Transcript Chapter 3 - Forecasting

BUAD306
Chapter 3 – Forecasting
Everyday Forecasting
Weather
 Time
 Traffic
 Other examples???

What is Forecasting?
Forecast: A statement about the
future
 Used to help managers:

Plan the system
 Plan the use of the system

Use of Forecasts
Accounting
Cost/profit estimates
Finance
Cash flow and funding
Human Resources
Hiring/recruiting/training
Marketing
Pricing, promotion, strategy
MIS
IT/IS systems, services
Operations
Schedules, MRP, workloads
Product/service design New products and services
Forecasting Basics
Assumes causal system past ==>
future
 Forecasts rarely perfect because of
randomness
 Forecasts more accurate for groups
vs. individuals
 Forecast accuracy decreases as time
horizon increases

Elements of a Good Forecast
Timely – feasible horizon
 Reliable – works consistently
 Accurate – degree should be stated
 Expressed in meaningful units
 Written – for consistency of usage
 Easy to Use - KISS

Approaches to Forecasting
Judgmental – subjective inputs
 Time Series – historical data
 Associative – explanatory variables

Judgmental Forecasts
Executive Opinions
 Accuracy??
 Outside Opinions
Industry experts
 Sales Force Feedback  Bias???
 Consumer Surveys  Guarantee???

What would you rather
evaluate?
60
50
40
30
20
10
0
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
Period
A
B
C
1
30
18
46
2
34
17
26
3
32
19
27
4
34
19
23
5
35
22
22
6
30
23
48
7
34
23
29
8
36
25
20
9
29
24
14
10
31
26
18
11
35
27
47
12
31
28
26
13
37
29
27
14
34
31
24
15
33
33
22
16
Time Series Forecasts
Based on observations over a period
of time
 Identifies:

Trend – LT movement in data
 Seasonality – ST variations
 Cycles – wavelike variations
 Irregular Variations – unusual events
 Random Variations – chance/residual

Forecast Variations
Irregular
variation
Random
variation
Trend
Cycles
90
89
88
Seasonal Variations
Naïve Forecasting
Simple to use
 Minimal to no cost
 Data analysis is almost nonexistent
 Easily understandable
 Cannot provide high accuracy
 Can be a standard for accuracy

RULE: “Whatever happened “yesterday”
is going to happen tomorrow as long as I
apply LOGIC.”
HW Problem 1
Day
Muffins
Buns
Cupcakes
1
30
18
46
2
34
17
26
3
32
19
27
4
34
19
23
5
35
22
22
6
30
23
48
7
34
23
29
8
36
25
20
9
29
24
14
10
31
26
18
11
35
27
47
12
31
28
26
13
37
29
27
14
34
31
24
15
33
33
22
16
HW Problem 1
Muffins
Buns
Cupcakes
60
50
40
30
20
10
0
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
Techniques for Averaging
Moving average
 Weighted moving average
 Exponential smoothing

Simple Moving Average
n
MAn =
Ai

i=1
n
Where:
i = index that corresponds to periods
n = number of periods (data points)
Ai = Actual value in time period I
MA = Moving Average
Ft = Forecast for period t
Example 1: Moving Average
Four period moving average for period 7:
Period
1
2
3
4
5
6
7
8
Sales
3520
2860
4005
3740
4310
5001
4890
??
Four period moving average for period 8:
Four period moving average for period 9 if
actual for 8 = 5025:
Weighted Moving Average
Similar to a moving average, but
assigns more weight to the most
recent observations.
 Total of weights must equal 1.

Example 2: Weighted Moving Average
Period Sales
1
2
3
4
3520
2860
4005
3740
5
6
7
4310
5001
4890
8
??
Compute a weighted moving average
forecast for period 8 using the
following weights: .40, .30, .20 and .10:
HW #2 – Let’s Discuss
Month
Feb
Sales
19
Mar
Apr
May
June
July
Aug
18
15
20
18
22
20
Calculating Error

Mathematically:
et = At - Ft
Let’s discuss examples on board…
Premise - Exponential
Smoothing
The most recent observations might
have the highest predictive value….
 And since all forecasts have error…
 We should give more weight to the
error in the more recent time periods
when forecasting.

Exponential Smoothing
Ft = Ft-1 + (At-1 - Ft-1)
Next forecast =
Previous forecast +  (Actual -Previous Forecast)
Smoothing Constant
About 
 = Smoothing constant selected by
forecaster
 It is a percentage of the forecast error
 The closer the value is to zero, the
slower the forecast will be to adjust to
forecast errors (greater smoothing)
 The higher the value is to 1.00, the
greater the responsiveness to errors
and the less smoothing

Example 3: Exponential Smoothing
Ft = Ft-1 + (At-1 - Ft-1)
Period Sales
1
3520
2
2860
3
4
5
6
4005
3740
4310
5001
7
8
4890
??


Assume a starting
forecast of 4030 for
period 3.
Given data at left and
 = .10, what would
the forecast be for
period 8?
Example 3: Exponential Smoothing
Period
Act
3
4005
4
3740
5
4310
6
5001
7
4890
8
Forecast Calc
Fore
4030
HW #2 – Let’s Discuss
Month
Feb
Sales
19
Mar
Apr
May
June
July
Aug
18
15
20
18
22
20
Techniques for Seasonality

Seasonal Variations – regularly
repeating movements in series values
that can be tied to recurring events
Computing Seasonal Relatives: Although we will
discuss how relatives are created in class, you do not
have to know this for exam – just how to apply the
relatives to a forecast.
Using Seasonal Relatives

Allows you to incorporate seasonality
or deseasonalize data
Incorporate: Adds seasonality into the
trend forecast so that you can see
peaks and valleys.
 Deseasonalize: Remove seasonal
components to get a clearer picture of
non-seasonal components (underlying
trend)

Example 4: Using Seasonal Relatives
A publisher wants to predict quarterly demand for a certain
book for periods 11 and 12, which happen to be in the 3rd and
4th quarters of a particular year. The data series consists of
both trend and seasonality. The trend portion of demand is
projected using the equation: yt=12,500 + 150.5t.
Quarter relatives are Q1= 1.3, Q2=.8, Q3=1.4, Q4=.9
Use this information to predict demand for periods 9 and 16.
The trend values:
Applying the relatives:
HW #11 – Let’s Discuss
The following equation summarizes the trend portion of
quarterly sales of condos over a long cycle. Prepare a
forecast for each Q of next year and the first quarter of the
following year.
Ft = 40 – 6.5t + 2t2
Ft = unit sales
t= 0 at 1Q of last year
Quarter
Relative
1
1.1
2
1
3
.6
4
1.3
Assoc. Forecasting Technique:
Simple Linear Regression
Predictor variables - used to predict
values of variable interest
 Regression - technique for fitting a
line to a set of points
 Least squares line - minimizes sum of
squared deviations around the line

Linear Regression
Assumptions

Variations around line are random

No patterns are apparent
Deviations around the line should be
normally distributed
 Predictions are being made only in the
range of observed values
 Should use minimum of 20
observations for best results

Suppose you analyze the following data...
X
7
2
6
4
14
15
16
12
14
20
15
7
Y
15
10
13
15
25
27
24
20
27
44
34
17
50
40
30
20
10
0
0
5
10
15
20
25
The regression line has the following equation:
y c = a + bx
Where:
y c = Predicted (dependent) variable
x = Predictor (independent) variable
b = slope of the line
a = Value of y c when x=0
b = n (xy) - (x)(y)
n(x2) - (x)2
a = y - b x
n
Example 5 - Linear Regression:
Suppose that a manufacturing company made batches of
a certain product. The accountant for the company wished
to determine the cost of a batch of product given the
following data:
Size of batch
20
30
40
50
70
80
100
120
150
Cost of batch (in 1000s)
$1.4
3.4
4.1
Question… which is
3.8
the dependent (y) and
6.7
which is the independent
6.6
(x) variable?
7.8
10.4
11.7
Batch (x) Cost (y)
20
1.4
30
3.4
40
4.1
50
3.8
70
6.7
80
6.6
100
7.8
120
10.4
150
11.7
SUM =
660
55.9
xy
28
102
164
190
469
528
780
1248
1755
5264
x^2
400
900
1600
2500
4900
6400
10000
14400
22500
63600
We are now ready to determine the
values of b and a:
SUM =
Batch (x) Cost (y)
660
55.9
xy
5264
x^2
63600
b = n (xy) - (x)(y) = 9 (5264) - (660)(55.9)
n(x2) - (x)2
= 47376-36894
572400-435600
9(63600) - (660)2
= 10482 =
136800
a = y - bx = 55.9 - .0766(660) =
n
9
Our linear regression equation:
y c = a + bx
yc=
What is the cost of a batch of 125 pieces?
yc=
Series1
Linear (Series1)
14
12
10
8
6
4
2
0
Cost
Batch (x) Cost (y)
20
1.4
30
3.4
40
4.1
50
3.8
70
6.7
80
6.6
100
7.8
120
10.4
150
11.7
0
100
Batch size
200
Problem #7
Freight car
loadings at a
busy port are as
follows:
Week
#
Week
#
1
220
10
380
2
245
11
420
3
280
12
450
4
275
13
460
5
300
14
475
6
310
15
500
7
350
16
510
8
360
17
525
9
400
18
541
Period (t) Loads (y)
1
220
2
245
3
280
4
275
5
300
6
310
7
350
8
360
9
400
10
380
11
420
12
450
13
460
14
475
15
500
16
510
17
525
18
541
SUM 171
7,001
t^2
1
4
9
16
25
36
49
64
81
100
121
144
169
196
225
256
289
324
2,109
Problem #7
t*y
220
490
840
1,100
1,500
1,860
2,450
2,880
3,600
3,800
4,620
5,400
5,980
6,650
7,500
8,160
8,925
9,738
75,713
b = n (xy) - (x)(y)
n(x2) - (x)2
a = y - bx
n
Correlation (r)

A measure of the relationship
between two variables
• Strength
• Direction (positive or negative)

Ranges from -1.00 to +1.00
• Correlation close to 0 signifies a weak
relationship – other variables may be at
play
• Correlation close to +1 or -1 signifies a
strong relationship
Calculating a Correlation
Coefficient
r =
n( xy) - ( x)( y)
n( x2)- ( x)2 * n( y2) - ( y)2
Example 6: Continued
Batch (x) Cost (y)
660
55.9
SUM =
r =
xy
5264
x^2
63600
y^2
439.11
n( xy) - ( x)( y)
n( x2)- ( x)2 * n( y2) - ( y)2
r =
r =
9 (5264) - (660)(55.9)
9(63600)- (660)2 * 9(439.11) - (55.9)2
47376 - 36894 =
10482 = .985
136800 * 827.18
369.86 * 28.76
Coefficient of Determination
(r2)
How well a regression line “fits” the
data
 Ranges from 0.00 to 1.00
 The closer to 1.0, the better the fit

Example 6: Continued
Cost
15
Series1
10
Linear
(Series1)
5
0
0
50
100
150
200
Batch size
r = .985
r2 = .9852 = .97
Conclusion of Example

R = .985


Positive, close to one
R2 = .9852 = .97

Closer to one, the better the fit to the
line
Forecast Accuracy
Error - difference between actual
value and predicted value
 Mean absolute deviation (MAD)



Average absolute error
Mean squared error (MSE)

Average of squared error
Why can’t we simply calculate error for each observed
period and then select the technique with the lowest error?
Error Example
Period
1
2
3
4
5
6
7
8
9
10
#errors?
Actual
55
60
75
58
80
90
70
92
100
3PMA
5PMA
3P WMA
.6, .3, .1 EX SM .2
63.33333
64.33333
71
76
80
84
87.33333
65.6
72.6
74.6
78
86.4
68.5
63.3
72.9
83.8
77
85.2
94.6
6
4
6
LR
55.69
55
60.66
56
65.63
59.8
70.6
59.44
75.57
63.552
80.54
68.8416
85.51
69.07328 90.48
73.65862 95.45
78.9269 100.42
8
9
Does the # of errors calculated impact the "accuracy" comparison???
Calculating Error

Mathematically:
et = At - Ft
What do the negative errors mean?
How do they affect total error?
Calculating MAD and MSE
 Actual forecast
MAD =
n
2
(Actual forecast)

MSE =
n -1
Conclusions with MAD & MSE
The MAD and MSE can be used as a
comparison tool for several
forecasting techniques.
 The forecasting technique that yields
the lowest MAD and MSE is the
preferred forecasting method.

MAD & MSE Comparison
Tech
MAD
MSE
3 Period MA
3.6
8.1
Exp Sm .02
2.2
5.6
Exp Sm .04
2.6
6.1
Which technique would you select for your forecast approach?