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Simple Exponential Smoothing New forecast = Old forecast +     (Actual – Old forecast) F t  = F t-1  +     (A t-1  – F t-1 ) Double Exponential Smoothing = S’ t  + ( S’ t  – S’’ t  ) + b t S’ t   = S’ t-1   +   ’ (A t  – S’ t-1 ) S’’ t   = S’’ t-1  +   ’’ (S’ t  – S’’ t-1 ) b t   = S’’ t  – S’’ t-1 Where F t+1  =  Single smoothed Single smoothed Double smoothed trend adjustment + + -
Linear trend Y t  =    +   t n   ty -   t   y    = n   t 2  - (  t) 2   y -   t n n  t  t 2 1 1 1 2 3 5 3 6 14 4 10 30 5 15 55 6 21 91 7 28 140 8 36 204 9 45 285 10 55 385 11 66 506 12 78 650 13 91 819 14 105 1,015 15 120 1,240 16 136 1,496 17 153 1,785 18 171 2,109 19 190 2,470 20 210 2,870
Parabolic trend Y t  =    +   t +   t 2 with t=0 at the  center  of the data  ty    =  t 2   y -   t 2 n    = n   t 4   -  (  t 2 ) 2 n   t 2 y -   t 2    y
Simple linear regression Y  = a + bx  a   y -   x n b   = n (  x 2 ) -  (  x) 2 n   xy  -  (  x) (  y)
a Σ   n + b Σ x 2  =  Σ xy The transposed formulae : r 2  is the ratio  Explained variation Total variation an + b Σ x  =  Σ y = a  = Σ y – b Σ x n b  =  n Σ xy –  Σ x Σ y n  Σ  x 2  - ( Σ x) 2 Σ  ( YE – Y ) 2 Σ   ( y - Y ) 2 where YE = estimate of Y given the regression equation for each value of x Y  = mean off actual values of y y  = individual actual value of y
r 2  = (n Σ xy -  Σ x Σ y) 2 {n Σ x 2  - ( Σ x) 2 } x {n  Σ   y 2  - (y) 2  } Σ  x 2 n Σ x 2  - ( Σ x) 2 S a  =  S e   = S e S b  = S e Σ x 2  - ( Σ x)  2 n An alternative formula for r 2  is : Standard error of regression = Standard errors of the intercept (a) and the gradient (b) The intercept  The gradient Σ  y 2  - a Σ y - b Σ xy n - 2
a  ±  t x S a b  ±  t x S b The confidence intervals for  α  and  β   may be established as follows : For the intercept  For the gradient For the intercept  For the gradient A test of significance for  α  and  β   The value of t is based upon n-2 degrees of freedom, and the chosen  confidence level α   β = = t  =  a -  α S a t  =  b -  β S b
Multiple Regression Y   = a + b  x 1  + b 2 x 2     x 1 y = a  x 1  + b 1    x 1 2  + b 2  x 1 x 2      x 2 y = a  x 2  + b 1    x 1 x 2  + b 2  x 2 2 a  y + b 1    x 1 y + b 2  x 2 y -     Y 2  -   R 2 =  y = an + b    x 1  + b 2  x 2   (  y) 2 n (  y) 2 n
Example :  586 – 2.338 (105) 14    = 14 (1,015) – 105 (105) 14 (4,927) -  105 (586) =  2.338 =  24.32 Y t  = 24.32 + 2.338 t  1. Year 2. Demand  (units) (1) X (2) 1 32 32 2 28 56 3 30 90 4 34 136 5 30 150 6 43 258 7 36 252 8 42 336 9 42 378 10 55 550 11 47 517 12 56 672 13 54 702 14 57 798  Y = 586  ty = 4,927
Assuming that the number of personnel does not change, a. What percentage of the volunteers will be working for each division in December 2007? b. Predict what percentage will be at the equilibrium state.  An organization dependent upon volunteer help is divided into three divisions and allows free  movement of personnel between any two.  Between January and April 2007, personnel movement was as indicated in the following table. Exercise   Gains Division  Jan. From 1 From 2 From 3 April 1 30 27   3   4 34 2 60   2 57   4 63 3 40   1   0 32 33
A Production Planning Department  is  considering  the production schedules  for  Period 9. In particular  they wish to calculate the time to be allocated for the man-  ufacture of a batch of 100 of a computer controlled machine tool called ROBO XI.  The first ROBO XI took  80  hours to make  and  it is known from past experience  that there is a learning effect. From past records the following information is avail-  able. They calculate that the cumulative production at the beginning of Period 9 will be 3,000 units. Required : a. What type of learning curve model do the records suggest? b. What value of learning curve do the records show? c. Calculate the learning coefficient. d. Calculate the time allowance necessary for the batch of 100 in  Period 9. ROBO XI Cumulative   Cumulative   Time per production (units)   time taken (hours)   unit (hours) 600  18,153.6  30.256 1,200  32,676  27.23
[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],Qualitative & Technological Forecasting
PT. Daya Utama runs a large machine that contains three identical vacuum tubes  which are a major cause of down time. The current practice is to replace these  tubes when they fail. A proposal has been made that all three tubes be replaced  whenever any one of them fails in order to reduce the frequency with which the  equipment must be shut down. The objective is to compare these alternatives on  a cost basis. The vacuum tubes themselves cost Rp.40,000,000.- each and installation costs are Rp.10,000,000.- for replacing one tube, Rp.16,000,000.- for two and Rp.20,000.000,- for three. Down time costs are estimated at Rp.90,000,000.- per hour and the time to replace one, two or three tubes is 30 mins, 50 mins and 60 mins respectively.  SIMULATION Historical data indicates that the time to failure of this type of tube has a distribution as shown below: a. Using the information above and by simulating the operation of this machine over a period of one year, estimate   the costs of maintenance under each of the following proposals: i)  replace each tube as it fails; ii)  replace all three after any one fails. Use the following random digits ,[object Object],[object Object],[object Object],Random digits Tube 1 8 0 0 1 7 2 1 9 9 7 Tube 2 8 8 8 4 8 5 2 3 7 0 Tube 3 5 7 2 0 2 8 6 2 8 5 Time to failure – weeks  7  8  9  10  11  12 Probability of failing 0.1 0.1 0.2 0.3 0.2 0.1

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Forecast2007

  • 1. Simple Exponential Smoothing New forecast = Old forecast +  (Actual – Old forecast) F t = F t-1 +  (A t-1 – F t-1 ) Double Exponential Smoothing = S’ t + ( S’ t – S’’ t ) + b t S’ t = S’ t-1 +  ’ (A t – S’ t-1 ) S’’ t = S’’ t-1 +  ’’ (S’ t – S’’ t-1 ) b t = S’’ t – S’’ t-1 Where F t+1 = Single smoothed Single smoothed Double smoothed trend adjustment + + -
  • 2. Linear trend Y t =  +  t n  ty -  t  y  = n  t 2 - (  t) 2   y -  t n n  t  t 2 1 1 1 2 3 5 3 6 14 4 10 30 5 15 55 6 21 91 7 28 140 8 36 204 9 45 285 10 55 385 11 66 506 12 78 650 13 91 819 14 105 1,015 15 120 1,240 16 136 1,496 17 153 1,785 18 171 2,109 19 190 2,470 20 210 2,870
  • 3. Parabolic trend Y t =  +  t +  t 2 with t=0 at the center of the data  ty  =  t 2   y -  t 2 n  = n  t 4 - (  t 2 ) 2 n  t 2 y -  t 2  y
  • 4. Simple linear regression Y = a + bx a   y -  x n b = n (  x 2 ) - (  x) 2 n  xy - (  x) (  y)
  • 5. a Σ n + b Σ x 2 = Σ xy The transposed formulae : r 2 is the ratio Explained variation Total variation an + b Σ x = Σ y = a = Σ y – b Σ x n b = n Σ xy – Σ x Σ y n Σ x 2 - ( Σ x) 2 Σ ( YE – Y ) 2 Σ ( y - Y ) 2 where YE = estimate of Y given the regression equation for each value of x Y = mean off actual values of y y = individual actual value of y
  • 6. r 2 = (n Σ xy - Σ x Σ y) 2 {n Σ x 2 - ( Σ x) 2 } x {n Σ y 2 - (y) 2 } Σ x 2 n Σ x 2 - ( Σ x) 2 S a = S e = S e S b = S e Σ x 2 - ( Σ x) 2 n An alternative formula for r 2 is : Standard error of regression = Standard errors of the intercept (a) and the gradient (b) The intercept The gradient Σ y 2 - a Σ y - b Σ xy n - 2
  • 7. a ± t x S a b ± t x S b The confidence intervals for α and β may be established as follows : For the intercept For the gradient For the intercept For the gradient A test of significance for α and β The value of t is based upon n-2 degrees of freedom, and the chosen confidence level α β = = t = a - α S a t = b - β S b
  • 8. Multiple Regression Y = a + b  x 1 + b 2 x 2  x 1 y = a  x 1 + b 1  x 1 2 + b 2  x 1 x 2  x 2 y = a  x 2 + b 1  x 1 x 2 + b 2  x 2 2 a  y + b 1  x 1 y + b 2  x 2 y -  Y 2 - R 2 =  y = an + b   x 1 + b 2  x 2 (  y) 2 n (  y) 2 n
  • 9. Example :  586 – 2.338 (105) 14  = 14 (1,015) – 105 (105) 14 (4,927) - 105 (586) = 2.338 = 24.32 Y t = 24.32 + 2.338 t  1. Year 2. Demand (units) (1) X (2) 1 32 32 2 28 56 3 30 90 4 34 136 5 30 150 6 43 258 7 36 252 8 42 336 9 42 378 10 55 550 11 47 517 12 56 672 13 54 702 14 57 798  Y = 586  ty = 4,927
  • 10. Assuming that the number of personnel does not change, a. What percentage of the volunteers will be working for each division in December 2007? b. Predict what percentage will be at the equilibrium state. An organization dependent upon volunteer help is divided into three divisions and allows free movement of personnel between any two. Between January and April 2007, personnel movement was as indicated in the following table. Exercise Gains Division Jan. From 1 From 2 From 3 April 1 30 27 3 4 34 2 60 2 57 4 63 3 40 1 0 32 33
  • 11. A Production Planning Department is considering the production schedules for Period 9. In particular they wish to calculate the time to be allocated for the man- ufacture of a batch of 100 of a computer controlled machine tool called ROBO XI. The first ROBO XI took 80 hours to make and it is known from past experience that there is a learning effect. From past records the following information is avail- able. They calculate that the cumulative production at the beginning of Period 9 will be 3,000 units. Required : a. What type of learning curve model do the records suggest? b. What value of learning curve do the records show? c. Calculate the learning coefficient. d. Calculate the time allowance necessary for the batch of 100 in Period 9. ROBO XI Cumulative Cumulative Time per production (units) time taken (hours) unit (hours) 600 18,153.6 30.256 1,200 32,676 27.23
  • 12.
  • 13.