Demand Forecasting
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Bangladesh University of Textiles**We aren't endorsed by this school
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TEXTILE EN 123
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Industrial Engineering
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Dec 18, 2024
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Demand Forecasting 521 3. Seasonal Variations (S): This seasonal variation occurs with some degree of regularity in a span of one year and are annually repetitive. These variation are caused by climatic conditions, sqcial customs, festivals, e.g., the sales of paint is highest during Diwali Festival, sale of umbrella and rain coat during monsoon, and warm clothes during winter. 4. Irregular Variations (R): These occur without any fixed pattern. These are chance variations that are not explained by trend, cyclic or seasonal variations. These may account for natural calamities like flood, draught, earthquake, etc. The most commonly used expression for a time series forecast is: Y =TCSR where Y = Forecasted value T = Secular trend C = Cyclic variations § = Seasonal variations R = Irregular fluctuations 21.8 LEAST SQUARE METHOD OF FORECASTING (Regression Analysis) This is the mathematical method of obtaining “the line of best fit between the dependent variable (usually demand) and an independent variable. This method is called least square method as the sum of the square of the deviations of the various points from the line of best fit is minimum or least. It gives the equation of the line for which the sum of the squares of vertical distances between the actual values and the line values are at minimum”. In a simple regression analysis, the relationship between the dependent variable y and some independent variable x can be represented by a straight line. y=a+bx where b is the slope of the line a is the y-intercept. The values of the constants a and b are determined by the two simultaneous equations. Zy = Na+bZx wi{3) Zxy = aZx +bEx? .(2) These two equations are called normal equations. To compute the values of a and b (i) Calculate the deviation (x) for each period and also the sum of deviations. (if) Find the value of £x? (iif) Find the value of Zxy (iv) Calculate the values of a and b (v) Make the sum of deviations Zx =0 Substituting the value of £x = 0 in equations (1) and (2) We get, Zy = Na Ixy = bix? which gives the values of a and b as a= Xy/N b= Zxy/ Ex® Note: (i) If the Time Series consists of odd number of years to make Ex = 0, the middle value of the time series is taken as the Origin.
522 Industrial Engineering and Production Management the midway period between two (ii) If the time series consists of even number of years, middle periods is taken as origin to make Zx = 0. E : Problem 1: The following data gives the sales of the company for various years. Fit the straight line. Forecast the sales for the year 1998 and 1999. Year Sales (000) 1989 13 1990 20 1991 20 1992 28 1993 30 1994 32 1995 33 1996 38 1997 43 Solution: Year Sale (y) Deviation (x) 2 xy 1 13 -4 16 -52 2 20 -3 9, - 60 3 20 -2 4 - 40 4 28 -1 1 -28 5 30 0 0 0 6 32 1 1 32 Z 33 2 4 66 8 38 3 9 114 9 43 4 14 172 N=9 Ty =257 Tx=0 2= 60 Txy = 204 Now, substituting the values in the equations to get a=Sy/n=257/9=2856 b=Sxy/x*=204/60=34 The equation of the straight line of best fit is y=2856 +34x (i) Sales for the year 1998 Yoy =28.56 +34 x5 =45.56 = 45560 (if) Sales for the year 1999 Yoo = 28.56 + 3.4 x 6 = 49000 Problem 2: The past data regarding the sales of SPMS for the last five years is given. Using the least square method, fit a straight line, estimate the sales for the year 1996 and 1997. Year 1991 1992 1993 1994 1995 Sales (‘00) 35 56 79 80 40
Demand Forecasting 523 Solution: Year Sales (y) Deviation (x) > xy 1991 35 -2 4 -70 1992 56 -1 1 - 56 1993 79 0 0 0 1994 80 1 1 80 1995 40 2 4 88 N=5 Ty =290 Ex=0 =0 Ixy =34 In this case, as the number of periods is odd, to make the Zx = 0, the deviations are calculated from the middle period, i.e., 1993. Now, to fit a straight line y = a + bx the values of a and b are to be computed using the formula a=Xy/n=290/5=58 b = Sxy/Ex?=34/10=34 y=58+34x (1) Deviation for 1996 is x = 3, and 1997 = 4 Substituting the values of x = 3 and x = 4 we get the forecast for 1996, 1997 Year 1996 = 58 + 3.4 x 3 = 68.2 Therefore Forecast for 1996 = 6820 units Year 1997 = 58 + 3.4 x 4=71.6 Therefore Forecast for 1997 = 7160 units Problem 3: The sales for the domestic water pumps manufactured by Ajit Manufacturing Company is given forecast the demand for the pumps for the next three years using least square method. Year 1986 1987 1988 1989 1990 1991 1992 1993 1994 1995 Sales ('000) 30 33 37 39 42 46 48 50 55 58 Solution: Year Sales (y) Deviation (x) Xt xy 1986 30 -4.5 20.25 -135 1987 33 -35 12.25 -455 1988 37 -25 6.25 =925 1989 39 -15 2.25 -58.5 1990 42 -5 0.25 -21 1991 46 0.5 0.25 23 1992 48 1.5 2.25 72 1993 50 2.5 6.25 125 1994 55 35 12.25 192.5 1995 58 4.5 20.25 261 N=10 438 Zx=0 =825 Txy =251 524 Industrial Engineering and Production Management As the number of periods is even, sum of the deviations is made equal to Zero (Ex =0) by making midway period between two middle periods is made as origin. The 1990 and 1991 are the two middle periods and midway period between 1990 and 1991 is selected as Origin. ie, 199005 1991 +0.5 and the deviations of the other periods are calculated as shown in the table. a = Zy/n=438/10=438 b = Zxy/¥x,=251/825=3.04 The regression line is y = 438 +3.04x Forecast for the pumps for next three years Yy = 438 +3.04 x5.5 60.52 = 60520 = 438+3.04 x65 63.56 = 63560 Yoy = 43.8+3.04x75 é‘< [
524 Industrial Engineering and Production Management As the number of periods is even, sum of the deviations is made equal to Zero (Ex =0) by making midway period between two middle periods is made as origin. The 1990 and 1991 are the two middle periods and midway period between 1990 and 1991 is selected as Origin. ie, 1990-05 s () 1991 + 0.5 and the deviations of the other periods are calculated as shown in the table. a = Zy/n=438/10=43.8 b = Zxy/Zx,=251/825=23.04 The regression line is y = 43.8 +3.04x Forecast for the pumps for next three years Yy = 438+3.04 %55 = 60.52 = 60520 Y,, = 43.8+3.04 x 65 = 63.56 = 63560 Y, = 438+3.04%7.5 = 66.6 = 66600 Least square method when the sum of the deviations is not zero (Ex = 0) For the straight line y = a + bx The constants a and b can be calculated from the expression— (zy.24%)~(2x-2xy) (N2et)-( ) o NExy - (Zx-2y) (N ~(zx)" or Alternatively The values of a and b can be computed by solving the following two equations: 2y = Na + bZx w (1) Zxy = aZx + bEx? (2 Problem 4: A company manufacturing washing machines establishes a fact that there is a relationship between sale of washing machines and population of the city. The market research carried out reveals the following information: Population (million) 5 7 15 22 27 36 No. of washing machines demanded (‘000) 28 40 65 80 96 130 Fit a linear regression equation and estimate the demand for washing machines for a city with a population of 45 million. Demand Forecasting 525 Solution: Population (x) | No. of Washing m/c’s demanded (y) e xy 5 28 25 140 7 40 49 280 15 65 225 975 22 80 484 1760 27 96 729 2592 36 130 1296 4680 Ix =112 Ty =439 Zx?=2808 | Zxw— 10407 To find the regression equation, the values of @ and b are computed as foll ? " s 7 (Ve (s _ (439x3808)-(112x 10427 ) ——taag0g (312
11:00 PM WZe sl Ll @3 55 & Industrial Engineering... [@ 6% Demand Forecasting 525 Solution: Population (x) | No. of Washing m/c's demanded (y) &2 xy ¥ 5 28 25 140 7 40 49 280 15 65 225 975 22 80 484 1760 27 96 729 2592 36 130 1296 4680 Zx =112 Ty =439 3x?=2808 | Zxy = 10427 To find the regression equation, the values of a and b are computed as follows: (zy-24%) - (2x-2ay) (N2 ~(zx)’ (439x3808)— (112 x 10427 ) 6x2808—(112)" g 1232712-1167824 —15.07 1684812544 Iy NExy—()‘.x- Ey) Nz ~(zx)" 6 x10427-112x439 _ 62562-49168 6x2808—(112)° T 16864-12544 13394 4304 | =311 (i) The regression equation is y = 15.07 +3.11x (i) The demand for the washing machine when population of city is 45 million, i.e., x=45 : 15.07 + 3.11 x 45 155.02 Therefore, the number of washing machines demanded when x = 45 is 155020 Nos. (Ans). y - 21.9 MOVING AVERAGE (MA) FORECASTING The past data of sales of a company can have fluctuations (high or low) because of the seasonal variations and random variations. Simple averaging of demand for previous periods is going to hide the trend and it is meaningless since trend is an important factor. Moving Average (MA) consists of series of arithmetic means calculated from overlapping groups of successive elements of time series. Moving average is a simple statistical method to extrapolate and establish trend of past sales. This method uses a past data and calculates a rolling average for a constant 526 Industrial Engineering and Production Management % period. At each period, fresh average is computed at the end of each period by a demand of the most recent period and deleting the data of the old-period since the aatan this method changes from period to period, it is called moving average method. ARt A Lo T
526 Industrial Engineering and Production Management period. At each period, fresh average is computed at the end of each period by adding the demand of the most recent period and deleting the data of the old-period since the data in this method changes from period to period, it is called moving average method. A simple moving average is calculated as follows: Sum of demands for periods Chosen number of periods For example, for the time series of values D, D,, D. the moving average for n periods, is given by, First value moving average = 1/n (D, + D, + Dy + .......Second value moving average = 1/ (D, + D, + D, + Third value of moving average = 1/n (D, + D, + +D,.q) Period n of moving average should be carefully selected. The wrongly selected period will distort the data and gives wrong picture of the trend. Larger the period of moving average, the greater is the smoothing effect. A three months MA has a weightage of 1/3" and 5 months MA has a weightage of 1/5". Larger the value of (period of moving average), the smaller is the effect of random variation and a higher smoothing effect. The value of 1 depends upon the speed at which the pattern of demand changes. If the demand pattern is stable a high value of n is selected. If the pattern is not stable, a small value of n should be selected. Centring of the Moving Average MA = . D, , etc,, for different periods, A three period moving average is located against the second period, a five months moving average is located against third month. il In general, n months moving average is located against =i Weighted Moving Average (WMA) Sometimes the forecaster wants to use a moving average but does not want all the n periods equally weighted as in simple MA method. But some organisations base their forecasts on a weighted moving average. In simple MA, equal weightage is given to 1* month, 2 month and 3 month in a three month moving average. But the organisation wants to attach more weightage to the third month and least to the first month. For example, depending upon the importance it assigns weightages, e.g., 0.2 to 1st period, 0.3 to second and 0.5 to the third. The sum of these weights should be equal to one. Problem 5: The past data on the load on the weaving machines is shown below: Month Load (HRS) May-96 June-96 585 JuJy-96 610 Aug.-96 675 Sep.-96 750 Oct.-96 860 Nov.-96 970
LG WEIBIIGEE 1UL EALIL UL WIS UEIIAHIUS L1 UIE Padt 15 WSLUWIIES Uy @ 1aLUs Ul {3~ uy. The last term is negligible for a very large value of period. The value of the a selected is small (0.05 to 0.1), if the demand pattern is smooth or stable and large value of « is used for the fluctuating demand. Problem 7: The demand for the disposable plastic tubing for a general hospital is 300 units and 350 units for September and October respectively. Using 200 units as demand for September, compute the forecast for the month of November. Assume the value of as 0.7. Solution: (i) Forecast (Ft) can be written as, F=a-D,_+(1-a)-F,_, In this case, D, , = Demand for September (300 units) F,_, = Forecast for September (given) 200 units. 2. Forecast for October can be computed as, (F) = 0.7 %300 +(1-0.7) x 200 = 210 + 60 = 270 units. (ii) Forecast for the month of November F,,,= a(Demand for Oct.) + (1 - a). Forecast for October = 0.7 x 350 + (1-0.7) x 270 = 326 units. Problem 8: The demand for the particular product is given for the last 8 periods. Compute the exponentially smoothed forecast for the periods taking a. = 0.1 and 0.3, Which of these forecast is better Period i 2 3 4 5 6 7z 8 Demand 10 18 29 15 30 12 16 8 Solution: The forecast for the initial period is not given and it is assumed equal to the demand for the first period. The forecasts with « = 0.1 and 0.3 are computed and the details are given in the table below. Period Demand Forecast Error Forecast Error Dt Fata=01 D,-F, | Fata=03 | D,-F, 1 10 15 =50 15 -5.0 2 18 14.5 3.5 13.5 4.5 3 29 14.85 14,15 14.85 14,15 4 15 16.26 -1.26 19.09 -4.09 5 30 16.14 13.86 17.86 12.14 6 12 17.52 -5.52 21.50 -9.50 7 16 16.97 =097, 18.65 -2.65 8 9 16.87 -8.87 17.85 -9.85 Problem 9: Estimate the sales forecast for the year 2000, using exponential smoothing forecastar. Take o = 0.5 and the forecast for the year 1995 as 160 x 10° units. Compare the forecast with least square method.
r:, : = Forecast for September (given) 200 units. . Forecast for October can be computed as, (F) = 0.7 %300+ (1-0.7) x 200 = 210 + 60 = 270 units. (if) Forecast for the month of November F,,; = a(Demand for Oct.) + (1 - a). Forecast for October 0.7 x 350 + (1-0.7) x 270 = 326 units. Problem 8: The demand for the particular product is given for the last 8 periods. Compute the exponentially smoothed forecast for the periods taking a = 0.1 and 0.3, Which of these forecast is better Period 1 2 3 4 5 6 7 8 Demand 10 18 29 15 30 12 16 8 Solution: The forecast for the initial period is not given and it is assumed equal to the demand for the first period. The forecasts with a = 0.1 and 0.3 are computed and the details are given in the table below. Period Demand Forecast Error Forecast Error Dt Foata=01 D,~F, Fata=03 D,-F, 1 10 15 -5.0 15 -5.0 2 18 14.5 35 13.5 45 3 29 14.85 14.15 14.85 14.15 4 15 16.26 ~1.26 19.09 -4.09 5 30 16.14 13.86 17.86 12.14 6 12 17.52 -5.52 21.50 -9.50 7 16 16.97 -097 18.65 -2.65 8 9 16.87 -8.87 17.85 -9.85 Problem 9: Estimate the sales forecast for the year 2000, using exponential smoothing Jorecastar. Take a. = 0.5 and the forecast for the year 1995 as 160 x 10° units. Compare the forecast with least square method. 530 Industrial Engineering and Production Management Year 1995 1996 1997 1998 1999 Sales T (x 10°) 180 168 159 170 188 Solution: The demand for the year 2000 can be computed using least square method as follows. Year Sales (x 10%) Deviation (x) » xy 1995 180 2 4 360 1996 168 -1 1 - 168 1997 159 0 0 0 1998 170 +1 1 170 1999 188 +2 4 376 N=5 Zy =505 Ex=0 Ex2=10 Ix2=18 Substituting these values in the equations, Iy _ 505 a= N = N =101 b= ZYI ' 10 The regression equation is y = 100 + 1.8¢ Now, the sales for year 2000 i.e,, for x = 3, we have, y=101+18x3=1064 The demand for year 2000 = 106.4 x 1052, The forecasted demand using the smoothing constant a = 0.5 is computed as shown in the table using the formula. F,=a-D, ,+(1-a).F,_, Period Demand Forecasted demand 5 Jora=05 1995 180 160 (Given) 1996 168 0.5 % 180 = (1 - 0.5) x 160 170.0 1997 159 05 x 168 = (0.5) x 170 169.0 1998 170 0.5 x 159 = (0.5) x 169 164.0 ? 1999 188 0.5 x 170 = (0.5) x 164 167.0 /’ 2000 0.5 x 188 = (0.5) x 167 1775 The forecast demand for the year 2000 with @ = 0.5 = 177.5. Trend Adjusted Exponential Smoothing
532 Industrial Engineering and Production Management 2. Weighted Moving Averages: Moving average with varying weights Weighted moving average = Z (weight for period IT) (demand in period) Z weights Fi_y = W, (D) + Wy (D,_,) + W, (Dy_p)..+ W, (Dy_y41) 3. Exponential Smoothing: A moving average with weights following an exponential distribution. New forecast = Last periods forecast + o (previous periods actual demand - previous periods forecast) Fi,,= F'+a(D,—F,) 4. Trend Adjusted Exponential Smoothing: An exponential smoothing model that can accommodate trend. Forecast with trend [Fy] = Exponentially smoothed trend (F) '+ Exponentially smoothed trend (T) F = o.(A,_l) +(1-a) (Fy_q+ T,y T, = B(Fi_Fl—l) +(1-p) T, 5. Trend Projection and Regression Analysis Y =a+bx (Zxy - nXy) bess % - nw? a=y-bx 6. Forecast Error (E) = D,-F, 7. Cumulative Sum of Forecast Errors (CFE) CFE = X E, 8. Mean Square Error (MSE) = CFE (EB)i=tes 9. Mean Square Standard Deviation (MSE) AE MSE = —- n 10. Mean Absolute Deviation (MAD) I|E| MAD = n 11. Mean Absolute Percent Deviation (MAPE) MAPE = ——(2 flEst l/nD' 2100) (Expressed as per~ ! CFE 2 12. Tracking Signal: MAD % 13. Exponentially smoothed error: MAD, = o|E|+(1-0) MAD,
simple to use. S 2. The method selected should give minimum forecast errors at the optimal cost. 3. The cost to make the forecast should be small. 4. The method selected should be stable in the sense that the changes should be minimum. Problem 12: There exists a relationship between expenditure on research and its annual profit. The details of the expenditure for the last six years is given below. Estimate the profit when the expenditure is 6 units. Year Expenditure for research (x) Annual Profit (y) 1994 2 20 1995 3 25 1996 5 34 1997 4 30 1998 11 40 1999 5 31 2000 6 ? (One unit corresponds to ¥ 1 Crore) Solution: Year Expenditure for Annual Profit Xy % research (x) 1994 2 20 40 4 1995 3 25 75 9 1996 5 34 170 25 1997 4 30 120 16 1998 11 40 440 121 1999 5 31 155 25 Total 30 180 1000 200 538 Industrial Engineering and Production Management X = % =5 — 180 ===% The values a and b are computed as follows: for a linear regression equation y=a+bx b= _ 1000-6x5x30 _ 1000-900 _ T 200-6x5x5 200-150 a = y-bx=30-2x5=20. Thus, the model is y = 20 + 2x The profit when the expenditure is 6 units is ¥ =20 +2 x 6 =32 units of X, Problem 13: Starwars Co. Ltd., uses simple exponential smoothing with smoothing constant o = 0.2 to forecast the demand. The forecast for the first week of March was 400 units and the actual demand turns out to be 450 units. (i) Estimate the demand for the second week of March. (ii) If the actual demand for the second week of March is 460 units, forecast the demand upto April second week. Assume that demand for subsequent weeks are 465, 434, 420, 498, 462 and 470 unit. Solution: (i) The forecast for the period (F)) is given by F=F_+a(D.,-F.,) The forecast demand for March second week is, F, =400 +0.2 (450 - 400) = 410 units. (if) The forecasted demand for the third week of March to April second week is worked out as shown in table below. Week Demand Old Forecast Correction New forecast D,_; | forecast error o(D,_;~F,_y) (Ft) E_y |DirFig F_y+a(D,_~F_y) March Week 2 460 410 50 10 420 Week 3 465 420 45 9 429 Week 434 429 5 1 430 April Week 1 420 430 -10 -2 428 Waal 2 = &, e - N AR
540 Industrial Engineering and Production Management Problem 15: The Forecast and the actual demand for the 6 months is given below. Compute the various Forecast errors and tracking signal and comment on the forecast accuracy. Month Actual Forecasted Deviation Cum deviation Demand (D) Demand (F) D-F, Z(D,~F) 1 71 78 =7 -7 2 80 75 5 =2 3 101 83 18 16 4 84 84 0 16 5 60 88 -28 -12 6 73 85 =12 -24 Solution: The deviations and cumulative deviations are shown in table above. (i) Mean Absolute Deviation (MAD) MAD is the mean of absolute deviation of forecast demands from actual demand values. MAD = i|u, -F| = where D, = actual demand for period ‘' F, = forecast for period ‘t’ n = number of time period used. 7+5+18+0+28+12 MAD = Ty — =117 (if) Cumulative deviations or Running Sum of Forecast Error (RSFE) This is the algebraic sum of the forecasting error i.e. both positive and negative signs are considered. RSFE = -7+5+18+0-28-12=~12 RSFE _ 24 MAD ~ 117 The tracking signal tells how well the forecast is predicting actual values, for it yields a measurement of the consistent difference between actual and forecasted values by expressing the cumulative deviation in terms of number of average deviations. (iv) Mean Square Error (MSE) It is the mean of the squares of the deviations of forecast demands from the actual demand values, (iii) Tracking signal = ~2.05=|205| MSE = i(u-;fi) (=7) +(5)" +(18)" +(-28)" +(-12)’ i = E:i’flzl. (1) Mean Forecast Error (MFE) Mean forecast errors is the mean of the deviations of the forecast demands from the actual demand. Demand Forecasting 541 (D,-F) n MSE = 3 ] n _ ~7+5+18+0-28-12 . -4 6 (vi) Mean Absolute Percentage Error (MAPE) Mean absolute percentage error is the mean of the percentage deviation of the forecast demands from the actual demands. mare= L3P Al 10002009 nia D, Problem 16: The number of people visiting on arts exhibition appears to depend upon (a) the [facilities available in the exhibition, (b) the distance of the location from the residential area (c) the entrance fee charged. Since these factors are difficult to measure directly, they are measured by means of a score for factors on 0-10 scale. The organisers believe that there is a direct and linear relationship bettween the number of people and these factors. The following data is collected on various exhibitions from cities. Estimate the relationship between the variables? Exhibition Facilities Distance Entrance Age of people/ No. score score Jee score week (000) 1 8 4 8 5 2 10 5 2 4 3 2 6 4 15 4 4 6 6 25 5 6 8 9 7 ?/ 6 3 10 7 5 ) 7 4 5 4 3 8 10 7 9 9 9 1 5 5 1 10 q 2 4 2