QUESTION 2

1. The information below relates to the Bimbila Supplies Ltd. As the operations manager of the firm you have been tasked to use the data below to forecast demand from February 2014 to March 2015. State the strengths and weaknesses of the methods used? Show how you arrived at the forecasted figures. 15 Marks

Period Month Demand

Exponential Smoothing

α= 0.50

Forecast Ft+1 Trend

Tt+1

Adjusted β = 0.30

Forecast AFt+1

1 January 37 37.002 February 40 37.003 March 41 38.504 April 40 39.755 May 41 39.886 June 37 40.447 July 50 38.728 August 47 44.369 September 56 45.6810 October 52 50.8411 November 55 51.4212 December 54 53.2113 January 50 53.6014 February 49 51.8015 March 50 50.40

ANSWER

α = 0.50

F2 = (0.50 x 37) + (1-0.50)37 F2 = 37.00

F3 = (0.50 x 40) + (1-0.50)37 F3 = 38.50

F4 = (0.50 x 41) + (1-0.50)38.5 F4 = 39.75

F5 = (0.50 x 40) + (1-0.50)39.75 F5 = 39.88

F6 = (0.50 x 41) + (1-0.50)39.88 F6 = 40.44

F7 = (0.50 x 37) + (1-0.50)40.44 F7 = 38.72

F8 = (0.50 x 50) + (1-0.50)38.72 F8 = 44.36

F t+1=α Dt+(1−α)F t

F t+1=t he forecast for t henext period ,D t = actual demand in the present period, F t = the previously determined forecast for the present period; α= smoothing constant

F9 = (0.50 x 47) + (1-0.50)44.36 F9 = 45.68

F10 = (0.50 x 56) + (1-0.50)45.68 F10 = 50.84

F11 = (0.50 x 52) + (1-0.50)50.84 F11 = 51.42

F12 = (0.50 x 55) + (1-0.50)51.42 F12 = 53.21

F13 = (0.50 x 54) + (1-0.50)53.21 F13 = 53.60

F14 = (0.50 x 50) + (1-0.50)53.60 F14 = 51.80

F15 = (0.50 x 49) + (1-0.50)51.31 F15 = 50.40

To calculate the Adjusted Exponential Smoothing;

T t+1=β (Ft+1−F t)+(1−β)T t

T t=t h e last period' s trend factor ; β = a smoothing constant for trend

T2 = 0.3 (37.00 – 37.00) + (0.7)(0 ) T2 = 0T3 = 0.3 (38.5 – 37.00) + (0.7)(0) T3 = 0.45T4 = 0.3 (39.75 – 38.5) + (0.7)(0.45) T4 = 0.69T5 = 0.3 (39.88 – 39.75) + (0.7)(0.69) T5 = 0.52T6 = 0.3 (40.44 – 39.88) + (0.7)(0.52) T6 = 0.53T7 = 0.3 (38.72 – 40.44) + (0.7)(0.53) T7 = -0.14T8 = 0.3 (44.36 – 38.72) + (0.7)(-0.14) T8 = 1.59T9 = 0.3 (45.68 – 44.36) + (0.7)(1.59) T9 = 1.51T10 = 0.3 (50.84 – 45.68) + (0.7)(1.51) T10 = 2.61T11 = 0.3 (51.42– 50.84) + (0.7)(2.61) T11 = 2.00T12 = 0.3 (53.21– 51.42) + (0.7)(2.00) T12 = 1.94T13 = 0.3 (53.60– 53.21) + (0.7)(1.94) T13 = 1.47T14 = 0.3 (51.80 – 53.60) + (0.7)(1.47) T14 = 0.49T15 = 0.3 (50.40 – 51.40) + (0.7)(0.49) T15 = - 0.08

AFt+1=F t+1+T t+1

AF1 = F1 + T1

AF1 = 37.00 + 0.00 AF1 = 37.00AF2 = 37.00 + 0.00 AF2 = 37.00AF3 = 38.50 + 0.45 AF3 = 38.95AF4 = 39.75 + 0.69 AF4 = 40.44AF5 = 39.88 + 0.52 AF5 = 40.40AF6 = 40.44 + 0.53 AF6 = 40.97AF7 = 38.72 + (-0.14) AF7 = 38.58AF8 = 44.36 + 1.59 AF8 = 45.95AF9 = 45.68 + 1.51 AF9 = 47.19AF10 = 50.84 + 2.61 AF10 = 53.45

AF11 = 51.42 + 2.00 AF11 = 53.42AF12 = 53.21 + 1.94 AF12 = 55.15AF13 = 53.60 + 1.47 AF13 = 55.07AF14 = 51.80 + 0.49 AF14 = 52.29AF15 = 50.40 + (-0.08)AF15 = 50.32