CH_15_FORECASTING

Time Series Forecasting
Solution with QM for Windows (1 of 2)

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Time Series Forecasting
Solution with QM for Windows (2 of 2)

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Regression Methods
Overview

Time series techniques relate a single variable being forecast to
time.

Regression is a forecasting technique that measures the
relationship of one variable to one or more other variables.

Simplest form of regression is linear regression.

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Regression Methods
Linear Regression

Linear regression relates demand (dependent variable ) to an
independent variable.

y = a+bx

a= y−bx

b = ∑ xy−nxy
∑x2 −nx2

where: ∑x
n
x = = mean of the x data

y = ∑y = mean of the y data
n

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Regression Methods
Linear Regression Example (1 of 3)

State University Athletic Department.

Wins Attendance x y xy x2
4 36,300 (wins) (attendance, 1,000s) 145.2 16
6 40,100 240.6 36
6 41,200 4 36.3 247.2 36
8 53,000 6 40.1 424.0 64
6 44,000 6 41.2 264.0 36
7 45,600 8 53.0 319.2 49
5 39,000 6 44.0 195.0 25
7 47,500 7 45.6 332.5 49
5 39.0 2,167.7 311
7 47.5
49 346.7

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Regression Methods
Linear Regression Example (2 of 3)

x = 49 = 6.125
8

y = 346.9 = 43.34
8

b = ∑ xy − nx y = (2,167.70 − (8)(6.125)(43.34) = 4.06
∑x2 −nx2 (311) − (8)(6.125)2

a = y −bx = 43.34−(.406)(6.125) =18.46

Therefore, y =18.46+4.06x

Attendance forecast for x = 7 wins is
y =18.46+4.06(7) = 46.88 or 46,880

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Regression Methods
Linear Regression Example (3 of 3)

Figure 15.6 15-57
Linear Regression Line

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Regression Methods
Correlation (1 of 2)

Correlation is a measure of the strength of the relationship

between independent and dependent variables.

Formula:

r = n∑ xy−∑x∑ y
( ) n∑ 2 
x2 −  ∑ x2 y2 − ∑ y 
n∑    



Value lies between +1 and -1.
Value of zero indicates little or no relationship between

variables.

Values near 1.00 and -1.00 indicate strong linear relationship.

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Regression Methods
Correlation (2 of 2)

Value for State University example:

r = (8)(2,167.7)−(49)(346.7) =.948
(8)(15,224.7) (346.7)2
(8)(311) − (49)(49) − 


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Regression Methods
Coefficient of Determination

The Coefficient of determination is the percentage of the
variation in the dependent variable that results from the

independent variable.

Computed by squaring the correlation coefficient, r.

For State University example:
r = .948, r2 = .899

This value indicates that 89.9% of the amount of variation in
attendance can be attributed to the number of wins by the team,
with the remaining 10.1% due to other, unexplained, factors.

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Regression Analysis with Excel (1 of 6)

Exhibit 15.8 15-61

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Regression Analysis with Excel (2 of 6)

Exhibit 15.9 15-62

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Regression Analysis with Excel (3 of 6)

Exhibit 15.10 15-63

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Regression Analysis with Excel (4 of 6)

Exhibit 15.11 15-64

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Regression Analysis with Excel (5 of 6)

Exhibit 15.12 15-65

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Regression Analysis with Excel (6 of 6)

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15-66

Multiple Regression with Excel (1 of 4)

Multiple regression relates demand to two or more independent

variables.
General form:

y = β0 + β 1x1 + β 2x2 + . . . + β kxk
where β 0 = the intercept
β 1 . . . β k = parameters representing

contributions of the independent
variables
x1 . . . xk = independent variables

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Multiple Regression with Excel (2 of 4)

State University example:

Wins Promotion ($) Attendance

4 29,500 36,300
6 55,700 40,100
6 71,300 41,200
8 87,000 53,000
6 75,000 44,000
7 72,000 45.600
5 55,300 39,000
7 81,600 47,500

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Multiple Regression with Excel (3 of 4)

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Multiple Regression with Excel (4 of 4)

Exhibit 15.15

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Example Problem Solution
Computer Software Firm (1 of 4)

Problem Statement:

For data below, develop an exponential smoothing forecast using α =
.40, and an adjusted exponential smoothing forecast using α = .40 and β
= .20.

Compare the accuracy of the forecasts using MAD and cumulative error.

Period Units
1 56
2 61
3 55
4 70
5 66
6 65
7 72
8 75

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Example Problem Solution
Computer Software Firm (2 of 4)

Step 1: Compute the Exponential Smoothing Forecast.
Ft+1 = α Dt + (1 - α)Ft

Step 2: Compute the Adjusted Exponential Smoothing
Forecast

AFt+1 = Ft +1 + Tt+1
Tt+1 = β(Ft +1 - Ft) + (1 - β)Tt

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Example Problem Solution
Computer Software Firm (3 of 4)

Period Dt Ft AFt Dt - Ft Dt - AFt

1 56    
2 61 56.00 56.00 5.00 5.00
3 55 58.00 58.40 -3.00 -3.40
4 70 56.80 56.88 13.20 13.12
5 66 62.08 63.20 3.92 2.80
6 65 63.65 64.86 1.35 0.14
7 72 64.18 65.26 7.81 6.73
8 75 67.31 68.80 7.68 6.20
9  70.39 72.19
 
35.97 30.60

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Example Problem Solution
Computer Software Firm (4 of 4)

Step 3: Compute the MAD Values

MAD(Ft ) = ∑ Dt − Ft = 41.97 =5.99
n 7

MAD( AFt ) = ∑ Dt − AFt = 37.39 = 5.34
n 7

Step 4: Compute the Cumulative Error.
E(Ft) = 35.97

E(AFt) = 30.60

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Example Problem Solution
Building Products Store (1 of 5)

For the following data:
Develop a linear regression model
Determine the strength of the linear relationship using
correlation.
Determine a forecast for lumber given 10 building permits
in the next quarter.

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Example Problem Solution
Building Products Store (2 of 5)

Quarter Building Lumber Sales
Permits, x (1,000s of bd ft), y
1
2 8 12.6
3 12 16.3
4
5 7 9.3
6 9 11.5
7 15 18.1
8 6
9 5 7.6
10 8 6.2
10 14.2
12 15.0
17.8

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Example Problem Solution
Building Products Store (3 of 5)

Step 1: Compute the Components of the Linear Regression Equation.

x = 92 = 92
10

y =12180.6 =12.86

b = ∑ xy − nx y = (1,290.3) − (10)(9.2)(12.86) =1.25
∑x2 −nx2 (932) − (10)(9.2)2

a = y −bx =12.86−(1.25)(9.2) =1.36

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Example Problem Solution
Building Products Store (4 of 5)

Step 2: Develop the Linear regression equation.
y = a + bx, y = 1.36 + 1.25x

Step 3: Compute the Correlation Coefficient.

r = n∑ xy−∑x∑ y
( ) n∑ 2 
x2 −  ∑ x2 y2 − ∑ y 
n∑    



r = (10)(1,170.3)−(92)(128.6) =.925
(92)(92)(10)(1,810.48) (128.6)2 
(10)(932) − − 

 

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Example Problem Solution
Building Products Store (5 of 5)

Step 4: Calculate the forecast for x = 10 permits.

Y = a + bx = 1.36 + 1.25(10) = 13.86 or 1,386 board ft

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