Chapter 5 - Forecasting 1 Chapter 5 Forecasting Introduction to Management Science 8th Edition by Bernard W. Taylor III

Chapter 5 - Forecasting 1

Chapter 5

Forecasting

Introduction to Management Science

8th Edition

by

Bernard W. Taylor III


Forecasting Components

Time Series Methods

Forecast Accuracy

Time Series Forecasting Using Excel

Time Series Forecasting Using QM for Windows

Regression Methods

Chapter Topics


A variety of forecasting methods are available for use depending on the time frame of the forecast and the existence of patterns.

Time Frames:

Short-range (one to two months) Medium-range (two months to one or two years) Long-range (more than one or two years)

Patterns:

TrendRandom variationsCyclesSeasonal pattern

Forecasting Components


Trend - A long-term movement of the item being forecast.

Random variations - movements that are not predictable and follow no pattern.

Cycle - A movement, up or down, that repeats itself over a lengthy time span.

Seasonal pattern - Oscillating movement in demand that occurs periodically in the short run and is repetitive.

Forecasting ComponentsPatterns (1 of 2)


Figure 5.1Forms of Forecast Movement: (a) Trend, (b) Cycle, (c) Seasonal

Pattern, (d) Trend with Seasonal Pattern

Forecasting ComponentsPatterns (2 of 2)

trend-line


Forecasting ComponentsForecasting Methods

Times Series - Statistical techniques that use historical data to predict future behavior.

Regression Methods - Regression (or causal ) methods that attempt to develop a mathematical relationship between the item being forecast and factors that cause it to behave the way it does.

Qualitative Methods - Methods using judgment, expertise and opinion to make forecasts.


Forecasting ComponentsQualitative Methods

Qualitative methods, the “jury of executive opinion,” is the most common type of forecasting method for long-term strategic planning.

Performed by individuals or groups within an organization, sometimes assisted by consultants and other experts, whose judgments and opinion are considered valid for the forecasting issue.

Usually includes specialty functions such as marketing, engineering, purchasing, etc. in which individuals have experience and knowledge of the forecasted item.

Supporting techniques include the Delphi Method, market research, surveys, etc.


Time Series MethodsOverview

Statistical techniques that make use of historical data collected over a long period of time.

Methods assume that what has occurred in the past will continue to occur in the future.

Forecasts based on only one factor - time.

Chapter 5 - Forecasting 9€

MAn=1n Dii=1

n∑

where:n= number of periods in the moving averageDi= data in period i

Time Series MethodsMoving Average (1 of 5)

Moving average uses values from the recent past to develop forecasts.

This dampens or smoothes out random increases and decreases.

Useful for forecasting relatively stable items that do not display any trend or seasonal pattern.

Formula for:


Example: Instant Paper Clip Supply Company forecast of orders for the next month.

Three-month moving average:

Five-month moving average:€

MA3=13Dii=1

3∑ =90+110+130

3=110 orders

€

MA3=15Dii=1

5∑ =90+110+130+75+50

5=91 orders



Figure 5.2Three- and Five-Month Moving Averages



Figure 5.2Three- and Five-Month Moving Averages




Longer-period moving averages react more slowly to changes in demand than do shorter-period moving averages.

The appropriate number of periods to use often requires trial-and-error experimentation.

Moving average does not react well to changes (trends, seasonal effects, etc.) but is easy to use and inexpensive.

Good for short-term forecasting.


In a weighted moving average, weights are assigned to the most recent data.Formula:

Time Series MethodsWeighted Moving Average (1 of 2)


Determining precise weights and number of periods requires trial-and-error experimentation.

Time Series MethodsWeighted Moving Average (2 of 2)


Exponential smoothing weights recent past data more strongly than more distant data.

Two forms: simple exponential smoothing and adjusted exponential smoothing.

Simple exponential smoothing:

Ft + 1 = Dt + (1 - )Ft

where: Ft + 1 = the forecast for the next period Dt = actual demand in the present period Ft = the previously determined forecast for the present period = a weighting factor (smoothing constant) use F1 = D1.

Time Series MethodsExponential Smoothing (1 of 11)


The most commonly used values of are between.10 and .50.

Determination of is usually judgmental and subjective and often based on trial-and -error experimentation.



Example: PM Computer Services (see Table 5.4).

Exponential smoothing forecasts using smoothing constant of .30.

Forecast for period 2 (February):

F2 = D1 + (1- )F1 = (.30)(37) + (.70)(37) = 37 units

Forecast for period 3 (March):

F3 = D2 + (1- )F2 = (.30)(40) + (.70)(37) = 37.9 units



Table 5.4Exponential Smoothing Forecasts, = .30 and = .50


Using

F1 = D1


The forecast that uses the higher smoothing constant (.50) reacts more strongly to changes in demand than does the forecast with the lower constant (.30).

Both forecasts lag behind actual demand.

Both forecasts tend to be consistently lower than actual demand.

Low smoothing constants are appropriate for stable data without trend; higher constants appropriate for data with trends.



Figure 5.3Exponential Smoothing Forecasts



Adjusted exponential smoothing: exponential smoothing with a trend adjustment factor added.

Formula:

AFt + 1 = Ft + 1 + Tt+1

where: Tt = an exponentially smoothed trend factor: Tt + 1 = (Ft + 1 - Ft) + (1 - )Tt

Tt = the last (previous) period’s trend factor = smoothing constant for trend ( a value between zero

and one).

• Reflects the weight given to the most recent trend data.

• Determined subjectively.



Example: PM Computer Services exponential smoothedforecasts with = .50 and = .30 (see Table 5.5).

Start with T2 = 0.00

Adjusted forecast for period 3:

T3 = (F3 - F2) + (1 - )T2

= (.30)(38.5 - 37.0) + (.70)(0) = 0.45

AF3 = F3 + T3 = 38.5 + 0.45 = 38.95



Table 5.5Adjusted Exponentially Smoothed Forecast Values


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


Adjusted forecast is consistently higher than the simple exponentially smoothed forecast.

It is more reflective of the generally increasing trend of the data.



Figure 5.4Adjusted Exponentially Smoothed Forecast



€

y=a+bx

where: a= intercept (at period 0) b= slope of the line x= the time period y= forecast for demand for period x

€

b= xy∑ −nxyx2∑ −nx

a=y−bx

where: n= number of periods x=1

n x∑ y=1

n y∑

When demand displays an obvious trend over time, a least squares regression line , or linear trend line, can be used to forecast.

Formula:

Time Series MethodsLinear Trend Line (1 of 5)


Example: PM Computer Services (see Table 5.6)

€

x=7812

=6.5 y=55712

=46.42

b= xy∑ −nxyx2∑ −nx2

=3,867−(12)(6.5)(46.42)650−12(6.5)2

=1.72

a=y−bx=46.42−(1.72)(6.5)=35.2

y=35.2+1.72x linear trend line

for period 13, x = 13, y=35.2+1.72(13)=57.56



Table 5.6Least Squares Calculations



A trend line does not adjust to a change in the trend as does the exponential smoothing method.

This limits its use to shorter time frames in which trend will not change.



Figure 5.5Linear Trend Line



A seasonal pattern is a repetitive up-and-down movement in demand.

Seasonal patterns can occur on a monthly, weekly, or daily basis.

A seasonally adjusted forecast can be developed by multiplying the normal forecast by a seasonal factor.

A seasonal factor can be determined by dividing the actual demand for each seasonal period by total annual demand:

Si =Di/D

Time Series MethodsSeasonal Adjustments (1 of 4)


Seasonal factors lie between zero and one and represent the portion of total annual demand assigned to each season.

Seasonal factors are multiplied by annual demand to provide adjusted forecasts for each period.



S1 = D1/ D = 42.0/148.7 = 0.28 S2 = D2/ D = 29.5/148.7 = 0.20 S3 = D3/ D = 21.9/148.7 = 0.15 S4 = D4/ D = 55.3/148.7 = 0.37

Table 5.7Demand for Turkeys at Wishbone Farms

Example: Wishbone Farms



Multiply forecasted demand for entire year by seasonal factors to determine quarterly demand.

Forecast for entire year (trend line for data in Table 5.7):

y = 40.97 + 4.30x = 40.97 + 4.30(4) = 58.17

Seasonally adjusted forecasts:

SF1 = (S1)(F5) = (.28)(58.17) = 16.28

SF2 = (S2)(F5) = (.20)(58.17) = 11.63

SF3 = (S3)(F5) = (.15)(58.17) = 8.73

SF4 = (S4)(F5) = (.37)(58.17) = 21.53


Note the potential for confusion: the Si are “seasonal factors” (fractions),

whereas the SFi are “seasonally adjusted forecasts” (commodity values)!


Forecasts will always deviate from actual values.

Difference between forecasts and actual values referred to as forecast error.

Would like forecast error to be as small as possible.

If error is large, either technique being used is the wrong one, or parameters need adjusting.

Measures of forecast errors:

Mean Absolute deviation (MAD)

Mean absolute percentage deviation (MAPD)

Cumulative error (E-bar)

Average error, or bias (E)

Forecast AccuracyOverview


MAD is the average absolute difference between the forecast and actual demand.

Most popular and simplest-to-use measures of forecast error.

Formula:

€

MAD=1n Dt−Ft∑

where:

t = the period number Dt = demand in period t Ft = the forecast for period t n = the total number of periods

Forecast AccuracyMean Absolute Deviation (1 of 7)


Example: PM Computer Services (see Table 5.8).

Compare accuracies of different forecasts using MAD:

€

MAD=1n Dt−Ft∑ = 53.41

11=4.85



Table 5.8Computational Values for MAD



The lower the value of MAD relative to the magnitude of the data, the more accurate the forecast.

When viewed alone, MAD is difficult to assess.

Must be considered in light of magnitude of the data.



Can be used to compare accuracy of different forecasting techniques working on the same set of demand data (PM Computer Services):

Exponential smoothing ( = .50): MAD = 4.04

Adjusted exponential smoothing ( = .50, = .30): MAD = 3.81

Linear trend line: MAD = 2.29

Linear trend line has lowest MAD; increasing from .30 to .50 improved smoothed forecast.



A variation on MAD is the mean absolute percent deviation (MAPD).

Measures absolute error as a percentage of demand rather than per period.

Eliminates problem of interpreting the measure of accuracy relative to the magnitude of the demand and forecast values.

Formula:

10.3%or 103.520

41.53 ==∑

∑ −=

tDtFtDMAPD



MAPD for other three forecasts:

Exponential smoothing ( = .50): MAPD = 8.5%

Adjusted exponential smoothing ( = .50, = .30): MAPD = 8.1%

Linear trend: MAPD = 4.9%



Cumulative error is the sum of the forecast errors (E =et).

A relatively large positive value indicates forecast is biased low, a large negative value indicates forecast is biased high.

If preponderance of errors are positive, forecast is consistently low; and vice versa.

Cumulative error for trend line is always almost zero, and is therefore not a good measure for this method.

Cumulative error for PM Computer Services can be read directly from Table 5.8.

E = et = 49.31 indicating forecasts are frequently below actual demand.

Forecast AccuracyCumulative Error (1 of 2)


Cumulative error for other forecasts:

Exponential smoothing ( = .50): E = 33.21

Adjusted exponential smoothing ( = .50, =.30):E = 21.14

Average error (bias) is the per period average of cumulative error.

Average error for exponential smoothing forecast:

A large positive value of average error indicates a forecast is biased low; a large negative error indicates it is biased high.

Forecast AccuracyCumulative Error (2 of 2)

€

E= et∑n =49.31

11=4.48


Results consistent for all forecasts:

Larger value of alpha is preferable.Adjusted forecast is more accurate than exponential

smoothing forecasts.Linear trend is more accurate than all the others.

Table 5.9Comparison of Forecasts for PM Computer Services

Forecast AccuracyExample Forecasts by Different Measures


Exhibit 5.1

Time Series Forecasting Using Excel (1 of 4)


Exhibit 5.2



Exhibit 5.3



Exhibit 5.4



Exhibit 5.5

Exponential Smoothing Forecast with Excel QM


Time Series ForecastingSolution with QM for Windows (1 of 2)

Exhibit 5.6


Exhibit 5.7

Time Series ForecastingSolution with QM for Windows (2 of 2)


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.

Regression MethodsOverview


y=a+bx

a=y−bx

b= xy−nxy∑x2−nx2

∑

where:

x= x∑n = mean of the x data

y= y∑n = mean of the y data

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

Regression MethodsLinear Regression

Linear function


State University athletic department.

Wins Attendance 4 6 6 8 6 7 5 7

36,300 40,100 41,200 53,000 44,000 45,600 39,000 47,500

x (wins)

y (attendance, 1,000s)

xy

x2

4 6 6 8 6 7 5 7

49

36.3 40.1 41.2 53.0 44.0 45.6 39.0 47.5

346.7

145.2 240.6 247.2 424.0 264.0 319.2 195.0 332.5

2,167.7

16 36 36 64 36 49 25 49

311

Regression MethodsLinear Regression Example (1 of 3)


x=498

=6.125

y=346.98

=43.34

b= xy∑ −nxy

x2−nx2∑

=(2,167.70−(8)(6.125)(43.34)(311)−(8)(6.125)2

=4.06

a=y−bx=43.34−(.406)(6.125)=18.46

Therefore, y=18.46+4.06x

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



Figure 5.6Linear Regression Line



Correlation is a measure of the strength of the relationship between independent and dependent variables.

Formula:

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: correlation and anti-correlation.

€

r =n xy − x∑ y∑∑

n x 2 − x 2∑( )∑[ ] n y 2 − y∑( )2

∑[ ]

Regression MethodsCorrelation (1 of 2)


€

r= (8)(2,167.7)−(49)(346.7)(8)(311)−(49)(49) ⎡

⎣ ⎢ ⎤

⎦ ⎥(8)(15,224.7)−(346.7)2 ⎡

⎣ ⎢

⎤

⎦ ⎥

=.948

Value for State University example:

Regression MethodsCorrelation (2 of 2)


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.

Regression MethodsCoefficient of Determination


Exhibit 5.8

Regression Analysis with Excel (1 of 7)


Exhibit 5.9



Exhibit 5.10



Exhibit 5.11



Exhibit 5.12



Exhibit 5.13



Exhibit 5.14



Exhibit 5.15

Regression Analysis with QM for Windows


Multiple Regression with Excel (1 of 4)

Multiple regression relates demand to two or more independent variables.

General form:

y = 0 + 1x1 + 2x2 + . . . + kxk

where 0 = the intercept

1 . . . k = parameters representing contributions of the independent

variables

x1 . . . xk = independent variables


State University example:

Wins Promotion ($) Attendance 4 6 6 8 6 7 5 7

29,500 55,700 71,300 87,000 75,000 72,000 55,300 81,600

36,300 40,100 41,200 53,000 44,000 45.600 39,000 47,500



Exhibit 5.16



Exhibit 5.17



Problem Statement:

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

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

Example Problem SolutionComputer Software Firm (1 of 4)


Step 1: Compute the Exponential Smoothing Forecast.

Ft+1 = Dt + (1 - )Ft

Step 2: Compute the Adjusted Exponential SmoothingForecast

AFt+1 = Ft +1 + Tt+1

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



Period Dt Ft AFt Dt - Ft Dt - AFt 1 2 3 4 5 6 7 8 9

56 61 55 70 66 65 72 75 ⏐

⏐ 56.00 58.00 56.80 62.08 63.65 64.18 67.31 70.39

⏐ 56.00 58.40 56.88 63.20 64.86 65.26 68.80 72.19

⏐ 5.00 -3.00 13.20 3.92 1.35 7.81 7.68 ⏐

35.97

⏐ 5.00 -3.40 13.12 2.80 0.14 6.73 6.20 ⏐

30.60



Step 3: Compute the MAD Values

Step 4: Compute the Cumulative Error.

E(Ft) = 35.97

E(AFt) = 30.60

€

MAD(Ft ) =1

nDt − Ft∑ =

41.97

7= 5.99

MAD(AFt ) =1

nDt − AFt∑ =

37.39

7= 5.34



Problem Statement:

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.

Example Problem SolutionBuilding Products Store (1 of 5)


Quarter

Building Permits

x

Lumber Sales (1,000s of bd ft)

y 1 2 3 4 5 6 7 8 9 10

8 12 7 9 15 6 5 8 10 12

12.6 16.3 9.3 11.5 18.1 7.6 6.2 14.2 15.0 17.8



Step 1: Compute the Components of the Linear RegressionEquation.

€

x =92

10= 92

y =128.6

10=12.86

b =xy − nxy∑

x 2∑ − nx2 =

(1,290.3) − (10)(9.2)(12.86)

(932) − (10)(9.2)2=1.25

a = y −bx =12.86 − (1.25)(9.2) =1.36



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 x 2 − x 2∑( )∑[ ] n y 2 − y∑( )2

∑[ ]

€

r =(10)(1,170.3) − (92)(128.6)

(10)(932) − (92)(92)[ ] (10)(1,810.48) − (128.6)2[ ]

= .925




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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Chapter 5 - Forecasting 1 Chapter 5 Forecasting Introduction to Management Science 8th Edition by Bernard W. Taylor III