13 Forecasting PowerPoint Slides by Jeff Heyl For Operations Management, 9e by Krajewski/Ritzman/Malhotra 2010 Pearson Education 13 1 Forecasting 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 13 2 1
Demand 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 13 3 Demand Patterns Figure 13.1 Patterns of Demand Quantity Time (a) Horizontal: Data cluster about a horizontal line 13 4 2
Demand Patterns Figure 13.1 Patterns of Demand Quantity Time (b) Trend: Data consistently increase or decrease 13 5 Demand Patterns Figure 13.1 Patterns of Demand Quantity 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 13 6 3
Demand Patterns Figure 13.1 Patterns of Demand Quantity 1 2 3 4 5 6 Years (d) Cyclical: Data reveal gradual increases and decreases over extended periods 13 7 Key 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 13 8 4
Judgment 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 13 9 Judgment Methods Market research is a systematic approach to determine external customer interest through data-gathering th 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 g forecasts and technological forecasting 13 10 5
Linear Regression A dependent variable is related to one or more independent variables by a linear equation The independent d 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 13 11 Linear Regression Dependent variable Y Estimate of Y from regression equation Deviation, or error Actual value of Y Value of X used to estimate Y Regression equation: Y = a + bx Independent variable X Figure 13.2 Linear Regression Line Relative to Actual Data 13 12 6
Linear 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, r 2 Measures the amount of variation in the dependent variable about its mean that is explained by the regression line The values of r 2 range from 0.00 r 2 1.00 The standard error of the estimate, s yx Measures how closely the data on the dependent variable cluster around the regression line 13 13 Using 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 10 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. 13 14 7
Using 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 = 8.135 b = 109.229X r = 0.980 r 2 = 0.960 s yx = 15.603 The regression equation is Y = 8.135 + 109.229X 13 15 Using 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, r 2, implies that 96 percent of the variation in sales is explained by advertising expenditures. Brass Door Hinge 250 X (000 units) Sales 200 150 100 50 X X X X X Data Forecasts 0 1.0 2.0 Advertising ($000) Figure 13.3 Linear Regression Line for the Sales and Advertising Data 13 16 8
Time Series Methods In a naive forecast the forecast for the next period equals the demand for the current period (Forecast = D t ) 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 13 17 Time Series Methods 450 Patient arrivals 430 410 390 370 350 0 5 10 15 20 25 30 Week Figure 13.4 Weekly Patient Arrivals at a Medical Clinic 13 18 9
Simple 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 Sum of last n demands F t+1 = = n D t + D t-1 + D t-2 + + D t-n+1 n where D t = actual demand in period t n = total number of periods in the average F t+1 = forecast for period t + 1 13 19 Simple 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 E t = D t F t where E t = forecast error for period t D t = actual demand in period t F t = forecast for period t 13 20 10
Using 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? 13 21 Using the Moving Average Method SOLUTION a. The moving average forecast at the end of week 3 is Week Patient Arrivals 1 400 2 380 3 411 411 + 380 + 400 F 4 = = 397.0 3 b. The forecast error for week 4 is E 4 = D 4 F 4 = 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 F 5 = = 402.0 415 + 411 + 380 3 13 22 11
Application 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 D 4 + D 3 + D 2 790 + 810 + 740 F 5 = = = 780 3 3 Forecast for month 5 is 780 customer arrivals 13 23 Application 13.1a If the actual number of arrivals in month 5 is 805, what is the forecast for month 6? Month Customer arrival 1 800 2 740 3 810 4 790 D 5 + D 4 + D 3 805 + 790 + 810 F 6 = = = 801.667 3 3 Forecast for month 6 is 802 customer arrivals 13 24 12
Application 13.1a Forecast error is simply the difference found by subtracting the forecast from actual demand for a given period, or E t = D t F t Given the three-month moving average forecast for month 5, and the number of patients that actually arrived (805), what is the forecast error? E 5 = 805 780 = 25 Forecast error for month 5 is 25 13 25 Weighted Moving Averages 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: F t+1 = W 1 D 1 + W 2 D 2 + + W n D t-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 F t+1 = 0.50D t + 0.30D t 1 + 0.20D t 2 13 26 13
Application 13.1b Revisiting the customer arrival data in Application 13.1a. Let W 1 = 0.50, W 2 = 0.30, and W 3 = 0.20. Use the weighted moving average method to forecast arrivals for month 5. F 5 = W 1 D 4 + W 2 D 3 + W 3 D 2 = 0.50(790) + 0.30(810) + 0.20(740) = 786 Forecast for month 5 is 786 customer arrivals Given the number of patients that actually arrived (805), what is the forecast error? E 5 = 805 786 = 19 Forecast error for month 5 is 19 13 27 Application 13.1b If the actual number of arrivals in month 5 is 805, compute the forecast for month 6 F 6 = W 1 D 5 + W 2 D 4 + W 3 D 3 = 0.50(805) + 0.30(790) + 0.20(810) = 801.5 Forecast for month 6 is 802 customer arrivals 13 28 14
Exponential 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 F t+1 = α(demand this period) + (1 α)(forecast calculated last period) = αd t + (1 α)f t or the equivalent F t+1 = F t + α(d t F t ) 13 29 Exponential 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 13 30 15
Exponential Smoothing and Moving Average 450 3-week MA 6-week MA forecast forecast 430 Pa atient arrivals 410 390 370 Exponential smoothing α = 0.10 0 5 10 15 20 25 30 Week 13 31 Using 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. 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? Week Patient Arrivals 1 400 2 380 3 411 4 415 13 32 16
Using 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 F 4 = 0.10(411) + 0.90(390) = 392.1 Thus, the forecast for week 4 would be 392 patients. 13 33 Using Exponential Smoothing b. The forecast error for week 4 is E 4 = 415 392 = 23 c. The new forecast for week 5 would be F 5 = 0.10(415) + 0.90(392.1) = 394.4 or 394 patients. Note that we used F 4, not the integer-value forecast for week 4, in the computation for F 5. In general, we round off (when it is appropriate) only the final result to maintain as much accuracy as possible in the calculations. 13 34 17
Application 13.1c Suppose the value of the customer arrival series average in month 3 was 783 customers (let it be F 4 ). Use exponential smoothing with α = 0.20 to compute the forecast for month 5. F t+1 = F t + α(d t F t ) = 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? E 5 = 805 784 = 21 Forecast error for month 5 is 21 13 35 Application 13.1c Given the actual number of arrivals in month 5, what is the forecast for month 6? F t+1 = F t + α(d t F t ) = 784.4 + 0.20(805 784.4) = 788.52 Forecast for month 6 is 789 customer arrivals 13 36 18
Including 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 13 37 Including a Trend For each period, we calculate the average and the trend: A t = α(demand this period) + (1 α)(average + Trend estimate last period) = αd t + (1 α)(a t 1 + T t 1 ) T t = β(average this period Average last period) + (1 β)(trend estimate last period) = β(a t A t 1 ) + (1 β)t t 1 F t+1 = A t + T t where A t = exponentially smoothed average of the series in period t T t = 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 F t+1 = forecast for period t + 1 13 38 19
Using Trend-Adjusted Exponential Smoothing EXAMPLE 13.4 Medanalysis, Inc., provides medical laboratory services Managers are interested t 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 d was for 27 blood tests t We use α = 0.20 and β = 0.20 to calculate the forecast for next week 13 39 Using Trend-Adjusted Exponential Smoothing SOLUTION A 0 = 28 patients and T 0 = 3 patients The forecast for week 2 (next week) is A 1 = 0.20(27) + 0.80(28 + 3) = 30.2 T 1 = 0.20(30.2 28) + 0.80(3) = 2.8 F 2 = 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 A 2 = 0.20(44) + 0.80(30.2 + 2.8) = 35.2 T 2 = 0.2(35.2 30.2) + 0.80(2.8) = 3.2 F 3 = 35.2 + 3.2 = 38.4 or 38 blood tests 13 40 20
Using Trend-Adjusted Exponential Smoothing TABLE 13.1 FORECASTS FOR MEDANALYSIS USING THE TREND-ADJUSTED EXPONENTIAL SMOOTHING MODEL Calculations to Forecast Arrivals for Next Week Week Arrivals Smoothed Trend Average 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 13 41 Using Trend-Adjusted Exponential Smoothing TABLE 13.1 FORECASTS FOR MEDANALYSIS USING THE TREND-ADJUSTED EXPONENTIAL SMOOTHING MODEL Calculations to Forecast Arrivals for Next Week Week Arrivals Smoothed Trend Average Average Forecast for This Week Forecast Error 0 28 28.00 3.00 1 27 30.20 2.84 28.00 + 3.00 = 31.00 4 2 44 35.23 3.28 30.20 + 2.84 = 33.04 10.96 3 37 38.21 3.22 35.23 + 3.28 = 38.51 1.51 4 35 40.14 2.96 38.21 + 3.22 = 41.43 6.43 5 53 45.08 3.36 40.14 + 2.96 = 43.10 9.90 6 38 46.35 2.94 45.08 + 3.36 = 48.44 10.44 7 57 50.83 3.25 46.35 + 2.94 = 49.29 7.71 8 61 55.46 3.52 35 50.83 + 3.25 = 54.08 6.92 9 39 54.99 2.72 55.46 + 3.52 = 58.98 19.98 10 55 57.17 2.62 54.99 + 2.72 = 57.71 2.71 11 54 58.63 2.38 57.17 + 2.62 = 59.79 5.79 12 52 59.21 2.02 58.63 + 2.38 = 61.01 9.01 13 60 60.99 1.97 59.21 + 2.02 = 61.23 1.23 14 60 62.37 1.86 60.99 + 1.97 = 62.96 2.96 15 75 66.38 2.29 62.37 + 1.86 = 64.23 10.77 13 42 21
Using Trend-Adjusted Exponential Smoothing 80 70 Trend-adjusted forecast Patient arrivals 60 50 40 Actual blood test requests 30 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 Week Figure 13.5 Trend-Adjusted Forecast for Medanalysis 13 43 Application 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 330,000 000 cases. What is the forecast for May, assuming α = 0.20 and β = 0.10? A Apr = αd t + (1 α)(a Mar + T Mar ) = 0.20(330,000) + 0.80(300,000 + 8,000) = 312,400 cases T Apr = β(a Apr A Mar ) + (1 β)t Mar = 0.10(312,400 300,000) + 0.90(8,000) = 8,440 cases Forecast for May = A Apr + pt Apr = 312,400 + (1)(8,440) = 320,840 cases 13 44 22
Application 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 t we want in the forecast and add the results to the current average. Forecast for July = A Apr + pt Apr = 312,400 + (3)(8,440) = 337,720 cases 13 45 Seasonal 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 13 46 23
Multiplicative Seasonal Method Multiplicative seasonal method, whereby seasonal factors are multiplied by an estimate of the average demand to arrive at a seasonal forecast 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 13 47 Using the Multiplicative Seasonal Method 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 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 13 48 24
Using the Multiplicative Seasonal Method 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). 13 49 Using the Multiplicative Seasonal Method Figure 13.6 Demand Forecasts Using the Seasonal Forecast Solver of OM Explorer 13 50 25
Application 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 Quarter Demand Index Demand Index 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 13 51 Application 13.3 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 Quarter Average Index 1 0.52 2 148 1.48 3 1.24 4 0.76 Forecast for Quarter 1 = 0.52(330) 172 units Forecast for Quarter 2 = 1.48(330) 488 units Forecast for Quarter 3 = 1.24(330) 409 units Forecast for Quarter 4 = 0.76(330) 251 units 13 52 26
Seasonal Patterns (a) Multiplicative pattern Demand 0 2 4 5 8 10 12 14 16 Period 13 53 Seasonal Patterns (b) Additive pattern Demand 0 2 4 5 8 10 12 14 16 Period 13 54 27
Choosing a Time-Series Method Forecast performance is determined by forecast errors Forecast errors detect t 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 13 55 Measures of Forecast Error Σ(E t E ) 2 CFE = ΣE t σ = n 1 E = CFE n MAD = Σ E t n MSE = ΣE t 2 n MAPE = (Σ E t /D t )(100) n 13 56 28
Calculating 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 D t Forecast F t Error E t Error 2 E 2 t Absolute Error E t 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% Absolute % Error ( E t /D t )(100) 13 57 Calculating 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 D t Forecast F t Error E t Error 2 E 2 t Absolute Error E t 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% Absolute % Error ( E t /D t )(100) 13 58 29
Calculating Forecast Errors SOLUTION Using the formulas for the measures, we get Cumulative forecast error (bias): CFE = 15 Average forecast error (mean bias): CFE E = n = 1.875 Mean squared error: MSE = ΣE t 2 n = 5,275 8 13 59 Calculating Forecast Errors Standard deviation: σ = Σ[E 1 t ( 1.875)] 2 n 1 = 27.4 Mean absolute deviation: MAD = Σ E t n 195 = = 24.4 8 Mean absolute percent error: (Σ E t / D t )(100) 81.3% MAPE = n = = 10.2% 8 13 60 30
Calculating 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 es of forecast error variability. ab 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. 13 61 Tracking 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 13 62 31
Tracking Signals The MAD can be calculated as a weighted average determined by the exponential smoothing method MAD t = α E t + (1 α)mad t-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 13 63 Tracking Signals Tracking signal +2.0 +1.5 +1.0 +0.5 0 0.5 1.0 1.5 Control limit Control limit Out of control 0 5 10 15 20 25 Observation number Figure 13.7 Tracking Signal 13 64 32
Criteria 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 13 65 Using Multiple Techniques Combination forecasts are forecasts that are produced by averaging g 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 13 66 33
Forecasting 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 13 67 Forecasting as a Process Adjust history file 1 Prepare initial forecasts 2 Consensus meetings and collaboration 3 Finalize and communicate 6 Review by Operating Committee 5 Revise forecasts 4 13 68 34
Forecasting 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 13 69 Solved 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 l 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 13 70 35
Solved Problem 1 SOLUTION We use the computer to calculate the best values of a, b, the correlation coefficient, and the coefficient of determination The regression line is a = 1,454.60 b = 277.63 r = 0.84 r 2 = 0.71 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 13 71 Solved 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 d for these special pizzas so that t 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. 13 72 36
Solved Problem 2 SOLUTION a. The simple moving average method and the weighted moving average method give the following results: Current Week June 16 June 23 June 30 July 7 Simple Moving Average Forecast for Next Week 52 + 65 + 50 3 56 + 52 + 65 3 55 + 56 + 52 3 60 + 55 + 56 3 = 55.7 or 56 = 57.7 or 58 = 54.3 or 54 = 57.0 or 57 Weighted Moving Average Forecast for Next Week [(0.5 52) + (0.3 65) + (0.2 50)] = 55.5 or 56 [(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 [(0.5 60) + (0.3 55) + (0.2 56)] = 57.7 or 58 13 73 Solved 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 E t Forecast for This Week Absolute Errors E t June 23 56 56 56 56 = 0 56 56 56 = 0 June 30 55 58 55 58 = 3 57 55 57 = 2 July 7 60 54 60 54 = 6 55 60 55 = 5 0 + 3 + 6 MAD = = 3 3 0 + 2 + 2 MAD = = 2.3 3 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. 13 74 37
Solved 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? 13 75 Solved Problem 3 SOLUTION a. Current Month, t May June July August September October November December Calculating Forecast for Next Month F t+1 = αd t + (1 α)f t Forecast for Month t + 1 0.2(100) + 0.8(105) = 104.0 or 104 June 0.2(80) + 0.8(104.0) = 99.2 or 99 July 0.2(110) + 0.8(99.2) = 101.4 or 101 August 0.2(115) + 0.8(101.4) = 104.1 or 104 September 0.2(105) + 0.8(104.1) = 104.3 or 104 October 0.2(110) + 0.8(104.3) = 105.4 or 105 November 0.2(125) + 0.8(105.4) = 109.3 or 109 December 0.2(120) + 0.8(109.3) = 111.4 or 111 January 13 76 38
Solved Problem 3 b. Actual Demand, Forecast, Month, t Dt Ft June 80 104 July 110 99 August 115 101 September 105 104 October 110 104 November 125 105 December 120 109 Total 765 Error, Et = Dt Ft Absolute Error, Et Absolute Percent Error, ( Et /Dt)(100) 24 24 30.0% 11 11 10.0 14 14 12.0 1 1 1.0 6 6 5.5 20 20 16.0 11 11 9.2 39 87 83.7% MAD = Σ E t n 87 = = 12.4 7 (Σ E t /D t )(100) 83.7% MAPE = n = = 11.96% 7 13 77 Solved 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: Tracking signal = CFE MAD = 39 12.4 = 3.14 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. 13 78 39
Solved 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. 13 79 Solved Problem 4 SOLUTION a. Calculate the average daily mail volume for each week. Then for each day of the week kdivide id 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. 13 80 40
Solved Problem 4 Week 1 Week 2 Mail Seasonal Factor Mail Seasonal Factor Day Volume (1) Volume (2) Sunday 5 5/32 = 0.15625 8 8/30 = 0.26667 Monday 20 20/32 = 0.62500 15 15/30 = 0.50000 Tuesday 30 30/32 = 0.93750 32 32/30 = 1.06667 Wednesday 35 35/32 = 1.09375 30 30/30 = 1.00000 Thursday 49 49/32 = 1.53125 45 45/30 = 1.50000 Friday 70 70/32 = 2.18750 70 70/30 = 2.33333 Saturday 15 15/32 = 0.46875 10 10/30 = 0.33333 Total 224 210 Average 224/7 = 32 210/7 = 30 Average Seasonal Factor [(1) + (2)]/2 0.21146 0.56250 1.00209 1.04688 1.51563 2.26042 0.40104 13 81 Solved 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: Day Calculations Forecast Sunday Monday Tuesday 0.21146(32,857) = 0.56250(32,857) = 1.00209(32,857) = 6,948 18,482 32,926 Wednesday 1.04688(32,857) = 34,397 Thursday 1.51563(32,857) = 49,799 Friday 2.26042(32,857) = 74,271 Saturday 0.40104(32,857) = 13,177 Total 230,000 13 82 41
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