Demand Forecasting
Dr. M. Shamim Uddin Khan
Introduction:
A forecast is a prediction of future events used for planning purposes. Forecasts are vital to
every business organization and for every significant management decision. Forecasting is the basis
of corporate long-run planning. In the functional areas finance and accounting, forecasts provide
the basis for budgetary planning and cost control. In human resources, forecasts provide hiring
activities including recruitment, interviewing, training, layoff planning including outplacement, and
counseling. MIS forecasting provides new (or revised) information systems and internet services.
Marketing relies on sales forecasting to plan new products, compensate sales personnel, and make
other key decisions. Production and operations personnel use forecasts to make periodic decisions
involving process selection, capacity planning and facility layout as well as for continual decisions
about production planning, scheduling and inventory.
At any organization, management needed accurate forecasts to ensure value chain success.
Changing business conditions resulting from global competition, rapid technological change, and
increasing environmental concerns exerts pressure on a firm’s capability to generate accurate
forecasts. There are two basic sources of demand: dependent demand and independent demand.
Dependent demand is the demand for a product or service caused by the demand for other products
or services.
Elements of a Good Forecast:
A properly prepared forecast should fulfill certain requirements:
1. The forecast should be timely.
2. The forecast should be accurate and the degree of accuracy should be stated.
3. The forecast should be reliable.
4. The forecast should be expressed in meaningful units.
5. The forecast should be in writing.
6. The forecasting technique should be simple to understand and use.
Steps in the Forecasting Process:
There are six basic steps in the forecasting process:
1. Determine the purpose of the forecast
2. Establish a time horizon
3. Select a forecasting technique
4. Gather and analyze relevant data
5. Prepare the forecast
6. Monitor the forecast.
Types of Forecasting:
Forecasting can be classified into four basic types: qualitative, time series analysis, causal
relationships and simulation.
A. Qualitative:
(i) Grass Roots:
(ii) Market Research:
(iii) Panel Consensus:
(iv) Historical Analogy:
(v) Delphi Method: This is a group process intended to achieve a consensus forecast. The
Delphi method was developed by the Rand Corporation in the 1950’s. The step by step
procedure is
Step-1: Choose the experts to participate. There should be a variety of knowledgeable
people in different areas.
Step-2: Through a questionnaire (or e-mail) obtain forecast (and any premises or
qualifications for the forecasts) from all participants.
Step-3: Summarize the results and redistribution them to the participants along with
appropriate new questions.
Step-4: Summarize again, refining forecasts and conditions and again develop new
questions.
Step-5: Repeat Step 4 if necessary. Distribute the final results to all participants.
The Delphi technique can usually achieve satisfactory results in three rounds. The time
required is a function of the number of participants, how much work is involved for
them to develop their forecasts and their speed in responding.
B. Quantitative
(a) Time series analysis:
(i) Simple Moving Average
(ii) Weighted Moving Average
(iii) Exponential Smoothing
(iv) Regression Analysis
(v) Box Jenkins Technique
(vi) Shiskin Time Series
(vii) Trend Projections
(b) Causal Relationships:
(i) Regression Analysis
(ii) Econometric Models
(iii) Input/Output Models
(iv) Leading Indicators
Example-1: (a) Compute a three-week moving average forecast for the arrival of medical clinic
patients in week 4. The numbers of arrivals for the past 3 weeks were as follows:
Week
Patient Arrivals
1
400
2
380
3
411
(b) If the actual number of patient arrivals in week 4 is 415, what is the forecast error for week 4?
(c) What is the forecast for week 5?
411 + 380 + 400
= 397
Solution: (a) The moving average forecast at the end of week 3 is F4 =
3
(b) The forecast error for week 4 is E4 = D4 − F4 = 415 − 397 = 18
(c) The forecast for week 5 requires the actual arrivals from weeks 2 through 4, the three
415 + 411 + 380
= 402
most recent weeks of data F5 =
3
Example-2: Forecasts based on averages. Given the following data:
Period
Number of complaints
1
60
2
65
3
55
4
58
5
64
Prepare a forecast using each of these approaches:
(a) The appropriate naive approach.
(b) A three-period moving average.
(c) A weighted average using weights of 0.50 (most recent), 0.30 and 0.20.
(d) Exponential smoothing with a smoothing constant of 0.40
Solution:
(a) The values are stable. Therefore, the most recent value of the series becomes the next
forecast: 64.
55 + 58 + 64
= 59
(b) MA3 =
3
(c) F = 0.50(64) + 0.30(58) + 0.20(55) = 60.4
(d)
Period
1
Number of complaints
60
2
3
4
5
6
65
55
58
64
Forecast
60
62
59.2
58.72
60.83
Calculations
[The previous value of series is
used as the starting forecast]
60 + 0.40(65 – 60) = 62
62 + 0.40(55 – 62) = 59.2
59.2 + 0.40(58 – 59.2) = 58.72
58.72 + 0.40(64 – 58.72) = 60.83
Example-3: The Polish General’s Pizza Parlor is a small restaurant catering to patrons with a taste
for European pizza. One of its specialties is Polish Prize pizza. The manager must forecast weekly
demand for these special pizzas so that he can order pizza shells weekly. Recently demand has been
as follows:
Week
Pizzas Week
Pizzas
June 2
50
June 23
56
June 9
65
June 30
55
June 16 52
July 7
60
(a) Forecast the demand for pizza for June 23 to July 14by using the simple moving average
method with n = 3. Then repeat the forecast by using the weighted moving average method
with n = 3 and weights of o.50, 0.30 and 0.20, with 0.50 applying to the most recent
demand.
(b) Calculate the MAD for each method.
Solution:
(a) The simple moving average method and weighted moving average method give the
following results.
Current Simple Moving Average
Weighted Moving Average
Week
Forecast for next week
Forecast for next week
June
[(0.5  52) + (0.3  65) + (0.2  50)] = 55.5 or 56
52 + 65 + 50
= 55.7 or 56
16
3
June
or
[(0.5  56) + (0.3  52) + (0.2  65)] = 56.6
56 + 52 + 65
= 57.7 or 58
23
57
3
June
or
[(0.5  55) + (0.3  56) + (0.2  52)] = 54.7
55 + 56 + 52
= 54.3 or 54
30
55
3
July 7
or
[(0.5  60) + (0.3  55) + (0.2  56)] = 57.7
60 + 55 + 56
= 57.0 or 57
58
3
(b) The mean absolute deviation is calculated as follows:
Simple Moving Average
Weighted Moving Average
Week
Actual
Forecast
Forecast
Absolute Errors Et
Absolute Errors Et
Demand
June 23
56
56
56
56 − 56 = 0
56 − 56 = 0
June 30
55
58
55 − 58 = 3
57
55 − 57 = 2
July 7
60
54
60 − 54 = 6
55
60 − 55 = 5
MAD =
0+ 3+ 6
=3
3
MAD =
0+2+5
= 2 .3
3
Example-4: The monthly demand for units manufactured by the Acme Rocket Company has been
as follows:
Month Units Month
Units
May
100
September 105
June
80
October
110
July
110
November 125
August 115
December 120
(a) Use the exponential smoothing method to forecast the number of units for June to January.
The initial forecast for May was 105 units,  = 0.2
(b) Calculate the absolute percentage error for each month from June through December and
the MAD and MAPE of forecast error as of the end of December.
(c) Calculate the tracking signal as of the end of December. What can you say about the
performance of your forecasting method?
Solution: (a)
Current Month, t Ft +1 = Dt + (1 −  ) Ft
Forecast for Month, t + 1
May
June
July
August
September
October
November
December
(b)
Current
Month, t
June
July
August
September
October
November
December
Total
0.2(100) + 0.8(105) = 104.0 or 104
0.2(80) + 0.8(104) = 99.2 or 99
0.2(110) + 0.8(99.2) = 101.4 or 101
0.2(115) + 0.8(101.4) = 104.1 or 104
0.2(105) + 0.8(104.1) =1 04.3 or 104
0.2(110) + 0.8(104.3) = 105.4 or 105
0.2(125) + 0.8(105.4) = 109.3 or 109
0.2(120) + 0.8(109.3) = 111.4 or 111
Actual Forecast, Ft
Demand
80
110
115
105
110
125
120
765
104
99
101
104
104
105
109
Error
Et = Dt – Ft
-24
11
14
1
6
20
11
39
June
July
August
September
October
November
December
January
Absolute
Error Et
24
11
14
1
6
20
11
87
Absolute Percentage Error,
Et
( )(100%)
Dt
30.0
10.0
12.2
0.9
5.4
16.0
9.2
83.7%
E
( t
Dt )100 83.7%
87
MAD =
=
= 12.4 and MAPE =
=
= 11.96
n
7
n
7
(c) As of the end of December, the cumulative sum of forecast errors (CFE) is 39. Using the
mean absolute deviation calculated in part (b), we calculate the tracking signal:
CFE
39
=
= 3.14
Tracking signal =
MAD 12.4
The probability that a tracking signal value of 3.14 could be generated completely by chance is
small. Consequently, we should revise our approach. The long string of forecasts lower than the
actual demand suggests use of a trend method.
 Et
Regression and Correlation Analysis
Dr M. Shamim Uddin Khan
Introduction:
Regression and correlation analyses are the techniques of studying how the variations
in one series are related to variations in another series. Measurement of the degree of relationship
between two or more variables is called correlation analysis and using the relationship between a
known variable and an unknown variable to estimate the unknown one is termed as regression
analysis. Thus, correlation measures the degree of relationship between the variables while
regression analysis shows how the variables are related.
According to Blair, ‘Regression is the measure of the average relationship between two or more
variables in terms of the original units of the data’. The technique of regression analysis is used to
determine the statistical relationship between two (or more variables) and to make prediction of one
variable on the basis of the other(s).
Assumptions of Regression Analysis:
While making use of the regression analysis for making predictions it is always assumed:
1. There is an actual relationship between the dependent and independent variables.
2. The values of the dependent variable are random but the values of the independent
variable are fixed quantities without error and are chosen by the users.
3. There is clear indication of direction of the relationship. This means that dependent
variable is a function of independent variable. (For example when we say that
advertising has an effect on sales, then we are not saying that sales has an effect on
advertising).
4. The analysis can be used to predict values within the range (and not for values
outside the range) for which it is valid.
Differences between Correlation and Regression:
Correlation
Correlation is the relationship between two or
more variables, which vary in sympathy with the
other in the same or opposite direction.
It finds out the degree of relationship between
two and not the cause and effect of the variables.
It is used for testing and verifying the relation
between two variables and gives limited
information.
It has limited application because it is confined
only to linear relationship between the variables.
It is not very useful for further mathematical
treatment.
Regression
Regression means going back and it is a
mathematical measure showing the average
relationship between two or more variables.
It indicates the cause and effect relationship
between the variables and establishes a function
relationship.
Besides verification, it is used for the prediction
of one value in relationship to the other given
value.
It has wide application as it studies linear and
nonlinear relationship between the variables.
It is widely used for further mathematical
treatment.
Example-1: For the following set of data:
(a) Plot the scatter diagram.
(b) Develop the estimating equation that best describes the data.
(c) Predict Y for X = 10, 15, 20.
X
13
16
14
11
17
9
13
17
Y 6.2 8.6 7.2 4.5 9.0 3.5 6.5 9.3
18
9.5
12
5.7
Solution: (b)
X
13
16
14
11
17
9
13
17
18
12
 x = 140
X =
140
= 14,
10
Y =
 x y
Y
6.2
8.6
7.2
4.5
9.0
3.5
6.5
9.3
9.5
5.7
 y = 70
XY
80.6
137.6
100.8
49.5
153.0
31.5
84.5
158.1
171.0
68.4
 xy = 1035
X2
169
256
196
121
289
81
169
289
324
144
2
 x = 2038
70
=7
10
(140)(70)
n
10
b=
=
= 0.7051
2
2
(
x
)
(
140
)

2038 −
 x2 − n
10
a = Y − bX = 7 − (0.7051)(14) = −2.8714
)
Thus, Y = −2.8714 + 0.7051X
)
X = 10, Y = −2.8714 + 0.7051(10) = 4.1796
)
(c) X = 15, Y = −2.8714 + 0.7051(15) = 7.7051
)
X = 20, Y = −2.8714 + 0.7051(20) = 11.2306
Example-2: Cost accountants often estimate overhead based on the level of production. At the
standard Knitting Co. they have collected information on overhead expenses and units produced at
different plants, and want to estimate a regression equation to predict future overhead.
Overheads 191 170 272 155 280 173 234 116 153 178
Units
40
42
53
35
56
39
48
30
37
40
(a) Develop the regression equation for the cost accountants.
(b) Predict overhead when 50 units are produced.
(c) Calculate the standard error of estimates.
Solution: (a) Let Y = overhead and X = units produced.
X
Y
XY
X2
Y2
40
191
7640
1600
36481
42
170
7140
1764
28900
53
272
14416
2809
73984
35
155
5425
1225
24025
56
280
15680
3136
78400
39
173
6747
1521
29929
48
234
11232
2304
54756
30
116
3480
900
13456
37
153
5661
1369
23409
40
178
7120
1600
31684
2
2
 x = 420  y = 1922  xy = 84541  x = 18228  y = 395024
 xy −
X =
420
= 42,
10
Y =
1035 −
1922
= 192.2
10
 x y
(420)(1922)
n
10
b=
=
= 6.4915
2
2
(
x
)
(
420
)

18228 −
 x2 − n
10
a = Y − bX = 192.2 − 6.4915(42) = −80.4430
)
Thus, Y = −80.4430 − 6.4915 X
)
(b) X = 50, Y = −80.4430 + 6.4915(50) = 244.1320
 xy −
Y
84541 −
− a Y − b XY
395024 − (−80.4430)(1922) − 6.4915(84541)
= 10.2320
n−2
8
Example-3: Campus stores has been selling the Believe It or Not: Wonders of Statistics Study
Guide for 12 semesters and would like to estimate the relationship between sales and number of
sections of elementary statistics taught in each semester. The following data have been collected:
Sales (units)
33 38 24 61 52 45 65 82 29 63 50 79
No. of Sections
3
7
6
6
10 12 12 13 12 13 14 15
(a) Develop the estimating equation that best fits the data.
(b) Calculate the sample co-efficient of determination and the sample coefficient of correlation.
Solution:
Let Y = sales and X = number of sections.
X
Y
XY
X2
Y2
3
33
99
9
1089
7
38
266
49
1444
6
24
144
36
576
6
61
366
36
3721
10
52
520
100
2704
12
45
540
144
2025
12
65
780
144
4225
13
82
1066
169
6724
12
29
348
144
841
13
63
819
169
3969
14
70
700
196
2500
15
79
1185
225
6241
2
2
 x = 123  y = 621  xy = 6833  x = 1421  y = 36059
(c) S e =
2
=
123
621
= 10.25, Y =
= 51.75
12
12
 x y 6833 − (123)(621)
 xy − n
12
b=
=
= 2.9189
2
2
(
x
)
(
123
)

1421 −
 x2 − n
12
a = Y − bX = 51.75 − 2.9189(10.25) = 21.8313
)
Thus, Y = 21.8313 + 2.9189 X
(c) Coefficient of correlation,
( x ) ( y )
(30)(63)
290 −
 xy −
n
8
r=
=
= 0.59
2
2
2
2
(
x
)
(
y
)
(
30
)
(
63
)
 }{ y 2 −  } {172 −
}{595 −
{ x 2 −

8
8
n
n
X =
Coefficient of determination, r 2 = 0.3481
Example-4: The person in charge of production scheduling for a company must prepare forecasts
of product demand in order to plan for appropriate production quantities. During a luncheon
meeting the marketing manager gives her information about the advertising budget for a brass door
hinge. The following are sales and advertising data for the past 5 months:
Month
Sales
Advertising
(thousands of units)
(Thousands of $)
1
264
2.5
2
116
1.3
3
165
1.4
4
101
1.0
5
209
2.0
The marketing manager says that next month the company will spend $1750 on advertising for the
product. Use linear regression to develop an equation and a forecast for this period.
Solution: We use the computer to calculate the best values of a and b the correlation coefficient
and the coefficient of determination.
a = −8.135, b = 109.229 r = 0.980, r 2 = 0.960
The regression equation is Y = −8.135 + 109.229 X
As the advertising expenditure will be $1750, the forecast for month 6 is
Y = −8.135 + 109.229(1.75) = 183.016 or 183016 units
Example-5: RC Cola is studying the effect of its latest advertising campaign. People chosen at
random were called and asked how many cans of RC Cola they bought in the past week and how
many RC Cola advertisement they had either read or seen in the past week.
X (Number of Ads)
4
9
0
1
6
2
3
5
Y(Cans purchased) 12 14
6
3
5
4
7
10
Required:
(a) Find the regression equation that best fits the data.
(b) Forecast the number of cans purchased when the number of advertisement seen or read in
the past week were 10.
Solution:
Requirement (a)
Let Y = Cans purchased and X = Number of Ads
X
Y
XY
X2
Y2
4
12
48
16
144
9
14
126
81
196
0
6
0
0
36
1
3
3
1
9
6
5
30
36
25
2
4
8
4
16
3
7
21
9
49
5
10
50
25
100
2
 x = 30  y = 61  x y = 286  x = 172
 y 2 = 575
61
= 7.63
8
 x  y 286 − (30)(61)
 xy − n
8
b=
=
= 0.96
2
2
(
x
)
(
30
)

172 −
 x2 − n
8
a = Y − bX = 7.63 − 0.96(3.75) = 4.03
)
Thus, Y = 4.03 + 0.96 X
Requirement (b) )
When X= 10, then Y = 4.03 + 0.96(10) = 13.25
X =
30
= 3.75,
8
Y =
Example-6: Zippy Cola is studying the effect of its latest advertising campaign. People chosen at
random were called and asked how many cans of Zippy Cola they bought in the past week and how
many Zippy Cola advertisement they had either read or seen in the past week.
X (Number of Ads)
4
9
3
0
1
6
2
5
Y(Cans purchased) 12 14
7
6
3
5
6
10
Required:
(a) Develop the estimating equation that best fits the data.
(b) Calculate the sample co-efficient of determination and interpret it.
(c) Forecast the number of cans purchased when the number of advertisement seen or read in
the past week were 10.
Solution:
Requirement (a)
Let Y = Cans purchased and X = Number of Ads
X
Y
XY
4
12
48
9
14
126
3
7
21
0
6
0
1
3
3
6
5
30
2
6
12
5
10
50
 x = 30  y = 63  x y = 290
30
= 3.75,
8
X2
16
81
9
0
1
36
4
25
 x 2 = 172
Y2
144
196
49
36
9
25
36
100
 y 2 = 595
63
= 7.88
8
 x  y 290 − (30)(63)
 xy − n
8
b=
=
= 0.90
2
(
x
)
(
30
)2

2
172 −
x − n
8
a = Y − bX = 7.88 − 0.90(3.75) = 4.51
)
Thus, Y = 4.51 + 0.90 X
Requirement (b)
(c) Coefficient of correlation,
( x) ( y)
(30)(63)
290 −
 xy −  n 
8
r=
=
= 0.70
2
2
2
2
(
x
)
(
y
)
(
30
)
(
63
)


{172 −
}{595 −
{ x 2 −
}{ y 2 −
}
8
8
n
n
2
Coefficient of determination, r = 0.49
The value of r2 is the proportion of the variation in the dependent variable Y explained by
regression on the independent variable X. Further, the co-efficient of determination (i.e. the square
of the correlation co-efficient) r2, which is used for judging the explanatory power of the linear
regression of Y on X. r2 is a measure of the goodness of fit of the regression line.
The co-efficient of correlation:
r = 0.49 = 0.7
There is a high degree of positive correlation between number of ads and cans purchased.
Requirement (c)
)
When X= 10, then Y = 4.51 + 0.90(10) = 13.51
X =
Y =
Example-7: The following data relate to advertising expenditure (Tk. in lakh) and their
corresponding sales (Tk. in crore):
Year
Advertising expenditure
Sales
2007
10
14
2008
12
17
2009
15
23
2010
23
25
2011
20
21
Requirements:
(a) Co-efficient of correlation
(b) Regression equations
(c) The sales corresponding to advertising expenditure of Tk.30 lakhs.
(d) The advertising expenditure for a sales target of Tk.35 crores.
Solution: Let advertising expenditure be denoted by x and sales by y.
X
Y
Y −Y = y
y2
X −X =x
x2
10
-6
36
14
-6
36
12
-4
16
17
-3
9
15
-1
1
23
3
9
23
7
49
25
5
25
20
4
16
21
1
1
2
2
 X = 80  x = 0  x = 118 Y = 100 y = 0  y = 80
(a) Co-efficient of correlation r =
 xy
x y
2
2
=
84
118  80
xy
36
12
-3
35
4
 xy = 84
= 0.86
(b) Regression equation of Y on X is Y = a1 + b1 X
 X = 80 = 16
X =
N
5
 Y = 100 = 20
Y =
N
5
 xy = 84 = 0.712
b1 =
 x 2 118
a1 = Y − b1 X = 20 − 0.712(16) = 8.608
Again the regression equation of X on Y is X = a2 + b2Y
 xy = 84 = 1.05
b2 =
 y 2 80
a2 = X − b2Y = 16 − 1.05(20) = −5
X = −5 + 1.05Y
The co-efficient of correlation is the geometric mean of the two regression co-efficients.
Symbolically r = byx  bxy = b1  b2 = 0.712  1.05 = 0.8646
(c) Y (30) = 8.608 + 0.712(30) = 29.968
(d) X (35) = −5 + 1.05(35) = 31.75
Example-8: Calculate the two regression equations and the co-efficient of correlation from the
given below:
Marks in Statistics (Out of 50)
40 38 35 42 30
Marks in Mathematics (Out of 50) 30 35 40 36 29
Solution: Let the marks in Statistics be denoted by X and marks in Mathematics be denoted by Y.
X
Y
xy
Y −Y = y
y2
X −X =x
x2
40
38
35
42
30
 X = 185
3
1
-2
5
-7
x = 0
9
1
4
25
49
 x 2 = 88
30
35
40
36
29
Y = 70
The regression equation of Y on X is Y − Y = r
y
 xy = 22 = 0.25
= b1 =
x
 x 2 88
Y − 34 = 0.25( X − 37)
Y = 24.75 + 0.25 X
y
(X − X )
x
-4
1
6
2
-5
y=0
16
1
36
4
25
 y 2 = 82
-12
1
-12
10
35
 xy = 22
r
The regression equation of X on Y is X − X = r
185
 Y = 170 = 34
= 37 Y =
N
5
N
5
xy
x
 = 22 = 0.268
r = b2 =
y
 y 2 82
X =
X
x
(Y − Y )
y
=
X − 37 = 0.268(Y − 34)
X = 27.88 + 0.268Y
The co-efficient of correlation is the geometric mean of the two regression co-efficients.
Symbolically r = bxy  byx = 0.268  0.25 = 0.259