Question

The manager of the Petroco Service Station wants to forecast the demand for unleaded gasoline next month so that the proper number of gallons can be ordered from the distributor. The owner has accumulated the following data on demand for unleaded gasoline from sales during the past 10 months: MAT540 Homework Week 4 Page 2 of 5 Month Gasoline Demanded (gal.) October 775 November 835 December 605 January 450 February 600 March 700 April 820 May 925 June July 1500 1200 a. Compute an exponential smoothed forecast, using an α value of 0.4 b. Compute the MAD.

Answer 1

The exponential smoothed forecast is obtained using the following equation

F = F:-1 + a(At-1-F:-1)

Forecast for month of October = 775

Forecast for November = 775 + 0.4 ( 775 - 775) = 775

Forecast for December = 775 + 0.4 ( 835 - 775 ) = 799

Forecast for January = 799 + 0.4 ( 605 - 799) = 721.4

Forecast for February = 721.4 + 0.4 ( 450 - 721.4) = 612.84

Forecast for March = 612.84 + 0.4 ( 600 - 612.84 ) = 607.70

Forecast for April = 607.704 + 0.4 ( 700 - 607.704 ) = 644.62

Forecast for May = 644.62 + 0.4 ( 820 - 644.62 ) = 714.77

Forecast for June = 714.77 + 0.4 ( 925 - 714.77 ) = 798.86

Forecast for July =  798.86 + 0.4 ( 1500 - 798.86 ) = 1079.32

Month	Gasoline demanded	Forecasted demand
October	775	775
November	835	775
December	605	799
January	450	721.4
February	600	612.84
March	700	607.7
April	820	644.62
May	925	714.77
June	1500	798.86
July	1200	1079.32

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To calculate the mean absolute deviation ( MAD)

60 271.4 Gasoline demanded Forecasted demand Deviation = Actual - forecast Absolute deviatic 775 775 835 775 60 605 799 -194 194 450 721.4 -271.4 600 612.84 -12.84 12.84 700 607.7 92.3 92.3 820 644.62 175.38 175.38 925 714.77 210.23 210.23 1500 798.86 701.14 701.14 1200 1079.32 120.68 120.68 Sum of absolute deviation 1837.97 Mean absolute deviation (MAD) -Sum of the absolute deviations

Mean absolute deviation = 1837.97 $\div$ 10

Mean absolute deviation = 183.80