1. IE5403 Facilities Design and Planning Instructor: Assistant Prof. Dr. Rıfat Gürcan Özdemir http://web.iku.edu.tr/~rgozdemir/IE551/index(IE551).htm

2. Course topics Chapter 1: Forecasting methods Chapter 2: Capacity planning Chapter 3: Facility location Chapter 4: Plant layout Chapter 5: Material handling and storage systems

3. Grading Participation5% Quizzes15% (4 quizzes) Assignment15%(every week) Midterm 130%(chapters 1 and 2) Final35%(all chapters) 3

4. 4 IE5403 - Chapter 1 Forecasting methods

5. 5 Forecasting Forecasting is the process of analyzing the past data of a time – dependent variable & predicting its future values by the help of a qualitative or quantitative method

6. 6 Why is forecasting important? Proper forecasting Better use of capacity Reduced inventory costs Lower overall personel costs Increased customer satisfaction Poor forecasting Decreased profitability Collapse of the firm

7. 7 Planning horizon demand time now past demand planning horizon actual demand? actual demand? Forcast demand

8. 8 Designing a forcasting system Forecast need Data avilable? Quantitative? Analyze data YES Causal factors? YES Causal approach YES Collect data? NO YES Qualitative approach NO Time series NO

9. 9 Regression Methods Model dependent variable independent variable Unknown parameters Random error component xtxt t Simple linear model x t = a + b t

10. 10 Estimating a and b parameters xtxt t x t = a + b t e1e1 e2e2 e5e5 e3e3 e4e4 T = 5 0 Such that sum squares of the errors (SSE) is minimized forecast error e t = ( x t – x t )

11. 11 Least squares normal equations Least squares normal equations

12. 12 (x t – x t ) 2 unexplained deviation = (x t – x t ) 2 = explained deviation xtxt Coefficient of determination (r 2 ) xtxt           xtxt t  (x t – x t ) 2 = total deviation, 0  r 2  1

13. 13 Coefficient of corelation (r) r =  coeff. of determination =  r 2 Sign of r,(– / +), shows the direction, of the relationship between x t and t ()or  r  shows the strength of relationship between x t and t – 1  r  1

14. 14 Example – 3.1 It is assumed that the monthly furniture sales in a city is directly proportional to the establishment of new housing in that month a)Determine regression parameters, a and b b)Determine and interpret r and r 2 c)Estimate the furniture sales, when expected establishment of new housing is 250

15. 15 Example – 3.1(continued) Month New Housing /month Furniture Sales /month ($1000) Jan.100461 Feb.110473 March96450 April114472 May120481 June160538 Month New Housing /month Furniture Sales /month ($1000) July150540 Aug.124517 Sep.93449 Oct.88452 Nov.104454 Dec.116495

16. 16 Example – 3.1(solution to a) T = 12  x t = 5782  tx t = 670,215  t = 1375  t 2 = 162,853 txtxt 100461 110473 96450 114472 120481 160538 150540 124517 93449 88452 104454 116495 t2t2 10,000 12,100 9216 12,996 14,400 25,600 22,500 15,376 8649 7744 10,816 13,456 tx t 46,100 52,030 43,200 53,808 57,720 86,080 81,000 64,108 41,757 39,776 47,216 57,420 a T + b  t =  x t a  t + b  t 2 =  tx t

17. 17 Example – 3.1(solution to a) 12 a + 1375 b = 5782 a 1375 + b 162,853 = 670,215 1375 x – 12 x b (1375 2 – 12 x 162,853) = (1375 x 5782 – 12 x 670,215) b (1375 2 – 12 x 162,853) (1375 x 5782 – 12 x 670,215) == 1.45

18. 18 Example – 3.1(solution to a) b= 1.45 12 a + 1375 b = 5782 a 1375 + b 162,853 = 670,215 a 12 (5782 – 1375 x 1.45) = = 315.5 x t = a + b t x t = 315.5 + 1.45 t

19. 19 Example – 3.1(solution to b) txtxt 100461 110473 96450 114472 120481 160538 150540 124517 93449 88452 104454 116495 T 12 xt =xt = xtxt 5782 ==482 x t = 315.5 + 1.45 (100) = 461 461 475 455 481 490 548 533 495 450 443 466 484 xtxt

20. 20 Example – 3.1(solution to b) txtxt 100461 -21 110473475-7-9 96450455-27-32 114472481-10 1204814908 1605385486656 1505405335158 1245174951335 93449450-32-33 88452443-39-30 104454466-16-28 116495484213 xtxt xtxt xtxt xtxt xtxt Explained deviation xtxt xtxt xtxt t         Total deviation xtxt xtxt xtxt t        

21. 21 Example – 3.1(solution to b) txtxt 100461 -21 441 110473475-7-94981 96450455-27-327291.024 114472481-101100 1204814908641 16053854866564.3563.136 15054053351582.6013.364 12451749513351691.225 93449450-32-331.0241.089 88452443-39-301.521900 104454466-16-28256784 1164954842134169 xtxt xtxt xtxt xtxt xtxt xtxt xtxt ()2)2 xtxt xtxt ()2)2

22. 22 Example – 3.1(solution to b) r2r2 = xtxt xtxt ()2)2  xtxt xtxt ()2)2  = 11.215 12.314 =0.91 Coefficient of determination: 91% of the deviation in the furniture sales can be explained by the establishment of new housing in the city

23. 23 Example – 3.1(solution to b) = = 0.95 rr2r2  = 0.91  Coefficient of corelation: a very strong (+) relationship (highly corelated)

24. 24 Example – 3.1(solution to b)  x t 2 = 2.798.274 r = 12 x 670,215 – 1375 x 5782 [12 x 162,853 – (1375) 2 ][12 x 2,798,274 – (5782) 2 ]  = 0.95 r2r2 =(0.95) 2 = 0.91

25. 25 Example – 3.1(solution to c) x t = a + b t x t = 315.5 + 1.45 t t = 250 x t = 315.5 + 1.45 (250) = 678 x t = $ 678,000 x $1000

26. 26 Components of a time series 1. Trend ( a continious long term directional movement, indicating growth or decline, in the data) 2. Seasonal variation ( a decrease or increase in the data during certain time intervals, due to calendar or climatic changes. May contain yearly, monthly or weekly cycles) 3. Cyclical variation (a temporary upturn or downturn that seems to follow no observable pattern. Usually results from changes in economic conditions such as inflation, stagnation) 4. Random effects (occasional and unpredictable effects due to chance and unusual occurances. They are the residual after the trend, seasonali and cyclical variations are removed)

27. 27 Components of a time series 0 xtxt t 12345678 a2a2 a1a1 seasonal variation trend slope random effect Year 1Year 2

28. 28 Simple Moving Average Model xtxt =a + tt xtxt Constant process t x t = a a Forecast error

29. Simple Moving Average Forecast is average of N previous observations or actuals X t : Note that the N past observations are equally weighted. Issues with moving average forecasts:  All N past observations treated equally;  Observations older than N are not included at all;  Requires that N past observations be retained.

30. Simple Moving Average Include N most recent observations Weight equally Ignore older observations weight today TT-1 T-2... T+1-N 1/N

31. 31 Parameter N for Moving Average If the process is relatively stable  choose a large N If the process is changing  choose a small N

32. 32 Example 3.2 WeekDemand 1650 2678 3720 4785 5859 6920 7850 8758 9892 10920 11789 12844 What are the 3-week and 6-week Moving Average Forecasts for demand of periods 11, 12 and 13?

33. Weighted Moving Average Include N most recent observations Weight decreases linearly when age of demand increases

34. 34 Weighted Moving Average WM T =  t=T-N+1 T w t x t  t=T-N+1 T wtwt wtwt = weight value for xtxt The value of wtwt is higher for more recent data

35. 35 Example 3.3 MonthSales Jan.10 Feb.12 March13 April? May? a) Use 3-month weighted moving average with the following weight values to predict the demand of april b) Assume demand of april is realized as 16, what is the demand of may? wTwT = 3 w T-1 = 2 w T-2 = 1

36. 36 Exponential Smoothing Method A moving average technique which places weights on past observations exponentially STST = xTxT  + S T-1  Smoothed valueSmoothing constant Realized demand at period T

37. Exponential Smoothing Include all past observations Weight recent observations much more heavily than very old observations: weight today Decreasing weight given to older observations

38. Exponential Smoothing Include all past observations Weight recent observations much more heavily than very old observations: weight today Decreasing weight given to older observations

39. Exponential Smoothing Include all past observations Weight recent observations much more heavily than very old observations: weight today Decreasing weight given to older observations

40. Exponential Smoothing Include all past observations Weight recent observations much more heavily than very old observations: weight today Decreasing weight given to older observations

41. Exponential Smoothing Include all past observations Weight recent observations much more heavily than very old observations: weight today Decreasing weight given to older observations

42. Exponential Smoothing

44. 44 The meaning of smoothing equation STST = xTxT  + S T-1  xTxT STST =  + S T-1  S T-1 STST = xTxT  +  () STST = x T+  New forecast for future periods S T-1 xTxT = Old forecast for the most recent period eTeT xTxT = xTxT – Forecast error

45. Exponential Smoothing Thus, new forecast is weighted sum of old forecast and actual demand Notes:  Only 2 values ( and ) are required, compared with N for moving average  Parameter a determined empirically (whatever works best)  Rule of thumb:  < 0.5  Typically,  = 0.2 or  = 0.3 work well

46. 46 Choice of  Small   Slower response Large   Quicker response Equivelance between  and N  = 2 N+ 1  = 2 N –  

47. 47 Example 3.4 WeekDemand 1820 2775 3680 Given the weekly demand data, what are the exponential smoothing forecasts for periods 3 and 4 using  = 0.1 and  = 0.6 ? Assume that S 1 = x 1 = 820

48. 48 Example 3.4 (solution for  = 0.1) S 1 = x 1 = 820 S2S2 = x2x2  + S1S1  = x2x2 820 txtxt StSt 1 2775 3680 4 xtxt S2S2 = 0.1(775) + 0.9(820) =815.5 = x3x3 820 815.5820 815.5 801.95

49. 49 Example 3.4 (solution for  = 0.6) S 1 = x 1 = 820 S2S2 = x2x2  + S1S1  = x2x2 820 txtxt StSt 1 2775 3680 4 xtxt S2S2 = 0.6(775) + 0.4(820) =793.0 = x3x3 820 793.0820 793.0 725.2

50. 50 Winters’ Method for Seasonal Variation Model xtxt =a + tt b t () ctct + Constant parameter Trend parameter Seasonal factor for period t Random error component t xtxt

51. 51 Initial values of ac b, and  t=1 L a0a0 = xtxt L b0b0 =0 ctct = xtxt a0a0  L ctct =L ctct values are valid ctct = L ctct  t=1 L ctct ctct values are normalized : YES NO for one year available demand data

52. 52 Smoothing equations  0 << 1  0 < < 1  0 < < 1

53. 53 Forecast Equation aTaT bTbT c (T+  -  L) x (T+  ) = ( )+   = the smallest integer ≥  L

54. 54 Example 3.6 Month(2005)123456789101112 Demand4258111318159654 a) Forecast the demand of Jan.’06 using Winters method with  = 0.2,  = 0.1,  = 0.5 b) Forecast the demand of Feb.’06 when Jan.’06 realizes as 5 using Winters method with  = 0.2,  = 0.1,  = 0.5 c) Forecast the demand of Mar.’06 and Mar.’07 when Feb.’06 realizes as 4 using Winters method with  = 0.2,  = 0.1,  = 0.5

55. 55 Example 3.6 (solution to a) b0b0 =0  t=1 L a0a0 = xtxt L = 100 12 = 8.3 c1c1 x (T+  ) = ( ) + 1 a0a0 b0b0 c1c1 = x1x1 a0a0 = 4 8.3 = 0.48 t 10.48 20.24 30.60 40.96 51.32 61.56 72.16 81.80 91.08 100.72 110.60 120.48  12 ctct 0.48 x (12+1) = ( ) + 8.30 = 4

56. 56 Example 3.6 (solution to b) x 13 = 5 JAN.’06 realizes as: =0.2 5 0.48 + 0.8 8.3+0=8.72 =0.1 8.72 – 8.3+ 0.9 0=0.043 =0.5 5 8.72 + 0.5 0.48=0.53 c2c2 x (T+  ) = ( ) + 1 a 13 b 13 0.24 x (13+1) = ( ) + 8.72 0.043= 2.11 x

57. Updated seasonal values should be normalized! 57

58. 58 Example 3.6 (solution to c) FEB.’06 realizes as: x 14 = 4 0.5 4 10.34 + 0.5 0.24=0.31= c 14 =0.2 4 0.24 + 0.8 8.72+0.043=10.34 a 14 =0.1 10.34 – 8.72+ 0.9 0.043=0.2 b 14 0.60 x (14+1) = ( ) + 10.34 0.2= 6.31 x 0.60 x (14+13) = ( ) + 10.34 0.2= 7.7613 x Forecast of Mar.’06 Forecast of Mar.’07

59. 59 Forecast accuracy Forecast accuracy shows the performance of the model for complying with the demand process, and is measured by using forecast error Forecast error is the difference between the actual demand and the forecast ForecastActual demand xtxt etet = xtxt – Forecast error

60. 60 Error measures Looking at the error for an isolated period does not provide much useful information Rather we will look at errors over the history of the forecasting system. There are several methods for this process, although each has different meaning 1.Cumulative (sum) error, E t 2.Mean error, ME 3.Mean square error, MSE 4.Mean absolute deviation, MAD 5.Mean absolute percentage error, MAPE

61. 61 Cumulative (sum) error, E t E t should be close to zero if the forecast is behaving properly. That is, sometimes it overestimates and sometimes it underestimates, but in the long run these should cancel out T ETET =  t=1 etet

62. 62 Mean error, ME ME should be interpreted same as sum error, E t, that is, it shows whether the model is biased toward certain direction or not n ME =  t=1 etet n 1 A forecast consistently larger than actual is called biased high A forecast consistently lower than actual is called biased low

63. 63 Example 3.9 txtxt 19 212 315 418 521 624 727 830 933 Validate the moving average (N=3) if it is suited to the given past data using ME and E t, and say if it is bias

64. 64 Example 3.9 (solution) txtxt MA (t-1) etet 19 212 315 418 126 521 156 624 186 727 216 830 246 933 276 E 9 = 6 + 6 +... + 6 = 36.00 ME = 1/n (  e t )= E 9 / 6 = 6.00 BIASED LOW !

65. 65 Mean square error, MSE n MSE =  t=1 etet n 1 2 MSE is mainly used to counteract the inefficiency in error measuring as negative errors (– e t ) cancel out the positive error terms (+ e t ) By squaring the error terms, the “penalty” is increased for large errors. Thus a single large error greatly increases MSE

66. 66 Mean absolute deviation, MAD n MAD =  t=1 etet n 1 MAD is another error measure for solving the neutralizing problem MAD measures the dispersion of the errors, and if MAD is small the forecast should be close to actual demand

67. 67 Mean absolute percentage error, MAPE n MAPE =  t=1  PE t  n 1 PE t = xtxt xtxt – xtxt (100) MAPE is mainly used to counteract the inefficiency in error measuring as the previously defined mesaures depend on the magnitude of the numbers being forecast If the numbers are large the error tends to be large. It may more meaningful to look at error relative to the magnitude of the forecasts, which is done by MAPE

68. 68 Example 3.11 txtxt 110 212 315 445 5130 6180 7170 8120 9125 10100 11125 12135 Compare a 3-period moving average model and a 6-period moving average model using given past data and show which suits better with respect to MSE, MAPE.

69. 69 Example 3.11 (solution) t xtxt MA[3] t-1 etet 1 10 2 12 3 15 4 45 12.33 5 130 24.00 6180 63.33 7170 118.33 51.67 8120 160.00 -40.00 9125 156.67 -31.67 10100 138.33 -38.33 11125 115.00 10.00 12135 116.67 18.33 t=7 6 12 MSE =  etet 1 2 1196.30 =

70. 70 Example 3.11 (solution) t xtxt MA[6] t-1 etet 1 10 2 12 3 15 4 45 5 130 6180 717065.33 104.67 812092.00 28.00 9125110.00 15.00 10100128.33 -28.33 11125137.50 -12.50 12135136.67 -1.67 t=7 6 12 MSE =  etet 1 2 2154.32 =

71. 71 Example 3.11 (solution) t xtxt MA[3] t-1 etet  Pe t  7 170 118.33 51.6730.39 8 120 160.00 -40.0033.33 9 125 156.67 -31.6725.33 10 100 138.33 -38.3338.33 11 125 115.00 10.008.00 12135 116.67 18.3313.58 MA[6] t-1 etet  Pe t  65.33 104.6761.57 92.00 28.0023.33 110.00 15.0012.00 128.33 -28.3328.33 137.50 -12.5010.00 136.67 -1.671.23 MA[3] model results MA[6] model results MAPE PetPet 12 =  t=7 6 1 24.83 = MAPE PetPet 12 =  t=7 6 1 22.74 =

72. 72 Example 3.11 (solution) Error measures Forecast models MA[3]MA[6] MSE2154.32 MAPE24.83 1196.30 22.74