In: Math
Demand data on "Service Orders" for a particular service enterprise for the previous 12 months is as follows: 550, 652, 673, 707, 725, 752, 780, 797, 815, 836, 850, and 872. Problem 3b) Use the following three methods and prepare three forecasting tables with errors for the given demand data. • a 3-month Weighted Moving Average with the weights 0.6, 0.3, and 0.1 with the maximum weight going for the most recent data point into the past • Exponential Smoothing with smoothing constant = 0.9 • Linear Trend Regression
a)
| Month | Service order | 3 month Weighted average | 
| 1 | 550 | |
| 2 | 652 | |
| 3 | 673 | |
| 4 | 707 | 654.4 | 
| 5 | 725 | 691.3 | 
| 6 | 752 | 714.4 | 
| 7 | 780 | 739.4 | 
| 8 | 797 | 766.1 | 
| 9 | 815 | 787.4 | 
| 10 | 836 | 806.1 | 
| 11 | 850 | 825.8 | 
| 12 | 872 | 842.3 | 
b) Exponential smoothing method
| Month | Service order | Exponential smoothing | 
| 1 | 550 | 550 | 
| 2 | 652 | 550 | 
| 3 | 673 | 641.8 | 
| 4 | 707 | 669.88 | 
| 5 | 725 | 703.288 | 
| 6 | 752 | 722.8288 | 
| 7 | 780 | 749.08288 | 
| 8 | 797 | 776.908288 | 
| 9 | 815 | 794.990829 | 
| 10 | 836 | 812.999083 | 
| 11 | 850 | 833.699908 | 
| 12 | 872 | 848.369991 | 
c) Linear trend
Sum of X = 78
Sum of Y = 9009
Mean X = 6.5
Mean Y = 750.75
Sum of squares (SSX) = 143
Sum of products (SP) = 2997.5
Regression Equation = ŷ = bX + a
b = SP/SSX = 2997.5/143 =
20.96154
a = MY - bMX = 750.75 -
(20.96*6.5) = 614.5
ŷ = 20.96154X + 614.5
| Month | Service order | Linear trend | 
| 1 | 550 | 635.46154 | 
| 2 | 652 | 656.42308 | 
| 3 | 673 | 677.38462 | 
| 4 | 707 | 698.34616 | 
| 5 | 725 | 719.3077 | 
| 6 | 752 | 740.26924 | 
| 7 | 780 | 761.23078 | 
| 8 | 797 | 782.19232 | 
| 9 | 815 | 803.15386 | 
| 10 | 836 | 824.1154 | 
| 11 | 850 | 845.07694 | 
| 12 | 872 | 866.03848 |