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- 1. 13 – 1 ForecastingForecasting 13 ForFor Operations Management, 9eOperations Management, 9e byby Krajewski/Ritzman/MalhotraKrajewski/Ritzman/Malhotra © 2010 Pearson Education© 2010 Pearson Education PowerPoint SlidesPowerPoint Slides by Jeff Heylby Jeff Heyl
- 2. 13 – 2 ForecastingForecasting Forecasts are critical inputs to business plans, annual plans, and budgets Finance, 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 capacities Forecasts are made on many different variables Forecasts are important to managing both processes and managing supply chains
- 3. 13 – 3 Demand PatternsDemand Patterns A time series is the repeated observations of demand for a service or product in their order of occurrence There are five basic time series patterns Horizontal Trend Seasonal Cyclical Random
- 4. 13 – 4 Demand PatternsDemand Patterns Quantity Time (a) Horizontal: Data cluster about a horizontal line Figure 13.1 – Patterns of Demand
- 5. 13 – 5 Demand PatternsDemand Patterns Quantity Time (b) Trend: Data consistently increase or decrease Figure 13.1 – Patterns of Demand
- 6. 13 – 6 Demand PatternsDemand Patterns Quantity | | | | | | | | | | | | J F M A M J J A S O N D Months (c) Seasonal: Data consistently show peaks and valleys Year 1 Year 2 Figure 13.1 – Patterns of Demand
- 7. 13 – 7 Demand PatternsDemand Patterns Quantity | | | | | | 1 2 3 4 5 6 Years (d) Cyclical: Data reveal gradual increases and decreases over extended periods Figure 13.1 – Patterns of Demand
- 8. 13 – 8 Key DecisionsKey Decisions Deciding what to forecast Level of aggregation Units of measure Choosing a forecasting system Choosing the type of forecasting technique Judgment and qualitative methods Causal methods Time-series analysis Key factor in choosing the proper forecasting approach is the time horizon for the decision requiring forecasts
- 9. 13 – 9 Judgment MethodsJudgment Methods Other methods (casual and time-series) require an adequate history file, which might not be available Judgmental forecasts use contextual knowledge gained through experience Salesforce estimates Executive opinion is a method in which opinions, experience, and technical knowledge of one or more managers are summarized to arrive at a single forecast Delphi method
- 10. 13 – 10 Judgment MethodsJudgment Methods Market research is a systematic approach to determine external customer interest through data-gathering surveys Delphi method is a process of gaining consensus from a group of experts while maintaining their anonymity Useful when no historical data are available Can be used to develop long-range forecasts and technological forecasting
- 11. 13 – 11 Linear RegressionLinear Regression A dependent variable is related to one or more independent variables by a linear equation The independent variables are assumed to “cause” the results observed in the past Simple linear regression model is a straight line Y = a + bX where Y = dependent variable X = independent variable a = Y-intercept of the line b = slope of the line
- 12. 13 – 12 Linear RegressionLinear Regression Dependentvariable Independent variable X Y Estimate of Y from regression equation Regression equation: Y = a + bX Actual value of Y Value of X used to estimate Y Deviation, or error Figure 13.2 – Linear Regression Line Relative to Actual Data
- 13. 13 – 13 Linear RegressionLinear Regression The sample correlation coefficient, r Measures the direction and strength of the relationship between 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
- 14. 13 – 14 Using Linear RegressionUsing Linear Regression EXAMPLE 13.1 The supply chain manager seeks a better way to forecast the demand for door hinges and believes that the demand is related to advertising expenditures. The following are sales and advertising data for the past 5 months: Month Sales (thousands of units) Advertising (thousands of $) 1 264 2.5 2 116 1.3 3 165 1.4 4 101 1.0 5 209 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.
- 15. 13 – 15 Using Linear RegressionUsing Linear Regression SOLUTION We 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 = The regression equation is Y = –8.135 + 109.229X –8.135 109.229X 0.980 0.960 15.603
- 16. 13 – 16 Using Linear RegressionUsing Linear Regression 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, r2 , implies that 96 percent of the variation in sales is explained by advertising expenditures. | | 1.0 2.0 Advertising ($000) 250 – 200 – 150 – 100 – 50 – 0 – Sales(000units) Brass Door Hinge X X X X X X Data Forecasts Figure 13.3 – Linear Regression Line for the Sales and Advertising Data
- 17. 13 – 17 Time Series MethodsTime Series Methods In a naive forecast the forecast for the next period equals the demand for the current 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 fluctuation Most useful when demand has no pronounced trend or seasonal influences The stability of the demand series generally determines how many periods to include
- 18. 13 – 18 | | | | | | 0 5 10 15 20 25 30 Week 450 – 430 – 410 – 390 – 370 – 350 – Patientarrivals Time Series MethodsTime Series Methods Figure 13.4 – Weekly Patient Arrivals at a Medical Clinic
- 19. 13 – 19 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) as Ft+1 = = Sum of last n demands n Dt + Dt-1 + Dt-2 + … + Dt-n+1 n where Dt = actual demand in period t n = total number of periods in the average Ft+1 = forecast for period t + 1
- 20. 13 – 20 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 subtracting the forecast from actual demand for a given period, or where Et = forecast error for period t Dt = actual demand in period t Ft = forecast for period t Et = Dt – Ft
- 21. 13 – 21 Using the Moving Average MethodUsing the Moving Average Method EXAMPLE 13.2 a. Compute a three-week moving average forecast for the arrival of medical clinic patients in week 4. The numbers of arrivals for the past three weeks were as follows: Week Patient Arrivals 1 400 2 380 3 411 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?
- 22. 13 – 22 Using the Moving Average MethodUsing the Moving Average Method SOLUTION a. The moving average forecast at the end of week 3 is Week Patient Arrivals 1 400 2 380 3 411 b. The forecast error for week 4 is F4 = = 397.0 411 + 380 + 400 3 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 = = 402.0 415 + 411 + 380 3
- 23. 13 – 23 Application 13.1aApplication 13.1a Estimating with Simple Moving Average using the following customer-arrival data Month Customer arrival 1 800 2 740 3 810 4 790 Use a three-month moving average to forecast customer arrivals for month 5 F5 = = 780 D4 + D3 + D2 3 790 + 810 + 740 3 = Forecast for month 5 is 780 customer arrivals
- 24. 13 – 24 Application 13.1aApplication 13.1a If the actual number of arrivals in month 5 is 805, what is the forecast for month 6? F6 = = 801.667 D5 + D4 + D3 3 805 + 790 + 810 3 = Forecast for month 6 is 802 customer arrivals Month Customer arrival 1 800 2 740 3 810 4 790
- 25. 13 – 25 Application 13.1aApplication 13.1a 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? Forecast error for month 5 is 25 Et = Dt – Ft E5 = 805 – 780 = 25
- 26. 13 – 26 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 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: Weighted Moving AveragesWeighted Moving Averages 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
- 27. 13 – 27 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 Forecast for month 5 is 786 customer arrivals Given the number of patients that actually arrived (805), what is the forecast error? Forecast error for month 5 is 19 E5 = 805 – 786 = 19
- 28. 13 – 28 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 Forecast for month 6 is 802 customer arrivals
- 29. 13 – 29 Exponential SmoothingExponential Smoothing A sophisticated weighted moving average that calculates the average of a time series by giving recent demands more weight than earlier demands Requires only three items of data The last period’s forecast The demand for this period A 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
- 30. 13 – 30 Exponential SmoothingExponential Smoothing The emphasis given to the most recent demand levels can be adjusted by changing the smoothing parameter Larger α values emphasize recent levels of demand and result in forecasts more responsive to changes in the underlying average Smaller α values treat past demand more uniformly and result in more stable forecasts Exponential smoothing is simple and requires minimal data When the underlying average is changing, results will lag actual changes
- 31. 13 – 31 450 – 430 – 410 – 390 – 370 – Patientarrivals Week | | | | | | 0 5 10 15 20 25 30 3-week MA forecast 6-week MA forecast Exponential smoothing α = 0.10 Exponential Smoothing andExponential Smoothing and Moving AverageMoving Average
- 32. 13 – 32 Using Exponential SmoothingUsing Exponential Smoothing EXAMPLE 13.3 a. Reconsider the patient arrival data in Example 13.2. It is now the end of week 3. Using α = 0.10, calculate the exponential smoothing forecast for week 4. Week Patient Arrivals 1 400 2 380 3 411 4 415 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?
- 33. 13 – 33 Using Exponential SmoothingUsing Exponential Smoothing SOLUTION a. The exponential smoothing method requires an initial forecast. Suppose that we take the demand data for the first two 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
- 34. 13 – 34 Using Exponential SmoothingUsing Exponential Smoothing b. The forecast error for week 4 is c. The new forecast for week 5 would be E4 = F5 = 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. 415 – 392 = 23 0.10(415) + 0.90(392.1) = 394.4
- 35. 13 – 35 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 Forecast for month 5 is 784 customer arrivals Given the number of patients that actually arrived (805), what is the forecast error? E5 = Forecast error for month 5 is 21 805 – 784 = 21
- 36. 13 – 36 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 Forecast for month 6 is 789 customer arrivals
- 37. 13 – 37 Including a TrendIncluding a Trend A trend in a time series is a systematic increase or decrease in the average of the series over time The forecast can be improved by calculating an estimate of the trend Trend-adjusted exponential smoothing requires two smoothing constants
- 38. 13 – 38 Including a TrendIncluding a Trend For each period, we calculate the average and the trend: 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 Ft+1 = At + Tt where At = exponentially smoothed average of the series in period t Tt = 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
- 39. 13 – 39 Using Trend-Adjusted ExponentialUsing Trend-Adjusted Exponential SmoothingSmoothing EXAMPLE 13.4 Medanalysis, Inc., provides medical laboratory services Managers are interested in forecasting the number of blood analysis requests per week There has been a national increase in requests for standard blood tests Medanalysis recently ran an average of 28 blood tests per week and the trend has been about three additional patients per week This week’s demand was for 27 blood tests We use α = 0.20 and β = 0.20 to calculate the forecast for next week
- 40. 13 – 40 30.2 + 2.8 = 33 blood tests Using Trend-Adjusted ExponentialUsing Trend-Adjusted Exponential SmoothingSmoothing SOLUTION If the actual number of blood tests requested in week 2 proved to be 44, the updated forecast for week 3 would be A0 = 28 patients and T0 = 3 patients The forecast for week 2 (next week) is A1 = T1 = F2 = 0.20(27) + 0.80(28 + 3) = 30.2 0.20(30.2 – 28) + 0.80(3) = 2.8 A2 = F3 = 35.2 + 3.2 = 38.4 or 38 blood tests 0.2(35.2 – 30.2) + 0.80(2.8) = 3.2 0.20(44) + 0.80(30.2 + 2.8) = 35.2 T2 =
- 41. 13 – 41 Using Trend-Adjusted ExponentialUsing Trend-Adjusted Exponential SmoothingSmoothing 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 0 28 28.00 3.00 1 27 2 44 3 37 4 35 5 53 6 38 7 57 8 61 9 39 10 55 11 54 12 52 13 60 14 60 15 75
- 42. 13 – 42 Using Trend-Adjusted ExponentialUsing Trend-Adjusted Exponential SmoothingSmoothing 30.20 + 2.84 = 33.04 –430.20 35.23 2.84 3.28 28.00 + 3.00 = 31.00 35.23 + 3.28 = 38.51 38.21 + 3.22 = 41.43 40.14 + 2.96 = 43.10 45.08 + 3.36 = 48.44 46.35 + 2.94 = 49.29 50.83 + 3.25 = 54.08 55.46 + 3.52 = 58.98 54.99 + 2.72 = 57.71 57.17 + 2.62 = 59.79 58.63 + 2.38 = 61.01 59.21 + 2.02 = 61.23 60.99 + 1.97 = 62.96 62.37 + 1.86 = 64.23 10.96 –1.51 –6.43 9.90 –10.44 7.71 6.92 –19.98 –2.71 –5.79 –9.01 –1.23 –2.96 10.77 38.21 40.14 45.08 46.35 50.83 55.46 54.99 57.17 58.63 59.21 60.99 62.37 66.38 3.22 2.96 3.36 2.94 3.25 3.52 2.72 2.62 2.38 2.02 1.97 1.86 2.29 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 0 28 28.00 3.00 1 27 2 44 3 37 4 35 5 53 6 38 7 57 8 61 9 39 10 55 11 54 12 52 13 60 14 60 15 75
- 43. 13 – 43 | | | | | | | | | | | | | | | | 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 80 – 70 – 60 – 50 – 40 – 30 – Patientarrivals Week Actual blood test requests Trend-adjusted forecast Using Trend-Adjusted ExponentialUsing Trend-Adjusted Exponential SmoothingSmoothing Figure 13.5 – Trend-Adjusted Forecast for Medanalysis
- 44. 13 – 44 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 sales 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 Forecast for May = AApr + pTApr = 0.20(330,000) + 0.80(300,000 + 8,000) = 312,400 cases = 0.10(312,400 – 300,000) + 0.90(8,000) = 8,440 cases = 312,400 + (1)(8,440) = 320,840 cases
- 45. 13 – 45 Application 13.2Application 13.2 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 periods 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
- 46. 13 – 46 Seasonal PatternsSeasonal Patterns Seasonal patterns are regularly repeated upward or downward movements in demand 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 season This approach accounts for seasonal effects but discards considerable information on past demand
- 47. 13 – 47 1. 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 3. Calculate the average seasonal index for each season using the results from Step 2 4. Calculate each season’s forecast for next year 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 forecast
- 48. 13 – 48 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 Using the Multiplicative SeasonalUsing the Multiplicative Seasonal MethodMethod EXAMPLE 13.5 The manager of the Stanley Steemer carpet cleaning company needs 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 4 1 45 70 100 100 2 335 370 585 725 3 520 590 830 1160 4 100 170 285 215 Total 1000 1200 1800 2200
- 49. 13 – 49 Using the Multiplicative SeasonalUsing the Multiplicative Seasonal MethodMethod SOLUTION Figure 13.6 shows the solution using the Seasonal Forecasting Solver 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 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).
- 50. 13 – 50 Using the Multiplicative SeasonalUsing the Multiplicative Seasonal MethodMethod Figure 13.6 – Demand Forecasts Using the Seasonal Forecast Solver of OM Explorer
- 51. 13 – 51 Application 13.3Application 13.3 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 0.52 2 400 1.60 408 1.36 1.48 3 300 1.20 384 1.28 1.24 4 200 0.80 216 0.72 0.76 Average 250 300
- 52. 13 – 52 Application 13.3Application 13.3 1320 units ÷ 4 quarters = 330 units Quarter Average Index 1 0.52 2 1.48 3 1.24 4 0.76 If the projected demand for Year 3 is 1320 units, what is the forecast for each quarter of that year? Forecast for Quarter 1 = Forecast for Quarter 2 = Forecast for Quarter 3 = Forecast for Quarter 4 = 0.52(330) ≈ 172 units 1.48(330) ≈ 488 units 1.24(330) ≈ 409 units 0.76(330) ≈ 251 units
- 53. 13 – 53 (a) Multiplicative pattern Seasonal PatternsSeasonal Patterns Period | | | | | | | | | | | | | | | | 0 2 4 5 8 10 12 14 16 Demand
- 54. 13 – 54 Seasonal PatternsSeasonal Patterns (b) Additive pattern Period | | | | | | | | | | | | | | | | 0 2 4 5 8 10 12 14 16 Demand
- 55. 13 – 55 Choosing a Time-Series MethodChoosing a Time-Series Method Forecast performance is determined by forecast errors Forecast errors detect when something is going wrong with the forecasting system Forecast errors can be classified as either bias errors or random errors Bias errors are the result of consistent mistakes Random error results from unpredictable factors that cause the forecast to deviate from the actual demand
- 56. 13 – 56 CFE = ΣEt Measures of Forecast ErrorMeasures of Forecast Error Σ(Et – E )2 n – 1 σ = ΣEt 2 nMSE = Σ|Et | n MAD = (Σ|Et |/Dt)(100) nMAPE = E = CFE n
- 57. 13 – 57 Calculating Forecast ErrorsCalculating Forecast Errors EXAMPLE 13.6 The following table shows the actual sales of upholstered chairs 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. Month t Demand Dt Forecast Ft Error Et Error2 Et 2 Absolute Error |Et| Absolute % Error (| Et|/Dt)(100) 1 200 225 –25 2 240 220 20 3 300 285 15 4 270 290 –20 5 230 250 –20 400 20 8.7 6 260 240 20 400 20 7.7 7 210 250 40 1,600 40 19.0 8 275 240 35 1,225 35 12.7 Total –15 5,275 195 81.3%
- 58. 13 – 58 Calculating Forecast ErrorsCalculating Forecast Errors EXAMPLE 13.6 The following table shows the actual sales of upholstered chairs 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. Month t Demand Dt Forecast Ft Error Et Error2 Et 2 Absolute Error |Et| Absolute % Error (| Et|/Dt)(100) 1 200 225 –25 625 25 12.5% 2 240 220 20 400 20 8.3 3 300 285 15 225 15 5.0 4 270 290 –20 400 20 7.4 5 230 250 –20 400 20 8.7 6 260 240 20 400 20 7.7 7 210 250 40 1,600 40 19.0 8 275 240 35 1,225 35 12.7 Total –15 5,275 195 81.3%
- 59. 13 – 59 SOLUTION Using the formulas for the measures, we get Cumulative forecast error (bias): Calculating Forecast ErrorsCalculating Forecast Errors CFE = –15 Average forecast error (mean bias): Mean squared error: MSE = ΣEt 2 n CFE n E = –1.875= 5,275 8 =
- 60. 13 – 60 Standard deviation: Calculating Forecast ErrorsCalculating Forecast Errors Mean absolute deviation: Mean absolute percent error: Σ[Et – (–1.875)]2 n – 1 σ = Σ|Et | nMAD = (Σ|Et |/ Dt)(100) nMAPE = = 27.4 = = 24.4 195 8 = = 10.2% 81.3% 8
- 61. 13 – 61 Calculating Forecast ErrorsCalculating Forecast Errors 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 the 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.
- 62. 13 – 62 Tracking SignalsTracking Signals A measure that indicates whether a method of forecasting is accurately predicting actual changes in demand Useful when forecast systems are computerized because it alerts analysts when forecast are getting far from desirable limits Tracking signal = CFE MAD Each period, the CFE and MAD are updated to reflect current error, and the tracking signal is compared to some predetermined limits
- 63. 13 – 63 Tracking SignalsTracking Signals 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 σ = ( π /2)(MAD) ≅ 1.25(MAD) MAD = 0.7978σ ≅ 0.8σ
- 64. 13 – 64 +2.0 – +1.5 – +1.0 – +0.5 – 0 – –0.5 – –1.0 – –1.5 – | | | | | 0 5 10 15 20 25 Observation number Trackingsignal Out of control Tracking SignalsTracking Signals Control limit Control limit Figure 13.7 – Tracking Signal
- 65. 13 – 65 Criteria for Selecting MethodsCriteria for Selecting Methods Criteria to use in making forecast method and parameter choices include 1. Minimizing bias 2. Minimizing MAPE, MAD, or MSE 3. Meeting managerial expectations of changes in the components of demand 4. 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
- 66. 13 – 66 Using Multiple TechniquesUsing Multiple Techniques Combination forecasts are forecasts that are produced by averaging independent forecasts based on different methods or different data or both Focus forecasting selects the best forecast from a group of forecasts generated by individual techniques
- 67. 13 – 67 Forecasting as a ProcessForecasting as a Process A typical forecasting process Step 1: Adjust history file Step 2: Prepare initial forecasts Step 3: Consensus meetings and collaboration Step 4: Revise forecasts Step 5: Review by operating committee Step 6: Finalize and communicate Forecasting is not a stand-alone activity, but part of a larger process
- 68. 13 – 68 Forecasting as a ProcessForecasting as a Process Finalize and communicate 6 Review by Operating Committee 5 Revise forecasts 4 Consensus meetings and collaboration 3 Prepare initial forecasts 2 Adjust history file 1
- 69. 13 – 69 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 competition Better forecasts result in better customer service and lower costs, as well as better relationships with suppliers and customers The forecast can and must make sense based on the big picture, economic outlook, market share, and so on The best way to improve forecast accuracy is to focus on reducing forecast error Bias is the worst kind of forecast error; strive for zero bias Whenever possible, forecast at more aggregate levels. Forecast in detail only where necessary Far more can be gained by people collaborating and communicating well than by using the most advanced forecasting technique or model
- 70. 13 – 70 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 calculations 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 $2.70 760 2 $3.50 510 3 $2.00 980 4 $4.20 250 5 $3.10 320 6 $4.05 480 Total $19.55 3,300 Average $3.258 550
- 71. 13 – 71 Solved Problem 1Solved Problem 1 SOLUTION We use the computer to calculate the best values of a, b, the correlation coefficient, and the coefficient of determination a = b = r = r 2 = 0.71 –0.84 –277.63 1,454.60 The regression line is 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
- 72. 13 – 72 Solved Problem 2Solved Problem 2 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 demand 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.
- 73. 13 – 73 Solved Problem 2Solved Problem 2 SOLUTION a. The simple moving average method and the weighted 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 June 30 July 7 = 55.7 or 56 52 + 65 + 50 3 [(0.5 × 52) + (0.3 × 65) + (0.2 × 50)] = 55.5 or 56 = 57.7 or 58 56 + 52 + 65 3 = 54.3 or 54 55 + 56 + 52 3 [(0.5 × 56) + (0.3 × 52) + (0.2 × 65)] = 56.6 or 57 [(0.5 × 55) + (0.3 × 56) + (0.2 × 52)] = 54.7 or 55 = 57.0 or 57 60 + 55 + 56 3 [(0.5 × 60) + (0.3 × 55) + (0.2 × 56)] = 57.7 or 58
- 74. 13 – 74 Solved Problem 2Solved Problem 2 b. The mean absolute deviation is calculated as follows: Simple Moving Average Weighted Moving Average Week Actual Demand Forecast 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 = = 3 0 + 3 + 6 3 MAD = = 2.3 0 + 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.
- 75. 13 – 75 Solved Problem 3Solved Problem 3 The monthly demand for units manufactured by the Acme Rocket Company has been as follows: Month Units Month Units May 100 September 105 June 80 October 110 July 110 November 125 August 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?
- 76. 13 – 76 Solved Problem 3Solved Problem 3 SOLUTION a. Current Month, t Calculating Forecast for Next Month Ft+1 = αDt + (1 – α)Ft Forecast for Month t + 1 May June June July July August August September September October October November November December December January 0.2(100) + 0.8(105) = 104.0 or 104 0.2(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) 0.2(125) + 0.8(105.4) 0.2(120) + 0.8(109.3) = 104.1 or 104 = 104.3 or 104 = 105.4 or 105 = 109.3 or 109 = 111.4 or 111
- 77. 13 – 77 Solved Problem 3Solved Problem 3 b. –24 24 30.0% 11 11 10.0 Month, t Actual Demand, Dt Forecast, Ft Error, Et = Dt – Ft Absolute Error, |Et| Absolute Percent Error, (| Et|/Dt)(100) June 80 104 July 110 99 August 115 101 September 105 104 October 110 104 November 125 105 December 120 109 Total 765 14 14 12.0 1 1 1.0 6 6 5.5 20 20 16.0 11 11 9.2 39 87 83.7% Σ|Et | nMAD = (Σ|Et |/Dt)(100) nMAPE = = = 11.96% 83.7% 7 = = 12.4 87 7
- 78. 13 – 78 Solved Problem 3Solved Problem 3 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 should revise our approach. The long string of forecasts lower than actual demand suggests use of a trend method. Tracking signal = CFE MAD = = 3.14 39 12.4
- 79. 13 – 79 Solved Problem 4Solved Problem 4 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 mail: Day Week 1 Week 2 Sunday 5 8 Monday 20 15 Tuesday 30 32 Wednesday 35 30 Thursday 49 45 Friday 70 70 Saturday 15 10 Total 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.
- 80. 13 – 80 Solved Problem 4Solved Problem 4 SOLUTION a. Calculate the average daily mail volume for each week. Then for 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.
- 81. 13 – 81 Solved Problem 4Solved Problem 4 Week 1 Week 2 Day Mail Volume Seasonal Factor (1) Mail Volume Seasonal Factor (2) Average Seasonal Factor [(1) + (2)]/2 Sunday 5 8 Monday 20 15 Tuesday 30 32 Wednesday 35 30 Thursday 49 45 Friday 70 70 Saturday 15 10 Total 224 210 Average 224/7 = 32 210/7 = 30 5/32 = 0.15625 20/32 = 0.62500 30/32 = 0.93750 8/30 = 0.26667 15/30 = 0.50000 32/30 = 1.06667 0.21146 0.56250 1.00209 35/32 = 1.09375 49/32 = 1.53125 70/32 = 2.18750 15/32 = 0.46875 30/30 = 1.00000 45/30 = 1.50000 70/30 = 2.33333 10/30 = 0.33333 1.04688 1.51563 2.26042 0.40104
- 82. 13 – 82 Solved Problem 4Solved Problem 4 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 following forecasts: 6,948 18,482 32,926 0.21146(32,857) = 0.56250(32,857) = 1.00209(32,857) = 34,397 49,799 74,271 13,177 230,000 1.04688(32,857) = 1.51563(32,857) = 2.26042(32,857) = 0.40104(32,857) = Day Calculations Forecast Sunday Monday Tuesday Wednesday Thursday Friday Saturday Total
- 83. 13 – 83

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