Forecasting Time Horizons

Forecasting Time Horizons  Short-range forecast  Up to 1 year, generally less than 3 months  Purchasing, job scheduling, workforce levels, job assignments, production levels  Medium-range forecast  3 months to 3 years  Sales and production planning, budgeting  Long-range forecast  3+ years  New product planning, facility location, research and development 4–1 Time Series Components Trend Cyclical Seasonal Random 4–2 Demand for product or service Components of Demand Trend component Seasonal peaks Actual demand Average demand over four years Random variation | 1 | 2 Year | 3 | 4 Figure 4.1 4–3 Shed Sales Graph of Moving Average 30 28 26 24 22 20 18 16 14 12 10 – – – – – – – – – – – Moving Average Forecast Actual Sales | J | F | M | A | M | J | J | A | S | O | N | D 4–4 Impact of Different α Demand 225 – Actual demand 200 – α = .5 175 – 150 – | 1 | 2 | 3 | 4 | 5 | 6 α = .1 | 7 | 8 | 9 Quarter 4–5 Values of Dependent Variable Least Squares Method Actual observation (y value) Deviation7 Deviation5 Deviation6 Deviation3 Deviation4 Deviation1 Deviation2 Trend line, y^ = a + bx Time period Figure 4.4 4–6 Values of Dependent Variable Least Squares Method Actual observation (y value) Deviation7 Deviation5 Deviation3 Deviation6 Least squares method minimizes the sum of the Deviation squared errors (deviations) 4 Deviation1 Deviation2 Trend line, y^ = a + bx Time period Figure 4.4 4–7 Least Squares Example Year 1999 2000 2001 2002 2003 2004 2005 Time Period (x) 1 2 3 4 5 6 7 ∑x = 28 x=4 Electrical Power Demand 74 79 80 90 105 142 122 ∑y = 692 y = 98.86 x2 xy 1 4 9 16 25 36 49 ∑x2 = 140 74 158 240 360 525 852 854 ∑xy = 3,063 3,063 - (7)(4)(98.86) ∑xy - nxy b= = = 10.54 2) 2 2 140 (7)(4 ∑x - nx a = y - bx = 98.86 - 10.54(4) = 56.70 4–8 Least Squares Example Year Time Period (x) Electrical Power Demand x2 xy 1999 1 74 1 2000 2 79 4 line is 80 2001The trend 3 9 2002 4 90 16 2003 105 25 y^ 5= 56.70 + 10.54x 2004 6 142 36 2005 7 122 49 Σx = 28 Σy = 692 Σx2 = 140 x=4 y = 98.86 74 158 240 360 525 852 854 Σxy = 3,063 3,063 - (7)(4)(98.86) Σxy - nxy b= = = 10.54 140 - (7)(42) Σx2 - nx2 a = y - bx = 98.86 - 10.54(4) = 56.70 4–9 Least Squares Example Power demand 160 150 140 130 120 110 100 90 80 70 60 50 – – – – – – – – – – – – Trend line, y^ = 56.70 + 10.54x | 1999 | 2000 | 2001 | 2002 | 2003 Year | 2004 | 2005 | 2006 | 2007 4 – 10 Associative Forecasting Forecasting an outcome based on predictor variables using the least squares technique y^ = a + bx ^ = computed value of the variable to where y be predicted (dependent variable) a = y-axis intercept b = slope of the regression line x = the independent variable though to predict the value of the dependent variable 4 – 11 Associative Forecasting Example Local Payroll ($000,000,000), x 1 3 4 4.0 – 2 1 3.0 – 7 Sales Sales ($000,000), y 2.0 3.0 2.5 2.0 2.0 3.5 2.0 – 1.0 – 0 | 1 | 2 | | | | 3 4 5 6 Area payroll | 7 4 – 12 Associative Forecasting Example Sales, y 2.0 3.0 2.5 2.0 2.0 3.5 ∑y = 15.0 Payroll, x 1 3 4 2 1 7 ∑x = 18 x = ∑x/6 = 18/6 = 3 y = ∑y/6 = 15/6 = 2.5 x2 1 9 16 4 1 49 ∑x2 = 80 xy 2.0 9.0 10.0 4.0 2.0 24.5 ∑xy = 51.5 51.5 - (6)(3)(2.5) ∑xy - nxy b= = 80 - (6)(32) = .25 ∑x2 - nx2 a = y - bx = 2.5 - (.25)(3) = 1.75 4 – 13 Associative Forecasting Example If payroll next year is estimated to be $600 million, then: Sales = 1.75 + .25(6) Sales = $325,000 Sales = 1.75 + .25(payroll) 4.0 – 3.25 3.0 – Sales y^ = 1.75 + .25x 2.0 – 1.0 – 0 | 1 | 2 | | | | 3 4 5 6 Area payroll | 7 4 – 14 Standard Error of the Estimate  A forecast is just a point estimate of a future value 4.0 – 3.25 3.0 – Sales  This point is actually the mean of a probability distribution 2.0 – 1.0 – 0 Figure 4.9 | 1 | 2 | | | | 3 4 5 6 Area payroll | 7 4 – 15 Standard Error of the Estimate Sy,x = ∑(y - yc)2 n-2 where y = y-value of each data point yc = computed value of the dependent variable, from the regression equation n = number of data points 4 – 16 Standard Error of the Estimate Computationally, this equation is considerably easier to use Sy,x = ∑y2 - a∑y - b∑xy n-2 We use the standard error to set up prediction intervals around the point estimate 4 – 17 Standard Error of the Estimate Sy,x = .306 The standard error of the estimate is $30,600 in sales 4.0 – 3.25 3.0 – Sales Sy,x = 39.5 - 1.75(15) - .25(51.5) 6-2 ∑y2 - a∑y - b∑xy = n-2 2.0 – 1.0 – 0 | 1 | 2 | | | | 3 4 5 6 Area payroll | 7 4 – 18 Correlation  How strong is the linear relationship between the variables?  Correlation does not necessarily imply causality!  Coefficient of correlation, r, measures degree of association  Values range from -1 to +1 4 – 19 Correlation Coefficient r= nΣxy - ΣxΣy [nΣx2 - (Σx)2][nΣy2 - (Σy)2] 4 – 20 Correlation Coefficient y y n∑xy - ∑x∑y r= 2 - (∑x)2][n∑y2 - (∑y)2] [n∑x (a) Perfect positive x (b) Positive correlation: 0