast errors)
Rounded Absolute Rounded Absolute
MSE = Actual Forecast Deviation Forecast Deviation
Tonnage
n
with for with for
Quarter Unloaded  = .10  = .10  = .50  = .50
1
For 180
= .10 175 5.00 175 5.00
2 = 1,526.54/8
168 175.5 = 190.82
7.50 177.50 9.50
3 159 174.75 15.75 172.75 13.75
4 For 175
= .50 173.18 1.82 165.88 9.12
5 190 173.36 16.64 170.44 19.56
6 = 1,561.91/8
205 175.02 = 195.24
29.98 180.22 24.78
7 180 178.02 1.98 192.61 12.61
8 182 178.22 3.78 186.30 4.30
82.45 98.62
MAD 10.31 12.33

4 - 39
 Comparison of Forecast
n Error
∑100|deviation |/actual i i
Rounded Absolute Rounded Absolute
MAPE = i=1
Actual Forecast Deviation Forecast Deviation
Tonnage with n for with for
Quarter Unloaded  = .10  = .10  = .50  = .50
1
 = .10 175
For 180 5.00 175 5.00
2 168 = 44.75/8
175.5 = 7.50
5.59% 177.50 9.50
3 159 174.75 15.75 172.75 13.75
4 =
For 175 .50 173.18 1.82 165.88 9.12
5 190 173.36 16.64 170.44 19.56
6 205 = 54.05/8
175.02 =29.98
6.76% 180.22 24.78
7 180 178.02 1.98 192.61 12.61
8 182 178.22 3.78 186.30 4.30
82.45 98.62
MAD 10.31 12.33
MSE 190.82 195.24
4 - 40
 Comparison of Forecast
Error
Rounded Absolute Rounded Absolute
Actual Forecast Deviation Forecast Deviation
Tonnage with for with for
Quarter Unloaded  = .10  = .10  = .50  = .50
1 180 175 5.00 175 5.00
2 168 175.5 7.50 177.50 9.50
3 159 174.75 15.75 172.75 13.75
4 175 173.18 1.82 165.88 9.12
5 190 173.36 16.64 170.44 19.56
6 205 175.02 29.98 180.22 24.78
7 180 178.02 1.98 192.61 12.61
8 182 178.22 3.78 186.30 4.30
82.45 98.62
MAD 10.31 12.33
MSE 190.82 195.24
MAPE 5.59% 6.76%
4 - 41
 Exponential Smoothing with
Trend Adjustment
When a trend is present, exponential
smoothing must be modified to respond
to trend
Forecast Exponentially Exponentially
including (FITt) = smoothed (Ft) + smoothed (Tt)
trend forecast trend

4 - 42
Exponential Smoothing with
Trend Adjustment

Ft = (At - 1) + (1 - )(Ft - 1 + Tt - 1)

Tt = b(Ft - Ft - 1) + (1 - b)Tt - 1

Step 1: Compute Ft
Step 2: Compute Tt
Step 3: Calculate the forecast FITt = Ft + Tt

4 - 43
 EXAMPLE

• A Portland manufacturer wants to

forecast the demand for a pollution-
control equipment.
• Past data shows that there is an
increasing trend.
• The company assumes the initial forecast
for month 1 was 11 units and the trend
over that period was 2 units.
• α = 0.2 β =0.4
4 - 44
 Exponential Smoothing with
Trend Adjustment Example
Forecast
Actual Smoothed Smoothed Including
Month(t) Demand (At) Forecast, Ft Trend, Tt Trend, FITt
1 12 11 2 13.00
2 17
3 20
4 19
5 24
6 21
7 31
8 28
9 36
10

Table 4.1
4 - 45
 Exponential Smoothing with
Trend Adjustment Example
Forecast
Actual Smoothed Smoothed Including
Month(t) Demand (At) Forecast, Ft Trend, Tt Trend, FITt
1 12 11 2 13.00
2 17
3 20
4 19
5 24 Step 1: Forecast for Month 2
6 21
7 31 F2 = A1 + (1 - )(F1 + T1)
8 28 F2 = (.2)(12) + (1 - .2)(11 + 2)
9 36
10 = 2.4 + 10.4 = 12.8 units
Table 4.1
4 - 46
 Exponential Smoothing with
Trend Adjustment Example
Forecast
Actual Smoothed Smoothed Including
Month(t) Demand (At) Forecast, Ft Trend, Tt Trend, FITt
1 12 11 2 13.00
2 17 12.80
3 20
4 19
5 24 Step 2: Trend for Month 2
6 21
7 31 T2 = b(F2 - F1) + (1 - b)T1
8 28 T2 = (.4)(12.8 - 11) + (1 - .4)(2)
9 36
10 = .72 + 1.2 = 1.92 units
Table 4.1
4 - 47
 Exponential Smoothing with
Trend Adjustment Example
Forecast
Actual Smoothed Smoothed Including
Month(t) Demand (At) Forecast, Ft Trend, Tt Trend, FITt
1 12 11 2 13.00
2 17 12.80 1.92
3 20
4 19
5 24 Step 3: Calculate FIT for Month 2
6 21
7 31 FIT2 = F2 + T2
8 28 FIT2 = 12.8 + 1.92
9 36
10 = 14.72 units
Table 4.1
4 - 48
 Exponential Smoothing with
Trend Adjustment Example
Forecast
Actual Smoothed Smoothed Including
Month(t) Demand (At) Forecast, Ft Trend, Tt Trend, FITt
1 12 11 2 13.00
2 17 12.80 1.92 14.72
3 20 15.18 2.10 17.28
4 19 17.82 2.32 20.14
5 24 19.91 2.23 22.14
6 21 22.51 2.38 24.89
7 31 24.11 2.07 26.18
8 28 27.14 2.45 29.59
9 36 29.28 2.32 31.60
10 32.48 2.68 35.16

Table 4.1
4 - 49
Exponential Smoothing with
Trend Adjustment Example
35 –

30 – Actual demand (At)

Product demand

25 –

20 –

15 –

10 – Forecast including trend (FITt)

with  = .2 and b = .4
5 –

0 – | | | | | | | | |
1 2 3 4 5 6 7 8 9
Figure 4.3
Time (month)
4 - 50
 Trend Projections
Fitting a trend line to historical data points
to project into the medium to long-range
Linear trends can be found using the least
squares technique

y^ = a + bx
^ = computed value of the variable to
where y
be predicted (dependent variable)
a = y-axis intercept
b = slope of the regression line
x = the independent variable

4 - 51
Values of Dependent Variable Least Squares Method

Actual observation Deviation7

(y-value)

Deviation5 Deviation6

Deviation3

Deviation4

Deviation1
(error) Deviation2
Trend line, y^ = a + bx

Time period Figure 4.4

4 - 52
Values of Dependent Variable Least Squares Method

Actual observation Deviation7

(y-value)

Deviation5 Deviation6

Deviation3 Least squares method

minimizes the sum of the
Deviation
squared errors (deviations)
4

Deviation1
(error) Deviation2
Trend line, y^ = a + bx

Time period Figure 4.4

4 - 53
 Least Squares Method
Equations to calculate the regression variables

y^ = a + bx

Sxy - nxy
b=
Sx2 - nx2

a = y - bx

4 - 54
 Least Squares Example
Time Electrical Power
Year Period (x) Demand (megawatt) x2 xy
2006 1 74 1 74
2007 2 79 4 158
2008 3 80 9 240
2009 4 90 16 360
2010 5 105 25 525
2011 6 142 36 852
2012 7 122 49 854
∑x = 28 ∑y = 692 ∑x2 = 140 ∑xy = 3,063
x=4 y = 98.86

∑xy - nxy 3,063 - (7)(4)(98.86)

b= = = 10.54
∑x - nx
2 2 140 - (7)(4 2)

a = y - bx = 98.86 - 10.54(4) = 56.70

4 - 55
 Least Squares Example
Time Electrical Power
Year Period (x) Demand x2 xy
2003 1 74 1 74
2004 2 79 4 158
2005The trend
3 line is 80 9 240
2006 4 90 16 360
2007 y^ 5= 56.70 + 10.54x
105 25 525
2008 6 142 36 852
2009 7 122 49 854
∑x = 28 ∑y = 692 ∑x2 = 140 ∑xy = 3,063
x=4 y = 98.86

∑xy - nxy 3,063 - (7)(4)(98.86)

b= = = 10.54
∑x - nx
2 2 140 - (7)(4 2)

a = y - bx = 98.86 - 10.54(4) = 56.70

4 - 56
 Least Squares Example
160 –
Trend line,
150 – y^ = 56.70 + 10.54x
140 –
Power demand

130 –
120 –
110 –
100 –
90 –
80 –
70 –
60 –
50 –
| | | | | | | | |
2006 2007 2008 2009 2010 2011 2012 2013 2014
Year
4 - 57
Seasonal Variations In Data

The multiplicative
seasonal model
can adjust trend
data for seasonal
variations in
demand (jet skis,
snow mobiles)

4 - 59
Seasonal Variations In Data
Steps in the process:
1. Find average historical demand for each season
2. Compute the average demand over all seasons
3. Compute a seasonal index for each season
4. Estimate next year’s total demand
5. Divide this estimate of total demand by the
number of seasons, then multiply it by the
seasonal index for that season

4 - 60
 Seasonal Index Example
Demand Average Average Seasonal
Month 2010 2011 2012 2010-2012 Monthly Index
Jan 80 85 105 90 94
Feb 70 85 85 80 94
Mar 80 93 82 85 94
Apr 90 95 115 100 94
May 113 125 131 123 94
Jun 110 115 120 115 94
Jul 100 102 113 105 94
Aug 88 102 110 100 94
Sept 85 90 95 90 94
Oct 77 78 85 80 94
Nov 75 72 83 80 94
Dec 82 78 80 80 94
4 - 61
 Seasonal Index Example
Demand Average Average Seasonal
Month 2010 2011 2012 2010-2012 Monthly Index
Jan 80 85 105 90 94 0.957
Feb 70 85 85 80 94
Mar 80 93 Average
82 85 monthly 94
2010-2012 demand
Seasonal90index95= 115
Apr 100 94
Average monthly demand
May 113 125 131 123 94
= 90/94 = .957
Jun 110 115 120 115 94
Jul 100 102 113 105 94
Aug 88 102 110 100 94
Sept 85 90 95 90 94
Oct 77 78 85 80 94
Nov 75 72 83 80 94
Dec 82 78 80 80 94
4 - 62
 Seasonal Index Example
Demand Average Average Seasonal
Month 2010 2011 2012 2010-2012 Monthly Index
Jan 80 85 105 90 94 0.957
Feb 70 85 85 80 94 0.851
Mar 80 93 82 85 94 0.904
Apr 90 95 115 100 94 1.064
May 113 125 131 123 94 1.309
Jun 110 115 120 115 94 1.223
Jul 100 102 113 105 94 1.117
Aug 88 102 110 100 94 1.064
Sept 85 90 95 90 94 0.957
Oct 77 78 85 80 94 0.851
Nov 75 72 83 80 94 0.851
Dec 82 78 80 80 94 0.851
4 - 63
 Seasonal Index Example
Demand Average Average Seasonal
Month 2010 2011 2012 2010-2012
Monthly Index
Jan 80 85 105 90 94 0.957
Feb 70 85 Forecast
85 for802013 94 0.851
Mar 80 93 82 85 94 0.904
Apr 90Expected
95 115 annual demand
100 = 1,200
94 1.064
May 113 125 131 123 94 1.309
Jun 110 115 120 1,200 115 94 1.223
Jul 100 102Jan 113 x .957 = 96 94
105 1.117
12
Aug 88 102 110 100 94 1.064
1,200
Sept 85 90
Feb 95 x90
.851 = 85 94 0.957
Oct 77 78 85 12 80 94 0.851
Nov 75 72 83 80 94 0.851
Dec 82 78 80 80 94 0.851
4 - 64
 Seasonal Index Example
2013 Forecast
140 – 2012 Demand
130 – 2011 Demand
2010 Demand
120 –
Demand

110 –
100 –
90 –
80 –
70 –
| | | | | | | | | | | |
J F M A M J J A S O N D
Time
4 - 65
 Associative Forecasting
Used when changes in one or more
independent variables can be used to predict
the changes in the dependent variable

Most common technique is linear

regression analysis

We apply this technique just as we did

in the time series example

4 - 66
 Associative Forecasting
Forecasting an outcome based on
predictor variables using the least squares
technique
y^ = a + bx
^
where y = computed value of the variable to
be predicted (dependent variable)
a = y-axis intercept
b = slope of the regression line
x = the independent variable though to
predict the value of the dependent
variable
4 - 67
 Associative Forecasting
Example
Sales Area Payroll
($ millions), y ($ billions), x
2.0 1
3.0 3
2.5 4 4.0 –
2.0 2
2.0 1 3.0 –
3.5 7 Sales
2.0 –

1.0 –

| | | | | | |
0 1 2 3 4 5 6 7
Area payroll
4 - 68
 Associative Forecasting
Example
Sales, y Payroll, x x2 xy
2.0 1 1 2.0
3.0 3 9 9.0
2.5 4 16 10.0
2.0 2 4 4.0
2.0 1 1 2.0
3.5 7 49 24.5
∑y = 15.0 ∑x = 18 ∑x2 = 80 ∑xy = 51.5

∑xy - nxy 51.5 - (6)(3)(2.5)

x = ∑x/6 = 18/6 = 3 b=
∑x2 - nx2
= 80 - (6)(32) = .25

y = ∑y/6 = 15/6 = 2.5 a = y - bx = 2.5 - (.25)(3) = 1.75

4 - 69
 Associative Forecasting
Example
y^ = 1.75 + .25x Sales = 1.75 + .25(payroll)

If payroll next year

4.0 –
is estimated to be
$6 billion, then: 3.25
3.0 –
Nodel’s sales
2.0 –
Sales = 1.75 + .25(6)
Sales = $3,250,000 1.0 –

| | | | | | |
0 1 2 3 4 5 6 7
Area payroll

4 - 70
The End
???

Thank You
4 - 71