06.Ch4-Forecasting_Notes

Short-Term Time Series Forecasting:Time series methods are based on the analysis of past data; it is cheap, easy, and timely, and it is generally a very accurate method. Therefore, it should always be attempted, and in practice one finds this to be true. A survey conducted by the Institute of Business Forecasting in 1999 revealed that over 60 percent of the companies used time series models. It also showed that moving averages and exponential smoothing are the most popular methods. As such, let us study the following methods:•Naive 1•Naive 2•Simple Moving Average•Weighted Moving Average•Simple Exponential Smoothing (corresponding to Graph I; only Level)•Exponential Smoothing including Trend (corresponding to Graph II, Level and Trend)•Other Models, including seasonality and the composite one, will not be studied in this courseSimple IllustrationSweet & Sourdough Bakery is a small bakery that is open 7 days a week. Until just recently, the owners forecasted the number of customers using their “gut feel.” However, they want to open another bakery and recognize the need to adopt a more formal method of forecasting that can be used at both locations. The actual sales for the previous few periods are provided below.Month ActualJan 120Feb 90March 101April 91May 115June 83JulyThe owners decide to analyze several forecasting methods. The different methods to be tested are:•Naïve 1 & Naïve 2•Simple Moving Average•Weighted Moving Average•Simple Exponential Smoothing (level only)Naïve Methods:1. Forecast for July = Actual for Juneor in general:F t+1 = A t where t is the time periodExampleMonth Actual Forecast120_____JanFeb 90 __________101MarchApril 91 __________May115June 83 __________JulyClearly using this Naïve 1 method, we only have F FEB = 120; F MAR = 90 and so on, so that F JULY = 832. Forecast for July = Some predetermined constant, Cor in general: F t+1 = C usually C is an average over a large number of past periodsIn the following example simply looking at the 6 periods one can safely assume C to be about 100.ExampleMonthActual ForecastJan 120 _____ Feb 90 _____ March 101 _____ April 91 _____ May 115 _____ June 83 _____ July _____Thus, the forecast according to the Naïve 2 method would be the same, namely 100, for all the months.Sweet & Sourdough Bakery Sales758595105115125JanFebMarchAprilMayJuneJulyAre these methods any good? Why or why not? How would you improve them?Simple Moving Averages (SMA):Simple Moving averages are effective if we can assume that demand will be fairly steady over time. Ex. 3-Month Moving Average Forecast for July = Average of Actual Demands in June, May, and April F July = ( A june + A may + A April ) / 3 Forecast for August = Average of Actual Demands in July, June, and May F August = ( A July + A June + A May ) / 3In general, n-period simple moving average is computed as;ExampleMonthActual ForecastJan 120 _____ Feb 90 _____ March 101 _____ April 91 _____ May 115 _____ June 83 _____ July _____Calculate 3 period simple moving average: F July =If Actual demand in July is 90, then F August =1. understand why this is called a “moving” average and what are its implications and rationale.2. understand the impact of using 3 periods versus, let us say, 12 period moving average on the forecasts. * Is this method better than the Naïve methods? * How do you determine the number of past periods (n)? * What if n =1 ? * What if n = is very large, say n=100 ? * What is the impact of using small n on the forecasts? * What is the impact of using large n on the forecasts? * What are the weights associated with the periods used in the forecast with this method?Any suggestions for improving this method?F A A nt t t +−=++11.....To see the impact of n, lets compare the forecasts for July through November when n = 2 and n = 5.Month Actual Forecast(n=2) Forecast(n =5) Jan 120 Feb 90 March 101 April 91 May 115 June 83 July 90 99 96 August 105 86.5 96September 108 97.5 96.8October 110 106.5 100.2 November 10999.2Sweet & Sourdough Bakery Sales758595105115125JanFebMarchAprilMayJuneJulyAugustSeptemberOctoberNovemberWhich value of n results in a more stable forecast?Which value of n results in a more responsive forecast?Weighted Moving Averages (WMA):If there is an upward or downward trend in demand, we can use weights to place more emphasis on recent value. WMA makes forecasting more responsive to changes if recent periods are more heavily weighted. There is no formula to determine weight. Choosing weights requires experience.Ex. 3-Month Weighted Moving Average W1 * A June + W2 * A May + W3 * A April Forecast for July = ------------------------------------------------------ W1 + W2 + W3 Where W1 > W2 > W3 are weightsIn general,ExampleMonthActual ForecastJan 120 _____ Feb 90 _____ March 101 _____ April 91 _____ May 115 _____ June 83 _____ July _____Find the 3 period weighted moving average using: W1 = 3, W2 = 2, W3 = 1F July =If Actual demand in July is 90, then F August =If the weights are in percentages, then the formula reduces to: ...A W A W F 1t 2t 11t ++=−+Here W 1 > W 2 > W 3 … and W 1 + W 2 + …W n = 1Example: n = 3 and W 1 = 0.5, W 2 = 0.3, and W 3 = 0.2.Note that if the weights are all equal, i.e., W 1 = W 2 … and W 1 + W 2 + … W n = 1, then the weighted moving average formula reduces to the simple moving average. * Is this method better than the previous methods? * How do you determine n? * How do you determine the corresponding weights? * Impact of changing n and W’s?The next method, “Exponential Smoothing,” accomplishes the weighted moving average rationale without the need for determining n and corresponding W’s.n211t 2t 11t ...W W W .....A W A W F ++++=−+Simple Exponential SmoothingRecall that we are dealing with demand patterns exhibited in Fig 1, page 5, where demand is assumed to have relatively constant mean, with random fluctuations around it. For this, the simple exponential smoothing model can be explained by:Forecast for June = Forecast for May + α (Error made in May)where α is a number chosen between 0 and 1, and error is measured as the difference between actualdemand and forecasted demand.Thus, error (e t) = (Actual demand for period t) - (Forecast for period t)= A t - F tForecast errors can be positive or negative (i.e., forecasts may underestimate or over estimate the demand)Thus, in notations,F June = F May + α ( A May - F May ) or F June = α A May + (1- α) F Mayor F June = F May + α e MayIn general, F t+1 = F t + α e t or F t+1 = αΑt + (1-α)F tExampleSuppose α = 0.3 for the following data. Note that exponential smoothing requires an initial forecast and therefore let us take the forecast for January as F Jan = 100.t A t F t e t = A t - F tα e t F t+1_________120 100Jan____________Feb 90____________101March____________April 91____________115May____________June 83___JulyThus, e Jan = 120 – 100 = 20; hence F Feb = 100 + 0.3 x 20 = 106; then, e Feb = 90 – 106 = -16; henceF Mar = 106 + 0.3 x (-16), and so on.Another example: suppose α = 0.5, F Jan = 100. What would be the forecast for February? What about July?When forecasts are used in inventory management, for example:could result in _________________errorsPositiveNegative errors could result in _________________Both errors are clearly undesirable, because one results in overages and the other in shortages, but could they be equally so in Inventory Management? We will deal with inventory management in a later module and discuss these issues in detail.Let us revisit the exponential smoothing formula:F June = F May + α ( A May - F May ) (1)or F June = αA May + (1 - α) F MayThe strength (and hence the wide applicability) of Exponential Smoothing lies in the fact that the forecasts are simplyobtained by using the above formula; however, embedded in the above formula is the following (this is obtained byputting the formula for F May, F April, F March, and so on)F June = α A May + α (1 - α) A April + α (1-α)2 A March + (2)This can be looked at as a weighted average by writingF June = W1 A May + W2 A April + W3 A March + …W1 = α, W2 = α(1−α), W3 = α(1−α)2 and so on,whereIn other words, embedded in formula (1) above, lots of past periods are automatically included with different anddecreasing weights. (See the table below to understand this further and to study the implications of different choices ofα.)Let us now study the implications of the choice of αThe following table shows the decreasing weights that result with different values of αMonth Weight* α = 0.1 α = 0.4 α = 0.9May α0.10 0.40 0.90 April α(1-α) 0.09 0.24 0.09 March α(1-α)20.081 0.144 0.009 February α(1-α)30.0729 0.0864 9 * 10-4January α(1-α)40.0656 0.0518 9 * 10-5December etc.* From equation 2Note that when α = 0.1 (small), several past weights are significant and hence several past periods impact the new forecast. Conversely, when α = 0.9 (large) the past weights reduce to almost zero quite rapidly; thus very few past periods impact the new forecast.Thus, a small α will result in smooth forecasts and a large α will result in erratic forecasts.What if α = 0 ? What if α = 1 ?Answers to these questions can be obtained by simply putting these values in the exponential smoothing formula:F t+1 = αA t + (1- α) F tPutting α =0 gives F t+1 = F t Forecasts will not change (naïve 2)Putting α =1 gives F t+1 = A t (naive 1)Measures of ErrorAs stated earlier, forecasts are inaccurate most of the time; no method can predict the future precisely. The accuracy of the forecast is determined by how exactly it predicts. Let us now study the measures of errors that are commonly used in practice to determine the “goodness” of forecasts. For this, suppose we are given the following table of actuals and their corresponding forecasts and errors. Note the necessity of also getting the absolute and squared errors in order to determine the “magnitude” of the errors.t A tF te t |e t | e t 2Jan 120 100 20 20 400 Feb 90 106256Mar 101 102 April 91 101 May 115 98 June 83 103The following are three popular measures for measuring the “magnitude” of the errors.MagnitudeFor the Example Above 1. Mean Absolute Deviation (MAD)neMAD nt∑=1____________2.a.Mean Squared Error (MSE)()MSE e nt n=∑21____________2.b.Root Mean Squared Error (RMSE)RMSE MSE = ____________An accurate forecasting system will have small MAD, MSE and RMSE; ideally equal to zero. A large error may indicate that either the forecasting method used or the parameters such as α used in the method is wrong.Note: In the above, n is the number of periods, which in our example is 6.The following two measures are to determine if a positive or negative “bias” is present in the forecasting methods so that a corrective action can be taken to remove it. BiasRunning Sum of Forecast Errors (RSFE)RSFE e t n=∑1____________Clearly, ideally RSFE should be around zero (indicating that the negative and positive errors nullify each other). However, to get a good relative measure for bias, the RSFE should be adjusted by its MAD. Thus, a popular measure called tracking signal (TS) is often used to measure “bias” which combines the RSFE and MAD information as:TS RSFEMAD=____________Depending on RSFE, TS will be positive or negative. Ideally an unbiased system will have TS around zero. TS values closer to zero from negative side or positive side will be desirable. The APICS dictionary permits TS of -4 to +4 or at the most -8 to +8 depending on the importance of the item.Implementation of Forecasting in Inventory Management: A simple exampleConsider only two items A and B whose demand are shown below; note both items have random fluctuations only around more or less the same mean.Item AItem B• If you had a choice between managing items A and B which one would you choose? and why? • Which of these two items will have a small MAD?• For which item should we keep a high level of "safety stock”?•Is there a relationship between MAD and Safety Stock? (we will learn more about this later)2.1 Forecasting, page 16Another Example:Company A, a personal computer producer, purchases generic parts and assembles them to final product. Even though most of the orders require customization, they have many common components. Thus, managers of Company A need a good forecast of demand so that they can purchase computer parts accordingly to minimize inventory cost while meeting acceptable service level.Demand data for its computers for the past 6 months is given in the following table.As mentioned earlier, the determination of α is based on trial and error experiments. In this example let’s consider exponential smoothing forecasts using α equal to 0.3 and 0.5.As the following figure shows, the forecast using the higher smoothing constant (α = 0.5) reacts more rapidly to changes in demand (in the next section we discuss several quantitative methods for forecast accuracy). When demand displays an increasing or decreasing trend, as is the case in this example, a larger value of α is better. On the other hand if the demand were relatively stable, α = 0.3 could be more appropriate.In the above which α will have a higher MAD? And why?758595JanFebMarchAprilMayJune。