HomeWork-TimeSeriesMining-Sol

Homework: Time Series Mining Question 1. A dry cleaner uses exponential smoothing to forecast equipment usage at its main plant. August usage was forecasted to be 88 percent of capacity; actual usage was 89.6 percent of capacity. A smoothing constant of .1 is used. a. Prepare a forecast for September. Ans: Exponential smoothing forecast for September with alpha = 0.10: 88 + 0.10(89.6 – 88) = 88.16 (round to two decimals) b. Assuming actual September usage of 92 percent, prepare a forecast for October usage. Ans: Exponential smoothing forecast for October with alpha = 0.10: 88.16 + 0.10(92 – 88.16) = 88.54 (round to two decimals) Question 2: An electrical contractor’s records during the last five weeks indicate the number of job requests: Predict the number of requests for week 6 using each of these methods: a. Naive. b. A four-period moving average. c. Exponential smoothing with .30. Use 20 for week 2 forecast. Ans: Given: Week Requests 1 20 2 22 3 18 4 21 5 22 a. Naïve approach forecast for Week 6 = Demand in Week 5 = 22 b. Four-period moving average forecast for Week 6: 22 + 18 + 21 + 22 4 = 20.75 (round to two decimals)

Homework: Time Series Mining c. Exponential smoothing with alpha = 0.30 and a Week 2 Forecast = 20 (round to two decimals): F 3 = 20 + 0.30(22 – 20) = 20.60 F 4 = 20.60 + 0.30(18 – 20.6) = 19.82 F 5 = 19.82 + 0.30(21 – 19.82) = 20.17 F 6 = 20.17 + 0.30(22 – 20.17) = 20.72 Question 3: From the following graph, determine the equation of the linear trend line for time- share sales for Glib Marketing, Inc. Ans: Slope of the line is estimated by Rise/Run = (300-500)/(10-0) = -200/10 = -20.00. The Y Intercept = 500.

Homework: Time Series Mining b = n ∑ tY − ∑ t ∑ Y n ∑ t 2 −( ∑ t ) 2 = 18 ( 75 , 713 )− 171 ( 7 , 001 ) 18 ( 2 , 109 )−( 171 ) 2 = 19.00 a = ∑ Y − b ∑ t n = 7 , 001 − 19.00 ( 171 ) 18 = 208.44 b. Linear Trend Forecast for Week 20: F = 208.44 + (19.00)(20) = 588.44 Linear Trend Forecast for Week 21: F = 208.44 + (19.00)(21) = 607.44 The forecasted demand for week 20 and 21 is 588.44 and 607.44 respectively. c. Set the trend equation = 800 and solve for t : 208.44 + 19.00 t = 800 19.00 t = 800 – 208.44 19.00 t = 591.96 t = 591.96 / 19.00 t = 31.13 weeks (during Week 32) Question 5: Solution: Y t = 70 + 5 t t = 0 (June of last year) t = 1 (July of last year) t = 7 (January of this year) t = 8 (February of this year) t = 9 (March of this year) t = 19 (January of next year) t = 20 (February of next year) t = 21 (March of next year) Y Jan. = 70 + (5)(19) = 165 Y Feb. = 70 + (5)(20) = 170 Y Mar. = 70+ (5)(21) = 175 Forecast = Trend * Seasonal Relative (round to two decimals): Month Trend * Seasonal Relative

Homework: Time Series Mining January 165 * 1.10 = 181.50 February 170 * 1.02 = 173.40 March 175 * 0.95 = 166.25 Question 6: c. Compute MAPE for each data set. Which forecast appears to be more accurate? Solution: Period Demand F1 e  e  e 2 (  e  /Demand) x 100 (%) F2 e  e  e 2 (  e  /Demand) x 100 (%) 1 68 66 2 2 4 2.94% 66 2 2 4 2.94% 2 75 68 7 7 49 9.33% 68 7 7 49 9.33% 3 70 72 –2 2 4 2.86% 70 0 0 0 0.00% 4 74 71 3 3 9 4.05% 72 2 2 4 2.70% 5 69 72 –3 3 9 4.35% 74 –5 5 25 7.25% 6 72 70 +2 2 4 2.78% 76 –4 4 16 5.56% 7 80 71 9 9 81 11.25% 78 2 2 4 2.50% 8 78 74 4 4 16 5.13% 80 –2 2 4 2.56% Sum 32 176 42.69% 24 106 32.84% a. MAD F1: 32/8 = 4.00 (round to two decimals) MAD F2: 24/8 = 3.00 (F2 appears to be more accurate) b. MSE F1: 176/(8-1) = 25.14 MSE F2: 106/(8-1) = 15.14 (F2 appears to be more accurate)