IE5403 Facilities Design and Planning

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 Participation 5% Quizzes 15% (4 quizzes) Assignment 15% (every week) Midterm 1 30% (chapters 1 and 2) Final 35% (all chapters)

4. IE5403 - Chapter 1 Forecasting methods

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

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

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

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

10. e5 xt = a + b t e3 Such that sum squares of the errors (SSE) is minimized e4 e1 e2 forecast error et = ( xt – xt ) T = 5 Estimating a and b parameters xt t 0

11. Least squares normal equations Least squares normal equations

12. unexplained deviation = (xt – xt)2 (xt – xt)2 = total deviation (xt – xt)2 = explained deviation xt , 0  r2  1 xt Coefficient of determination (r2)  xt            t

13. r =  coeff. of determination =  r2 ( or ) Sign of r ,(– / +), shows the direction, of the relationship betweenxtand t – 1  r  1 Coefficient of corelation (r) r shows the strength of relationship betweenxtand t

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 r2 c) Estimate the furniture sales, when expected establishment of new housing is 250

15. Example – 3.1(continued)

16. a T + bt = xt at + b t2= txt Example – 3.1(solution to a) T = 12 t = 1375 xt = 5782 t2 = 162,853 txt = 670,215

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

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

19. xt 5782 xt xt= = = 482 T 12 xt = 315.5 + 1.45 (100) = 461 Example – 3.1(solution to b)

20.  xt    xt xt   xt xt   t Explained deviation xt  xt xt xt        t Total deviation xt xt Example – 3.1(solution to b)

21. xt ( )2 xt xt xt xt xt xt xt ( )2 xt Example – 3.1(solution to b)

22. xt ( )2  Coefficient of determination: xt 11.215 xt xt r2 = 0.91 = = 12.314  ( )2 Example – 3.1(solution to b) 91% of the deviation in the furniture sales can be explained by the establishment of new housing in the city

23. r r2 = = 0.95 = Coefficient of corelation: 0.91   Example – 3.1(solution to b) a very strong (+) relationship (highly corelated)

24. 2.798.274 xt2 = 12 x 670,215 – 1375 x 5782 r = = 0.95  [12 x 162,853 – (1375)2 ][12 x 2,798,274 – (5782)2] r2 = (0.95)2 = 0.91 Example – 3.1(solution to b)

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

26. Components of a time series • Trend ( a continious long term directional movement, indicating growth or decline, in the data) • 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) • 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) • 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. seasonal variation a2 trend slope a1 random effect Components of a time series xt t 0 1 2 3 4 5 6 7 8 Year 1 Year 2

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

29. Simple Moving Average • Forecast is average of N previous observations or actuals Xt: • 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 1/N ... T+1-N T-2 T-1 T today

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. Example 3.2 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. T S wt xt t=T-N+1 T The value of is higher wt S wt for more recent data t=T-N+1 Weighted Moving Average xt wt weight value for = WMT =

35. 3 wT = 2 wT-1 = wT-2 1 = Example 3.3 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?

36. Realized demand at period T xT a (1-a) ST = ST-1 + Smoothed value Smoothing constant Exponential Smoothing Method A moving average technique which places weights on past observations exponentially

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

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

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

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

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

42. Exponential Smoothing

43. Exponential Smoothing

44. xT a - a ST = ST-1 ST-1 + xT a (1-a) xT a ST = ST-1 - + ( ) ST = ST-1 + ST-1 xT xT xT+t ST = ST-1 = xT eT – = New forecast for future periods Old forecast for the most recent period Forecast error The meaning of smoothing equation

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

47. Example 3.4 Given the weekly demand data, what are the exponential smoothing forecasts for periods 3and 4 using a = 0.1 and a = 0.6 ? Assume that S1= x1 = 820

48. = 820 x2 x3 a x2 (1-a) S2 = S1 + = S2 + = 815.5 0.1(775) 0.9(820) = 815.5 xt Example 3.4 (solution for a = 0.1) S1= x1 = 820 820 815.5 820 815.5 801.95 801.95

49. = 820 x2 x3 a x2 (1-a) S2 = S1 + = S2 + = 793.0 0.6(775) 0.4(820) = 793.0 xt Example 3.4 (solution for a = 0.6) S1= x1 = 820 820 793.0 820 793.0 725.2 725.2

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