Exponential Smoothing in Forecasting
Chapter 4
Class 2
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
Exponential Smoothing
Exponential Smoothing
Exponential Smoothing
Exponential Smoothing
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F
F
t
t
= F
= F
t
t
– 1
– 1
+
+
(
(
A
A
t
t
– 1
– 1
-
-
F
F
t
t
– 1
– 1
)
)
where
where
F
F
t
t
=
=
new forecast
new forecast
F
F
t
t
– 1
– 1
=
=
previous forecast
previous forecast
=
=
smoothing (or weighting)
smoothing (or weighting)
constant
constant
(0
(0
1)
1)
Exponential Smoothing Example
Exponential Smoothing Example
Predicted demand
Predicted demand
= 142
= 142
Ford Mustangs
Ford Mustangs
Actual demand
Actual demand
= 153
= 153
Smoothing constant
Smoothing constant
= .20
= .20
New forecast
New forecast
= 142 + .2(153 – 142)
= 142 + .2(153 – 142)
= 142 + 2.2
= 142 + 2.2
= 144.2 ≈ 144 cars
= 144.2 ≈ 144 cars
Problem 4.4
Problem 4.4
A check-processing center uses exponential
A check-processing center uses exponential
smoothing to forecast the number of incoming
smoothing to forecast the number of incoming
checks each month. The number of checks received
checks each month. The number of checks received
in
in
June was 40 million
June was 40 million
, while the
, while the
forecast was 42
forecast was 42
million
million
. A
. A
smoothing constant of .2
smoothing constant of .2
is used.
is used.
A) What is the forecast for
A) What is the forecast for
July
July
?
?
B) If the center received 45 million checks in July,
B) If the center received 45 million checks in July,
what would be the forecast for
what would be the forecast for
August
August
?
?
C) Why might this be an inappropriate forecasting
C) Why might this be an inappropriate forecasting
method for this situation?
method for this situation?
Problem 4.4
Problem 4.4
A) What is the forecast for
A) What is the forecast for
July
July
?
?
B) If the center received 45 million checks in July, what would be the
B) If the center received 45 million checks in July, what would be the
forecast for
forecast for
August
August
?
?
C) Why might this be an inappropriate forecasting method for this
C) Why might this be an inappropriate forecasting method for this
situation?
situation?
Problem 4.18
Problem 4.18
Consider the following actual (At) and forecast (Ft) demand
levels for a product.
The first forecast, F 1 , was derived by observing A 1 and
setting equal to A 1 . Subsequent forecast averages were
derived by exponential smoothing. Using the exponential
smoothing method, find the forecast for time period 5
Problem 4.18
Problem 4.18
We need to find the smoothing constant
.
We know in general that
F
t
= F
t–
1
+
(
A
t–
1
– F
t–
1
);
t =
2, 3, 4
.
Choose either
t =
3
or
t =
4
(
t =
2
won’t let us find
because
F
2
=
50 = 50 +
(50 – 50)
holds for
any
).
Let’s pick
t
= 3
.
Then
F
3
= 48 = 50 +
(42 – 50)
or 48 = 50 + 42
– 50
or –2 = –8
So,
.25 =
Now we can find
F
5
:
F
5
= 50 +
(46 – 50)
F
5
= 50 + 46
– 50
= 50 – 4
For
= .25,
F
5
=
50 – 4(.25) = 49
The forecast for time period 5 = 49 units.
Common Measures of Error
Common Measures of Error
Common Measures of Error
Common Measures of Error
Comparison of Forecast Error
Comparison of Forecast Error
Comparison of Forecast Error
Comparison of Forecast Error
Comparison of Forecast Error
Comparison of Forecast Error
Comparison of Forecast Error
Comparison of Forecast Error
Comparison of Forecast Error
Comparison of Forecast Error
Problem 4.14
Problem 4.14
following are two weekly forecasts made by two different
following are two weekly forecasts made by two different
methods for the number of gallons of gasoline, in thousands,
methods for the number of gallons of gasoline, in thousands,
demanded at a local gasoline station. Also shown are actual
demanded at a local gasoline station. Also shown are actual
demand levels in thousands of gallons:
demand levels in thousands of gallons:
What are the MAD and MSE for each method?
What are the MAD and MSE for each method?
Problem 4.14
Problem 4.14
What are the MAD and MSE for each method?
What are the MAD and MSE for each method?
Method 1:
MAD: (0.20 + 0.05 + 0.05 + 0.20)/4 = .125
MSE : (0.04 + 0.0025 + 0.0025 + 0.04)/4 = .021
Method 2:
MAD: (0.1 + 0.20 + 0.10 + 0.11) / 4 = .1275
MSE : (0.01 + 0.04 + 0.01 + 0.0121) / 4 = .018
Exponential Smoothing with
Exponential Smoothing with
Trend Adjustment
Trend Adjustment
When a trend is present, exponential
When a trend is present, exponential
smoothing must be modified
smoothing must be modified
F
F
t
t
=
=
(
(
A
A
t
t
- 1
- 1
) + (1 -
) + (1 -
)(
)(
F
F
t
t
- 1
- 1
+
+
T
T
t
t
- 1
- 1
)
)
T
T
t
t
=
=
(
(
F
F
t
t
-
-
F
F
t
t
- 1
- 1
) + (1 -
) + (1 -
)
)
T
T
t
t
- 1
- 1
Step 1: Compute F
Step 1: Compute F
t
t
Step 2: Compute T
Step 2: Compute T
t
t
Step 3: Calculate the forecast FIT
Step 3: Calculate the forecast FIT
t
t
=
=
F
F
t
t
+
+
T
T
t
t
Exponential Smoothing with
Exponential Smoothing with
Trend Adjustment
Trend Adjustment
Exponential Smoothing with
Exponential Smoothing with
Trend Adjustment Example
Trend Adjustment Example
Table 4.1
Table 4.1
Table 4.1
Table 4.1
F
2
=
A
1
+ (1 -
)(
F
1
+
T
1
)
F
2
= (.2)(12) + (1 - .2)(11 + 2)
= 2.4 + 10.4 = 12.8 units
Step 1: Forecast for Month 2
Exponential Smoothing with
Exponential Smoothing with
Trend Adjustment Example
Trend Adjustment Example
Table 4.1
Table 4.1
T
2
=
(
F
2
-
F
1
) + (1 -
)
T
1
T
2
= (.4)(12.8 - 11) + (1 - .4)(2)
= .72 + 1.2 = 1.92 units
Step 2: Trend for Month 2
Exponential Smoothing with
Exponential Smoothing with
Trend Adjustment Example
Trend Adjustment Example
Table 4.1
Table 4.1
FIT
2
= F
2
+ T
1
FIT
2
= 12.8 + 1.92
= 14.72 units
Step 3: Calculate FIT for Month 2
Exponential Smoothing with
Exponential Smoothing with
Trend Adjustment Example
Trend Adjustment Example
Table 4.1
Table 4.1
15.18
15.18
2.10
2.10
17.28
17.28
17.82
17.82
2.32
2.32
20.14
20.14
19.91
19.91
2.23
2.23
22.14
22.14
22.51
22.51
2.38
2.38
24.89
24.89
24.11
24.11
2.07
2.07
26.18
26.18
27.14
27.14
2.45
2.45
29.59
29.59
29.28
29.28
2.32
2.32
31.60
31.60
32.48
32.48
2.68
2.68
35.16
35.16
Exponential Smoothing with
Exponential Smoothing with
Trend Adjustment Example
Trend Adjustment Example
Figure 4.3
Figure 4.3
Exponential Smoothing with
Exponential Smoothing with
Trend Adjustment Example
Trend Adjustment Example
Problem 4.19
Problem 4.19
Income at the law firm of Smith and Wesson for the period
Income at the law firm of Smith and Wesson for the period
February to July was as follows:
February to July was as follows:
Month
Month
FEB
FEB
MAR
MAR
APR
APR
MAY
MAY
JUNE
JUNE
JUL
JUL
Income
Income
70.0
70.0
68.5
68.5
64.8
64.8
71.7
71.7
71.3
71.3
72.8
72.8
(In $ thousand)
(In $ thousand)
Use trend-adjusted exponential smoothing to forecast the
Use trend-adjusted exponential smoothing to forecast the
law firm’s
law firm’s
August income
August income
. Assume that the initial forecast for
. Assume that the initial forecast for
February is
February is
$65,000
$65,000
and the initial trend adjustment is
and the initial trend adjustment is
0
0
.The
.The
smoothing constants selected are
smoothing constants selected are
α = 0.1 and β =0.2
α = 0.1 and β =0.2
Problem 4.19
Problem 4.19
Trend Projections
Trend Projections
Fitting a trend line to historical data points to
Fitting a trend line to historical data points to
project into the medium-to-long-range
project into the medium-to-long-range
Linear trends can be found using the least
Linear trends can be found using the least
squares technique
squares technique
Least Squares Method
Least Squares Method
Figure 4.4
Figure 4.4
Figure 4.4
Figure 4.4
Least squares method
minimizes the sum of the
squared errors (deviations)
Least Squares Method
Least Squares Method
Equations to calculate the regression variables
Equations to calculate the regression variables
Least Squares Method
Least Squares Method
Least Squares Example
Least Squares Example
Least Squares Example
Least Squares Example
Least Squares Example
Least Squares Example
Least Squares Requirements
Least Squares Requirements
1.
We always plot the data to insure a
We always plot the data to insure a
linear relationship
linear relationship
2.
We do not predict time periods far
We do not predict time periods far
beyond the database
beyond the database
3.
Deviations around the least squares line
Deviations around the least squares line
are assumed to be random
are assumed to be random
Problem 4.25
Problem 4.25
The following gives the number of accidents that occurred
The following gives the number of accidents that occurred
on Florida State Highway 101 during the last 4 months:
on Florida State Highway 101 during the last 4 months:
Forecast the number of accidents that will occur in
Forecast the number of accidents that will occur in
May
May
,
,
using
using
least squares regression
least squares regression
to derive a trend equation.
to derive a trend equation.
Problem 4.25
Problem 4.25
Problem 4.25
Problem 4.25
Exponential smoothing is a popular method used in forecasting, where the most recent data is weighted the most. It involves a smoothing constant that ranges from 0 to 1, with little record-keeping of past data required. The method calculates new forecasts based on previous forecasts and actual demand. However, it may not be suitable for all situations, as seen in a check-processing center's case. Learn more about exponential smoothing through practical examples and problems provided.
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Presentation Transcript
Chapter 4 Class 2
Exponential Smoothing 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
Exponential Smoothing last period s forecast + (last period s actual demand last period s forecast) New forecast = Ft = Ft 1 + (At 1 - Ft 1) where Ft = new forecast Ft 1 = = previous forecast smoothing (or weighting) constant (0 1)
Exponential Smoothing Example Predicted demand = 142 Ford Mustangs Actual demand = 153 Smoothing constant = .20 New forecast = 142 + .2(153 142) = 142 + 2.2 = 144.2 144 cars
Problem 4.4 A check-processing center uses exponential smoothing to forecast the number of incoming checks each month. The number of checks received in June was 40 million, while the forecast was 42 million. A smoothing constant of .2 is used. A) What is the forecast for July? B) If the center received 45 million checks in July, what would be the forecast for August? C) Why might this be an inappropriate forecasting method for this situation?
Problem 4.4 A) What is the forecast for July? B) If the center received 45 million checks in July, what would be the forecast for August? C) Why might this be an inappropriate forecasting method for this situation?
Problem 4.18 Consider the following actual (At) and forecast (Ft) demand levels for a product. The first forecast, F 1 , was derived by observing A 1 and setting equal to A 1 . Subsequent forecast averages were derived by exponential smoothing. Using the exponential smoothing method, find the forecast for time period 5
Problem 4.18 We need to find the smoothing constant . We know in general that Ft = Ft 1 + (At 1 Ft 1); t = 2, 3, 4. Choose either t = 3or t = 4(t = 2won t let us find because F2 = 50 = 50 + (50 50) holds for any ). Let s pick t = 3. Then F3 = 48 = 50 + (42 50) or 48 = 50 + 42 50 or 2 = 8 So, .25 = Now we can find F5 :F5 = 50 + (46 50) F5 = 50 + 46 50 = 50 4 For = .25, F5 = 50 4(.25) = 49 The forecast for time period 5 = 49 units.
Common Measures of Error Mean Absolute Deviation (MAD) |actual - forecast| MAD = n Mean Squared Error (MSE) (forecast errors)2 MSE = n
Common Measures of Error Mean Absolute Percent Error (MAPE) n 100 |actuali - forecasti|/actuali i = 1 MAPE = n
Comparison of Forecast Error Rounded Forecast with = .10 Absolute Deviation for = .10 Rounded Forecast with = .50 Absolute Deviation for = .50 Quarter Actual Tonnage Unloaded 1 2 3 4 5 6 7 8 180 168 159 175 190 205 180 182 175 176 175 173 173 175 178 178 5 8 175 178 173 166 170 180 193 186 5 10 14 16 2 17 30 2 4 84 9 20 25 13 4 100
Comparison of Forecast Error |deviations| Rounded Forecast with = .10 Absolute Deviation for = .10 Rounded Forecast with = .50 Absolute Deviation for = .50 MAD = Quarter Actual Tonage Unloaded For = .10 n 1 2 3 4 5 6 7 8 180 168 159 175 190 205 180 182 175 176 175 173 173 175 178 178 5 8 175 178 173 166 170 180 193 186 5 = 84/8 = 10.50 10 14 16 2 17 30 2 4 84 For = .50 9 20 25 13 = 100/8 = 12.50 4 100
Comparison of Forecast Error (forecast errors)2 Rounded Forecast with = .10 Absolute Deviation for = .10 Rounded Forecast with = .50 Absolute Deviation for = .50 MSE = Quarter Actual Tonage Unloaded For = .10 n 1 2 3 4 5 6 7 8 180 168 159 175 190 205 180 182 175 176 175 173 173 175 178 178 5 8 175 178 173 166 170 180 193 186 5 = 1,558/8 = 194.75 10 14 16 2 17 30 2 4 84 For = .50 9 20 25 13 = 1,612/8 = 201.50 4 100 MAD 10.50 12.50
Comparison of Forecast Error 100 |deviationi|/actuali i = 1 n Rounded Forecast with = .10 Absolute Deviation for = .10 Rounded Forecast with = .50 Absolute Deviation for = .50 MAPE = Quarter Actual Tonage Unloaded For = .10 n 1 2 3 4 5 6 7 8 180 168 159 175 190 205 180 182 175 176 175 173 173 175 178 178 5 8 175 178 173 166 170 180 193 186 5 = 45.62/8 = 5.70% 10 14 16 2 17 30 2 4 84 For = .50 9 20 25 13 = 54.8/8 = 6.85% 4 100 MAD MSE 10.50 194.75 12.50 201.50
Comparison of Forecast Error Rounded Forecast with = .10 Absolute Deviation for = .10 Rounded Forecast with = .50 Absolute Deviation for = .50 Quarter Actual Tonnage Unloaded 1 2 3 4 5 6 7 8 180 168 159 175 190 205 180 182 175 176 175 173 173 175 178 178 5 8 175 178 173 166 170 180 193 186 5 10 14 16 2 17 30 2 4 84 9 20 25 13 4 100 MAD MSE MAPE 10.50 194.75 5.70% 12.50 201.50 6.85%
Problem 4.14 following are two weekly forecasts made by two different methods for the number of gallons of gasoline, in thousands, demanded at a local gasoline station. Also shown are actual demand levels in thousands of gallons: What are the MAD and MSE for each method?
Problem 4.14 What are the MAD and MSE for each method? Method 1: MAD: (0.20 + 0.05 + 0.05 + 0.20)/4 = .125 MSE : (0.04 + 0.0025 + 0.0025 + 0.04)/4 = .021 Method 2: MAD: (0.1 + 0.20 + 0.10 + 0.11) / 4 = .1275 MSE : (0.01 + 0.04 + 0.01 + 0.0121) / 4 = .018
Exponential Smoothing with Trend Adjustment When a trend is present, exponential smoothing must be modified Forecast including (FITt) = trend exponentially smoothed (Ft) + forecast (Tt) smoothed trend exponentially
Exponential Smoothing with Trend Adjustment Ft = (At - 1) + (1 - )(Ft - 1 + Tt - 1) Tt = (Ft - Ft - 1) + (1 - )Tt - 1 Step 1: Compute Ft Step 2: Compute Tt Step 3: Calculate the forecast FITt = Ft + Tt
Exponential Smoothing with Trend Adjustment Example Forecast Including Trend, FITt 13.00 Month(t) 1 2 3 4 5 6 7 8 9 10 Actual Demand (At) 12 17 20 19 24 21 31 28 36 Smoothed Forecast, Ft 11 Smoothed Trend, Tt 2 Table 4.1
Exponential Smoothing with Trend Adjustment Example Forecast Including Trend, FITt 13.00 Month(t) 1 2 3 4 5 6 7 8 9 10 Actual Demand (At) 12 17 20 19 24 21 31 28 36 Smoothed Forecast, Ft 11 Smoothed Trend, Tt 2 Step 1: Forecast for Month 2 F2 = A1 + (1 - )(F1 + T1) F2 = (.2)(12) + (1 - .2)(11 + 2) = 2.4 + 10.4 = 12.8 units Table 4.1
Exponential Smoothing with Trend Adjustment Example Forecast Including Trend, FITt 13.00 Month(t) 1 2 3 4 5 6 7 8 9 10 Actual Demand (At) 12 17 20 19 24 21 31 28 36 Smoothed Forecast, Ft 11 12.80 Smoothed Trend, Tt 2 Step 2: Trend for Month 2 T2 = (F2 - F1) + (1 - )T1 T2 = (.4)(12.8 - 11) + (1 - .4)(2) = .72 + 1.2 = 1.92 units Table 4.1
Exponential Smoothing with Trend Adjustment Example Forecast Including Trend, FITt 13.00 Month(t) 1 2 3 4 5 6 7 8 9 10 Actual Demand (At) 12 17 20 19 24 21 31 28 36 Smoothed Forecast, Ft 11 12.80 Smoothed Trend, Tt 2 1.92 Step 3: Calculate FIT for Month 2 FIT2 = F2 + T1 FIT2 = 12.8 + 1.92 = 14.72 units Table 4.1
Exponential Smoothing with Trend Adjustment Example Forecast Including Trend, FITt 13.00 14.72 17.28 20.14 22.14 24.89 26.18 29.59 31.60 35.16 Month(t) 1 2 3 4 5 6 7 8 9 10 Actual Demand (At) 12 17 20 19 24 21 31 28 36 Smoothed Forecast, Ft 11 12.80 15.18 17.82 19.91 22.51 24.11 27.14 29.28 32.48 Smoothed Trend, Tt 2 1.92 2.10 2.32 2.23 2.38 2.07 2.45 2.32 2.68 Table 4.1
Exponential Smoothing with Trend Adjustment Example 35 Actual demand (At) 30 25 Product demand 20 15 Forecast including trend (FITt) 10 5 0 | | | | | | | | | 1 2 3 4 5 6 7 8 9 Time (month) Figure 4.3
Problem 4.19 Income at the law firm of Smith and Wesson for the period February to July was as follows: Month Income (In $ thousand) FEB 70.0 MAR APR 68.5 MAY JUNE JUL 71.7 71.3 64.8 72.8 Use trend-adjusted exponential smoothing to forecast the law firm s August income. Assume that the initial forecast for February is $65,000 and the initial trend adjustment is 0.The smoothing constants selected are = 0.1 and =0.2
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 ^ 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
Least Squares Method Actual observation (y value) Deviation7 Values of Dependent Variable Deviation5 Deviation6 Deviation3 Deviation4 Deviation1 Deviation2 ^ Trend line, y = a + bx Time period Figure 4.4
Least Squares Method Actual observation (y value) Values of Dependent Variable Deviation7 Deviation5 Deviation6 Least squares method minimizes the sum of the Deviation3 Deviation4 squared errors (deviations) Deviation1 Deviation2 ^ Trend line, y = a + bx Time period Figure 4.4
Least Squares Method Equations to calculate the regression variables ^ y = a + bx xy - nxy x2 - nx2 b = a = y - bx
Least Squares Example Time Period (x) Electrical Power Demand x2 Year xy 1999 2000 2001 2002 2003 2004 2005 1 2 3 4 5 6 7 74 79 80 90 105 142 122 1 4 9 74 158 240 360 525 852 854 16 25 36 49 x2 = 140 x = 28 x = 4 y = 692 y = 98.86 xy = 3,063 3,063 - (7)(4)(98.86) xy - nxy b = x2 - nx2 = 140 - (7)(42) = 10.54 a = y - bx = 98.86 - 10.54(4) = 56.70
Least Squares Example Time Period (x) Electrical Power Demand x2 Year xy 1999 2000 2001 2002 2003 2004 2005 1 2 3 4 5 6 7 74 79 80 90 105 142 122 1 4 9 74 158 240 360 525 852 854 The trend line is 16 25 36 49 ^ y = 56.70 + 10.54x x2 = 140 x = 28 x = 4 y = 692 y = 98.86 xy = 3,063 3,063 - (7)(4)(98.86) xy - nxy b = x2 - nx2 = 140 - (7)(42) = 10.54 a = y - bx = 98.86 - 10.54(4) = 56.70
Least Squares Example Trend line, y = 56.70 + 10.54x ^ 160 150 140 130 120 110 100 90 80 70 60 50 Power demand | | | | | | | | | 1999 2000 2001 2002 2003 Year 2004 2005 2006 2007
Least Squares Requirements 1. We always plot the data to insure a linear relationship 2. We do not predict time periods far beyond the database 3. Deviations around the least squares line are assumed to be random
Problem 4.25 The following gives the number of accidents that occurred on Florida State Highway 101 during the last 4 months: Month Number of Accidents January 30 February 40 March 60 April 90 Forecast the number of accidents that will occur in May, using least squares regression to derive a trend equation.
