5, 6 & 7 - demand management
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Demand Management
2
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
3
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)
4
Methods of Forecasting
• Qualitative Techniques Market research Panel Consensus History Analogy Delphi Method
• Casual Methods Linear Regression Multiple Regression Input Output Analysis End use analysis
• Simulation – Monte Carlo Simulation
5
Methods of Forecasting . . contd
• Time Series Extrapolation – Used when past data is linear Moving Average – Used when data is cyclic
o Simple – Average of specified past period is consideredo Weighted – different weights are assigned to past data
Exponential Smoothing – weightage decreased exponentially
o Simple Exponential Smoothingo Trend adjusted o Seasonality considered
6
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
7
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
8
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
9
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
10
Uses of Technology Forecasting
In planning of future discoveries & technologies
Government IndividualsUniversitiesIndustry
Planning
Policy formation for the allocation of resources
Corporate Planning
Investment in production
Investment in research & development
Future Academic Roles
Selection of fertile areas of research
11
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
12
Forecast Error Monitoring - MAD
•Mean Absolute Deviation (MAD)Absolute means the positive & negative signs are ignoredDeviation is difference between forecast & actuals
Period Forecast Demand Actual Demand Deviation
1 900 1000 100
2 1000 1100 100
3 1050 1000 -50
4 1010 960 -50
5 980 970 - 15
MAD = ∑I Deviation I = 315 = 63
N 5
13
Forecast Error Monitoring - RSFE
•Running sum of forecast errors (RSFE)This is algebraic sum of forecasting errors (deviation)
Period Forecast Demand Actual Demand Deviation
1 900 1000 100
2 1000 1100 100
3 1050 1000 -50
4 1010 960 -50
5 980 970 -15
RSFE = ∑Deviation = 85
14
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
15
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
16
Problem
Forecast Demand
100 110
90 85
80 88
85 95
75 65
85 80
65 52
Table shows actual & forecasted demand. Calculate:
- MAD
- MSE
- MAPE
- Tracking Signal
17
Problem
Forecast Demand Deviation
100 110 10
90 85 -5
80 88 8
85 95 10
75 65 -10
85 80 -5
65 52 -13
18
Problem
Forecast Demand Deviation (Deviation)²
100 110 10 100
90 85 -5 25
80 88 8 64
85 95 10 100
75 65 -10 100
85 80 -5 25
65 52 -13 169
19
Problem
Forecast Demand Deviation (Deviation)² Percentage Error
100 110 10 100 9.09
90 85 -5 25 -5.88
80 88 8 64 9.09
85 95 10 100 10.52
75 65 -10 100 -15.38
85 80 -5 25 -6.25
65 52 -13 169 -25
20
Problem
MAD =∑ I Deviation I = 61 = 8.71
N 7
MSE = ∑ (Deviation)² = 583 = 83.29
7 7
MAPE = ∑ I PE I = 81.23 = 11.6%
N 7
Tracking Signal = RSFE = -5 = -0.57
MAD 8.71
21
Practice Problem
Forecast Demand
150 160
125 130
130 135
145 150
180 160
170 165
165 145
155 150
155 155
150 160
150 165
160 160
Table shows actual & forecasted demand. Calculate:
- MAD
- MSE
- MAPE
- Tracking Signal
22
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
23
Problem – Simple Moving Average Method
Month Demand (D)
Jan 450
Feb 440
Mar 460
Apr 510
May 520
Jun 495
Jul 475
Aug 560
Sep 510
Oct 520
Nov 540
Dec 550
Forecast using 3 month and 6 month moving average and determine which is a better forecast
24
Problem – Simple Moving Average Method
Month Demand (D) Moving Average
(3 mth)
Jan 450 -
Feb 440 -
Mar 460 450
Apr 510 470
May 520 497
Jun 495 508
Jul 475 497
Aug 560 510
Sep 510 515
Oct 520 530
Nov 540 523
Dec 550 537
25
Problem – Simple Moving Average Method
Month Demand (D) Moving Average (3 mth)
Moving Average Forecast (f)(3 mth)
Jan 450 - -
Feb 440 - -
Mar 460 450 -
Apr 510 470 450
May 520 497 470
Jun 495 508 497
Jul 475 497 508
Aug 560 510 497
Sep 510 515 510
Oct 520 530 515
Nov 540 523 530
Dec 550 537 523
26
Problem – Simple Moving Average Method
Month Demand (D) Moving Average (3 mth)
Moving Average Forecast (f)(3 mth)
Deviation
Jan 450 - - -
Feb 440 - - -
Mar 460 450 - -
Apr 510 470 450 60
May 520 497 470 50
Jun 495 508 497 -2
Jul 475 497 508 -33
Aug 560 510 497 63
Sep 510 515 510 0
Oct 520 530 515 5
Nov 540 523 530 10
Dec 550 537 523 27
27
Problem – Simple Moving Average Method
Month Demand (D) MA (3 mth) F (3 mth) Deviation
Jan 450 - - -
Feb 440 - - -
Mar 460 450 - -
Apr 510 470 450 60
May 520 497 470 50
Jun 495 508 497 -2
Jul 475 497 508 -33
Aug 560 510 497 63
Sep 510 515 510 0
Oct 520 530 515 5
Nov 540 523 530 10
Dec 550 537 523 27
MAD = ∑I Deviation I
N
MAD = 250 / 9
MAD = 27.78
RSFE = ∑ Deviation
RSFE = 180
Tracking Signal = RSFE / MAD
= 6.48
28
Problem – Simple Moving Average Method
Month Demand (D) Moving Average
(6 mth)
Jan 450 -
Feb 440 -
Mar 460 -
Apr 510 -
May 520 -
Jun 495 479
Jul 475 483
Aug 560 503
Sep 510 512
Oct 520 513
Nov 540 517
Dec 550 526
29
Problem – Simple Moving Average Method
Month Demand (D) Moving Average
(6 mth)
Moving Average Forecast(6 mth)
Jan 450 - -
Feb 440 - -
Mar 460 - -
Apr 510 - -
May 520 - -
Jun 495 479 -
Jul 475 483 479
Aug 560 503 483
Sep 510 512 503
Oct 520 513 512
Nov 540 517 513
Dec 550 526 517
30
Problem – Simple Moving Average Method
Month Demand (D) Moving Average
(6 mth)
Moving Average Forecast(6 mth)
Deviation
Jan 450 - - -
Feb 440 - - -
Mar 460 - - -
Apr 510 - - -
May 520 - - -
Jun 495 479 - -
Jul 475 483 479 -4
Aug 560 503 483 77
Sep 510 512 503 7
Oct 520 513 512 8
Nov 540 517 513 27
Dec 550 526 517 33
31
Problem – Simple Moving Average Method
Month Demand (D) MA (6mth)
F (6mth) Deviation
Jan 450 - - -
Feb 440 - - -
Mar 460 - - -
Apr 510 - - -
May 520 - - -
Jun 495 479 - -
Jul 475 483 479 -4
Aug 560 503 483 77
Sep 510 512 503 7
Oct 520 513 512 8
Nov 540 517 513 27
Dec 550 526 517 33
MAD = ∑I Deviation I
N
MAD = 156 / 6
MAD = 26
RSFE = ∑ Deviation
RSFE = 148
Tracking Signal = RSFE / MAD
= 5.7
32
Problem – Simple Moving Average Method
• Since Tracking Signal of 6 month moving average is more closer to zero it is a better forecasting technique
33
Weighted Moving Average - Formula
• Weighted Moving Average
= ∑ Ct Dt
Where,
Ct = Fraction used as weight for period t
0 ≤ Ct ≤1
t = 1
n
34
Problem – Weighted Moving Average
Month Demand (D) Moving
Average
(3 mth)
Moving Average Forecast
(3 mth)
Jan 450 - -
Feb 440 - -
Mar 460 450 -
Apr 510 470 450
May 520 497 470
Jun 495 508 497
Jul 475 497 508
Aug 560 510 497
Sep 510 515 510
Oct 520 530 515
Nov 540 523 530
Dec 550 537 523
Since it is a 3 month moving average, assume values of:
C1 = 0.25
C2 = 0.25
C3 = 0.5
35
Problem – Weighted Moving Average
Month Demand (D) MA
(3 mth)
MA Forecast
(3 mth)
3 mnth WMA
Jan 450 - - -
Feb 440 - - -
Mar 460 450 - 453
Apr 510 470 450 480
May 520 497 470 503
Jun 495 508 497 505
Jul 475 497 508 491
Aug 560 510 497 523
Sep 510 515 510 514
Oct 520 530 515 528
Nov 540 523 530 528
Dec 550 537 523 541
36
Problem – Weighted Moving Average
Month Demand (D) MA
(3 mth)
MA Forecast
(3 mth)
3 mnth WMA
3 mnth WMA forecast
Jan 450 - - - -
Feb 440 - - - -
Mar 460 450 - 453 -
Apr 510 470 450 480 453
May 520 497 470 503 480
Jun 495 508 497 505 503
Jul 475 497 508 491 505
Aug 560 510 497 523 491
Sep 510 515 510 514 523
Oct 520 530 515 528 514
Nov 540 523 530 528 528
Dec 550 537 523 541 528
37
Practice Problem
Month Demand (D)
Jan 125
Feb 135
Mar 130
Apr 120
May 115
Jun 140
Jul 135
Aug 110
Sep 120
Oct 120
Nov 140
Dec 145
Use Simple Moving Average and Weighted Moving average method for 2 months. Forecast and compare two methods. Assume appropriate values
38
Simple Exponential Smoothing - Formula
Ft = F t-1 + α(Dt - Ft-1) OR α (Dt) + (1 – α ) Ft-1
ft = Ft-1
Where,
F = Simple Exponential average
f = Forecast for time t
D = Demand
α = Smoothing constant between 0 to 1
39
Problem – Simple Exponential Smoothing
Month Demand
Jan 97
Feb 93
Mar 110
Apr 98
May 104
Jun 103
Jul 99
Aug 108
Sep 106
Oct 94
Nov 109
Dec 95
A firm uses exponential smoothing method for forecasting. Try α = 0.1 & F0 = 100
40
Problem – Simple Exponential Smoothing
Month Demand Exponential Avg (Ft)
100
Jan 97 99.7
Feb 93 99.03
Mar 110 100.73
Apr 98 99.91
May 104 100.32
Jun 103 100.60
Jul 99 100.44
Aug 108 101.20
Sep 106 101.68
Oct 94 82.11
Nov 109 84.8
Dec 95 85.82
41
Problem – Simple Exponential Smoothing
Month Demand Exponential Avg (Ft)
Forecast (ft)
100
Jan 97 99.7 100
Feb 93 99.03 99.7
Mar 110 100.73 99.03
Apr 98 99.91 100.73
May 104 100.32 99.91
Jun 103 100.60 100.32
Jul 99 100.44 100.60
Aug 108 101.20 100.44
Sep 106 101.68 101.20
Oct 94 82.11 101.68
Nov 109 84.8 82.11
Dec 95 85.82 84.8
42
Practice Problem
• A firm uses exponential smoothing method for forecasting, with α = 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 = α (Dt) + (1 – α ) Ft-1
FMAR = α (DMAR) + (1 – α ) FFEB
F MAR = 0.2 (460) + (1-0.2) 500
F MAR = 492
43
Practice Problem
Month Demand
Jan 122
Feb 127
Mar 125
Apr 126
May 139
Jun 127
Jul 134
Aug 128
Sep 134
Oct 136
Nov 132
Dec 131
Given in table is the data for 1994. F0 = 150. Try α = 0.1 and 0.3, which is a better value?
44
Exponential Smoothing with Trend (Winter’s)
Ft = α (Dt) + (1 – α ) (Ft-1 + Tt-1)
Tt = β (Ft – Ft-1) + (1 – β) Tt-1
ft = Ft-1 + Tt-1
Where,
F = Winters Exponential average
f = Forecast for time t
D = Demand
α = Smoothing constant between 0 to 1
T = Trend estimate at time t
Β = Averaging fraction
45
Problem – Exponential smoothing with trend
Month Demand
Jan 460
Feb 510
Mar 520
Apr 495
May 475
Jun 560
Jul 510
Aug 520
Sep 540
Oct 550
Nov 555
Dec 569
Calculate forecast with:
α = 0.2
β = 0.2
F0 = 480
T0 = 9
46
Problem – Exponential smoothing with trend
Month Demand Winter’s exp avg (Ft)
Trend (Tt)
480 9
Jan 460 483.20 7.84
Feb 510 494.83 8.60
Mar 520 506.74 9.26
Apr 495 511.80 8.42
May 475 511.18 6.61
Jun 560 526.23 8.30
Jul 510 529.62 7.32
Aug 520 533.55 6.64
Sep 540 540.16 6.63
Oct 550 547.43 6.76
Nov 555 554.35 6.79
Dec 569 562.72 7.11
47
Problem – Exponential smoothing with trend
Month Demand Winter’s exp avg (Ft)
Trend (Tt) Winter’s Forecast (ft)
480 9
Jan 460 483.20 7.84 489.00
Feb 510 494.83 8.60 491.04
Mar 520 506.74 9.26 503.43
Apr 495 511.80 8.42 516.00
May 475 511.18 6.61 520.22
Jun 560 526.23 8.30 517.79
Jul 510 529.62 7.32 534.53
Aug 520 533.55 6.64 536.94
Sep 540 540.16 6.63 540.20
Oct 550 547.43 6.76 546.79
Nov 555 554.35 6.79 554.19
Dec 569 562.72 7.11 561.15
48
Practice problem
Month Demand
Jan 128
Feb 136
Mar 137
Apr 141
May 157
Jun 148
Jul 158
Aug 155
Sep 164
Oct 169
Nov 168
Dec 160
Given data is for year 1994. Calculate forecast with:
α = 0.2
β = 0.05
F0 = 130
T0 = 0
49
Exponential smoothing with seasonality
Ft = α Dt + (1-α) Ft-1
It-m
It = γ Dt + (1-γ) It-m
Ft
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
γ = smoothing constant, normally ≤ 0.05
50
Prob - Exponential smoothing with seasonality
Month 1993 1994
Jan 80 100
Feb 75 85
Mar 80 90
Apr 90 110
May 115 131
Jun 110 120
Jul 100 110
Aug 90 110
Sep 85 95
Oct 75 85
Nov 75 85
Dec 80 80
The table shows demand data of 1993 & 1994. Forecast for the year 1995.
Other data:
α = 0.1 , γ = 0.05 , FDEC94 = 94
Next Step: Calculate average demand of 1993 & 1994 and average monthly demand
Demand
51
Prob - Exponential smoothing with seasonality
Month 1993 1994 Average Demand
Jan 80 100 90
Feb 75 85 80
Mar 80 90 85
Apr 90 110 100
May 115 131 123
Jun 110 120 115
Jul 100 110 105
Aug 90 110 100
Sep 85 95 90
Oct 75 85 80
Nov 75 85 80
Dec 80 80 80
Demand
Avg Monthly Demand = 1128 / 12
= 94
Next Step: Calculate Seasonal Index It = Dt / Ft
52
Prob - Exponential smoothing with seasonality
Month 1993 1994 Average Demand
Seasonality Index (It)
Jan 80 100 90 0.957
Feb 75 85 80 0.851
Mar 80 90 85 0.904
Apr 90 110 100 1.064
May 115 131 123 1.309
Jun 110 120 115 1.223
Jul 100 110 105 1.117
Aug 90 110 100 1.064
Sep 85 95 90 0.957
Oct 75 85 80 0.851
Nov 75 85 80 0.851
Dec 80 80 80 0.851
Demand
The demand for 1995 is given as:
53
Prob - Exponential smoothing with seasonality
Month Seasonality Index (It-m)
Demand
(Dt) 1995
Jan 0.957 95
Feb 0.851 75
Mar 0.904 90
Apr 1.064 105
May 1.309 120
Jun 1.223 117
Jul 1.117 102
Aug 1.064 98
Sep 0.957 95
Oct 0.851 75
Nov 0.851 85
Dec 0.851 75
Next Step: Calculate Ft for 1995 using formula
Ft = α Dt + (1-α) Ft-1
It-m
54
Prob - Exponential smoothing with seasonality
Month Seasonality Index (It-m)
Demand
(Dt) 1995
Average
(Ft)
Jan 0.957 95 94.50
Feb 0.851 75 93.86
Mar 0.904 90 94.43
Apr 1.064 105 94.86
May 1.309 120 94.54
Jun 1.223 117 94.65
Jul 1.117 102 94.32
Aug 1.064 98 94.10
Sep 0.957 95 94.62
Oct 0.851 75 93.37
Nov 0.851 85 94.56
Dec 0.851 75 93.91
Next Step: Calculate Forecast values for 1995 using formula:
ft+1 = Ft x It+1-m
55
Prob - Exponential smoothing with seasonality
Month Seasonality Index (It-m)
Demand
(Dt) 1995
Average
(Ft)
Forecast
(ft)
Jan 0.957 95 94.50 89.96
Feb 0.851 75 93.86 80.42
Mar 0.904 90 94.43 84.45
Apr 1.064 105 94.86 100.47
May 1.309 120 94.54 124.17
Jun 1.223 117 94.65 115.62
Jul 1.117 102 94.32 105.72
Aug 1.064 98 94.10 100.36
Sep 0.957 95 94.62 90.05
Oct 0.851 75 93.37 80.52
Nov 0.851 85 94.56 79.97
Dec 0.851 75 93.91 80.47
Next Step: Calculate It using formula
It = γ Dt + (1-γ) It-m
Ft
56
Prob - Exponential smoothing with seasonality
Month Seasonality Index (It-m)
Demand
(Dt) 1995
Average
(Ft)
Forecast
(ft)
New Seasonality Index (It)
Jan 0.957 95 94.50 89.96 0.959
Feb 0.851 75 93.86 80.42 0.848
Mar 0.904 90 94.43 84.45 0.906
Apr 1.064 105 94.86 100.47 1.066
May 1.309 120 94.54 124.17 1.307
Jun 1.223 117 94.65 115.62 1.224
Jul 1.117 102 94.32 105.72 1.115
Aug 1.064 98 94.10 100.36 1.063
Sep 0.957 95 94.62 90.05 0.959
Oct 0.851 75 93.37 80.52 0.849
Nov 0.851 85 94.56 79.97 0.853
Dec 0.851 75 93.91 80.47 0.848
57
Practice Problem
Month 2003 2004 2005
Jan 120 130 145
Feb 115 121 127
Mar 117 125 132
Apr 122 122 127
May 125 122 118
Jun 127 120 115
Jul 125 125 128
Aug 122 130 135
Sep 120 133 140
Oct 115 130 140
Nov 117 127 135
Dec 120 127 130
The table shows demand data of 2003, 2004 & 2005. Forecast for the year 2005.
Other data:
α = 0.2 , γ = 0.05 , F0 = 130
Demand
58
Exponential smoothing with seasonality & trend (Winter’s complete model)
Ft = α Dt + (1-α) (Ft-1 + Tt-1)
It-m
Tt = β (Ft – Ft-1) + (1 – β) Tt-1
It = γ Dt + (1-γ) It-m
Ft
ft+1 = (Ft + Tt) x It+1-m
Consider values of α = 0.2, β = 0.05 and γ = 0.01
59
Prob – Smoothing with trend & seasonality
MonthDemand
‘95Seasonality
Index ‘95Demand
‘96
Jan 95 0.959 80
Feb 75 0.848 85
Mar 90 0.906 90
Apr 105 1.066 95
May 120 1.307 100
Jun 117 1.224 100
Jul 102 1.115 95
Aug 98 1.063 95
Sep 95 0.959 90
Oct 75 0.849 95
Nov 85 0.853 85
Dec 75 0.848 80
Forecast the demand for 1996 where:
F0 = 80
T0 = 4.5
α = 0.2
β = 0.05
γ = 0.01
Step 1: Calculate smoothing average and trend for each month of 1996
60
Prob – Smoothing with trend & seasonality
Month Demand
‘95
It-m Demand
‘96
Average
Ft
Trend
Tt
80 4.5
Jan 95 0.959 80 84.284 4.489
Feb 75 0.848 85 91.066 4.604
Mar 90 0.906 90 96.403 4.641
Apr 105 1.066 95 98.659 4.521
May 120 1.307 100 97.846 4.255
Jun 117 1.224 100 98.020 4.051
Jul 102 1.115 95 98.697 3.882
Aug 98 1.063 95 99.937 3.750
Sep 95 0.959 90 101.719 3.651
Oct 75 0.849 95 106.676 3.717
Nov 85 0.853 85 108.243 3.609
Dec 75 0.848 80 108.350 3.434
Now do forecasting for 1996 and calculate new seasonal index
61
Prob – Smoothing with trend & seasonality
Month Demand
‘95
It-m Demand
‘96
Average
Ft
Trend
Tt
Forecast’96 (ft)
New Index (It)
80 4.5
Jan 95 0.959 80 84.284 4.489 81.036 0.959
Feb 75 0.848 85 91.066 4.604 75.280 0.849
Mar 90 0.906 90 96.403 4.641 86.677 0.906
Apr 105 1.066 95 98.659 4.521 107.713 1.065
May 120 1.307 100 97.846 4.255 134.856 1.304
Jun 117 1.224 100 98.020 4.051 124.971 1.222
Jul 102 1.115 95 98.697 3.882 113.809 1.113
Aug 98 1.063 95 99.937 3.750 109.041 1.062
Sep 95 0.959 90 101.719 3.651 99.436 0.958
Oct 75 0.849 95 106.676 3.717 89.460 0.849
Nov 85 0.853 85 108.243 3.609 94.165 0.852
Dec 75 0.848 80 108.350 3.434 94.851 0.847
62
Problem – Smoothing with trend & seasonality
1992 1993 1994
Quarter 1 146 192 272
Quarter 2 96 127 155
Quarter 3 59 79 98
Quarter 4 133 186 219
Estimate forecast for 1995 using winter’s complete model with α = 0.2 , β = 0.1 and γ = 0.05
63
Problem – Smoothing with trend & seasonality
Ft = α Dt + (1-α) (Ft-1 + Tt-1)
It-m
Tt = β (Ft – Ft-1) + (1 – β) Tt-1
It = γ Dt + (1-γ) It-m
Ft
ft+1 = (Ft + Tt) x It+1-m
Here, F0, T0 and It-m (i.e I -3) are unknown.
Lets calculate it first.
64
Problem – Smoothing with trend & seasonality
F0 = D – T0 (2.5) for quarterly data
And
F0 = D – T0 (6.5) 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) = 9.38So, F0 = 108.5 – 9.38 (2.5) = 85.05
65
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 1993
Quarter 1 94.43 131.95
Quarter 2 103.81 141.33
Quarter 3 113.19 150.71
Quarter 4 122.57 160.09
From these trend estimates (table) we can develop initial seasonal indices as:
Index = Demand / Estimate
66
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 1993
Quarter 1 1.55 1.46
Quarter 2 0.92 0.90
Quarter 3 0.52 0.52
Quarter 4 1.09 1.13
Lets check our indices are correct or not
To check, take average of indices for 1992 and 1993 and calculate ∑ average
67
Problem – Smoothing with trend & seasonality
1992 1993 Average
Quarter 1 1.55 1.46 1.51
Quarter 2 0.92 0.90 0.91
Quarter 3 0.52 0.52 0.52
Quarter 4 1.09 1.13 1.13
∑ Average 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
68
Problem – Smoothing with trend & seasonality
Correction factor = (4 / 4.07) = 0.983. Now recalculate the indices
1992 1993 Average I t-m
Quarter 1 1.55 1.46 1.51 1.51 x 0.983 = 1.48
Quarter 2 0.92 0.90 0.91 0.91 x 0.983 = 0.89
Quarter 3 0.52 0.52 0.52 0.52 x 0.983 = 0.51
Quarter 4 1.09 1.13 1.13 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
69
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
70
Problem – Smoothing with trend & seasonality
Ft Tt It
0 85.05 9.38 1.11
1 95.27 9.46 1.48
2 105.36 9.53 0.89
3 115.05 9.54 0.51
4 123.64 9.45 1.11
5 132.37 9.38 1.48
6 141.90 9.39 0.89
7 152.01 9.46 0.51
8 162.74 9.59 1.11
9 174.60 9.82 1.48
10 182.31 9.61 0.90
11 191.92 9.61 0.52
12 199.92 9.48 1.11Lets forecast for 1995
71
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
72
Practice Problem
Month Demand’ 93 Forecast ‘93 Demand ‘94
Jan 97 100 78
Feb 93 100 0
Mar 110 100 55
Apr 98 100 75
May 130 102 87
Jun 133 104 0
Jul 129 106 73
Aug 138 108 0
Sep 136 110 0
Oct 124 112 0
Nov 139 114 0
Dec 125 116 53
Calculate Winter’s trend ratio and seasonality index.
What is the forecast for Q1 of 1995?
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