ME 403 Production Planning and Control

Tutorial Sheet 2: Forecasting (contd.)
By: Gunaranjan Chaudhry

 

Answer 1

 

Exponential Smoothing:             S_t = a D_t + (1-a) S_t-1

                                                            F_t,t+1 = S_t

F_0,1 ( S₀) is given as 25

 

(a) The value of a is 0.15

 

S₁=0.15D₁ + 0.85S₀

   = 0.15 X 23.3 + 0.85 X 25

   = 24.745

F_1,2 = 24.745 (forecast for February made in January)

 

F_2,3 = S₂ = 0.15 D₂ + 0.85 S_­1

              = 0.15 X 72.3 + 0.85 X 24.745

   = 31.878

 

F_3,4 = S₃ = 0.15 D₃ + 0.85 S_­2

              = 0.15 X 30.3 + 0.85 X 31.878

   = 31.641

 

F_4,5 = S₄ = 0.15 D₄ + 0.85 S_­3

              = 0.15 X 15.5 + 0.85 X 31.641

   = 29.220

 

(b) The value of a is 0.40. The calculations are similar.

 

F_­1,2 = 24.320

F_2,3 = 43.512

F_3,4 = 38.227

F_4,5 = 29.136

 

            The 1^st set of forecasts is more stable (lower value of a)

            The 2^nd set of forecasts has greater fluctuations (more responsive to demand)

 

(c) Calculation of MSEs and MADs

 

The following values are obtained (on the basis of 3 months, Feb to Apr)

	a = 0.15	a = 0.4
MAD	21.76	27.97
MSE	841.65	997.74

On the basis of these, a = 0.15 gives a more accurate forecast

Answer 2

 

(a) This method makes exponential smoothing more responsive to change because it allows changes in the value of a. When there are changes in the pattern of data that make the initial value of a no longer appropriate, the value of a changes so that it is better suited to the data.

 

(b) Applying this technique to the data of problem 1….

 

            Relevant equations:            E_t = b e_t + (1-b) E_t-1

                                                M_t = b |e_t| + (1-b) M_t-1

                                                a_t = | E_t / M_t |

                                               

                                                S_t = a_t D_t + (1-a_t) S­_t-1

                                                F_t+1 = S_t

 

            Starting values, E₁ = e₁, M₁ = |e₁|, where e₁ is the error in January

            e₁ = D₁ – F₁ = 23.3 – 25 = -1.7

            b=0.1

           

            Forecast for February

E₁ = e₁ = -1.7

            M₁ = |e₁| = 1.7

            a₁ = | E₁ / M₁ | = 1

            S₁ = 1 X D₁ + 0 X S₀ = 23.3

            F_1,2 = S₁ = 23.3

 

            For 2^nd period, error = e₂ = D₂ – F₂ = 72.3 – 23.3 = 49

           

Forecast for March

E₂ = 0.1 X e₂ + 0.9 X E₁= 0.1 X 49 + 0.9 X –1.7 = 3.37

            M₁ = 0.1 X e₂ + 0.9 X E₁= 0.1 X 49 + 0.9 X 1.7 = 6.43

            a₁ = | E₁ / M₁ | = 0.524

            S₁ = 0.524 X D₂ + 0.476 X S₁ = 48.98

            F_2,3 = S₂ = 48.98

 

            Similarly, the forecast for April can be determined. The relevant values are:

            E₃ = 1.165

            M₃ = 7.655

            a₃ = 0.152

            S₃ = 46.14

            F_3,4 = 46.14

 

The MAD for this method can now be calculated. It is equal to 32.77, which is greater than the MAD obtained in Problem 1.

 

Answer 3

 

Both R and S use the same method (exponential smoothing) with the same parameter. S updates his forecasts weekly.

R updates his forecasts monthly.

 

(a) Given the recent drop, S is more likely to have an accurate forecast. Both R and S give less weightage to the demand than the previous best guess of the demand (a=0.2). However, S updates his forecasts more regularly and it is quite likely that his forecasts would have had a greater opportunity to adapt to the recent change in demand.

 

(b) If S used an a=0.1, his model gives less weightage to the demand than R’s. However, it is updated more regularly. It is hard to say whether the greater frequency of updation will offset the effect of a lower a. If it does, S will have a more accurate forecast; if not, R’s will be more accurate.