CH.8 Forecasting Learning objectives:

1. CH.8 Forecasting Learning objectives: After completing this chapter, you should be able to: 1.Explain the importance of forecasting in organization. 2.Describe the three major approaches to forecasting. 3.Measure the accuracy of a forecast over time.

2. Summary Decision maker rely on forecasting to reduce uncertainty about the future, for this reason forecasts play a vital role in planning. Decision makers have a wide variety of forecasting models to choose from, these can be classified as qualitative, projection of historical pattern, and explanatory models.

3. Glossary Time series: A time- ordered sequence of observation. Trend: A long- term movement in time series data, often linear, but not necessarily.

4. CH.8 Forecasting If there is considerable uncertainty , it is much more difficult to formulate plans that will produce the desired results than when there is little or no uncertainty involved. That is Forecasts provide decision makers with an improved picture of probable future events. So, forecasts are important because they, can help to reduce some of the uncertainty?

5. In fact, forecasts are vital inputs for almost all planning processes. Forecasts are used planning the system Product and service design, process design, capacity planning, and equipment investment decision. As well as for planning the use of system Advertising production and inventory planning and scheduling, purchasing, planning the size of work force, budgeting and cost estimation.

6. Good forecasts are more likely to result using this process: 1. Determine the purpose of the forecast. 2. Determine the time horizon. 3. Select an appropriate technique. 4. Identify the necessary date. 5. Make the forecast. 6. Monitor forecast errors. For forecasting we use Qualitative forecasts Quantitative forecasts We will use this approach

7. Quantitative models Associative models 1. Simple regression. Time series models : 1. Naïve 2. Moving averages. 3. Exponential smoothing. 4. Trend. Let’s go to example to see how we could apply each type

8. Forecasts that use Time series DATA * Time series data are historical values of a variable that have been intervals. Daily demand, weekly sales, quarterly revenues, annual demand. Forecasting techniques that use time series data typically involve the assumption that past experience reflects probable future experience. So, what are the observed patterns in historical data? These are : Trends, seasonal variation, cyclical variations, and random variations.

9. Look at figures: Cyclical variation Random variation around an average average Demand Demand 0 0 1 2 3 1 2 3 4 month year Trend with random Seasonal variation Demand Demand 0 0 4 1 2 3 1 2 3 Year Year

10. Techniques for averaging 1. Naïve forecasts. 2. Moving averages. 3. Exponential smoothing. Naïve forecasts It’s is simple , for any period the previous periods actual value . How? If demand last week was 50 units the forecasts for the upcoming week is 50 units. If the demand in the upcoming week turns to be 54 units, the forecast for the following week would be 54 units.

11. Moving averages Uses a number of the most recent actual data values in generating forecast. The moving average forecast can be computed using the following equation: MAn = Ai n M i= 1 n Where : i = the ‘age’ of the data I = 1,2,3,……….. n = the number of periods in the moving average. Ai = actual value with age i. Go to example

12. Example 1. Compute a three-period moving average forecast given demand for shopping product for the last 5 years: Period Age Demand 1 5 40 2 4 44 3 3 36 MA3 = 36 + 42 + 40 = 38.33 4 2 42 3 5 1 40 If actual demand in period 6 turns out to be 41, the moving average forecast for period 7 would be : MA3 = 42 + 40 + 41 = 41.00 3 If we want to make a weight for each data used in computing moving average ? Go to the example

13. Example 1. Weighted MA. WMA = Ai (w) M Weight determined base number of period MA. w M Period Weight Demand If we use: WMA3 = 36(1) + 42(2) + 40(3) = 40.00 1 40 2 44 6 3 3 36 If actual demand in period 6 turns to be 40,the weighted MA forecast for period 7 would be : 4 2 42 5 1 40 Weight WMA3 = 42(1) + 40(2) + 40(3) = 40.33 For 7 6 Note :- as we move forward we change the weighted ( it’s arrangement ).

14. Exponential smoothing model Is an averaging technique that reduces difficulties associated the previous model . The exponentially smoothed forecast can be computed using this equation: Ft = F + X ( A – F ) Where : Ft = the forecast for period (t) . F t-1 = the forecast for period ( t-1 ) . X = smoothing constraint . A t-1 = actual demand or sales for period ( t-1 ) . Go to understanding smoothing constant, and how to use it ? t-1 t-1 t-1

15. The smoothing constant , represent a percentage of the forecast error . Each new forecast is equal to the previous forecast plus a percentage of the previous error. For example: suppose the previous forecast was 100 units, actual demand was 90 units, and = .10 . The new forecast would be computed as follows Ft = 100 + .10 ( 90- 100 ) = 99 Then, if the next actual demand turns out to be 102, the next forecast would be: Ft = 99 + .10 ( 102- 99 ) =99.3 For forecasting which a percentage to use low or high?

16. The sensitivity of forecast adjustment to error is determined by the smoothing constant, alpha . The closer its value ( ) is to zero , the slower the forecast will be to adjust to forecast errors ( the greater the smoothing ) conversely , the closer the value of is to 1.00 the greater the sensitivity and the less the smoothing . Selecting a smoothing constant is, basically, a matter of judgment and trial and error. If you want to be sure about that use .05 and .50 Go to the example

17. = .05 = .50 Period A Ft Error A Ft Error 1 42 - - - - - 40 2 43 3 40 4 5 41 6 39 7 46 44 8 9 45 38 10 40 11 12 Ft = F - ( A t-1 – F t-1 ) t-1 Error = A – Ft

18. For accuracy use mean absolute deviation ( MAD ) equation : MAD = e = ( A – Ft ) M M n n value with MAD less is the best for forecasting? Suppose you have these info . Alternative MAD 1 26.3 Best accuracy 2 18.2 3 37.0 Complete the previous example to see that works.

19. Techniques for trend Trend is a persistent upward or downward movement in a time series. The trend may be linear or nonlinear. We will focus on linear trends. There are two important techniques that can be used to develop forecasts when trend present. 1. Trend equation. 2. Trend- adjusted exponential smoothing. Go to examples

20. Trend equation : a linear trend equation : Yt = a + bt Where : t = a specified number of time periods from t=0 Yt = the forecast for period t . a = value of Yt at t=0. b = slope of the line . The coefficient of the line, a and b , can be computed from historical data using these two equation: b = n Ty - t y M M M M M n t - ( t) 2 2 M M a = Y - b t n Where n the number of periods

21. Note :- Consider the trend equation Yt = 45 + 5t . The value of Yt when t = 0 is 45 , and the slope of the line is 5 , which means that the value of Yt will increase by 5 units for each one unit increase in t. If t = 10, the forecast, Yt, is 45 +5(10) = 95 units. Example : Month Actual demand Mar 112 Apr 125 May 120 Jun 133 Jul 136 Auq 146 Sept 140 Oct 155 Nov 152 1. Plot the data to determine if linear trend equation is appropriate. 2. Obtain a trend equation . 3. Forecast demand for the next two months.

22. Solution: 1. The data seem to show an upward, roughly linear trend . 160 150 Demand 140 130 120 110 1 2 3 4 5 6 7 8 9 Period Month t y ty Mar 112 1 112 Apr 125 2 250 May 120 3 360 Jun 4 133 532 Jul 136 5 660 Auq 146 6 876 Sept 140 7 980 Oct 155 8 1240 Nov 152 9 1360 M M M y t = 45 = 1218 Ty = 6398

23. From the table n = 9 , t = 45 , t = 285 2 M M b = 9 ( 6398 ) – 45 ( 1219 ) = 2727 = t 5.05 9 (285) – 45 ( 45 ) 540 a = 1219 – 5,05 ( 45 ) = 110.19 9 Yt = 110.19 + 5.05 t 3. The next two months, t= 10 , t = 1 . Y dec = 110.19 + 5.05 ( 10 ) = 160.69 . Y jan = 110.19 + 5.05 ( 11 ) = 165.74 .

24. Trend- adjusted exponential smoothing Note: we apply this technique when time series includes trend. The equation for trend adjusted forecast is: TAFt+1 = St + Tt Where St = smoothed error Tt = current trend estimate. And St = TAF t + ( A t – TAF t ) Tt = T t-1 + ( TAF t – TAF t-1 – T t-1 ) 2

25. Suppose a manager estimates a trend of + 10 units, based on observed changes in the first four data, and uses a starting forecast of 250, and =.5 and = .4 if the next actual value is 255, TAF for the following period would be computed in this manager: St = 250 + .5 ( 255 – 250 ) = 252.5 Tt = 10 + .4 (0)* * Since this is the initial forecast, no previous error is available, so the value zero is used. TAF = St + Tt = 252.5 + 10 = 262.5 If the next actual value is 265 the next TAF would be St = 262.5 + 05 ( 265 – 262.5 ) = 263.75 Tt = 10+ .4 ( 262.5 – 250 – 10 ) = 11.00 TAF = 263.75 + 11.00 = 274.75 We can use another equation for TAF 2

26. FITL = F t + T t Where : FIT t forecast including trend F t = new forecast simple smoothing . T t = trend for period For F t = F t-1 + ( A t-1 – F t-1 ) For T t = ( 1 – B ) T t-1 + B ( F t – F t-1 ) To compute FIT t follow these steps : 1. Find F t . 2. Compute T t 3. Find FIT t = 1 + 2 . Go to example

27. Example You have the following info: F 2 = 11 + .2 ( 12 – 11 ) = 11.2 T 2 = ( 1 - .4) 0 + .4 ( 11.2 – 11.0 ) = .08 FIT 2 = 11.2 + .08 = 11.28 And so on for the other periods. Month demand 1 12 2 17 3 20 4 19 5 24 6 26 Assume the initial forecast for month first was 11 unit 7 31 8 32 9 36 = .2 B = .4 Tt M d Ft FITt 1 12 11 0 .08 .51 .92 .96 1.30 1.52 1.91 2.02 - 11.28 12.87 14.81 15.87 18.03 20.10 22.98 25.27 2 17 11.2 3 20 12.56 13.89 14.91 16.75 18.58 21.07 23.25 4 19 5 24 6 26 7 31 8 32 9 36

28. Techniques for seasonality Seasonal variation in time series data are regularly upward or downward movements. Weather variation and it’s impact on travel sports product. To compute seasonal relative, use this model : Centered moving average, which based on Identical to a moving average forecast as are the resulting values. However the values are not projected as in a forecast, but, instead, they are positioned in the middle of the periods to compute average. Go to example

29. Period Demand Three-period centered average 1 40 40 + 46 + 42 2 46 Average = = 42.67 3 3 42 It would be positioned of period 2, the average representative of the series at that point. The ratio of demand at period 2 to this centered average at period 2 is an estimate of the seasonal relative at that point. Because the ratio is 46 = 1.08 the series is about 8% above average at this point. What if both trend and seasonality appear in a series? 42.67

30. Trend + seasonality Suppose a furniture manufacturer wants to predict quarterly demand for a certain seat for period 15 and 16, which happen to be the second and third quarters of a particular year. The series consists of both trend and seasonality. The trend portion of demand can be projected the equation Yt= 124 + 7.5t . Quarter relatives are : Q1 = 1.20 Q2 = 1.10 Q3 = .75 Q4 = .95 Use the info. To predict demand for period 15 and 16. Solution : The trend values at t = 15 and t = 16 Y15 = 124 + 7.5 ( 15 ) = 235.5 Y 16 = 124 + 7.5 ( 16 ) = 244.0 Period 15 235.5 ( 1.10 ) = 259.05 Period 16 244.0 ( .75 ) = 183.00 Quarter relative

31. Forecasts that use explanatory model These model incorporate one or more variables that are related to the variable of interest and, therefore, they can be used to predict future values of that variable. Are called regression models We are going to use the simple linear regression. * For multiple regression go to the computer?

32. Simple linear regression Consists two variable The variable to be forecasted is dependent variable ? The other value used to explain or predict the dependent variable is called independent variable ? To make forecast , the two variables must be expressed in a linear equation of the form : Y = a + bx Where : Y = the dependent variable . X = the independent variable. B = slope of the line . b = n x y - x y M M M M M A = y - b x = y – bx 2 2 M M N x - ( x ) n

33. Note : In order to successfully use this approach , it is necessary to be able to : 1. Identify an appropriate independent variable . 2. Obtain a sample of at least 10 observations. 3. Develop an equation . 4. Identify any restriction on prediction . 5. Measure accuracy in a given forecast . Go to the example ??

34. Conceder, for example , 10 paired observations about sales and ads. ( millions of $ ) observation Ads expenditure, x Sales , y 7 1 1.1 8 2 1.4 3 1.4 10 4 2 10 5 0.9 7 6 1.6 10 7 2 11 8 1.7 11 9 1.2 9 10 0.8 6 You can plot the pairs to see if a linear relationship is appropriate,?? It appear / linear relationship

35. Solution : Calculations for regression coefficients 2 2 observation Ads expenditure, x x y x y Sales , y 1 7 7.7 1.21 49 1.1 2 1.4 8 11.2 1.96 64 3 1.4 10 14.0 1.69 100 4 2 10 20.0 4.00 100 5 0.9 7 6.3 0.81 49 6 1.6 10 16.0 2.56 100 7 2 11 22.0 4.00 121 8 1.7 11 18.7 2.89 121 9 1.2 9 10.8 1.44 81 10 0.8 6 4.8 0.64 36 14.1 89 131.5 21.47 821 b = 10 ( 131.5 ) – 14.1 ( 89 ) = 60.1 = + 3.78 15.89 10 ( 21.47 ) – 14.1 ( 14.1 ) a = 89 – 3.78 ( 14.1 ) = 3.57 10 y = 3.57 + 3.87

36. Before we go in prediction process ,we must measure the strength of the relationship between the two variable . For that, we use correlation coefficient ? M M M r = n x y - x y 2 2 M M M M 2 n ( x ) - ( x ) X n ( y ) - ( y ) 2 r = 10 ( 131.5 ) - ( 14.1 ) ( 89 ) 10 (21.47 ) - ( 14.1 ) X 10 ( 821 ) - ( 89 ) = + .887 The same as the sign of b . Relationship is good to use for prediction .

37. 2 Square of the correlation coefficient, r , can be used as a measure of goodness of fit : the closer the value is to 1.00 the better the fit, for Ads data r = 0.787 this means that .79 of the sum of the squared deviation of Y value around the mean of the Y . We say that the independent variable explain .79 of the variation . There are a number of important questions that must be answered if a regression equation is to be used for forecasting ?? 2

38. 1. Is the relationship significant ? 2. How accurate will a particular forecast probably be ? These questions and other depend on the extend to which the points scatter around the regression line . This is measured by standard error of estimate, se, Se = n – 2 / because we have two coefficient a and b M M M y - a y - b x y 2 n – 2

39. For our example Se = 821 – 3.57 (89) – 3.78 ( 131.5 ) = .880 10 – 2 2 Note : if the value of r is less than , say .40 the computed relationship will not be strong enough to yield acceptable forecast accuracy. To test for the existence of relationship ( than the slope is non zero ) Compute the ration T test = b S b Where b = slope of regression line . S b = Se 1 M M 2 2 X - ( x ) n

40. For our example S b = .880 1 .629 = .698 = .880 2 21.47 – ( 14.1 ) 10 Compare this value with value of t .025 for n – 2 / from the table of T test T test = 3.78 = 5.42 .698 n 10 11 12 13 14 15 16 17 18 19 20 21 22 t 2.31 2.26 2.23 2.20 2.18 2.16 2.15 2.13 2.12 2.11 2.10 2.09 2.08 If the test is greater than the table value, as it is in our case ( 5.42 ), it can be concluded that the relationship is significant. Not due to the chance ??

41. How accurate will a given predication be ? Here we should use the regression equation to compute expected value of the dependent variable for a selected or given value of X the construct a 95% confidence interval around that value . Suppose expected value of X is 1.3 ? Y + t S reg Where Y = a + b x t = from the T test table S reg = 2 ( x g – x ) 1 + 1 + n 2 2 x - ( x ) M M n X g = value of x to be used

42. X = 1.3 , X = 14.1 = 1.41 10 S reg = .880 2 ( 1.3 – 1.41 ) 1 + 1 + = .926 2 10 21.47 - (14.1) 10 For our case : X g = 1.3 , 95% confidence interval For the predicted value of y is Y = a + b x = 3.57 + 3.78 (1.3) = 8.48 95% interval = 8.48 + 2.31 ( .926 ) = 8.48 + 2.14 or 6.34 to 10.62 for test table .