1. ISQA 459/559 Advanced Forecasting Mellie Pullman

2. 30 Recall Forecast Error Measurements MFE: mean forecast error MAD: mean absolute deviation

3. Best Error Measurement (What it the problem with the MAD calculation as an error measurement for long histories?) 365 days Averaged ?

4. Solution?  Smoothed MAD  Phi () is a smoothing parameter, which is set in advance.  It is important that we fix (set) phi BEFORE we try to find the best forecasting method. Why?

5. Phi  Phi controls the period of time over which we are evaluating forecast accuracy--the smaller the value of phi, the larger the number of historical periods that are considered in the measurement of the "average" forecast error.  What effect would changing phi have while you are trying to compare the accuracy of two different forecasting methods?

6. Suggested Values for Phi Forecasting IntervalGood Values of Phi Daily.02 (149 days).03 (99 days).04 (74 days).05 (59 days).10 (29 days) Weekly.05 (59 weeks).10 (29 weeks).15 (19 weeks).20 (14 weeks) Monthly.10 (29 months).15 (19 months).20 (14 months).25 (11 months).30 (9 months)

7. Phi0.3 MonthDemandForecast ErrorMAD ----- 1200200.00.0 2134200.0-66.019.8 3157180.2-23.220.8 4165173.2-8.217.0 5177170.86.213.8 6125172.6-47.624.0 7146158.3-12.320.5 8150154.6-4.615.7 9182153.228.819.6 10197161.935.124.3 11136172.4-36.427.9 12163161.51.520.0 13157161.9-4.915.5 14169160.58.513.4 --------- TOTALS2258.02381.3-123.3252.3

8. QuarterYear 1Year 2Year 3Year 4 14570100100 2335370585725 35205908301160 4100170285215 Total1000120018002200 Average250300450550 Seasonal Index/Factor We estimate 2600 for Year 5 but need to know how many to make each quarter.

9. Seasonal Factor Method

10. QuarterYear 1Year 2Year 3Year 4 145/250 = 0.1870100100 2335370585725 35205908301160 4100170285215 Total1000120018002200 Average250300450550 Seasonal Index = = 0.18 45 250 Seasonal Index/Factor

11. Quarter Year 1 Year 2 Year 3 Year 4 145/250 = 0.1870/300 = 0.23100/450 = 0.22100/550 = 0.18 2335/250 = 1.34370/300 = 1.23585/450 = 1.30725/550 = 1.32 3520/250 = 2.08590/300 = 1.97830/450 = 1.841160/550 = 2.11 4100/250 = 0.40170/300 = 0.57285/450 = 0.63215/550 = 0.39 QuarterAverage Seasonal Index 1(0.18 + 0.23 + 0.22 + 0.18)/4 = 0.20 2(1.34 + 1.23 + 1.30 + 1.32)/4 = 1.30 3(2.08 + 1.97 + 1.84 + 2.11)/4 = 2.00 4(0.40 + 0.57 + 0.63 + 0.39)/4 = 0.50 Seasonal Index/Factor

12. Quarter Year 1 Year 2 Year 3 Year 4 145/250 = 0.1870/300 = 0.23100/450 = 0.22100/550 = 0.18 2335/250 = 1.34370/300 = 1.23585/450 = 1.30725/550 = 1.32 3520/250 = 2.08590/300 = 1.97830/450 = 1.841160/550 = 2.11 4100/250 = 0.40170/300 = 0.57285/450 = 0.63215/550 = 0.39 QuarterAverage Seasonal IndexForecast 1(0.18 + 0.23 + 0.22 + 0.18)/4 = 0.20650(0.20) =130 2(1.34 + 1.23 + 1.30 + 1.32)/4 = 1.30650(1.30) =845 3(2.08 + 1.97 + 1.84 + 2.11)/4 = 2.00650(2.00) =1300 4(0.40 + 0.57 + 0.63 + 0.39)/4 = 0.50650(0.50) =325 Seasonal Influences

13. In- Class Problem: Forecast Year 3 (Overall forecast = 1500) Qtr Year 1Year 2 Average Index DemandIndexDemandIndex 1100192 2400408 3300384 4200216 Avg

14. Decomposition of Season & Trend  Decompose the data into components Find seasonal component Deseasonalize demand Find Trend component  Forecast future values of each component Project Trend component into future Multiply trend component by seasonal component

15. Example of Deseasonalized Data

16. Project Future and Re-seasonalize

17. Options for Brewery Case that use regression and/or seasonal adjustment?  Using Yearly Data to start?  Using Monthly data to start?

18. Trend-Adjusted Exponential Smoothing ||||||||||||||| 0123456789101112131415 80 — 70 — 60 — 50 — 40 — 30 — Guest arrivals Week Actual room requests

19. Trend-Adjusted Exponential Smoothing

20. ||||||||||||||| 0123456789101112131415 80 — 70 — 60 — 50 — 40 — 30 — Guest arrivals Week Guest Arrivals A t =  D t + (1 -  )(A t-1 + T t-1 ) T t =  (A t - A t-1 ) + (1 -  )T t-1

21. Trend-Adjusted Exponential Smoothing ||||||||||||||| 0123456789101112131415 80 — 70 — 60 — 50 — 40 — 30 — Guest arrivals Week A 1 = 0.2(27) + 0.80(28 + 3)= 30.2 T 1 = 0.2(30.2 - 28) + 0.80(3)= 2.8 Guest Arrivals A 0 = 28 g D 1 = 27 g T 0 = 3 g  = 0.20  = 0.20 A t =  D t + (1 -  )(A t-1 + T t-1 ) T t =  (A t - A t-1 ) + (1 -  )T t-1

22. Trend-Adjusted Exponential Smoothing ||||||||||||||| 0123456789101112131415 80 — 70 — 60 — 50 — 40 — 30 — Guest arrivals Week A 1 = 30.2 T 1 = 2.8 Guest Arrivals A 0 = 28 guests T 0 = 3 guests  = 0.20  = 0.20 A t =  D t + (1 -  )(A t-1 + T t-1 ) T t =  (A t - A t-1 ) + (1 -  )T t-1 Forecast 2 = 30.2 + 2.8 = 33

23. Trend-Adjusted Exponential Smoothing ||||||||||||||| 0123456789101112131415 80 — 70 — 60 — 50 — 40 — 30 — Guest arrivals Week Guest Arrivals A 1 = 30.2 D 2 = 44 T 1 = 2.8  = 0.20  = 0.20 A t =  D t + (1 -  )(A t-1 + T t-1 ) T t =  (A t - A t-1 ) + (1 -  )T t-1 A 2 = T 2 = Forecast =

24. Trend-Adjusted Exponential Smoothing ||||||||||||||| 0123456789101112131415 80 — 70 — 60 — 50 — 40 — 30 — Guest arrivals Week Guest Arrivals A 1 = 30.2 D 2 = 44 T 1 = 2.8  = 0.20  = 0.20 A t =  D t + (1 -  )(A t-1 + T t-1 ) T t =  (A t - A t-1 ) + (1 -  )T t-1 A 2 = 35.2 T 2 = 3.2 Forecast = 35.2 + 3.2 = 38.4

25. Trend-Adjusted Exponential Smoothing ||||||||||||||| 0123456789101112131415 80 — 70 — 60 — 50 — 40 — 30 — Guest arrivals Week Trend-adjusted forecast Actual guest arrivals

26. In Class Exercise  A mar = 300,000 cases; T mar = +8,000 cases D apr = 330,000 cases; = 0.20 =.10  What are the forecasts for May and July?

27. The End