Description
Demand Management and it contains topics like what is demand, what is forecasting, methods of forecasting, typical forecasting program, error monitoring.
Demand Management
Demand
• Planning as most important function of management • Demand Management deals with both consumer needs and supplier coordination • Purpose of demand management is to coordinate and control all resources to efficiently utilise systems
• Demands has to be forecasted to run businesses profitably
2
Forecasting
• Many business decisions depend on some sort of forecasting • Forecasting is scientifically calculated guess, it forms the basis of planning • Levels of forecasting - Short term (upto 1 yr) Medium term (1 to 3 yrs) and Long term (over 5 yrs) • Extrinsic forecast & intrinsic forecast • Elements of forecasting
? ?
Internal factors (past, present and future) External factors (controllable & uncontrollable)
3
Methods of Forecasting
• Qualitative Techniques
? ? ? ?
Market research Panel Consensus History Analogy Delphi Method
Linear Regression Multiple Regression Input Output Analysis End use analysis
4
• Casual Methods
? ? ? ?
• Simulation – Monte Carlo Simulation
Methods of Forecasting . . contd
• Time Series
? ?
Extrapolation – Used when past data is linear Moving Average – Used when data is cyclic
o o
Simple – Average of specified past period is considered Weighted – different weights are assigned to past data
?
Exponential Smoothing – weightage decreased exponentially
o o
o
Simple Exponential Smoothing Trend adjusted Seasonality considered
5
Forecasting programme for any company
• Observing, listing and studying external factors (cultural, social, political, technological) nationally & internationally • Gather info on internal company policies (changes in design, quality, sales) & their effect on demand • Analyse data to establish various relationships and their relative effects of each factors on final demand • Create various scenarios assuming certain happenings in external environment & alternative internal policies • Operationally apply the forecast by breaking it down on the number of product lines. Do Break Even analysis • Regularly monitor forecast errors & update method
6
Prerequisites & Pitfalls in Forecasting
• Define the purpose of forecasting, this will help to decide the accuracy & type of technique to be chosen • It should be combined effort organisationally • Forecasting technique will vary depending on the Product Life Cycle and stage of the product
? ?
New product development stage – Delphi, PDMT Steady state of PLC – Time series with trend & seasonality
• Quite often wrong things are forecasted • People go for minute forecasting of odd products • No timely tracking of forecasting
7
Range & Precision of Forecast
• Forecast may be in terms of ranges
? ?
Higher range – low value, low turnover items in inventory Lower range – Capital intensive machineries
• It will also guide the company form its strategic posture • Precision in forecasting is not as important as proper use of available data • It is better to use the available data according to situational demands
8
Technology Forecasting
• • • • • It deals with estimation of future trends in technology Helps making current decisions examining future choices Guides wide range of long term planning process Forecasting tool at micro level corporate planning also Very effective for high range technology products as gestation time to become productive is quite long • Exploratory technology forecasting – Predicting future with present trends & capabilities • Normative Forecasting – Set goals & objectives of future technology and take necessary current action
9
Uses of Technology Forecasting
In planning of future discoveries & technologies
Government
Planning
Industry
Corporate Planning
Universities
Future Academic Roles
Individuals
Selection of fertile areas of research
Policy formation for the allocation of resources
Investment in production
Investment in research & development
10
Selecting a forecasting method
• Choice depends on
? ? ? ? ?
Purpose of forecasting Type and amount of data available Time horizon of forecast Degree of accuracy required Cost involved
11
Forecast Error Monitoring - MAD
•Mean Absolute Deviation (MAD)
Absolute means the positive & negative signs are ignored ?Deviation is difference between forecast & actuals
?
Period 1 2 3 4 5
Forecast Demand 900 1000 1050 1010 980
Actual Demand 1000 1100 1000 960 970
Deviation 100 100 -50 -50 - 15
MAD = ?I Deviation I = N
315 = 63 5
12
Forecast Error Monitoring - RSFE
•Running sum of forecast errors (RSFE) ?This is algebraic sum of forecasting errors (deviation)
Period 1 2 Forecast Demand 900 1000 Actual Demand 1000 1100 Deviation 100 100
3
4 5
1050
1010 980
1000
960 970
-50
-50 -15
RSFE = ?Deviation = 85
13
Forecast Error Monitoring - Formulas
Mean Square Error (MSE) = ? (Deviation)² N Percentage Error (PE) = (Deviation / Demand) x 100 Mean Absolute Percentage Error = ? I PE I
N
Tracking Signal = RSFE MAD
14
Forecast Error Monitoring . . contd
• RSFE is calculated to determine whether or not the forecast has positive or negative bias • MAD indicates the volume or amplitude of deviation from actuals • Both the bias and amplitude of forecast errors are important • It is important to monitor MAD & Tracking signal for any modifications to be made in original forecasting model • A good forecast should have approximately as much positive as negative deviation
15
Problem
Forecast 100 Demand 110
Table shows actual & forecasted demand. Calculate: - MAD - MSE - MAPE - Tracking Signal
90
80 85 75 85 65
85
88 95 65 80 52
16
Problem
Forecast 100 Demand 110 Deviation 10
90
80 85 75 85 65
85
88 95 65 80 52
-5
8 10 -10 -5 -13
17
Problem
Forecast 100 Demand 110 Deviation 10 (Deviation)² 100
90
80 85 75 85 65
85
88 95 65 80 52
-5
8 10 -10 -5 -13
25
64 100 100 25 169
18
Problem
Forecast 100 90 80 85 75 85 Demand 110 85 88 95 65 80 Deviation 10 -5 8 10 -10 -5 (Deviation)² 100 25 64 100 100 25 Percentage Error 9.09 -5.88 9.09 10.52 -15.38 -6.25
65
52
-13
169
-25
19
Problem
MAD =? I Deviation I N MSE = ? (Deviation)² = 61 7 = 8.71
= 583
= 83.29
7
7
MAPE = ? I PE I = 81.23 = 11.6% N 7
Tracking Signal = RSFE
MAD
= -5
8.71
= -0.57
20
Practice Problem
Forecast 150 125 130 145 Demand 160 130 135 150
Table shows actual & forecasted demand. Calculate: - MAD
180
170 165 155
160
165 145 150
- MSE
- MAPE - Tracking Signal
155
150 150 160
155
160 165 160
21
Simple Moving Average - Formula
( MA ) t = Dt + Dt-1 + Dt-2 + . . . . Dt-n+1
n
( MA ) t = ( f ) t+1 Where; MA = Moving Average f = Moving Avg Forecast t = time
22
Problem – Simple Moving Average Method
Month Demand (D) Jan Feb Mar Apr May Jun Jul Aug Sep Oct 450 440 460 510 520 495 475 560 510 520
Forecast using 3 month and 6 month moving average and determine which is a better forecast
Nov
Dec
540
550
23
Problem – Simple Moving Average Method
Month Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec Demand (D) 450 440 460 510 520 495 475 560 510 520 540 550 Moving Average (3 mth) 450 470 497 508 497 510 515 530 523 537
24
Problem – Simple Moving Average Method
Month Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec Demand (D) 450 440 460 510 520 495 475 560 510 520 540 550 Moving Average (3 mth) 450 470 497 508 497 510 515 530 523 537 Moving Average Forecast (f)(3 mth) 450 470 497 508 497 510 515 530 523
25
Problem – Simple Moving Average Method
Month Demand (D) Moving Average (3 mth) 450 470 497 508 Moving Average Forecast (f)(3 mth) 450 470 497 Deviation
Jan Feb Mar Apr May Jun
450 440 460 510 520 495
60 50 -2
Jul
Aug Sep Oct
475
560 510 520
497
510 515 530
508
497 510 515
-33
63 0 5
Nov Dec
540 550
523 537
530 523
10 27
26
Problem – Simple Moving Average Method
Month Jan Feb Mar Apr May Jun Jul Aug Sep Demand (D) 450 440 460 510 520 495 475 560 510 MA (3 mth) 450 470 497 508 497 510 515 F (3 mth) 450 470 497 508 497 510 Deviation 60 50 -2 -33 63 0
MAD = ?I Deviation I
N
MAD = 250 / 9 MAD = 27.78 RSFE = ? Deviation RSFE = 180
Oct
Nov
520
540
530
523
515
530
5
10
Tracking Signal = RSFE / MAD
= 6.48
27
Dec
550
537
523
27
Problem – Simple Moving Average Method
Month Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec Demand (D) 450 440 460 510 520 495 475 560 510 520 540 550 Moving Average (6 mth) 479 483 503 512 513 517 526
28
Problem – Simple Moving Average Method
Month Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec Demand (D) 450 440 460 510 520 495 475 560 510 520 540 550 Moving Average (6 mth) 479 483 503 512 513 517 526 Moving Average Forecast(6 mth) 479 483 503 512 513 517
29
Problem – Simple Moving Average Method
Month Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec Demand (D) 450 440 460 510 520 495 475 560 510 520 540 550 Moving Average (6 mth) 479 483 503 512 513 517 526 Moving Average Forecast(6 mth) 479 483 503 512 513 517 Deviation -4 77 7 8 27 33
30
Problem – Simple Moving Average Method
Month Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec Demand (D) 450 440 460 510 520 495 475 560 510 520 540 550 MA (6mth) 479 483 503 512 513 517 526 F (6mth) 479 483 503 512 513 517 Deviation -4 77 7 8 27 33
31
MAD = ?I Deviation I N MAD = 156 / 6 MAD = 26
RSFE = ? Deviation
RSFE = 148
Tracking Signal = RSFE / MAD = 5.7
Problem – Simple Moving Average Method
• Since Tracking Signal of 6 month moving average is more closer to zero it is a better forecasting technique
32
Weighted Moving Average - Formula
• Weighted Moving Average = ? Ct Dt
n
t=1
Where, Ct = Fraction used as weight for period t 0 ? Ct ?1
33
Problem – Weighted Moving Average
Month Demand (D) Moving Average (3 mth) 450 470 497 508 497 510 Moving Average Forecast (3 mth) 450 470 497 508 497
Jan Feb Mar Apr May Jun Jul Aug
450 440 460 510 520 495 475 560
Since it is a 3 month moving average, assume values of:
C1 = 0.25
C2 = 0.25 C3 = 0.5
Sep
Oct Nov Dec
510
520 540 550
515
530 523 537
510
515 530 523
34
Problem – Weighted Moving Average
Month Demand (D) MA (3 mth) 450 MA Forecast (3 mth) 3 mnth WMA 453
Jan Feb Mar
450 440 460
Apr
May Jun Jul Aug
510
520 495 475 560
470
497 508 497 510
450
470 497 508 497
480
503 505 491 523
Sep
Oct Nov Dec
510
520 540 550
515
530 523 537
510
515 530 523
514
528 528 541
35
Problem – Weighted Moving Average
Month Demand (D) MA (3 mth) 450 MA Forecast (3 mth) 3 mnth WMA 453 3 mnth WMA forecast -
Jan Feb Mar
450 440 460
Apr
May Jun Jul Aug
510
520 495 475 560
470
497 508 497 510
450
470 497 508 497
480
503 505 491 523
453
480 503 505 491
Sep
Oct Nov Dec
510
520 540 550
515
530 523 537
510
515 530 523
514
528 528 541
523
514 528 528
36
Practice Problem
Month Jan Feb Mar Apr May Jun Jul Aug Sep Demand (D) 125 135 130 120 115 140 135 110 120
Use Simple Moving Average and Weighted Moving average method for 2 months. Forecast and compare two methods. Assume appropriate values
Oct
Nov
120
140
Dec
145
37
Simple Exponential Smoothing - Formula
Ft = F t-1 + ?(Dt - Ft-1)
ft = Ft-1
OR
? (Dt) + (1 – ? ) Ft-1
Where, F = Simple Exponential average f = Forecast for time t D = Demand ? = Smoothing constant between 0 to 1
38
Problem – Simple Exponential Smoothing
Month Jan Feb Demand 97 93
A firm uses exponential smoothing method for forecasting. Try ? = 0.1 & F0 = 100
Mar
Apr May Jun Jul
110
98 104 103 99
Aug
Sep Oct Nov Dec
108
106 94 109 95
39
Problem – Simple Exponential Smoothing
Month Demand Exponential Avg (Ft) 100 Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec 97 93 110 98 104 103 99 108 106 94 109 95 99.7 99.03 100.73 99.91 100.32 100.60 100.44 101.20 101.68 82.11 84.8 85.82
40
Problem – Simple Exponential Smoothing
Month Demand Exponential Avg (Ft) 100 Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec 97 93 110 98 104 103 99 108 106 94 109 95 99.7 99.03 100.73 99.91 100.32 100.60 100.44 101.20 101.68 82.11 84.8 85.82 100 99.7 99.03 100.73 99.91 100.32 100.60 100.44 101.20 101.68 82.11 84.8
41
Forecast (ft)
Practice Problem
• A firm uses exponential smoothing method for forecasting, with ? = 0.2 The forecast for month of March was 500 units but actual demand turned out to be 460. Forecast demand in April. ft = F t-1 i.e f APR = F MAR Ft = ? (Dt) + (1 – ? ) Ft-1 FMAR = ? (DMAR) + (1 – ? ) FFEB F MAR = 0.2 (460) + (1-0.2) 500 F MAR = 492
42
Practice Problem
Month Jan Demand 122
Feb
Mar Apr May Jun Jul Aug Sep Oct Nov Dec
127
125 126 139 127 134 128 134 136 132 131
Given in table is the data for 1994. F0 = 150. Try ? = 0.1 and 0.3, which is a better value?
43
Exponential Smoothing with Trend (Winter’s)
Ft = ? (Dt) + (1 – ? ) (Ft-1 + Tt-1) Tt = ? (Ft – Ft-1) + (1 – ?) Tt-1 ft = Ft-1 + Tt-1
Where, F = Winters Exponential average f = Forecast for time t D = Demand ? = Smoothing constant between 0 to 1 T = Trend estimate at time t ? = Averaging fraction
44
Problem – Exponential smoothing with trend
Month Demand
Calculate forecast with: ? = 0.2 ? = 0.2 F0 = 480 T0 = 9
Jan
Feb Mar Apr May
460
510 520 495 475
Jun
Jul Aug Sep Oct
560
510 520 540 550
Nov
Dec
555
569
45
Problem – Exponential smoothing with trend
Month Demand Winter’s exp avg (Ft) Trend (Tt)
480
Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec 460 510 520 495 475 560 510 520 540 550 555 569 483.20 494.83 506.74 511.80 511.18 526.23 529.62 533.55 540.16 547.43 554.35 562.72
9
7.84 8.60 9.26 8.42 6.61 8.30 7.32 6.64 6.63 6.76 6.79 7.11
46
Problem – Exponential smoothing with trend
Month Demand Winter’s exp avg (Ft) Trend (Tt) Winter’s Forecast (ft) 489.00 491.04 503.43 516.00 520.22 517.79 534.53 536.94 540.20 546.79 554.19 561.15
47
480
Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec 460 510 520 495 475 560 510 520 540 550 555 569 483.20 494.83 506.74 511.80 511.18 526.23 529.62 533.55 540.16 547.43 554.35 562.72
9
7.84 8.60 9.26 8.42 6.61 8.30 7.32 6.64 6.63 6.76 6.79 7.11
Practice problem
Month Jan Feb Mar Apr May Jun Jul Aug Sep Oct Demand 128 136 137 141 157 148 158 155 164 169
Given data is for year 1994. Calculate forecast with: ? = 0.2 ? = 0.05 F0 = 130 T0 = 0
Nov
Dec
168
160
48
Exponential smoothing with seasonality
Ft = ? It = ? Dt It-m Dt Ft + (1-?) Ft-1 + (1-?) It-m
ft+1 = Ft x It+1-m Where; It-m = Index calculated m=12 months ago for monthly forecast, m=4 quarters ago for quarterly forecast ? = smoothing constant, normally ? 0.05
49
Prob - Exponential smoothing with seasonality
Demand Month Jan Feb Mar Apr May Jun Jul Aug Sep Oct 1993 80 75 80 90 115 110 100 90 85 75 1994 100 85 90 110 131 120 110 110 95 85
The table shows demand data of 1993 & 1994. Forecast for the year 1995. Other data: ? = 0.1 , ? = 0.05 , FDEC94 = 94 Next Step: Calculate average demand of 1993 & 1994 and average monthly demand
Nov
Dec
75
80
85
80
50
Prob - Exponential smoothing with seasonality
Demand Month Jan Feb Mar Apr 1993 80 75 80 90 1994 100 85 90 110 Average Demand 90 80 85 100
Avg Monthly Demand = 1128 / 12 = 94
May
Jun Jul Aug Sep Oct Nov Dec
115
110 100 90 85 75 75 80
131
120 110 110 95 85 85 80
123
115 105 100 90 80 80 80
Next Step: Calculate Seasonal Index It = Dt / Ft
51
Prob - Exponential smoothing with seasonality
Demand Month Jan Feb Mar Apr 1993 80 75 80 90 1994 100 85 90 110 Average Demand 90 80 85 100 Seasonality Index (It) 0.957 0.851 0.904 1.064
The demand for 1995 is given as:
May
Jun Jul Aug Sep Oct Nov Dec
115
110 100 90 85 75 75 80
131
120 110 110 95 85 85 80
123
115 105 100 90 80 80 80
1.309
1.223 1.117 1.064 0.957 0.851 0.851 0.851
52
Prob - Exponential smoothing with seasonality
Month Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec Seasonality Index (It-m) 0.957 0.851 0.904 1.064 1.309 1.223 1.117 1.064 0.957 0.851 0.851 0.851 Demand (Dt) 1995 95 75 90 105 120 117 102 98 95 75 85 75
53
Next Step: Calculate Ft for 1995 using formula Ft = ? Dt It-m + (1-?) Ft-1
Prob - Exponential smoothing with seasonality
Month Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec Seasonality Index (It-m) 0.957 0.851 0.904 1.064 1.309 1.223 1.117 1.064 0.957 0.851 0.851 0.851 Demand (Dt) 1995 95 75 90 105 120 117 102 98 95 75 85 75 Average (Ft) 94.50 93.86 94.43 94.86 94.54 94.65 94.32 94.10 94.62 93.37 94.56 93.91
54
Next Step: Calculate Forecast values for 1995 using formula: ft+1 = Ft x It+1-m
Prob - Exponential smoothing with seasonality
Month Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec Seasonality Index (It-m) 0.957 0.851 0.904 1.064 1.309 1.223 1.117 1.064 0.957 0.851 0.851 0.851 Demand (Dt) 1995 95 75 90 105 120 117 102 98 95 75 85 75 Average (Ft) 94.50 93.86 94.43 94.86 94.54 94.65 94.32 94.10 94.62 93.37 94.56 93.91 Forecast (ft) 89.96 80.42 84.45 100.47 124.17 115.62 105.72 100.36 90.05 80.52 79.97 80.47
55
Next Step: Calculate It using formula It = ? Dt + (1-?) It-m Ft
Prob - Exponential smoothing with seasonality
Month Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec Seasonality Index (It-m) 0.957 0.851 0.904 1.064 1.309 1.223 1.117 1.064 0.957 0.851 0.851 0.851 Demand (Dt) 1995 95 75 90 105 120 117 102 98 95 75 85 75 Average (Ft) 94.50 93.86 94.43 94.86 94.54 94.65 94.32 94.10 94.62 93.37 94.56 93.91 Forecast (ft) 89.96 80.42 84.45 100.47 124.17 115.62 105.72 100.36 90.05 80.52 79.97 80.47 New Seasonality Index (It) 0.959 0.848 0.906 1.066 1.307 1.224 1.115 1.063 0.959 0.849 0.853 0.848
56
Practice Problem
Demand Month Jan Feb Mar Apr May Jun Jul Aug Sep Oct 2003 120 115 117 122 125 127 125 122 120 115 2004 130 121 125 122 122 120 125 130 133 130 2005 145 127 132 127 118 115 128 135 140 140
The table shows demand data of 2003, 2004 & 2005. Forecast for the year 2005. Other data: ? = 0.2 , ? = 0.05 , F0 = 130
Nov
Dec
117
120
127
127
135
130
57
Exponential smoothing with seasonality & trend (Winter’s complete model)
Ft = ? Dt It-m + (1-?) (Ft-1 + Tt-1)
Tt = ? (Ft – Ft-1) + (1 – ?) Tt-1 It = ? Dt Ft + (1-?) It-m
ft+1 = (Ft + Tt) x It+1-m
Consider values of ? = 0.2, ? = 0.05 and ? = 0.01
58
Prob – Smoothing with trend & seasonality
Demand ‘95 95 75 90 105 Seasonality Index ‘95 0.959 0.848 0.906 1.066 Demand ‘96 80 85 90 95
Month Jan Feb Mar Apr
Forecast the demand for 1996 where: F0 = 80 T0 = 4.5 ? = 0.2 ? = 0.05 ? = 0.01 Step 1: Calculate smoothing average and trend for each month of 1996
59
May
Jun Jul Aug Sep
120
117 102 98 95
1.307
1.224 1.115 1.063 0.959
100
100 95 95 90
Oct
Nov Dec
75
85 75
0.849
0.853 0.848
95
85 80
Prob – Smoothing with trend & seasonality
Month Demand ‘95 95 75 90 It-m Demand ‘96 80 85 90 Average Ft 80 Jan Feb Mar 0.959 0.848 0.906 84.284 91.066 96.403 Trend Tt 4.5 4.489 4.604 4.641
Now do forecasting for 1996 and calculate new seasonal index
Apr
May Jun Jul Aug Sep Oct Nov Dec
105
120 117 102 98 95 75 85 75
1.066
1.307 1.224 1.115 1.063 0.959 0.849 0.853 0.848
95
100 100 95 95 90 95 85 80
98.659
97.846 98.020 98.697 99.937 101.719 106.676 108.243 108.350
4.521
4.255 4.051 3.882 3.750 3.651 3.717 3.609 3.434
60
Prob – Smoothing with trend & seasonality
Month Demand ‘95 95 75 90 It-m Demand ‘96 80 85 90 Average Ft 80 Jan Feb Mar 0.959 0.848 0.906 84.284 91.066 96.403 Trend Tt 4.5 4.489 4.604 4.641 81.036 75.280 86.677 0.959 0.849 0.906 Forecast ’96 (ft) New Index (It)
Apr
May Jun Jul Aug Sep Oct Nov Dec
105
120 117 102 98 95 75 85 75
1.066
1.307 1.224 1.115 1.063 0.959 0.849 0.853 0.848
95
100 100 95 95 90 95 85 80
98.659
97.846 98.020 98.697 99.937 101.719 106.676 108.243 108.350
4.521
4.255 4.051 3.882 3.750 3.651 3.717 3.609 3.434
107.713
134.856 124.971 113.809 109.041 99.436 89.460 94.165 94.851
1.065
1.304 1.222 1.113 1.062 0.958 0.849 0.852
61 0.847
Problem – Smoothing with trend & seasonality
1992 1993 1994
Quarter 1
Quarter 2 Quarter 3 Quarter 4
146
96 59 133
192
127 79 186
272
155 98 219
Estimate forecast for 1995 using winter’s complete model with ? = 0.2 , ? = 0.1 and ? = 0.05
62
Problem – Smoothing with trend & seasonality
Ft = ? Dt It-m + (1-?) (Ft-1 + Tt-1)
Tt = ? (Ft – Ft-1) + (1 – ?) Tt-1 It = ? Dt Ft + (1-?) It-m
ft+1 = (Ft + Tt) x It+1-m
Here, F0, T0 and It-m (i.e I -3) are unknown. Lets calculate it first.
63
Problem – Smoothing with trend & seasonality
F0 = D – T0 (2.5) And F0 = D – T0 (6.5) for quarterly data for monthly data
Lets Calculate D for year 1992 and 1993 D1992 = 108.5 and D1993 = 146 Here we see the upward trend movement from 1992 to 1993 is = 146 – 108.5 = 37.5, hence quarterly movement (T0) = 9.38 So, F0 = 108.5 – 9.38 (2.5) = 85.05
64
Problem – Smoothing with trend & seasonality
Now, lets calculate the trend line sales estimate for 1992 & 1993 1992 Q1 = F0 + T0(1) = 85.05 + 9.38 = 94.43 1992 Q2 = F0 + T0(2) = 85.05 + 9.38 (2) = 103.81 and so on
1992
Quarter 1 Quarter 2 Quarter 3 Quarter 4 94.43 103.81 113.19 122.57
1993
131.95 141.33
From these trend estimates (table) we can develop initial seasonal indices as:
Index = Demand / Estimate
150.71 160.09
65
Problem – Smoothing with trend & seasonality
Index for Q1 of 1992 = 146 / 94.43 = 1.55
Q2 of 1992 = 96 / 103.81 = 0.92 and so on
1992 Quarter 1 Quarter 2 Quarter 3 Quarter 4 1.55 0.92 0.52 1.09 1993 1.46 0.90 0.52 1.13
Lets check our indices are correct or not
To check, take average of indices for 1992 and 1993 and calculate ? average
66
Problem – Smoothing with trend & seasonality
1992 Quarter 1 Quarter 2 Quarter 3 1.55 0.92 0.52 1993 1.46 0.90 0.52 Average 1.51 0.91 0.52
Quarter 4
1.09
1.13 ? Average
1.13 4.07
? Average = 4.07. It should have been 4, so there is a mistake in calculated indices. Lets introduce a correction factor and recalculate the indices
67
Problem – Smoothing with trend & seasonality
Correction factor = (4 / 4.07) = 0.983. Now recalculate the indices
1992 Quarter 1 Quarter 2 1.55 0.92 1993 1.46 0.90 Average 1.51 0.91 I t-m 1.51 x 0.983 = 1.48 0.91 x 0.983 = 0.89
Quarter 3
Quarter 4
0.52
1.09
0.52
1.13
0.52
1.13
0.52 x 0.983 = 0.51
1.13 x 0.983 = 1.11
Now we have values of all the unknowns F0, T0 and It-m (i.e I-3) and we can calculate Ft, Tt, It and also forecast for 1995
68
Problem – Smoothing with trend & seasonality
F1 = 0.2 (146 / 1.48) + (1 – 0.8)(85.05 + 9.38) = 95.27 T1 = 0.1 (95.27 – 85.05) + (1 – 0.1) 9.38 = 9.46
I1 = 0.05 (146 / 95.27) + (1 – 0.05) 1.48 = 1.48
. . . And so on till the value of F12, T12 and I12
69
Problem – Smoothing with trend & seasonality
Ft 0 1 2 85.05 95.27 105.36 Tt 9.38 9.46 9.53 It 1.11 1.48 0.89
3
4 5 6 7 8 9 10 11 12
115.05
123.64 132.37 141.90 152.01 162.74 174.60 182.31 191.92 199.92
9.54
9.45 9.38 9.39 9.46 9.59 9.82 9.61 9.61 9.48
0.51
1.11 1.48 0.89 0.51 1.11 1.48 0.90 0.52 1.11
Lets forecast for 1995
70
Problem – Smoothing with trend & seasonality
Forecast for Q1 of 1995 = f13 = (199.92 + 9.48) * 1.49 = 312 Forecast for Q2 of 1995 = f14 = (199.92 + (2 x 9.48)) * 0.90 = 197 Forecast for Q3 of 1995 = f15 = (199.92 + (3 x 9.48)) * 0.52 = 119 Forecast for Q4 of 1995 = f16 = (199.92 + (4 x 9.48)) * 1.11 = 264
71
Practice Problem
Month Jan Feb Demand’ 93 Forecast ‘93 97 93 110 98 130 133 129 138 136 124 100 100 100 100 102 104 106 108 110 112 Demand ‘94 78 0 55 75 87 0 73 0 0 0
Calculate Winter’s trend ratio and seasonality index. What is the forecast for Q1 of 1995?
Mar
Apr May Jun Jul
Aug
Sep Oct
Nov
Dec
139
125
114
116
0
53
72
doc_938724223.pptx
Demand Management and it contains topics like what is demand, what is forecasting, methods of forecasting, typical forecasting program, error monitoring.
Demand Management
Demand
• Planning as most important function of management • Demand Management deals with both consumer needs and supplier coordination • Purpose of demand management is to coordinate and control all resources to efficiently utilise systems
• Demands has to be forecasted to run businesses profitably
2
Forecasting
• Many business decisions depend on some sort of forecasting • Forecasting is scientifically calculated guess, it forms the basis of planning • Levels of forecasting - Short term (upto 1 yr) Medium term (1 to 3 yrs) and Long term (over 5 yrs) • Extrinsic forecast & intrinsic forecast • Elements of forecasting
? ?
Internal factors (past, present and future) External factors (controllable & uncontrollable)
3
Methods of Forecasting
• Qualitative Techniques
? ? ? ?
Market research Panel Consensus History Analogy Delphi Method
Linear Regression Multiple Regression Input Output Analysis End use analysis
4
• Casual Methods
? ? ? ?
• Simulation – Monte Carlo Simulation
Methods of Forecasting . . contd
• Time Series
? ?
Extrapolation – Used when past data is linear Moving Average – Used when data is cyclic
o o
Simple – Average of specified past period is considered Weighted – different weights are assigned to past data
?
Exponential Smoothing – weightage decreased exponentially
o o
o
Simple Exponential Smoothing Trend adjusted Seasonality considered
5
Forecasting programme for any company
• Observing, listing and studying external factors (cultural, social, political, technological) nationally & internationally • Gather info on internal company policies (changes in design, quality, sales) & their effect on demand • Analyse data to establish various relationships and their relative effects of each factors on final demand • Create various scenarios assuming certain happenings in external environment & alternative internal policies • Operationally apply the forecast by breaking it down on the number of product lines. Do Break Even analysis • Regularly monitor forecast errors & update method
6
Prerequisites & Pitfalls in Forecasting
• Define the purpose of forecasting, this will help to decide the accuracy & type of technique to be chosen • It should be combined effort organisationally • Forecasting technique will vary depending on the Product Life Cycle and stage of the product
? ?
New product development stage – Delphi, PDMT Steady state of PLC – Time series with trend & seasonality
• Quite often wrong things are forecasted • People go for minute forecasting of odd products • No timely tracking of forecasting
7
Range & Precision of Forecast
• Forecast may be in terms of ranges
? ?
Higher range – low value, low turnover items in inventory Lower range – Capital intensive machineries
• It will also guide the company form its strategic posture • Precision in forecasting is not as important as proper use of available data • It is better to use the available data according to situational demands
8
Technology Forecasting
• • • • • It deals with estimation of future trends in technology Helps making current decisions examining future choices Guides wide range of long term planning process Forecasting tool at micro level corporate planning also Very effective for high range technology products as gestation time to become productive is quite long • Exploratory technology forecasting – Predicting future with present trends & capabilities • Normative Forecasting – Set goals & objectives of future technology and take necessary current action
9
Uses of Technology Forecasting
In planning of future discoveries & technologies
Government
Planning
Industry
Corporate Planning
Universities
Future Academic Roles
Individuals
Selection of fertile areas of research
Policy formation for the allocation of resources
Investment in production
Investment in research & development
10
Selecting a forecasting method
• Choice depends on
? ? ? ? ?
Purpose of forecasting Type and amount of data available Time horizon of forecast Degree of accuracy required Cost involved
11
Forecast Error Monitoring - MAD
•Mean Absolute Deviation (MAD)
Absolute means the positive & negative signs are ignored ?Deviation is difference between forecast & actuals
?
Period 1 2 3 4 5
Forecast Demand 900 1000 1050 1010 980
Actual Demand 1000 1100 1000 960 970
Deviation 100 100 -50 -50 - 15
MAD = ?I Deviation I = N
315 = 63 5
12
Forecast Error Monitoring - RSFE
•Running sum of forecast errors (RSFE) ?This is algebraic sum of forecasting errors (deviation)
Period 1 2 Forecast Demand 900 1000 Actual Demand 1000 1100 Deviation 100 100
3
4 5
1050
1010 980
1000
960 970
-50
-50 -15
RSFE = ?Deviation = 85
13
Forecast Error Monitoring - Formulas
Mean Square Error (MSE) = ? (Deviation)² N Percentage Error (PE) = (Deviation / Demand) x 100 Mean Absolute Percentage Error = ? I PE I
N
Tracking Signal = RSFE MAD
14
Forecast Error Monitoring . . contd
• RSFE is calculated to determine whether or not the forecast has positive or negative bias • MAD indicates the volume or amplitude of deviation from actuals • Both the bias and amplitude of forecast errors are important • It is important to monitor MAD & Tracking signal for any modifications to be made in original forecasting model • A good forecast should have approximately as much positive as negative deviation
15
Problem
Forecast 100 Demand 110
Table shows actual & forecasted demand. Calculate: - MAD - MSE - MAPE - Tracking Signal
90
80 85 75 85 65
85
88 95 65 80 52
16
Problem
Forecast 100 Demand 110 Deviation 10
90
80 85 75 85 65
85
88 95 65 80 52
-5
8 10 -10 -5 -13
17
Problem
Forecast 100 Demand 110 Deviation 10 (Deviation)² 100
90
80 85 75 85 65
85
88 95 65 80 52
-5
8 10 -10 -5 -13
25
64 100 100 25 169
18
Problem
Forecast 100 90 80 85 75 85 Demand 110 85 88 95 65 80 Deviation 10 -5 8 10 -10 -5 (Deviation)² 100 25 64 100 100 25 Percentage Error 9.09 -5.88 9.09 10.52 -15.38 -6.25
65
52
-13
169
-25
19
Problem
MAD =? I Deviation I N MSE = ? (Deviation)² = 61 7 = 8.71
= 583
= 83.29
7
7
MAPE = ? I PE I = 81.23 = 11.6% N 7
Tracking Signal = RSFE
MAD
= -5
8.71
= -0.57
20
Practice Problem
Forecast 150 125 130 145 Demand 160 130 135 150
Table shows actual & forecasted demand. Calculate: - MAD
180
170 165 155
160
165 145 150
- MSE
- MAPE - Tracking Signal
155
150 150 160
155
160 165 160
21
Simple Moving Average - Formula
( MA ) t = Dt + Dt-1 + Dt-2 + . . . . Dt-n+1
n
( MA ) t = ( f ) t+1 Where; MA = Moving Average f = Moving Avg Forecast t = time
22
Problem – Simple Moving Average Method
Month Demand (D) Jan Feb Mar Apr May Jun Jul Aug Sep Oct 450 440 460 510 520 495 475 560 510 520
Forecast using 3 month and 6 month moving average and determine which is a better forecast
Nov
Dec
540
550
23
Problem – Simple Moving Average Method
Month Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec Demand (D) 450 440 460 510 520 495 475 560 510 520 540 550 Moving Average (3 mth) 450 470 497 508 497 510 515 530 523 537
24
Problem – Simple Moving Average Method
Month Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec Demand (D) 450 440 460 510 520 495 475 560 510 520 540 550 Moving Average (3 mth) 450 470 497 508 497 510 515 530 523 537 Moving Average Forecast (f)(3 mth) 450 470 497 508 497 510 515 530 523
25
Problem – Simple Moving Average Method
Month Demand (D) Moving Average (3 mth) 450 470 497 508 Moving Average Forecast (f)(3 mth) 450 470 497 Deviation
Jan Feb Mar Apr May Jun
450 440 460 510 520 495
60 50 -2
Jul
Aug Sep Oct
475
560 510 520
497
510 515 530
508
497 510 515
-33
63 0 5
Nov Dec
540 550
523 537
530 523
10 27
26
Problem – Simple Moving Average Method
Month Jan Feb Mar Apr May Jun Jul Aug Sep Demand (D) 450 440 460 510 520 495 475 560 510 MA (3 mth) 450 470 497 508 497 510 515 F (3 mth) 450 470 497 508 497 510 Deviation 60 50 -2 -33 63 0
MAD = ?I Deviation I
N
MAD = 250 / 9 MAD = 27.78 RSFE = ? Deviation RSFE = 180
Oct
Nov
520
540
530
523
515
530
5
10
Tracking Signal = RSFE / MAD
= 6.48
27
Dec
550
537
523
27
Problem – Simple Moving Average Method
Month Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec Demand (D) 450 440 460 510 520 495 475 560 510 520 540 550 Moving Average (6 mth) 479 483 503 512 513 517 526
28
Problem – Simple Moving Average Method
Month Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec Demand (D) 450 440 460 510 520 495 475 560 510 520 540 550 Moving Average (6 mth) 479 483 503 512 513 517 526 Moving Average Forecast(6 mth) 479 483 503 512 513 517
29
Problem – Simple Moving Average Method
Month Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec Demand (D) 450 440 460 510 520 495 475 560 510 520 540 550 Moving Average (6 mth) 479 483 503 512 513 517 526 Moving Average Forecast(6 mth) 479 483 503 512 513 517 Deviation -4 77 7 8 27 33
30
Problem – Simple Moving Average Method
Month Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec Demand (D) 450 440 460 510 520 495 475 560 510 520 540 550 MA (6mth) 479 483 503 512 513 517 526 F (6mth) 479 483 503 512 513 517 Deviation -4 77 7 8 27 33
31
MAD = ?I Deviation I N MAD = 156 / 6 MAD = 26
RSFE = ? Deviation
RSFE = 148
Tracking Signal = RSFE / MAD = 5.7
Problem – Simple Moving Average Method
• Since Tracking Signal of 6 month moving average is more closer to zero it is a better forecasting technique
32
Weighted Moving Average - Formula
• Weighted Moving Average = ? Ct Dt
n
t=1
Where, Ct = Fraction used as weight for period t 0 ? Ct ?1
33
Problem – Weighted Moving Average
Month Demand (D) Moving Average (3 mth) 450 470 497 508 497 510 Moving Average Forecast (3 mth) 450 470 497 508 497
Jan Feb Mar Apr May Jun Jul Aug
450 440 460 510 520 495 475 560
Since it is a 3 month moving average, assume values of:
C1 = 0.25
C2 = 0.25 C3 = 0.5
Sep
Oct Nov Dec
510
520 540 550
515
530 523 537
510
515 530 523
34
Problem – Weighted Moving Average
Month Demand (D) MA (3 mth) 450 MA Forecast (3 mth) 3 mnth WMA 453
Jan Feb Mar
450 440 460
Apr
May Jun Jul Aug
510
520 495 475 560
470
497 508 497 510
450
470 497 508 497
480
503 505 491 523
Sep
Oct Nov Dec
510
520 540 550
515
530 523 537
510
515 530 523
514
528 528 541
35
Problem – Weighted Moving Average
Month Demand (D) MA (3 mth) 450 MA Forecast (3 mth) 3 mnth WMA 453 3 mnth WMA forecast -
Jan Feb Mar
450 440 460
Apr
May Jun Jul Aug
510
520 495 475 560
470
497 508 497 510
450
470 497 508 497
480
503 505 491 523
453
480 503 505 491
Sep
Oct Nov Dec
510
520 540 550
515
530 523 537
510
515 530 523
514
528 528 541
523
514 528 528
36
Practice Problem
Month Jan Feb Mar Apr May Jun Jul Aug Sep Demand (D) 125 135 130 120 115 140 135 110 120
Use Simple Moving Average and Weighted Moving average method for 2 months. Forecast and compare two methods. Assume appropriate values
Oct
Nov
120
140
Dec
145
37
Simple Exponential Smoothing - Formula
Ft = F t-1 + ?(Dt - Ft-1)
ft = Ft-1
OR
? (Dt) + (1 – ? ) Ft-1
Where, F = Simple Exponential average f = Forecast for time t D = Demand ? = Smoothing constant between 0 to 1
38
Problem – Simple Exponential Smoothing
Month Jan Feb Demand 97 93
A firm uses exponential smoothing method for forecasting. Try ? = 0.1 & F0 = 100
Mar
Apr May Jun Jul
110
98 104 103 99
Aug
Sep Oct Nov Dec
108
106 94 109 95
39
Problem – Simple Exponential Smoothing
Month Demand Exponential Avg (Ft) 100 Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec 97 93 110 98 104 103 99 108 106 94 109 95 99.7 99.03 100.73 99.91 100.32 100.60 100.44 101.20 101.68 82.11 84.8 85.82
40
Problem – Simple Exponential Smoothing
Month Demand Exponential Avg (Ft) 100 Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec 97 93 110 98 104 103 99 108 106 94 109 95 99.7 99.03 100.73 99.91 100.32 100.60 100.44 101.20 101.68 82.11 84.8 85.82 100 99.7 99.03 100.73 99.91 100.32 100.60 100.44 101.20 101.68 82.11 84.8
41
Forecast (ft)
Practice Problem
• A firm uses exponential smoothing method for forecasting, with ? = 0.2 The forecast for month of March was 500 units but actual demand turned out to be 460. Forecast demand in April. ft = F t-1 i.e f APR = F MAR Ft = ? (Dt) + (1 – ? ) Ft-1 FMAR = ? (DMAR) + (1 – ? ) FFEB F MAR = 0.2 (460) + (1-0.2) 500 F MAR = 492
42
Practice Problem
Month Jan Demand 122
Feb
Mar Apr May Jun Jul Aug Sep Oct Nov Dec
127
125 126 139 127 134 128 134 136 132 131
Given in table is the data for 1994. F0 = 150. Try ? = 0.1 and 0.3, which is a better value?
43
Exponential Smoothing with Trend (Winter’s)
Ft = ? (Dt) + (1 – ? ) (Ft-1 + Tt-1) Tt = ? (Ft – Ft-1) + (1 – ?) Tt-1 ft = Ft-1 + Tt-1
Where, F = Winters Exponential average f = Forecast for time t D = Demand ? = Smoothing constant between 0 to 1 T = Trend estimate at time t ? = Averaging fraction
44
Problem – Exponential smoothing with trend
Month Demand
Calculate forecast with: ? = 0.2 ? = 0.2 F0 = 480 T0 = 9
Jan
Feb Mar Apr May
460
510 520 495 475
Jun
Jul Aug Sep Oct
560
510 520 540 550
Nov
Dec
555
569
45
Problem – Exponential smoothing with trend
Month Demand Winter’s exp avg (Ft) Trend (Tt)
480
Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec 460 510 520 495 475 560 510 520 540 550 555 569 483.20 494.83 506.74 511.80 511.18 526.23 529.62 533.55 540.16 547.43 554.35 562.72
9
7.84 8.60 9.26 8.42 6.61 8.30 7.32 6.64 6.63 6.76 6.79 7.11
46
Problem – Exponential smoothing with trend
Month Demand Winter’s exp avg (Ft) Trend (Tt) Winter’s Forecast (ft) 489.00 491.04 503.43 516.00 520.22 517.79 534.53 536.94 540.20 546.79 554.19 561.15
47
480
Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec 460 510 520 495 475 560 510 520 540 550 555 569 483.20 494.83 506.74 511.80 511.18 526.23 529.62 533.55 540.16 547.43 554.35 562.72
9
7.84 8.60 9.26 8.42 6.61 8.30 7.32 6.64 6.63 6.76 6.79 7.11
Practice problem
Month Jan Feb Mar Apr May Jun Jul Aug Sep Oct Demand 128 136 137 141 157 148 158 155 164 169
Given data is for year 1994. Calculate forecast with: ? = 0.2 ? = 0.05 F0 = 130 T0 = 0
Nov
Dec
168
160
48
Exponential smoothing with seasonality
Ft = ? It = ? Dt It-m Dt Ft + (1-?) Ft-1 + (1-?) It-m
ft+1 = Ft x It+1-m Where; It-m = Index calculated m=12 months ago for monthly forecast, m=4 quarters ago for quarterly forecast ? = smoothing constant, normally ? 0.05
49
Prob - Exponential smoothing with seasonality
Demand Month Jan Feb Mar Apr May Jun Jul Aug Sep Oct 1993 80 75 80 90 115 110 100 90 85 75 1994 100 85 90 110 131 120 110 110 95 85
The table shows demand data of 1993 & 1994. Forecast for the year 1995. Other data: ? = 0.1 , ? = 0.05 , FDEC94 = 94 Next Step: Calculate average demand of 1993 & 1994 and average monthly demand
Nov
Dec
75
80
85
80
50
Prob - Exponential smoothing with seasonality
Demand Month Jan Feb Mar Apr 1993 80 75 80 90 1994 100 85 90 110 Average Demand 90 80 85 100
Avg Monthly Demand = 1128 / 12 = 94
May
Jun Jul Aug Sep Oct Nov Dec
115
110 100 90 85 75 75 80
131
120 110 110 95 85 85 80
123
115 105 100 90 80 80 80
Next Step: Calculate Seasonal Index It = Dt / Ft
51
Prob - Exponential smoothing with seasonality
Demand Month Jan Feb Mar Apr 1993 80 75 80 90 1994 100 85 90 110 Average Demand 90 80 85 100 Seasonality Index (It) 0.957 0.851 0.904 1.064
The demand for 1995 is given as:
May
Jun Jul Aug Sep Oct Nov Dec
115
110 100 90 85 75 75 80
131
120 110 110 95 85 85 80
123
115 105 100 90 80 80 80
1.309
1.223 1.117 1.064 0.957 0.851 0.851 0.851
52
Prob - Exponential smoothing with seasonality
Month Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec Seasonality Index (It-m) 0.957 0.851 0.904 1.064 1.309 1.223 1.117 1.064 0.957 0.851 0.851 0.851 Demand (Dt) 1995 95 75 90 105 120 117 102 98 95 75 85 75
53
Next Step: Calculate Ft for 1995 using formula Ft = ? Dt It-m + (1-?) Ft-1
Prob - Exponential smoothing with seasonality
Month Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec Seasonality Index (It-m) 0.957 0.851 0.904 1.064 1.309 1.223 1.117 1.064 0.957 0.851 0.851 0.851 Demand (Dt) 1995 95 75 90 105 120 117 102 98 95 75 85 75 Average (Ft) 94.50 93.86 94.43 94.86 94.54 94.65 94.32 94.10 94.62 93.37 94.56 93.91
54
Next Step: Calculate Forecast values for 1995 using formula: ft+1 = Ft x It+1-m
Prob - Exponential smoothing with seasonality
Month Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec Seasonality Index (It-m) 0.957 0.851 0.904 1.064 1.309 1.223 1.117 1.064 0.957 0.851 0.851 0.851 Demand (Dt) 1995 95 75 90 105 120 117 102 98 95 75 85 75 Average (Ft) 94.50 93.86 94.43 94.86 94.54 94.65 94.32 94.10 94.62 93.37 94.56 93.91 Forecast (ft) 89.96 80.42 84.45 100.47 124.17 115.62 105.72 100.36 90.05 80.52 79.97 80.47
55
Next Step: Calculate It using formula It = ? Dt + (1-?) It-m Ft
Prob - Exponential smoothing with seasonality
Month Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec Seasonality Index (It-m) 0.957 0.851 0.904 1.064 1.309 1.223 1.117 1.064 0.957 0.851 0.851 0.851 Demand (Dt) 1995 95 75 90 105 120 117 102 98 95 75 85 75 Average (Ft) 94.50 93.86 94.43 94.86 94.54 94.65 94.32 94.10 94.62 93.37 94.56 93.91 Forecast (ft) 89.96 80.42 84.45 100.47 124.17 115.62 105.72 100.36 90.05 80.52 79.97 80.47 New Seasonality Index (It) 0.959 0.848 0.906 1.066 1.307 1.224 1.115 1.063 0.959 0.849 0.853 0.848
56
Practice Problem
Demand Month Jan Feb Mar Apr May Jun Jul Aug Sep Oct 2003 120 115 117 122 125 127 125 122 120 115 2004 130 121 125 122 122 120 125 130 133 130 2005 145 127 132 127 118 115 128 135 140 140
The table shows demand data of 2003, 2004 & 2005. Forecast for the year 2005. Other data: ? = 0.2 , ? = 0.05 , F0 = 130
Nov
Dec
117
120
127
127
135
130
57
Exponential smoothing with seasonality & trend (Winter’s complete model)
Ft = ? Dt It-m + (1-?) (Ft-1 + Tt-1)
Tt = ? (Ft – Ft-1) + (1 – ?) Tt-1 It = ? Dt Ft + (1-?) It-m
ft+1 = (Ft + Tt) x It+1-m
Consider values of ? = 0.2, ? = 0.05 and ? = 0.01
58
Prob – Smoothing with trend & seasonality
Demand ‘95 95 75 90 105 Seasonality Index ‘95 0.959 0.848 0.906 1.066 Demand ‘96 80 85 90 95
Month Jan Feb Mar Apr
Forecast the demand for 1996 where: F0 = 80 T0 = 4.5 ? = 0.2 ? = 0.05 ? = 0.01 Step 1: Calculate smoothing average and trend for each month of 1996
59
May
Jun Jul Aug Sep
120
117 102 98 95
1.307
1.224 1.115 1.063 0.959
100
100 95 95 90
Oct
Nov Dec
75
85 75
0.849
0.853 0.848
95
85 80
Prob – Smoothing with trend & seasonality
Month Demand ‘95 95 75 90 It-m Demand ‘96 80 85 90 Average Ft 80 Jan Feb Mar 0.959 0.848 0.906 84.284 91.066 96.403 Trend Tt 4.5 4.489 4.604 4.641
Now do forecasting for 1996 and calculate new seasonal index
Apr
May Jun Jul Aug Sep Oct Nov Dec
105
120 117 102 98 95 75 85 75
1.066
1.307 1.224 1.115 1.063 0.959 0.849 0.853 0.848
95
100 100 95 95 90 95 85 80
98.659
97.846 98.020 98.697 99.937 101.719 106.676 108.243 108.350
4.521
4.255 4.051 3.882 3.750 3.651 3.717 3.609 3.434
60
Prob – Smoothing with trend & seasonality
Month Demand ‘95 95 75 90 It-m Demand ‘96 80 85 90 Average Ft 80 Jan Feb Mar 0.959 0.848 0.906 84.284 91.066 96.403 Trend Tt 4.5 4.489 4.604 4.641 81.036 75.280 86.677 0.959 0.849 0.906 Forecast ’96 (ft) New Index (It)
Apr
May Jun Jul Aug Sep Oct Nov Dec
105
120 117 102 98 95 75 85 75
1.066
1.307 1.224 1.115 1.063 0.959 0.849 0.853 0.848
95
100 100 95 95 90 95 85 80
98.659
97.846 98.020 98.697 99.937 101.719 106.676 108.243 108.350
4.521
4.255 4.051 3.882 3.750 3.651 3.717 3.609 3.434
107.713
134.856 124.971 113.809 109.041 99.436 89.460 94.165 94.851
1.065
1.304 1.222 1.113 1.062 0.958 0.849 0.852
61 0.847
Problem – Smoothing with trend & seasonality
1992 1993 1994
Quarter 1
Quarter 2 Quarter 3 Quarter 4
146
96 59 133
192
127 79 186
272
155 98 219
Estimate forecast for 1995 using winter’s complete model with ? = 0.2 , ? = 0.1 and ? = 0.05
62
Problem – Smoothing with trend & seasonality
Ft = ? Dt It-m + (1-?) (Ft-1 + Tt-1)
Tt = ? (Ft – Ft-1) + (1 – ?) Tt-1 It = ? Dt Ft + (1-?) It-m
ft+1 = (Ft + Tt) x It+1-m
Here, F0, T0 and It-m (i.e I -3) are unknown. Lets calculate it first.
63
Problem – Smoothing with trend & seasonality
F0 = D – T0 (2.5) And F0 = D – T0 (6.5) for quarterly data for monthly data
Lets Calculate D for year 1992 and 1993 D1992 = 108.5 and D1993 = 146 Here we see the upward trend movement from 1992 to 1993 is = 146 – 108.5 = 37.5, hence quarterly movement (T0) = 9.38 So, F0 = 108.5 – 9.38 (2.5) = 85.05
64
Problem – Smoothing with trend & seasonality
Now, lets calculate the trend line sales estimate for 1992 & 1993 1992 Q1 = F0 + T0(1) = 85.05 + 9.38 = 94.43 1992 Q2 = F0 + T0(2) = 85.05 + 9.38 (2) = 103.81 and so on
1992
Quarter 1 Quarter 2 Quarter 3 Quarter 4 94.43 103.81 113.19 122.57
1993
131.95 141.33
From these trend estimates (table) we can develop initial seasonal indices as:
Index = Demand / Estimate
150.71 160.09
65
Problem – Smoothing with trend & seasonality
Index for Q1 of 1992 = 146 / 94.43 = 1.55
Q2 of 1992 = 96 / 103.81 = 0.92 and so on
1992 Quarter 1 Quarter 2 Quarter 3 Quarter 4 1.55 0.92 0.52 1.09 1993 1.46 0.90 0.52 1.13
Lets check our indices are correct or not
To check, take average of indices for 1992 and 1993 and calculate ? average
66
Problem – Smoothing with trend & seasonality
1992 Quarter 1 Quarter 2 Quarter 3 1.55 0.92 0.52 1993 1.46 0.90 0.52 Average 1.51 0.91 0.52
Quarter 4
1.09
1.13 ? Average
1.13 4.07
? Average = 4.07. It should have been 4, so there is a mistake in calculated indices. Lets introduce a correction factor and recalculate the indices
67
Problem – Smoothing with trend & seasonality
Correction factor = (4 / 4.07) = 0.983. Now recalculate the indices
1992 Quarter 1 Quarter 2 1.55 0.92 1993 1.46 0.90 Average 1.51 0.91 I t-m 1.51 x 0.983 = 1.48 0.91 x 0.983 = 0.89
Quarter 3
Quarter 4
0.52
1.09
0.52
1.13
0.52
1.13
0.52 x 0.983 = 0.51
1.13 x 0.983 = 1.11
Now we have values of all the unknowns F0, T0 and It-m (i.e I-3) and we can calculate Ft, Tt, It and also forecast for 1995
68
Problem – Smoothing with trend & seasonality
F1 = 0.2 (146 / 1.48) + (1 – 0.8)(85.05 + 9.38) = 95.27 T1 = 0.1 (95.27 – 85.05) + (1 – 0.1) 9.38 = 9.46
I1 = 0.05 (146 / 95.27) + (1 – 0.05) 1.48 = 1.48
. . . And so on till the value of F12, T12 and I12
69
Problem – Smoothing with trend & seasonality
Ft 0 1 2 85.05 95.27 105.36 Tt 9.38 9.46 9.53 It 1.11 1.48 0.89
3
4 5 6 7 8 9 10 11 12
115.05
123.64 132.37 141.90 152.01 162.74 174.60 182.31 191.92 199.92
9.54
9.45 9.38 9.39 9.46 9.59 9.82 9.61 9.61 9.48
0.51
1.11 1.48 0.89 0.51 1.11 1.48 0.90 0.52 1.11
Lets forecast for 1995
70
Problem – Smoothing with trend & seasonality
Forecast for Q1 of 1995 = f13 = (199.92 + 9.48) * 1.49 = 312 Forecast for Q2 of 1995 = f14 = (199.92 + (2 x 9.48)) * 0.90 = 197 Forecast for Q3 of 1995 = f15 = (199.92 + (3 x 9.48)) * 0.52 = 119 Forecast for Q4 of 1995 = f16 = (199.92 + (4 x 9.48)) * 1.11 = 264
71
Practice Problem
Month Jan Feb Demand’ 93 Forecast ‘93 97 93 110 98 130 133 129 138 136 124 100 100 100 100 102 104 106 108 110 112 Demand ‘94 78 0 55 75 87 0 73 0 0 0
Calculate Winter’s trend ratio and seasonality index. What is the forecast for Q1 of 1995?
Mar
Apr May Jun Jul
Aug
Sep Oct
Nov
Dec
139
125
114
116
0
53
72
doc_938724223.pptx