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ForecastingForecasting13

13 – 1Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.

For For Operations Management, 9eOperations Management, 9e by by Krajewski/Ritzman/Malhotra Krajewski/Ritzman/Malhotra © 2010 Pearson Education© 2010 Pearson Education

PowerPoint Slides PowerPoint Slides by Jeff Heylby Jeff Heyl

ForecastingForecasting

Forecasts are critical inputs to business plans, annual plans, and budgetsFinance, human resources, marketing, operations, and supply chain managers need forecasts to plan: output levels, purchases of services and materials, workforce and output schedules, inventories, and long-term capacitiesForecasts are made on many different variables


Forecasts are important to managing both processes and managing supply chains

2

Demand PatternsDemand Patterns

A time series is the repeated observations of demand for a service or product in their porder of occurrenceThere are five basic time series patterns

HorizontalTrendSeasonal


SeasonalCyclicalRandom

Demand PatternsDemand PatternsFigure 13.1 – Patterns of Demand

Qua

ntity


Time

(a) Horizontal: Data cluster about a horizontal line

3


Qua

ntity


Time

(b) Trend: Data consistently increase or decrease


Qua

ntity

Year 1

Year 2


| | | | | | | | | | | |J F M A M J J A S O N D

Months

(c) Seasonal: Data consistently show peaks and valleys

4


Qua

ntity


| | | | | |1 2 3 4 5 6

Years

(d) Cyclical: Data reveal gradual increases and decreases over extended periods

Key DecisionsKey Decisions

Deciding what to forecastLevel of aggregationU i fUnits of measure

Choosing a forecasting systemChoosing the type of forecasting technique

Judgment and qualitative methodsCausal methods


Time-series analysisKey factor in choosing the proper forecasting approach is the time horizon for the decision requiring forecasts

5

Judgment MethodsJudgment Methods

Other methods (casual and time-series) require an adequate history file, which might not be availableJudgmental forecasts use contextual knowledge gained through experienceSalesforce estimatesExecutive opinion is a method in which opinions, experience, and technical knowledge of one or more managers are summarized to arrive at a


gsingle forecastDelphi method

Judgment MethodsJudgment Methods

Market research is a systematic approach to determine external customer interest through d t th idata-gathering surveysDelphi method is a process of gaining consensus from a group of experts while maintaining their anonymityUseful when no historical data are available Can be used to develop long-range forecasts and


p g gtechnological forecasting

6

Linear RegressionLinear Regression

A dependent variable is related to one or more independent variables by a linear equationTh i d d t i bl d tThe independent variables are assumed to “cause” the results observed in the pastSimple linear regression model is a straight line

Y = a + bX

where


Y = dependent variableX = independent variablea = Y-intercept of the lineb = slope of the line


Y

Estimate of

Regressionequation:Y = a + bX

Deviation,or error

Dep

ende

nt v

aria

ble

Estimate ofY fromregressionequation

Actualvalueof Y

Value of X used


Independent variableX

Value of X usedto estimate Y

Figure 13.2 – Linear Regression Line Relative to Actual Data

7


The sample correlation coefficient, rMeasures the direction and strength of the relationship between the independent variable and the dependentbetween the independent variable and the dependent variable.The value of r can range from –1.00 ≤ r ≤ 1.00

The sample coefficient of determination, r2

Measures the amount of variation in the dependent variable about its mean that is explained by the regression line


The values of r2 range from 0.00 ≤ r2 ≤ 1.00

The standard error of the estimate, syx

Measures how closely the data on the dependent variable cluster around the regression line

Using Linear RegressionUsing Linear Regression

EXAMPLE 13.1The supply chain manager seeks a better way to forecast the demand for door hinges and believes that the demand is related gto advertising expenditures. The following are sales and advertising data for the past 5 months:

Month Sales (thousands of units) Advertising (thousands of $)1 264 2.52 116 1.33 165 1.44 101 1 0


4 101 1.05 209 2.0

The company will spend $1,750 next month on advertising for the product. Use linear regression to develop an equation and a forecast for this product.

8


SOLUTIONWe used POM for Windows to determine the best values of a, b, the correlation coefficient, the coefficient of determination, and , ,the standard error of the estimate

a =b =r =

r2 =syx =

–8.135109.229X0.9800.96015.603


The regression equation is Y = –8.135 + 109.229X


The regression line is shown in Figure 13.3. The r of 0.98 suggests an unusually strong positive relationship between sales and advertising expenditures. The coefficient of determination 2 implies that 96 percent of the variation indetermination, r2, implies that 96 percent of the variation in sales is explained by advertising expenditures.

250 –

200 –

150 –

(000

uni

ts)

Brass Door Hinge

X

X

X

XData


| |1.0 2.0

Advertising ($000)

100 –

50 –

0 –

Sale

s

XXData

Forecasts

Figure 13.3 – Linear Regression Line for the Sales and Advertising Data

9

Time Series MethodsTime Series Methods

In a naive forecast the forecast for the next period equals the demand for the current period (Forecast = D )period (Forecast = Dt)Estimating the average: simple moving averages

Used to estimate the average of a demand time series and thereby remove the effects of random fluctuationMost useful when demand has no pronounced


Most useful when demand has no pronounced trend or seasonal influencesThe stability of the demand series generally determines how many periods to include

450 –

430

Time Series MethodsTime Series Methods

430 –

410 –

390 –

370 –

350 –

Patie

nt a

rriv

als


| | | | | |

0 5 10 15 20 25 30Week

Figure 13.4 – Weekly Patient Arrivals at a Medical Clinic

10

Simple Moving AveragesSimple Moving Averages

Specifically, the forecast for period t + 1 can be calculated at the end of period t (after the actual demand for period t is known) asdemand for period t is known) as

Ft+1 = =Sum of last n demands

nDt + Dt-1 + Dt-2 + … + Dt-n+1

n

where


Dt = actual demand in period tn = total number of periods in the average

Ft+1 = forecast for period t + 1

Simple Moving AveragesSimple Moving Averages

For any forecasting method, it is important to measure the accuracy of its forecasts. Forecast error is simply the difference found by subtractingerror is simply the difference found by subtracting the forecast from actual demand for a given period, or

where

Et = Dt – Ft


whereEt = forecast error for period tDt = actual demand in period tFt = forecast for period t

11

Using the Moving Average MethodUsing the Moving Average Method

EXAMPLE 13.2a. Compute a three-week moving average forecast for the

arrival of medical clinic patients in week 4. The numbers ofarrival of medical clinic patients in week 4. The numbers of arrivals for the past three weeks were as follows:

Week Patient Arrivals1 4002 3803 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?

Using the Moving Average MethodUsing the Moving Average Method

SOLUTIONa. The moving average forecast at

the end of week 3 is

Week Patient Arrivals1 4002 3803 411the end of week 3 is

b. The forecast error for week 4 is

F4 = = 397.0411 + 380 + 4003

E4 = D4 – F4 = 415 – 397 = 18


c. The forecast for week 5 requires the actual arrivals from weeks 2 through 4, the three most recent weeks of data

F5 = = 402.0415 + 411 + 3803

12

Application 13.1aApplication 13.1a

Estimating with Simple Moving Average using the following customer-arrival data

Month Customer arrival1 8002 7403 8104 790

Use a three-month moving average to forecast customer arrivals for month 5


arrivals for month 5

F5 = = 780D4 + D3 + D2

3790 + 810 + 740

3=

Forecast for month 5 is 780 customer arrivals


If the actual number of arrivals in month 5 is 805, what is the forecast for month 6?

F = = 801 667D5 + D4 + D3 805 + 790 + 810

=

Month Customer arrival1 8002 7403 8104 790


F6 = = 801.6673 3=


13


Forecast error is simply the difference found by subtracting the forecast from actual demand for a given period, or

Given the three-month moving average forecast for month 5, and the number of patients that actually arrived (805), what is the forecast error?

Et = Dt – Ft

E5 = 805 – 780 = 25


Forecast error for month 5 is 25

In the weighted moving average method, each historical demand in the average can have its own weight provided that the sum of the weights equals

Weighted Moving AveragesWeighted Moving Averages

weight, provided that the sum of the weights equals 1.0. The average is obtained by multiplying the weight of each period by the actual demand for that period, and then adding the products together:

Ft+1 = W1D1 + W2D2 + … + WnDt-n+1

A three-period weighted moving average model with


the most recent period weight of 0.50, the second most recent weight of 0.30, and the third most recent might be weight of 0.20

Ft+1 = 0.50Dt + 0.30Dt–1 + 0.20Dt–2

14

Application 13.1bApplication 13.1b

Revisiting the customer arrival data in Application 13.1a. Let W1 = 0.50, W2 = 0.30, and W3 = 0.20. Use the weighted moving average method to forecast arrivals for month 5.

= 0.50(790) + 0.30(810) + 0.20(740)F5 = W1D4 + W2D3 + W3D2

= 786


Given the number of patients that actually arrived (805), what is the forecast error?


is the forecast error?


E5 = 805 – 786 = 19

Application 13.1bApplication 13.1b

If the actual number of arrivals in month 5 is 805, compute the forecast for month 6

= 0.50(805) + 0.30(790) + 0.20(810)F6 = W1D5 + W2D4 + W3D3

= 801.5



15

Exponential SmoothingExponential Smoothing

A sophisticated weighted moving average that calculates the average of a time series by giving recent demands more weight than earlier demandsrecent demands more weight than earlier demandsRequires only three items of data

The last period’s forecastThe demand for this periodA smoothing parameter, alpha (α), where 0 ≤ α ≤ 1.0

The equation for the forecast is


Ft+1 = α(Demand this period) + (1 – α)(Forecast calculated last period)= αDt + (1 – α)Ft

Ft+1 = Ft + α(Dt – Ft)or the equivalent

Exponential SmoothingExponential Smoothing

The emphasis given to the most recent demand levels can be adjusted by changing the smoothing parameterparameterLarger α values emphasize recent levels of demand and result in forecasts more responsive to changes in the underlying averageSmaller α values treat past demand more uniformly and result in more stable forecastsE ti l thi i i l d i


Exponential smoothing is simple and requires minimal dataWhen the underlying average is changing, results will lag actual changes

16

450 – 3-week MA 6-week MAforecast

Exponential Smoothing and Exponential Smoothing and Moving AverageMoving Average

430 –

410 –

390 –

370 –

atie

nt a

rriv

als

forecast forecast

Exponential


Pa

Week

| | | | | |0 5 10 15 20 25 30

Exponential smoothingα = 0.10

Using Exponential SmoothingUsing Exponential Smoothing

EXAMPLE 13.3a. Reconsider the patient arrival data in Example 13.2. It is

now the end of week 3. Using α = 0.10, calculate thenow the end of week 3. Using α 0.10, calculate the exponential smoothing forecast for week 4.

Week Patient Arrivals1 400

b. What was the forecast error for week 4 if the actual demand turned out to be 415?

c. What is the forecast for week 5?


2 3803 4114 415

17


SOLUTIONa. The exponential smoothing method requires an initial

forecast. Suppose that we take the demand data for the first pptwo weeks and average them, obtaining (400 + 380)/2 = 390 as an initial forecast. (POM for Windows and OM Explorer simply use the actual demand for the first week as a default setting for the initial forecast for period 1, and do not begin tracking forecast errors until the second period). To obtain the forecast for week 4, using exponential smoothing with and the initial forecast of 390, we calculate the average at the end of week 3 as


F4 =

Thus, the forecast for week 4 would be 392 patients.

0.10(411) + 0.90(390) = 392.1


b. The forecast error for week 4 is

E 415 392 23

c. The new forecast for week 5 would be

E4 =

F5 =

or 394 patients Note that we used F not the integer value

415 – 392 = 23

0.10(415) + 0.90(392.1) = 394.4


or 394 patients. Note that we used F4, not the integer-value forecast for week 4, in the computation for F5. In general, we round off (when it is appropriate) only the final result to maintain as much accuracy as possible in the calculations.

18

Application 13.1cApplication 13.1c

Suppose the value of the customer arrival series average in month 3 was 783 customers (let it be F4). Use exponential smoothing with α = 0.20 to compute the forecast for month 5.

Ft+1 = Ft + α(Dt – Ft) = 783 + 0.20(790 – 783) = 784.4


Given the number of patients that actually arrived (805), what is the forecast error?


E5 =


805 – 784 = 21

Application 13.1cApplication 13.1c

Given the actual number of arrivals in month 5, what is the forecast for month 6?

Ft+1 = Ft + α(Dt – Ft) = 784.4 + 0.20(805 – 784.4) = 788.52



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Including a TrendIncluding a Trend

A trend in a time series is a systematic increase or decrease in the average of theincrease or decrease in the average of the series over timeThe forecast can be improved by calculating an estimate of the trendTrend-adjusted exponential smoothing requires two smoothing constants


requires two smoothing constants

Including a TrendIncluding a Trend

For each period, we calculate the average and the trend:

A = α(Demand this period)At = α(Demand this period) + (1 – α)(Average + Trend estimate last period)

= αDt + (1 – α)(At–1 + Tt–1)

Tt = β(Average this period – Average last period)+ (1 – β)(Trend estimate last period)

= β(At – At–1) + (1 – β)Tt–1

F A + T


Ft+1 = At + Tt

whereAt = exponentially smoothed average of the series in period tTt = exponentially smoothed average of the trend in period t

= smoothing parameter for the average, with a value between 0 and 1= smoothing parameter for the trend, with a value between 0 and 1

Ft+1 = forecast for period t + 1

20

Using TrendUsing Trend--Adjusted Exponential Adjusted Exponential SmoothingSmoothing

EXAMPLE 13.4Medanalysis, Inc., provides medical laboratory services M i t t d i f ti th b f bl dManagers are interested in forecasting the number of blood analysis requests per weekThere has been a national increase in requests for standard blood testsMedanalysis recently ran an average of 28 blood tests per week and the trend has been about three additional patients per weekThi k’ d d f 27 bl d t t


This week’s demand was for 27 blood testsWe use α = 0.20 and β = 0.20 to calculate the forecast for next week


SOLUTIONA0 = 28 patients and T0 = 3 patients

30.2 + 2.8 = 33 blood tests

If the actual number of blood tests requested in week 2 proved to be 44, the updated forecast for week 3 would be

The forecast for week 2 (next week) is

A1 =T1 =F2 =

0.20(27) + 0.80(28 + 3) = 30.20.20(30.2 – 28) + 0.80(3) = 2.8


p , p

A2 =

F3 = 35.2 + 3.2 = 38.4 or 38 blood tests0.2(35.2 – 30.2) + 0.80(2.8) = 3.20.20(44) + 0.80(30.2 + 2.8) = 35.2

T2 =

21


TABLE 13.1 | FORECASTS FOR MEDANALYSIS USING THE TREND-ADJUSTED EXPONENTIAL| SMOOTHING MODEL

Calculations to Forecast Arrivals for Next Week

Week Arrivals Smoothed Average

Trend Average Forecast for This Week Forecast Errorg g

0 28 28.00 3.001 272 443 374 355 536 387 578 61


9 3910 5511 5412 5213 6014 6015 75


TABLE 13.1 | FORECASTS FOR MEDANALYSIS USING THE TREND-ADJUSTED EXPONENTIAL| SMOOTHING MODEL

Calculations to Forecast Arrivals for Next Week

Week Arrivals Smoothed Average

Trend Average Forecast for This Week Forecast Error

30.20 + 2.84 = 33.04–430.20

35.232.843.28

28.00 + 3.00 = 31.00

35.23 + 3.28 = 38.5138.21 + 3.22 = 41.4340.14 + 2.96 = 43.1045.08 + 3.36 = 48.4446.35 + 2.94 = 49.2950.83 + 3.25 = 54.08

10.96–1.51–6.43

9.90–10.44

7.716.92

38.2140.1445.0846.3550.8355.46

3.222.963.362.943.253.52

g g0 28 28.00 3.001 272 443 374 355 536 387 578 61


55.46 + 3.52 = 58.9854.99 + 2.72 = 57.7157.17 + 2.62 = 59.7958.63 + 2.38 = 61.0159.21 + 2.02 = 61.2360.99 + 1.97 = 62.9662.37 + 1.86 = 64.23

–19.98–2.71–5.79–9.01–1.23–2.9610.77

55 654.9957.1758.6359.2160.9962.3766.38

3 52.722.622.382.021.971.862.29

9 3910 5511 5412 5213 6014 6015 75

22

80 –

70 –Trend-adjusted

forecast


70

60 –

50 –

40 –

30

Patie

nt a

rriv

als

Actual blood test requests

forecast


| | | | | | | | | | | | | | | |0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

30 –

WeekFigure 13.5 – Trend-Adjusted Forecast for Medanalysis

Application 13.2Application 13.2

The forecaster for Canine Gourmet dog breath fresheners estimated (in March) the sales average to be 300,000 cases sold per month and the trend to be +8,000 per month. The actual

l f A il 330 000 Wh t i th f t f Msales for April were 330,000 cases. What is the forecast for May, assuming α = 0.20 and β = 0.10?

AApr = αDt + (1 – α)(AMar + TMar)

TApr = β(AApr – AMar) + (1 – β)TMar

= 0.20(330,000) + 0.80(300,000 + 8,000) = 312,400 cases


Forecast for May = AApr + pTApr

= 0.10(312,400 – 300,000) + 0.90(8,000) = 8,440 cases

= 312,400 + (1)(8,440) = 320,840 cases

23


Suppose you also wanted the forecast for July, three months ahead. To make forecasts for periods beyond the next period, we multiply the trend estimate by the number of additional

i d th t t i th f t d dd th lt t thperiods that we want in the forecast and add the results to the current average.

Forecast for July = AApr + pTApr

= 312,400 + (3)(8,440) = 337,720 cases


Seasonal PatternsSeasonal Patterns

Seasonal patterns are regularly repeated upward or downward movements in demand measured in periods of less thandemand measured in periods of less than one year Account for seasonal effects by using one of the techniques already described but to limit the data in the time series to those periods in the same seasonThi h t f l


This approach accounts for seasonal effects but discards considerable information on past demand

24

Multiplicative Seasonal MethodMultiplicative Seasonal Method

Multiplicative seasonal method, whereby seasonal factors are multiplied by an estimate of the average demand to arrive at a seasonal forecast1. For each year, calculate the average demand for

each season by dividing annual demand by the number of seasons per year

2. For each year, divide the actual demand for each season by the average demand per season, resulting in a seasonal index for each season

demand to arrive at a seasonal forecast


resulting in a seasonal index for each season3. Calculate the average seasonal index for each

season using the results from Step 24. Calculate each season’s forecast for next year

Using the Multiplicative Seasonal Using the Multiplicative Seasonal Method Method

EXAMPLE 13.5The manager of the Stanley Steemer carpet cleaning company needs a quarterly forecast of the number of customersneeds a quarterly forecast of the number of customers expected next year. The carpet cleaning business is seasonal, with a peak in the third quarter and a trough in the first quarter. Following are the quarterly demand data from the past 4 years:

Quarter Year 1 Year 2 Year 3 Year 41 45 70 100 1002 335 370 585 7253 520 590 830 1160


The manager wants to forecast customer demand for each quarter of year 5, based on an estimate of total year 5 demand of 2,600 customers

3 520 590 830 11604 100 170 285 215

Total 1000 1200 1800 2200

25


SOLUTIONFigure 13.6 shows the solution using the Seasonal Forecasting Solver in OM Explorer. For the Inputs the forecast for the totalSolver in OM Explorer. For the Inputs the forecast for the total demand in year 5 is needed. The annual demand has been increasing by an average of 400 customers each year (from 1,000 in year 1 to 2,200 in year 4, or 1,200/3 = 400). The computed forecast demand is found by extending that trend, and projecting an annual demand in year 5 of 2,200 + 400 = 2,600 customers. The Results sheet shows quarterly forecasts by multiplying the seasonal factors by the average demand per quarter For


seasonal factors by the average demand per quarter. For example, the average demand forecast in year 5 is 650 customers (or 2,600/4 = 650). Multiplying that by the seasonal index computed for the first quarter gives a forecast of 133 customers (or 650 × 0.2043 = 132.795).



Figure 13.6 – Demand Forecasts Using the Seasonal Forecast Solver of OM Explorer

26


Suppose the multiplicative seasonal method is being used to forecast customer demand. The actual demand and seasonal indices are shown below.

Year 1 Year 2 Average IndexQuarter Demand Index Demand Index

1 100 0.40 192 0.64 0.522 400 1.60 408 1.36 1.483 300 1.20 384 1.28 1.244 200 0.80 216 0.72 0.76


Average 250 300


Quarter Average Index1 0.522 1 48

If the projected demand for Year 3 is 1320 units, what is the forecast for each quarter of that year?

1320 units ÷ 4 quarters = 330 units

2 1.483 1.244 0.76

y

Forecast for Quarter 1 =


0.52(330) ≈ 172 units

1.48(330) ≈ 488 units




1.24(330) ≈ 409 units

0.76(330) ≈ 251 units

27

(a) Multiplicative pattern


Dem

and


Period

| | | | | | | | | | | | | | | |0 2 4 5 8 10 12 14 16


(b) Additive pattern

Dem

and


Period

| | | | | | | | | | | | | | | |0 2 4 5 8 10 12 14 16

28

Choosing a TimeChoosing a Time--Series MethodSeries Method

Forecast performance is determined by forecast errorsF t d t t h thi i iForecast errors detect when something is going wrong with the forecasting systemForecast errors can be classified as either bias errors or random errorsBias errors are the result of consistent mistakesRandom error results from unpredictable factors


pthat cause the forecast to deviate from the actual demand

CFE = ΣE

Measures of Forecast ErrorMeasures of Forecast Error

Σ(Et – E )2σ =CFE = ΣEt n – 1σ =

Σ|Et |nMAD =E =

CFEn


ΣEt2

nMSE =(Σ|Et |/Dt)(100)

nMAPE =

29

Calculating Forecast ErrorsCalculating Forecast Errors

EXAMPLE 13.6The following table shows the actual sales of upholstered chairs for a furniture manufacturer and the forecasts made forchairs for a furniture manufacturer and the forecasts made for each of the last eight months. Calculate CFE, MSE, σ, MAD, and MAPE for this product.

Montht

DemandDt

ForecastFt

ErrorEt

Error2

Et2

Absolute Error |Et|

Absolute % Error (|Et|/Dt)(100)

1 200 225 –252 240 220 203 300 285 154 270 290 20


4 270 290 –205 230 250 –20 400 20 8.76 260 240 20 400 20 7.77 210 250 40 1,600 40 19.08 275 240 35 1,225 35 12.7

Total –15 5,275 195 81.3%


EXAMPLE 13.6The following table shows the actual sales of upholstered chairs for a furniture manufacturer and the forecasts made forchairs for a furniture manufacturer and the forecasts made for each of the last eight months. Calculate CFE, MSE, σ, MAD, and MAPE for this product.

Montht

DemandDt

ForecastFt

ErrorEt

Error2

Et2

Absolute Error |Et|

Absolute % Error (|Et|/Dt)(100)

1 200 225 –25 625 25 12.5%2 240 220 20 400 20 8.33 300 285 15 225 15 5.04 270 290 20 400 20 7 4


4 270 290 –20 400 20 7.45 230 250 –20 400 20 8.76 260 240 20 400 20 7.77 210 250 40 1,600 40 19.08 275 240 35 1,225 35 12.7

Total –15 5,275 195 81.3%

30

SOLUTIONUsing the formulas for the measures, we getC l ti f t (bi )


Cumulative forecast error (bias):

CFE = –15

Average forecast error (mean bias):

CFEnE = –1.875=


Mean squared error:

MSE =ΣEt

2

n5,275

8=

Standard deviation:


Σ[E ( 1 875)]2

Mean absolute deviation:

Σ[Et – (–1.875)]2

n – 1σ =

Σ|Et |nMAD =

= 27.4

= = 24.41958


Mean absolute percent error:

(Σ|Et |/ Dt)(100)nMAPE = = = 10.2%

81.3%8

31


A CFE of –15 indicates that the forecast has a slight bias to overestimate demand. The MSE, σ, and MAD statistics provide measures of forecast error variability. A MAD of 24.4 means that easu es o o ecast e o a ab ty o ea s t atthe average forecast error was 24.4 units in absolute value. The value of σ, 27.4, indicates that the sample distribution of forecast errors has a standard deviation of 27.4 units. A MAPE of 10.2 percent implies that, on average, the forecast error was about 10 percent of actual demand. These measures become more reliable as the number of periods of data increases.


Tracking SignalsTracking Signals

A measure that indicates whether a method of forecasting is accurately predicting actual changes in demandchanges in demandUseful when forecast systems are computerized because it alerts analysts when forecast are getting far from desirable limits

Tracking signal =CFEMAD


MAD

Each period, the CFE and MAD are updated to reflect current error, and the tracking signal is compared to some predetermined limits

32


The MAD can be calculated as a weighted average determined by the exponential smoothing method

MADt = α|Et| + (1 – α)MADt-1

If forecast errors are normally distributed with a mean of 0, the relationship between σ and MAD is simple

( /2)(MAD) 1 25(MAD)


σ = ( π /2)(MAD) ≅ 1.25(MAD)

MAD = 0.7978σ ≅ 0.8σ

+2.0 –

+1 5

Out of control


Control limit+1.5 –

+1.0 –

+0.5 –

0 –

–0.5 –

–1.0 –Trac

king

sig

nal


–1.5 –| | | | |

0 5 10 15 20 25Observation number

Control limit

Figure 13.7 – Tracking Signal

33

Criteria for Selecting MethodsCriteria for Selecting Methods

Criteria to use in making forecast method and parameter choices include

1 Minimizing bias1. Minimizing bias2. Minimizing MAPE, MAD, or MSE3. Meeting managerial expectations of changes in the

components of demand4. Minimizing the forecast error last period

Statistical performance measures can be used


1. For projections of more stable demand patterns, use lower α and β values or larger n values

2. For projections of more dynamic demand patterns try higher α and β values or smaller n values

Using Multiple TechniquesUsing Multiple Techniques

Combination forecasts are forecasts that are produced by averaging independent p y g g pforecasts based on different methods or different data or bothFocus forecasting selects the best forecast from a group of forecasts generated by individual techniques


34

Forecasting as a ProcessForecasting as a Process

A typical forecasting processStep 1: Adjust history fileStep 1: Adjust history fileStep 2: Prepare initial forecastsStep 3: Consensus meetings and collaborationStep 4: Revise forecastsStep 5: Review by operating committeeStep 6: Finalize and communicate


Step 6: Finalize and communicate

Forecasting is not a stand-alone activity, but part of a larger process

Forecasting as a ProcessForecasting as a Process

Finalize Review by R i

Consensus meetings and collaboration

3

Prepare initial

forecasts2

Adjust history

file1


Finalize and

communicate6

Review by Operating Committee

5

Revise forecasts

4

35

Forecasting PrinciplesForecasting Principles

TABLE 13.2 | SOME PRINCIPLES FOR THE FORECASTING PROCESS

Better processes yield better forecasts

Demand forecasting is being done in virtually every company, either formally or informally. The challenge is to do it well—better than the competitionBetter forecasts result in better customer service and lower costs, as well as better relationships with suppliers and customersThe forecast can and must make sense based on the big picture, economic outlook, market share, and so onThe best way to improve forecast accuracy is to focus on reducing forecast error

Bi i th t ki d f f t t i f bi


Bias is the worst kind of forecast error; strive for zero bias

Whenever possible, forecast at more aggregate levels. Forecast in detail only where necessaryFar more can be gained by people collaborating and communicating well than by using the most advanced forecasting technique or model

Solved Problem 1Solved Problem 1

Chicken Palace periodically offers carryout five-piece chicken dinners at special prices. Let Y be the number of dinners sold and X be the price. Based on the historical observations and

l l ti i th f ll i t bl d t i th icalculations in the following table, determine the regression equation, correlation coefficient, and coefficient of determination. How many dinners can Chicken Palace expect to sell at $3.00 each?

Observation Price (X) Dinners Sold (Y)1 $2.70 7602 $3.50 5103 $2 00 980


3 $2.00 9804 $4.20 2505 $3.10 3206 $4.05 480

Total $19.55 3,300Average $3.258 550

36


SOLUTIONWe use the computer to calculate the best values of a, b, the correlation coefficient, and the coefficient of determinationcorrelation coefficient, and the coefficient of determination

a =b =r =

r 2 = 0.71–0.84–277.631,454.60

The regression line isY = a + bX = 1 454 60 277 63X


Y = a + bX = 1,454.60 – 277.63X

For an estimated sales price of $3.00 per dinner

Y = a + bX = 1,454.60 – 277.63(3.00)

= 621.71 or 622 dinners


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 d d f th i l i th t h d idemand 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 14 by using the simple moving average method with n = 3 then using the weighted moving average method with and weights of 0.50, 0.30, and 0.20, with 0.50.

b. Calculate the MAD for each method.

37


SOLUTIONa. The simple moving average method and the weighted

moving average method give the following results:moving average method give the following results:

Current Week

Simple Moving Average Forecast for Next Week

Weighted Moving Average Forecast for Next Week

June 16

June 23

= 55.7 or 5652 + 65 + 50

3[(0.5 × 52) + (0.3 × 65) + (0.2 × 50)] = 55.5 or 56

= 57.7 or 5856 + 52 + 65

3

55 + 56 + 52

[(0.5 × 56) + (0.3 × 52) + (0.2 × 65)] = 56.6 or 57


June 30

July 7

= 54.3 or 5455 56 52

3 [(0.5 × 55) + (0.3 × 56) + (0.2 × 52)] = 54.7 or 55

= 57.0 or 5760 + 55 + 56

3 [(0.5 × 60) + (0.3 × 55) + (0.2 × 56)] = 57.7 or 58


b. The mean absolute deviation is calculated as follows:

Simple Moving Average Weighted Moving AverageSimple Moving Average Weighted Moving Average

WeekActual

DemandForecast for This Week Absolute Errors |Et|

Forecast for This Week Absolute Errors |Et|

June 23 56 56 56

June 30 55 58 57

July 7 60 54 55

|56 – 56| = 0

|55 – 58| = 3

|60 – 54| = 6

MAD = = 30 + 3 + 6

3 MAD = = 2.30 + 2 + 2

3

|56 – 56| = 0|55 – 57| = 2

|60 – 55| = 5


For this limited set of data, the weighted moving average method resulted in a slightly lower mean absolute deviation. However, final conclusions can be made only after analyzing much more data.

38


The monthly demand for units manufactured by the Acme Rocket Company has been as follows:

Month Units Month UnitsMonth Units Month UnitsMay 100 September 105June 80 October 110July 110 November 125August 115 December 120

a. Use the exponential smoothing method to forecast June to January. The initial forecast for May was 105 units; α = 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?


SOLUTIONa.

Current Month, tCalculating Forecast for Next Month Ft+1 = αDt + (1 – α)Ft Forecast for Month t + 1

May JuneJune JulyJuly AugustAugust SeptemberSeptember OctoberOctober November

0.2(100) + 0.8(105) = 104.0 or 1040.2(80) + 0.8(104.0)0.2(110) + 0.8(99.2)

= 99.2 or 99= 101.4 or 101

0.2(115) + 0.8(101.4)0.2(105) + 0.8(104.1)0 2(110) + 0 8(104 3)

= 104.1 or 104= 104.3 or 104

105 4 105


October NovemberNovember DecemberDecember January

0.2(110) + 0.8(104.3)0.2(125) + 0.8(105.4)0.2(120) + 0.8(109.3)

= 105.4 or 105= 109.3 or 109= 111.4 or 111

39


b.

ActualAbsolute Percent

–24 24 30.0%11 11 10.0

Month, t

Actual Demand,

DtForecast,

FtError,

Et = Dt – FtAbsolute Error, |Et|

Percent Error,

(|Et|/Dt)(100)June 80 104July 110 99August 115 101September 105 104October 110 104November 125 105

14 14 12.01 1 1.06 6 5.5

20 20 16 0


November 125 105December 120 109

Total 765

20 20 16.011 11 9.239 87 83.7%

Σ|Et |nMAD =

(Σ|Et |/Dt)(100)nMAPE = = = 11.96%

83.7%7= = 12.487

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:

The probability that a tracking signal value of 3.14 could be generated completely by chance is small. Consequently, we

Tracking signal = CFEMAD = = 3.1439

12.4


g p y y q y,should revise our approach. The long string of forecasts lower than actual demand suggests use of a trend method.

40


The Northville Post Office experiences a seasonal pattern of daily mail volume every week. The following data for two representative weeks are expressed in thousands of pieces of

ilmail:Day Week 1 Week 2Sunday 5 8Monday 20 15Tuesday 30 32Wednesday 35 30Thursday 49 45Friday 70 70S t d 15 10


Saturday 15 10Total 224 210

a. Calculate a seasonal factor for each day of the week.b. If the postmaster estimates 230,000 pieces of mail to be

sorted next week, forecast the volume for each day.


SOLUTIONa. Calculate the average daily mail volume for each week. Then

f h d f th k di id th il l b thfor each day of the week divide the mail volume by the week’s average to get the seasonal factor. Finally, for each day, add the two seasonal factors and divide by 2 to obtain the average seasonal factor to use in the forecast.


41


Week 1 Week 2

Mail Seasonal Factor Mail Seasonal FactorAverage

Seasonal FactorDay

Mail Volume

Seasonal Factor (1)

Mail Volume

Seasonal Factor (2)

Seasonal Factor[(1) + (2)]/2

Sunday 5 8Monday 20 15Tuesday 30 32Wednesday 35 30Thursday 49 45Friday 70 70Saturday 15 10

5/32 = 0.1562520/32 = 0.6250030/32 = 0.93750

8/30 = 0.2666715/30 = 0.5000032/30 = 1.06667

0.211460.562501.00209

35/32 = 1.0937549/32 = 1.5312570/32 = 2.1875015/32 = 0.46875

30/30 = 1.0000045/30 = 1.5000070/30 = 2.3333310/30 = 0.33333

1.046881.515632.260420.40104


Total 224 210Average 224/7 = 32 210/7 = 30


b. The average daily mail volume is expected to be 230,000/7 = 32,857 pieces of mail. Using the average seasonal factors calculated in part (a), we obtain the f ll i f tfollowing forecasts:

6,94818,482

32,926

0.21146(32,857) =

0.56250(32,857) =

1.00209(32,857) =

34,3971.04688(32,857) =

Day Calculations Forecast

SundayMonday

Tuesday

Wednesday


49,799

74,271

13,177

230,000

1.51563(32,857) =

2.26042(32,857) =

0.40104(32,857) =

y

ThursdayFriday

Saturday

Total

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