Description
various forecasting methods like moving average methods, exponential method, Linear trend equation trend adjusted, Time-Series Methods with seasonality.
Forecasting
Forecasting _ Anchor
1. Averaging methods: Moving Averages smoothing Weighted Moving Averages smoothing Exponential Smoothing 2. Trend analysis : Linear trend equation, Trend adjusted Exponential smoothing 2. 2 Causal methods: Simple linear regression 3. Time Series Decomposition seasonality
SUMMARIZING:
Product Demand Charted over 4 Years with Trend and Seasonality
Seasonal peaks Demand for product or servic ce Trend component
Actual demand line Average demand g over four years
Random variation
Year 1 Year 2
Year 3
Year 4
Naive Approach
• Assumes demand in next ssu es de a d e t period is the same as demand in most recent period
– e.g., If May sales were 48, then June sales will be 48
• Sometimes cost effective & efficient
© 1995 Coel Corp.
Moving Average Method
• • • •
MA is a series of arithmetic means Used little U d if li l or no trend d Used often for smoothing
– Provides overall impression of data over time
Equation Demand in Previous n Periods MA = ? n
Moving Average Example
You’re manager of a museum store that sells historical replicas You want to replicas. forecast sales (000) for 2003 using a 3period moving average. 1998 4 1999 6 2000 5 2001 3 2002 7
Moving Average Solution
Time 1998 1999 2000 2001 2002 2003
Response Yi 4 6 5 3 7 NA
Moving Total (n=3) ( 3) NA NA NA 4+6+5=15 4 6 5 15
Moving Average (n=3) ( 3) NA NA NA 15/3 = 5
Moving Average Solution
Time 1998 1999 2000 2001 2002 2003
Response Yi 4 6 5 3 7 NA
Moving Moving Total Average (n=3) (n=3) NA NA NA NA NA NA 4+6+5=15 1 /3 = 5 46 1 15/3 6+5+3=14 14/3=4 2/3
Moving Average Solution
Time 1998 1999 2000 2001 2002 2003
Response Yi 4 6 5 3 7 NA
Moving Total (n=3) NA NA NA 4+6+5=15 6+5+3=14 5+3+7=15
Moving Average g (n=3) NA NA NA 15/3=5.0 14/3=4.7 15/3=5.0
Moving Average Graph
Sales 8 6 4 2 95 96 97 98 Year 99
Actual Forecast
00
TimeTime-Series Methods Simple Moving Averages
450 — 430 — Pat tient arrivals 410 — 390 — 370 — | 5 Actual patient arrivals | 10 | 15 Week | 20 | 25 | 30
0
TimeTime-Series Methods Simple Moving Averages
450 — 430 — Pat tient arrivals 410 — 390 — 370 — | 5 Actual patient arrivals | 10 | 15 Week | 20 | 25 | 30
0
TimeTime-Series Methods Simple Moving Averages
450 — 430 — Pat tient arrivals 410 — 390 — 370 — | 5 Actual patient arrivals | 10 | 15 Week | 20 | 25 | 30
Week 1 2 3
Patient Arrivals 400 380 411
0
TimeTime-Series Methods Simple Moving Averages
450 — 430 — Pat tient arrivals 410 — 390 — 370 — | 5 Actual patient arrivals | 10 | 15 Week | 20 | 25 | 30
Week 1 2 3
Patient Arrivals 400 380 411
0
TimeTime-Series Methods Simple Moving Averages
450 — 430 — Pat tient arrivals 410 — 390 — 370 — | 5
Week 1 2 3
Patient Arrivals 400 380 411
Actual patient arrivals | 10
F4 =
411 + 380 + 400 3
| 20 | 25 | 30
0
| 15 Week
TimeTime-Series Methods Simple Moving Averages
450 — 430 — Pat tient arrivals 410 — 390 — 370 — | 5 Actual patient arrivals | 10
Week 1 2 3
Patient Arrivals 400 380 411
F4 = 397.0
0
| 15 Week
| 20
| 25
| 30
TimeTime-Series Methods Simple Moving Averages
450 — 430 — Pat tient arrivals 410 — 390 — 370 — | 5 Actual patient arrivals | 10
Week 1 2 3
Patient Arrivals 400 380 411
F4 = 397.0
0
| 15 Week
| 20
| 25
| 30
TimeTime-Series Methods Simple Moving Averages
450 — 430 — Pat tient arrivals 410 — 390 — 370 — | 5 Actual patient arrivals | 10
Week 2 3 4
Patient Arrivals 380 411 415
F5 =
415 + 411 + 380 3
| 20 | 25 | 30
0
| 15 Week
TimeTime-Series Methods Simple Moving Averages
450 — 430 — Pat tient arrivals 410 — 390 — 370 — | 5 Actual patient arrivals | 10
Week 2 3 4
Patient Arrivals 380 411 415
F5 = 402.0
0
| 15 Week
| 20
| 25
| 30
TimeTime-Series Methods Simple Moving Averages
450 — 430 — Pat tient arrivals 410 — 390 — 370 — | 5 Actual patient arrivals | 10 | 15 Week | 20 | 25 | 30
0
TimeTime-Series Methods Simple Moving Averages
450 — 430 — Pat tient arrivals 410 — 390 — 370 — | 5 Actual patient arrivals | 10 | 15 Week | 20 | 25 | 30 3-week MA 3 k forecast
0
TimeTime-Series Methods Simple Moving Averages
450 — 430 — Pat tient arrivals 410 — 390 — 370 — | 5 Actual patient arrivals | 10 | 15 Week | 20 | 25 | 30 3-week MA 3 k forecast 6-week MA forecast
0
Disadvantages of Moving Average Methods
• Increasing n makes forecast less sensitive to changes • Does not forecast trend well • Require much historical data
Weighted Moving Average Method • Used when trend is present
– Older data usually less important, so recent values get higher weights
• Weights based on intuition
– Often lie between 0 & 1, & sum to 1.0
• Equation
WMA =
?(Weight for period n) (Demand in period n) ?Weights
Actual Demand, Moving Average, Weighted Moving Average
35 30 Sales Demand 25 20 15 10 5 0 Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec Month
Weighted moving average Actual sales
Moving average
TimeTime-Series Methods Weighted Moving Average
450 — 430 — Pat tient arrivals 410 — 390 — 370 — | 5 Actual patient arrivals | 10 | 15 Week | 20 | 25 | 30 3-week MA 3 k forecast
Weighted Moving Average forecast
Assigned weights t t-1 t-2 0.70 0.20 0.10
6-week MA
0
TimeTime-Series Methods Weighted Moving Average
450 — 430 — Pat tient arrivals 410 — 390 — 370 — | 5 Actual patient arrivals | 10 3-week MA 3 k forecast
Weighted Moving Average forecast
Assigned weights t t-1 t-2 0.70 0.20 0.10
6-week MA
F4 = 0.70(411) + 0.20(380) + 0.10(400) 0 10(400)
0
| 15 Week
| 20
| 25
| 30
TimeTime-Series Methods Weighted Moving Average
450 — 430 — Pat tient arrivals 410 — 390 — 370 — | 5 Actual patient arrivals | 10 | 15 Week | 20 | 25 | 30 3-week MA 3 k forecast
Weighted Moving Average forecast
Assigned weights t t-1 t-2 F4 = 403.7 0.70 0.20 0.10
6-week MA
0
TimeTime-Series Methods Weighted Moving Average
450 — 430 — Pat tient arrivals 410 — 390 — 370 — | 5 Actual patient arrivals | 10 | 15 Week | 20 | 25 | 30 3-week MA 3 k forecast
Weighted Moving Average forecast
Assigned weights t t-1 t-2 F4 = 404 0.70 0.20 0.10 F5 = 410.7
6-week MA
0
TimeTime-Series Methods Weighted Moving Average
450 — 430 — Pat tient arrivals 410 — 390 — 370 — | 5 Actual patient arrivals | 10 | 15 Week | 20 | 25 | 30 3-week MA 3 k forecast
Weighted Moving Average forecast
Assigned weights t t-1 t-2 F4 = 404 0.70 0.20 0.10 F5 = 411
6-week MA
0
Time series decomposition, Anchor
Exponential Smoothing Method
• Form of weighted moving average g g g
– Weights decline exponentially – Most recent data weighted most
• Requires smoothing constant (?)
– Ranges from 0 to 1 – Subjectively chosen
Exponential Smoothing Equations
• Ft+1 = Ft + ?(At - Ft)
– Use for computing forecast – Above equation can also be written as
• Ft+1 = ? At + (1- ?) Ft
Ft +1 = ? At + ?(1- ?)At - 1 + ?(1- ?)2At - 2 + ...
Exponential Smoothing Example
During the past 8 quarters, the Port of Baltimore has unloaded large quantities of grain. (? = .10). The first quarter forecast was 175 175.. Quarter Actual 1 2 3 4 5 6 7 8 9 180 168 159 175 190 205 180 182 ?
Find the forecast for the 9th quarter.
Forecast Effects of Smoothing Constant ?
•Ft+1 = ? At + (1- ? ) Ft
Ft +1 = ? At + ?(1- ?)At - 1 + ?(1- ?)2At - 2 + ... ( ) (
Weights ?=
?= 0.10 ?= 0.90 Prior Period ? 2 periods ago 3 periods ago ?(1 - ?) ?(1 - ?)2
10%
Forecast Effects of Smoothing Constant ? Ft = ? At - 1 + ?(1- ?) At - 2 + ?(1- ?)2At - 3 + ... ( (
Weights ?=
Prior Period ? ?= 0.10 ?= 0.90 2 periods ago 3 periods ago ?(1 - ?) ?(1 - ?)2
10% 90%
9% 9%
8.1% 0.9%
TimeTime-Series Methods Exponential Smoothing
450 — 430 — Pat tient arrivals 410 — 390 — 370 — | 5 | 10 | 15 Week | 20 | 25 | 30
Exponential Smoothing ? = 0.10 Ft +1 = Ft + ? (At – Ft )
0
TimeTime-Series Methods Exponential Smoothing
450 — 430 — Pat tient arrivals 410 — 390 — 370 — | 5 | 10 | 15 Week
Exponential Smoothing ? = 0.10 Ft +1 = Ft + ? (At – Ft ) F t+1 = ? A t + (1- ?) Ft F3 = (400 + 380)/2 A3 = 411 F4 = 0.10(411) + 0.90(390)
0
| 20
| 25
| 30
TimeTime-Series Methods Exponential Smoothing
450 — 430 — Pat tient arrivals 410 — 390 — 370 — | 5 | 10 | 15 Week
Exponential Smoothing ? = 0.10 Ft +1 = Ft + ? (At – Ft ) F3 = (400 + 380)/2 A3 = 411 F4 = 392.1
0
| 20
| 25
| 30
TimeTime-Series Methods Exponential Smoothing
450 — 430 — Pat tient arrivals 410 — 390 — 370 — | 5 | 10 | 15 Week
Exponential Smoothing ? = 0.10 Ft +1 = Ft + ? (At – Ft ) F4 = 392.1 A4 = 415 F4 = 392.1 F5 = 394.4
| 25 | 30
0
| 20
TimeTime-Series Methods Exponential Smoothing
450 — 430 — Pat tient arrivals 410 — 390 — 370 — | 5 | 10 | 15 Week | 20 | 25 | 30
0
TimeTime-Series Methods Exponential Smoothing
450 — 430 — Pat tient arrivals 410 — 390 — 370 — | 5 | 10 Exponential smoothing ? = 0.10 | | 15 20 Week
0
| 25
| 30
TimeTime-Series Methods Exponential Smoothing
450 — 430 — Pat tient arrivals 410 — 390 — 370 — | 5 | 10 Exponential smoothing ? = 0.10 | | 15 20 Week 3-week MA 3 k forecast 6-week MA forecast
0
| 25
| 30
Trend Line fitting
Y Deman variable nd
Line equation: Yt = a + bt
a = y - bt
b=
t Time intervals
? ty - n
(t) ? t - n(t)
2 2
TimeTime-Series Methods
TrendTrend-Adjusted Exponential Smoothing
Also called: Holt’s Linear Trend Model
Trend adjusted ES Ft = ??At + (1 – ??)(Ft-1 + bt-1) bt = ??(Ft – Ft-1) + (1 – ?)bt-1 Ft+1= Ft + bt
TimeTime-Series Methods
TrendTrend-Adjusted Exponential Smoothing
80 — 70 — Pa atient arrivals 60 — 50 — 40 — 30 — | 1 | 2 | 3 | 4 | 5 | 6 | | 7 8 Week | 9 | | | | 10 11 12 13 | | 14 15 Actual blood test requests
0
TimeTime-Series Methods
TrendTrend-Adjusted Exponential Smoothing
80 — 70 — Pa atient arrivals 60 — 50 — 40 — 30 — | 1 | 2 | 3 | 4 | 5 | 6 | | 7 8 Week | 9 | | | | 10 11 12 13 | | 14 15
Medanalysis, Inc. Demand for blood analysis Ft = ? At + (1 – ?)(Ft-1 + bt-1) bt = ? (Ft – Ft-1) + (1 – ?)bt-1
Ft+1 = Ft + b t
0
TimeTime-Series Methods
TrendTrend-Adjusted Exponential Smoothing
80 — 70 — Pa atient arrivals 60 — 50 — 40 — 30 — | 1 | 2 | 3 | 4 | 5
Medanalysis, Inc. Demand for blood analysis Ft = ? At + (1 – ?)(Ft-1 + bt-1) bt = ? (Ft – Ft-1) + (1 – ?)bt-1 F0 = 28 patients ? = 0 20 0.20 b0 = 3 patients ? = 0 20 0.20
F1 = 0.2(27) + 0.80(28 + 3) b1 = 0.2(30.2 |- 28) + |0.80(3) | | | | | | |
6 7 8 Week 9 10 11 12 13
0
| 14 15
TimeTime-Series Methods
TrendTrend-Adjusted Exponential Smoothing
80 — 70 — Pa atient arrivals 60 — 50 — 40 — 30 — | 1 | 2 | 3 | 4 | 5
Medanalysis, Inc. Demand for blood analysis Ft = ? At + (1 – ?)(Ft-1 + bt-1) bt = ? (Ft – Ft-1) + (1 – ?)bt-1 F0 = 28 patients ? = 0 20 0.20 b0 = 3 patients ? = 0 20 0.20 Forecast F2 = 30.2 + 2.8 = 33
F1 = 30.2 b1 = 2.8 | | | |
6 7 8 Week 9
0
| | | | 10 11 12 13
| | 14 15
TimeTime-Series Methods
TrendTrend-Adjusted Exponential Smoothing
80 — 70 — Pa atient arrivals 60 — 50 — 40 — 30 — | 1 | 2 | 3 | 4 | 5
Medanalysis, Inc. Demand for blood analysis Ft = ? At + (1 – ?)(Ft-1 + bt-1) bt = ? (Ft – Ft-1) + (1 - ?)bt-1 F2 = 30.2 ? = 0 20 0.20 A2 = 44 b1 = 2.8 ? = 0 20 0.20
F2 = 0.2(44) + 0.80(30.2 + 2.8) b | | | | 2 = 0.2(35.2 |- 30.2) |+ 0.80(2.8) | | | |
6 7 8 Week 9 10 11 12 13 14 15
0
TimeTime-Series Methods
TrendTrend-Adjusted Exponential Smoothing
80 — 70 — Pa atient arrivals 60 — 50 — 40 — 30 — | 1 | 2 | 3 | 4 | 5
Medanalysis, Inc. Medanalysis Inc Demand for blood analysis Ft = ? At + (1 – ?)(Ft-1 + bt-1) bt = ? (Ft – Ft-1) + (1 - ?)bt-1 F2 = 30.2 ? = 0.20 0 20 A2 = 44 b1 = 2.8 ? = 0.20 0 20 Forecast F 3 = 35.2 + 3.2 = 38.4
F2 = 35.2 b | | 3.2| | 2=
6 7 8 Week 9
0
| | | | 10 11 12 13
| | 14 15
TimeTime-Series Methods
TrendTrend-Adjusted Exponential Smoothing
80 — 70 — Pa atient arrivals 60 — 50 — 40 — 30 — | 1 | 2 | 3 | 4 | 5 | 6 | | 7 8 Week | 9 | | | | 10 11 12 13 | | 14 15 Actual blood test requests
0
TimeTime-Series Methods
TrendTrend-Adjusted Exponential Smoothing
80 — 70 — Pa atient arrivals 60 — 50 — 40 — 30 — | 1 | 2 | 3 | 4 | 5 | 6 | | 7 8 Week | 9 | | | | 10 11 12 13 | | 14 15 Actual blood test requests Trend-adjusted forecast
0
TimeTime-Series Methods
TrendTrend-Adjusted Exponential Smoothing
80 — 70 — Pa atient arrivals 60 — 50 — 40 — 30 — | 1 | 2 | 3 | 4 | 5 | 6 | | 7 8 Week | 9 | | | | 10 11 12 13 | | 14 15 Actual blood test requests Trend-adjusted forecast
0
Forecasting _Simple Linear Regression
Simple Linear Regression
1. Fit a line such that the sum of squared errors is minimum. (least squares estimation) 2. Used only where there is a causal link between the two variables. 3. X and Y ( the two variables) correlated to the extent of correlation coefficient ‘r’. (naturally occurring phenomenon are correlated t th extent of r = 0.3) h l t d to the t t f 0 3) 4.If assumption of linear variation fails, the method of linear regression is not very good. (if n is small then unstable results, n>30 desirable)
Causal Methods Linear Regression
Y Depen ndent variable
X Independent variable
Linear Regression
Y Depen ndent variable Regression equation: Y = a + bX
X Independent variable
Linear Regression
Y Depen ndent variable Regression equation: Y = a + bX
Actual value of Y Value of X used to estimate Y
X Independent variable
Linear Regression
Y Depen ndent variable
Estimate f E ti t of Y from regression equation Actual value of Y Value of X used to estimate Y
Regression equation: Y = a + bX
X Independent variable
Linear Regression
Y Depen ndent variable
Deviation, Estimate f E ti t of or error Y from regression equation
Regression equation: Y = a + bX
Actual value of Y Value of X used to estimate Y
{
X Independent variable
Linear Regression
Yt = a + bx
a = y - bx
? xy - n(x) ? x - n(x )
2 2
b=
Linear Regression
Solved example _ Advertising
Month 1 2 3 4 5
Sales (000 units) 264 116 165 101 209
Advertising (000 $) 2.5 1.3 1.4 1.0 2.0
Linear Regression
300 — Sales (thousands of units) s 250 — 200 —
Month 1 2 150 — 3 4 100 — 5
50
Sales (000 units)
Advertising (000 $) 2.5 1.3 1.4 1.0 2.0 + 109.229X
264 116 165 101 209 – 8.136 Y=
a = b = r = r2 = syx =
– 8.136 109.229X 0.98 0 96 0.96 15.61
| | | | 1.0 1.5 2.0 2.5 Advertising (thousands of dollars)
Linear Regression
300 — Sales (thousands of units) s 250 — 200 —
Month 1 2 150 — 3 4 100 — 5
50
Sales (000 units)
Advertising (000 $) 2.5 1.3 1.4 1.0 2.0 + 109.229X
|
264 116 165 101 209 – 8.136 Y=
| | |
a = b = r = r2 = syx =
– 8.136 109.229X 0.98 0 96 0.96 15.61
Forecast for Month 2.5 6 1.0 1.5 2.0
Advertising (thousands of dollars)
X = $1750, Y = – 8.136 + 109.229(1.75)
Linear Regression
300 — Sales (thousands of units) s 250 — 200 —
Month 1 2 150 — 3 4 100 — 5
50
Sales (000 units)
Advertising (000 $) 2.5 1.3 1.4 1.0 2.0 + 109.229X
|
264 116 165 101 209 – 8.136 Y=
| | |
a = b = r = r2 = syx =
– 8.136 109.229X 0.98 0 96 0.96 15.61
Forecast for Month 2.5 6 1.0 1.5 2.0
Advertising (thousands of dollars)
X = $1750, Y = 183.015, or 183,015 units
Linear Regression
300 — Sales (thousands of units) s 250 — 200 —
Month 1 2 150 — 3 4 100 — 5
50
Sales (000 units)
Advertising (000 $) 2.5 1.3 1.4 1.0 2.0 + 109.229X
264 116 165 101 209 – 8.136 Y=
a = b = r = r2 = syx =
– 8.136 109.229X 0.98 0 96 0.96 15.61
| | | | 1.0 1.5 2.0 2.5 Advertising (thousands of dollars)
Linear Regression
300 — Sales (thousands of units) s 250 — 200 —
Month 1 2 150 — 3 4 100 — 5
50
Sales (000 units)
Advertising (000 $) 2.5 1.3 1.4 1.0 2.0 + 109.229X
|
264 116 165 101 209 – 8.136 Y=
| | |
a = b = r = r2 = syx =
– 8.136 109.229X 0.98 0 96 0.96 15.61
If 1.0 current stock = 62,500 units, 1.5 2.0 2.5
Advertising (thousands of dollars)
Production = 183,015 – 62,500 = 120,015 units
Summarizing • Exponential smoothing
F t +1 = Ft + ?(At – Ft )
• Trend adjusted ES
• Ft+1= Ft + b t • Ft = ? ?At + (1 – ??)(Ft-1 + bt-1) • b t = ?? (Ft – Ft-1) + (1 – ?)bt-1
• Seasonality (multiplicative)
– Follows using time series decomposition method – If seasonality is additive, simply subtract this from the series and proceed with new series. When making prediction, add back seasonal value for period.
TimeTime-Series Methods
Seasonal Influences
Quarter 1 2 3 4 Total Average Year 1 45 335 520 100 1000 250 Year 2 70 370 590 170 1200 300 Year 3 100 585 830 285 1800 450 Year 4 100 725 1160 215 2200 550
TimeTime-Series Methods
Seasonal Influences
Quarter 1 2 3 4 Total Average Year 1 45 335 520 100 1000 250 Year 2 70 370 590 170 1200 300 Year 3 100 585 830 285 1800 450 Year 4 100 725 1160 215 2200 550
Seasonal Index =
Actual Demand Average Demand
TimeTime-Series Methods
Seasonal Influences
Quarter 1 2 3 4 Total Average Year 1 45 335 520 100 1000 250 Year 2 70 370 590 170 1200 300 Year 3 100 585 830 285 1800 450 Year 4 100 725 1160 215 2200 550
Seasonal Index =
45 250
= 0.18
TimeTime-Series Methods
Seasonal Influences
Quarter 1 2 3 4 Year 1 45/250 = 0.18 335 520 100 1000 250 Year 2 70 370 590 170 1200 300 Year 3 100 585 830 285 1800 450 Year 4 100 725 1160 215 2200 550
Total Average
Seasonal Index =
45 250
= 0.18
TimeTime-Series Methods
Seasonal Influences
Quarter 1 2 3 4 Year 1 Year 2 70/300 = 0.23 370/300 = 1.23 590/300 = 1.97 170/300 = 0.57 Year 3 Year 4 45/250 = 0.18 335/250 = 1.34 520/250 = 2.08 100/250 = 0.40 100/450 = 0.22 100/550 = 0.18 585/450 = 1.30 725/550 = 1.32 830/450 = 1.84 1160/550 = 2.11 285/450 = 0.63 215/550 = 0.39
TimeTime-Series Methods
Seasonal Influences
Quarter 1 2 3 4 Year 1 Year 2 70/300 = 0.23 370/300 = 1.23 590/300 = 1.97 170/300 = 0.57 Year 3 Year 4 45/250 = 0.18 335/250 = 1.34 520/250 = 2.08 100/250 = 0.40 100/450 = 0.22 100/550 = 0.18 585/450 = 1.30 725/550 = 1.32 830/450 = 1.84 1160/550 = 2.11 285/450 = 0.63 215/550 = 0.39
Quarter 1 2 3 4
Average Seasonal Index (0.18 0.23 0.22 0.18)/4 0.20 (0 18 + 0 23 + 0 22 + 0 18)/4 = 0 20
TimeTime-Series Methods
Seasonal Influences
Quarter 1 2 3 4 Year 1 Year 2 70/300 = 0.23 370/300 = 1.23 590/300 = 1.97 170/300 = 0.57 Year 3 Year 4 45/250 = 0.18 335/250 = 1.34 520/250 = 2.08 100/250 = 0.40 100/450 = 0.22 100/550 = 0.18 585/450 = 1.30 725/550 = 1.32 830/450 = 1.84 1160/550 = 2.11 285/450 = 0.63 215/550 = 0.39
Quarter 1 2 3 4
Average Seasonal Index (0.18 0.23 0.22 0.18)/4 0.20 (0 18 + 0 23 + 0 22 + 0 18)/4 = 0 20 (1.34 + 1.23 + 1.30 + 1.32)/4 = 1.30 (2.08 + 1.97 + 1.84 + 2.11)/4 = 2.00 (0.40 + 0.57 + 0.63 + 0.39)/4 = 0.50
TimeTime-Series Methods
Seasonal Influences
Quarter 1 2 3 4 Year 1 Year 2 Year 3 Year 4 45/250 = 0.18 70/300 = 0.23 100/450 = 0.22 Projected Annual Demand = 2600 100/550 = 0.18 335/250 = 1.34 370/300 = 1.23 585/450 = 1.30 725/550 = 1.32 Average 590/300 = 1.97 830/450 2600/4 = 650 520/250 = 2.08 Quarterly Demand = = 1.84 1160/550 = 2.11 100/250 = 0.40 170/300 = 0.57 285/450 = 0.63 215/550 = 0.39
Quarter 1 2 3 4
Average Seasonal Index (0.18 0.23 0.22 0.18)/4 0.20 (0 18 + 0 23 + 0 22 + 0 18)/4 = 0 20 (1.34 + 1.23 + 1.30 + 1.32)/4 = 1.30 (2.08 + 1.97 + 1.84 + 2.11)/4 = 2.00 (0.40 + 0.57 + 0.63 + 0.39)/4 = 0.50
Forecast
TimeTime-Series Methods
Seasonal Influences
Quarter 1 2 3 4 Year 1 Year 2 Year 3 Year 4 45/250 = 0.18 70/300 = 0.23 100/450 = 0.22 Projected Annual Demand = 2600 100/550 = 0.18 335/250 = 1.34 370/300 = 1.23 585/450 = 1.30 725/550 = 1.32 Average 590/300 = 1.97 830/450 2600/4 = 650 520/250 = 2.08 Quarterly Demand = = 1.84 1160/550 = 2.11 100/250 = 0.40 170/300 = 0.57 285/450 = 0.63 215/550 = 0.39
Quarter 1 2 3 4
Average Seasonal Index (0.18 0.23 0.22 0.18)/4 0.20 (0 18 + 0 23 + 0 22 + 0 18)/4 = 0 20 (1.34 + 1.23 + 1.30 + 1.32)/4 = 1.30 (2.08 + 1.97 + 1.84 + 2.11)/4 = 2.00 (0.40 + 0.57 + 0.63 + 0.39)/4 = 0.50
Forecast 650(0.20) 650(0 20) = 130
TimeTime-Series Methods
Seasonal Influences
Quarter 1 2 3 4 Year 1 Year 2 70/300 = 0.23 370/300 = 1.23 590/300 = 1.97 170/300 = 0.57 Year 3 Year 4 45/250 = 0.18 335/250 = 1.34 520/250 = 2.08 100/250 = 0.40 100/450 = 0.22 100/550 = 0.18 585/450 = 1.30 725/550 = 1.32 830/450 = 1.84 1160/550 = 2.11 285/450 = 0.63 215/550 = 0.39
Quarter 1 2 3 4
Average Seasonal Index (0.18 0.23 0.22 0.18)/4 0.20 (0 18 + 0 23 + 0 22 + 0 18)/4 = 0 20 (1.34 + 1.23 + 1.30 + 1.32)/4 = 1.30 (2.08 + 1.97 + 1.84 + 2.11)/4 = 2.00 (0.40 + 0.57 + 0.63 + 0.39)/4 = 0.50
Forecast 650(0.20) 650(0 20) = 650(1.30) = 650(2.00) = 650(0.50) = 130 845 1300 325
Anchor
Steps in forecasting with seasonal data • Find seasonal index • Divide term with appropriate seasonal index and pp p find de-seasonalized series • Use ES or MA or some similar method for forecasting the next term’s figure • Multiply by seasonal index for term • Modify forecast using judgment.
Time Series Decomposition
The oil supply problem (seasonality)
• Mr Ram Kumar is a manager in a major oil company. He is responsible for supplies to the ‘Lensdown area’, and is planning supply arrangements for the next year. The demand for oil (petrol etc.) is seasonal and is less in the winter months with a peak during the summer tourist season. Quarterly demand for the last two years is given below. Can you help Mr Ram Kumar forecast the supplies for the next year? Does he need more data?
Demand in KL Qtr 1 2 3 4 Yr 1 40 350 290 210 Yr 2 60 440 320 280 1100
Total 890
Solution to Ram Kumar’s Oil Supply problem
Basic_Oil_data_Ram_kumar 500 400 300 200 100 0 1 2 3 4 5 6 7 8 Time Period
Demand
DATA DeSeason
Guide to forecast method selection _JBF_1992
Forecast method Simple exponential smoothen Amnt of historical data 5-10 obs Data Pattern Stationary ForeDev. cast time horizon Short short Personnel background
Moderate sophistication Moderate sophistication
Trend Adjust 10 – 15 obs exponential smoothen Trend models
Trend
Short mediu m Short mediu m Short mediu m Short, med., long
short
Trend 10 -20 obs for seasonal > 5 per season
Enough to see
short
Little sophistication
seasonality
two peaks and troughs Causal regression models
Cyclical and seasonal patterns
Short - Little sophistication moderate Long Considerable sophistication
10 obs per Complex independent patterns variable
Demand Forecast Applications
Time Horizon Short Term (0–3 months) Individual products or services Inventory management Final assembly scheduling Workforce scheduling Master production scheduling Time series Causal Judgment Medium Term (3 months– 2 years) Total sales Groups or families of products or services Staff planning Production planning Master production scheduling g Purchasing Distribution Causal Judgment Long Term (more than 2 years) Total sales Application Forecast quantity
Decision area
Facility location Capacity planning Process management g
Forecasting technique
Causal Judgment
Thank You
doc_239183969.pdf
various forecasting methods like moving average methods, exponential method, Linear trend equation trend adjusted, Time-Series Methods with seasonality.
Forecasting
Forecasting _ Anchor
1. Averaging methods: Moving Averages smoothing Weighted Moving Averages smoothing Exponential Smoothing 2. Trend analysis : Linear trend equation, Trend adjusted Exponential smoothing 2. 2 Causal methods: Simple linear regression 3. Time Series Decomposition seasonality
SUMMARIZING:
Product Demand Charted over 4 Years with Trend and Seasonality
Seasonal peaks Demand for product or servic ce Trend component
Actual demand line Average demand g over four years
Random variation
Year 1 Year 2
Year 3
Year 4
Naive Approach
• Assumes demand in next ssu es de a d e t period is the same as demand in most recent period
– e.g., If May sales were 48, then June sales will be 48
• Sometimes cost effective & efficient
© 1995 Coel Corp.
Moving Average Method
• • • •
MA is a series of arithmetic means Used little U d if li l or no trend d Used often for smoothing
– Provides overall impression of data over time
Equation Demand in Previous n Periods MA = ? n
Moving Average Example
You’re manager of a museum store that sells historical replicas You want to replicas. forecast sales (000) for 2003 using a 3period moving average. 1998 4 1999 6 2000 5 2001 3 2002 7
Moving Average Solution
Time 1998 1999 2000 2001 2002 2003
Response Yi 4 6 5 3 7 NA
Moving Total (n=3) ( 3) NA NA NA 4+6+5=15 4 6 5 15
Moving Average (n=3) ( 3) NA NA NA 15/3 = 5
Moving Average Solution
Time 1998 1999 2000 2001 2002 2003
Response Yi 4 6 5 3 7 NA
Moving Moving Total Average (n=3) (n=3) NA NA NA NA NA NA 4+6+5=15 1 /3 = 5 46 1 15/3 6+5+3=14 14/3=4 2/3
Moving Average Solution
Time 1998 1999 2000 2001 2002 2003
Response Yi 4 6 5 3 7 NA
Moving Total (n=3) NA NA NA 4+6+5=15 6+5+3=14 5+3+7=15
Moving Average g (n=3) NA NA NA 15/3=5.0 14/3=4.7 15/3=5.0
Moving Average Graph
Sales 8 6 4 2 95 96 97 98 Year 99
Actual Forecast
00
TimeTime-Series Methods Simple Moving Averages
450 — 430 — Pat tient arrivals 410 — 390 — 370 — | 5 Actual patient arrivals | 10 | 15 Week | 20 | 25 | 30
0
TimeTime-Series Methods Simple Moving Averages
450 — 430 — Pat tient arrivals 410 — 390 — 370 — | 5 Actual patient arrivals | 10 | 15 Week | 20 | 25 | 30
0
TimeTime-Series Methods Simple Moving Averages
450 — 430 — Pat tient arrivals 410 — 390 — 370 — | 5 Actual patient arrivals | 10 | 15 Week | 20 | 25 | 30
Week 1 2 3
Patient Arrivals 400 380 411
0
TimeTime-Series Methods Simple Moving Averages
450 — 430 — Pat tient arrivals 410 — 390 — 370 — | 5 Actual patient arrivals | 10 | 15 Week | 20 | 25 | 30
Week 1 2 3
Patient Arrivals 400 380 411
0
TimeTime-Series Methods Simple Moving Averages
450 — 430 — Pat tient arrivals 410 — 390 — 370 — | 5
Week 1 2 3
Patient Arrivals 400 380 411
Actual patient arrivals | 10
F4 =
411 + 380 + 400 3
| 20 | 25 | 30
0
| 15 Week
TimeTime-Series Methods Simple Moving Averages
450 — 430 — Pat tient arrivals 410 — 390 — 370 — | 5 Actual patient arrivals | 10
Week 1 2 3
Patient Arrivals 400 380 411
F4 = 397.0
0
| 15 Week
| 20
| 25
| 30
TimeTime-Series Methods Simple Moving Averages
450 — 430 — Pat tient arrivals 410 — 390 — 370 — | 5 Actual patient arrivals | 10
Week 1 2 3
Patient Arrivals 400 380 411
F4 = 397.0
0
| 15 Week
| 20
| 25
| 30
TimeTime-Series Methods Simple Moving Averages
450 — 430 — Pat tient arrivals 410 — 390 — 370 — | 5 Actual patient arrivals | 10
Week 2 3 4
Patient Arrivals 380 411 415
F5 =
415 + 411 + 380 3
| 20 | 25 | 30
0
| 15 Week
TimeTime-Series Methods Simple Moving Averages
450 — 430 — Pat tient arrivals 410 — 390 — 370 — | 5 Actual patient arrivals | 10
Week 2 3 4
Patient Arrivals 380 411 415
F5 = 402.0
0
| 15 Week
| 20
| 25
| 30
TimeTime-Series Methods Simple Moving Averages
450 — 430 — Pat tient arrivals 410 — 390 — 370 — | 5 Actual patient arrivals | 10 | 15 Week | 20 | 25 | 30
0
TimeTime-Series Methods Simple Moving Averages
450 — 430 — Pat tient arrivals 410 — 390 — 370 — | 5 Actual patient arrivals | 10 | 15 Week | 20 | 25 | 30 3-week MA 3 k forecast
0
TimeTime-Series Methods Simple Moving Averages
450 — 430 — Pat tient arrivals 410 — 390 — 370 — | 5 Actual patient arrivals | 10 | 15 Week | 20 | 25 | 30 3-week MA 3 k forecast 6-week MA forecast
0
Disadvantages of Moving Average Methods
• Increasing n makes forecast less sensitive to changes • Does not forecast trend well • Require much historical data
Weighted Moving Average Method • Used when trend is present
– Older data usually less important, so recent values get higher weights
• Weights based on intuition
– Often lie between 0 & 1, & sum to 1.0
• Equation
WMA =
?(Weight for period n) (Demand in period n) ?Weights
Actual Demand, Moving Average, Weighted Moving Average
35 30 Sales Demand 25 20 15 10 5 0 Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec Month
Weighted moving average Actual sales
Moving average
TimeTime-Series Methods Weighted Moving Average
450 — 430 — Pat tient arrivals 410 — 390 — 370 — | 5 Actual patient arrivals | 10 | 15 Week | 20 | 25 | 30 3-week MA 3 k forecast
Weighted Moving Average forecast
Assigned weights t t-1 t-2 0.70 0.20 0.10
6-week MA
0
TimeTime-Series Methods Weighted Moving Average
450 — 430 — Pat tient arrivals 410 — 390 — 370 — | 5 Actual patient arrivals | 10 3-week MA 3 k forecast
Weighted Moving Average forecast
Assigned weights t t-1 t-2 0.70 0.20 0.10
6-week MA
F4 = 0.70(411) + 0.20(380) + 0.10(400) 0 10(400)
0
| 15 Week
| 20
| 25
| 30
TimeTime-Series Methods Weighted Moving Average
450 — 430 — Pat tient arrivals 410 — 390 — 370 — | 5 Actual patient arrivals | 10 | 15 Week | 20 | 25 | 30 3-week MA 3 k forecast
Weighted Moving Average forecast
Assigned weights t t-1 t-2 F4 = 403.7 0.70 0.20 0.10
6-week MA
0
TimeTime-Series Methods Weighted Moving Average
450 — 430 — Pat tient arrivals 410 — 390 — 370 — | 5 Actual patient arrivals | 10 | 15 Week | 20 | 25 | 30 3-week MA 3 k forecast
Weighted Moving Average forecast
Assigned weights t t-1 t-2 F4 = 404 0.70 0.20 0.10 F5 = 410.7
6-week MA
0
TimeTime-Series Methods Weighted Moving Average
450 — 430 — Pat tient arrivals 410 — 390 — 370 — | 5 Actual patient arrivals | 10 | 15 Week | 20 | 25 | 30 3-week MA 3 k forecast
Weighted Moving Average forecast
Assigned weights t t-1 t-2 F4 = 404 0.70 0.20 0.10 F5 = 411
6-week MA
0
Time series decomposition, Anchor
Exponential Smoothing Method
• Form of weighted moving average g g g
– Weights decline exponentially – Most recent data weighted most
• Requires smoothing constant (?)
– Ranges from 0 to 1 – Subjectively chosen
Exponential Smoothing Equations
• Ft+1 = Ft + ?(At - Ft)
– Use for computing forecast – Above equation can also be written as
• Ft+1 = ? At + (1- ?) Ft
Ft +1 = ? At + ?(1- ?)At - 1 + ?(1- ?)2At - 2 + ...
Exponential Smoothing Example
During the past 8 quarters, the Port of Baltimore has unloaded large quantities of grain. (? = .10). The first quarter forecast was 175 175.. Quarter Actual 1 2 3 4 5 6 7 8 9 180 168 159 175 190 205 180 182 ?
Find the forecast for the 9th quarter.
Forecast Effects of Smoothing Constant ?
•Ft+1 = ? At + (1- ? ) Ft
Ft +1 = ? At + ?(1- ?)At - 1 + ?(1- ?)2At - 2 + ... ( ) (
Weights ?=
?= 0.10 ?= 0.90 Prior Period ? 2 periods ago 3 periods ago ?(1 - ?) ?(1 - ?)2
10%
Forecast Effects of Smoothing Constant ? Ft = ? At - 1 + ?(1- ?) At - 2 + ?(1- ?)2At - 3 + ... ( (
Weights ?=
Prior Period ? ?= 0.10 ?= 0.90 2 periods ago 3 periods ago ?(1 - ?) ?(1 - ?)2
10% 90%
9% 9%
8.1% 0.9%
TimeTime-Series Methods Exponential Smoothing
450 — 430 — Pat tient arrivals 410 — 390 — 370 — | 5 | 10 | 15 Week | 20 | 25 | 30
Exponential Smoothing ? = 0.10 Ft +1 = Ft + ? (At – Ft )
0
TimeTime-Series Methods Exponential Smoothing
450 — 430 — Pat tient arrivals 410 — 390 — 370 — | 5 | 10 | 15 Week
Exponential Smoothing ? = 0.10 Ft +1 = Ft + ? (At – Ft ) F t+1 = ? A t + (1- ?) Ft F3 = (400 + 380)/2 A3 = 411 F4 = 0.10(411) + 0.90(390)
0
| 20
| 25
| 30
TimeTime-Series Methods Exponential Smoothing
450 — 430 — Pat tient arrivals 410 — 390 — 370 — | 5 | 10 | 15 Week
Exponential Smoothing ? = 0.10 Ft +1 = Ft + ? (At – Ft ) F3 = (400 + 380)/2 A3 = 411 F4 = 392.1
0
| 20
| 25
| 30
TimeTime-Series Methods Exponential Smoothing
450 — 430 — Pat tient arrivals 410 — 390 — 370 — | 5 | 10 | 15 Week
Exponential Smoothing ? = 0.10 Ft +1 = Ft + ? (At – Ft ) F4 = 392.1 A4 = 415 F4 = 392.1 F5 = 394.4
| 25 | 30
0
| 20
TimeTime-Series Methods Exponential Smoothing
450 — 430 — Pat tient arrivals 410 — 390 — 370 — | 5 | 10 | 15 Week | 20 | 25 | 30
0
TimeTime-Series Methods Exponential Smoothing
450 — 430 — Pat tient arrivals 410 — 390 — 370 — | 5 | 10 Exponential smoothing ? = 0.10 | | 15 20 Week
0
| 25
| 30
TimeTime-Series Methods Exponential Smoothing
450 — 430 — Pat tient arrivals 410 — 390 — 370 — | 5 | 10 Exponential smoothing ? = 0.10 | | 15 20 Week 3-week MA 3 k forecast 6-week MA forecast
0
| 25
| 30
Trend Line fitting
Y Deman variable nd
Line equation: Yt = a + bt
a = y - bt
b=
t Time intervals
? ty - n
(t) ? t - n(t)2 2
TimeTime-Series Methods
TrendTrend-Adjusted Exponential Smoothing
Also called: Holt’s Linear Trend Model
Trend adjusted ES Ft = ??At + (1 – ??)(Ft-1 + bt-1) bt = ??(Ft – Ft-1) + (1 – ?)bt-1 Ft+1= Ft + bt
TimeTime-Series Methods
TrendTrend-Adjusted Exponential Smoothing
80 — 70 — Pa atient arrivals 60 — 50 — 40 — 30 — | 1 | 2 | 3 | 4 | 5 | 6 | | 7 8 Week | 9 | | | | 10 11 12 13 | | 14 15 Actual blood test requests
0
TimeTime-Series Methods
TrendTrend-Adjusted Exponential Smoothing
80 — 70 — Pa atient arrivals 60 — 50 — 40 — 30 — | 1 | 2 | 3 | 4 | 5 | 6 | | 7 8 Week | 9 | | | | 10 11 12 13 | | 14 15
Medanalysis, Inc. Demand for blood analysis Ft = ? At + (1 – ?)(Ft-1 + bt-1) bt = ? (Ft – Ft-1) + (1 – ?)bt-1
Ft+1 = Ft + b t
0
TimeTime-Series Methods
TrendTrend-Adjusted Exponential Smoothing
80 — 70 — Pa atient arrivals 60 — 50 — 40 — 30 — | 1 | 2 | 3 | 4 | 5
Medanalysis, Inc. Demand for blood analysis Ft = ? At + (1 – ?)(Ft-1 + bt-1) bt = ? (Ft – Ft-1) + (1 – ?)bt-1 F0 = 28 patients ? = 0 20 0.20 b0 = 3 patients ? = 0 20 0.20
F1 = 0.2(27) + 0.80(28 + 3) b1 = 0.2(30.2 |- 28) + |0.80(3) | | | | | | |
6 7 8 Week 9 10 11 12 13
0
| 14 15
TimeTime-Series Methods
TrendTrend-Adjusted Exponential Smoothing
80 — 70 — Pa atient arrivals 60 — 50 — 40 — 30 — | 1 | 2 | 3 | 4 | 5
Medanalysis, Inc. Demand for blood analysis Ft = ? At + (1 – ?)(Ft-1 + bt-1) bt = ? (Ft – Ft-1) + (1 – ?)bt-1 F0 = 28 patients ? = 0 20 0.20 b0 = 3 patients ? = 0 20 0.20 Forecast F2 = 30.2 + 2.8 = 33
F1 = 30.2 b1 = 2.8 | | | |
6 7 8 Week 9
0
| | | | 10 11 12 13
| | 14 15
TimeTime-Series Methods
TrendTrend-Adjusted Exponential Smoothing
80 — 70 — Pa atient arrivals 60 — 50 — 40 — 30 — | 1 | 2 | 3 | 4 | 5
Medanalysis, Inc. Demand for blood analysis Ft = ? At + (1 – ?)(Ft-1 + bt-1) bt = ? (Ft – Ft-1) + (1 - ?)bt-1 F2 = 30.2 ? = 0 20 0.20 A2 = 44 b1 = 2.8 ? = 0 20 0.20
F2 = 0.2(44) + 0.80(30.2 + 2.8) b | | | | 2 = 0.2(35.2 |- 30.2) |+ 0.80(2.8) | | | |
6 7 8 Week 9 10 11 12 13 14 15
0
TimeTime-Series Methods
TrendTrend-Adjusted Exponential Smoothing
80 — 70 — Pa atient arrivals 60 — 50 — 40 — 30 — | 1 | 2 | 3 | 4 | 5
Medanalysis, Inc. Medanalysis Inc Demand for blood analysis Ft = ? At + (1 – ?)(Ft-1 + bt-1) bt = ? (Ft – Ft-1) + (1 - ?)bt-1 F2 = 30.2 ? = 0.20 0 20 A2 = 44 b1 = 2.8 ? = 0.20 0 20 Forecast F 3 = 35.2 + 3.2 = 38.4
F2 = 35.2 b | | 3.2| | 2=
6 7 8 Week 9
0
| | | | 10 11 12 13
| | 14 15
TimeTime-Series Methods
TrendTrend-Adjusted Exponential Smoothing
80 — 70 — Pa atient arrivals 60 — 50 — 40 — 30 — | 1 | 2 | 3 | 4 | 5 | 6 | | 7 8 Week | 9 | | | | 10 11 12 13 | | 14 15 Actual blood test requests
0
TimeTime-Series Methods
TrendTrend-Adjusted Exponential Smoothing
80 — 70 — Pa atient arrivals 60 — 50 — 40 — 30 — | 1 | 2 | 3 | 4 | 5 | 6 | | 7 8 Week | 9 | | | | 10 11 12 13 | | 14 15 Actual blood test requests Trend-adjusted forecast
0
TimeTime-Series Methods
TrendTrend-Adjusted Exponential Smoothing
80 — 70 — Pa atient arrivals 60 — 50 — 40 — 30 — | 1 | 2 | 3 | 4 | 5 | 6 | | 7 8 Week | 9 | | | | 10 11 12 13 | | 14 15 Actual blood test requests Trend-adjusted forecast
0
Forecasting _Simple Linear Regression
Simple Linear Regression
1. Fit a line such that the sum of squared errors is minimum. (least squares estimation) 2. Used only where there is a causal link between the two variables. 3. X and Y ( the two variables) correlated to the extent of correlation coefficient ‘r’. (naturally occurring phenomenon are correlated t th extent of r = 0.3) h l t d to the t t f 0 3) 4.If assumption of linear variation fails, the method of linear regression is not very good. (if n is small then unstable results, n>30 desirable)
Causal Methods Linear Regression
Y Depen ndent variable
X Independent variable
Linear Regression
Y Depen ndent variable Regression equation: Y = a + bX
X Independent variable
Linear Regression
Y Depen ndent variable Regression equation: Y = a + bX
Actual value of Y Value of X used to estimate Y
X Independent variable
Linear Regression
Y Depen ndent variable
Estimate f E ti t of Y from regression equation Actual value of Y Value of X used to estimate Y
Regression equation: Y = a + bX
X Independent variable
Linear Regression
Y Depen ndent variable
Deviation, Estimate f E ti t of or error Y from regression equation
Regression equation: Y = a + bX
Actual value of Y Value of X used to estimate Y
{
X Independent variable
Linear Regression
Yt = a + bx
a = y - bx
? xy - n
(x) ? x - n(x )2 2
b=
Linear Regression
Solved example _ Advertising
Month 1 2 3 4 5
Sales (000 units) 264 116 165 101 209
Advertising (000 $) 2.5 1.3 1.4 1.0 2.0
Linear Regression
300 — Sales (thousands of units) s 250 — 200 —
Month 1 2 150 — 3 4 100 — 5
50
Sales (000 units)
Advertising (000 $) 2.5 1.3 1.4 1.0 2.0 + 109.229X
264 116 165 101 209 – 8.136 Y=
a = b = r = r2 = syx =
– 8.136 109.229X 0.98 0 96 0.96 15.61
| | | | 1.0 1.5 2.0 2.5 Advertising (thousands of dollars)
Linear Regression
300 — Sales (thousands of units) s 250 — 200 —
Month 1 2 150 — 3 4 100 — 5
50
Sales (000 units)
Advertising (000 $) 2.5 1.3 1.4 1.0 2.0 + 109.229X
|
264 116 165 101 209 – 8.136 Y=
| | |
a = b = r = r2 = syx =
– 8.136 109.229X 0.98 0 96 0.96 15.61
Forecast for Month 2.5 6 1.0 1.5 2.0
Advertising (thousands of dollars)
X = $1750, Y = – 8.136 + 109.229(1.75)
Linear Regression
300 — Sales (thousands of units) s 250 — 200 —
Month 1 2 150 — 3 4 100 — 5
50
Sales (000 units)
Advertising (000 $) 2.5 1.3 1.4 1.0 2.0 + 109.229X
|
264 116 165 101 209 – 8.136 Y=
| | |
a = b = r = r2 = syx =
– 8.136 109.229X 0.98 0 96 0.96 15.61
Forecast for Month 2.5 6 1.0 1.5 2.0
Advertising (thousands of dollars)
X = $1750, Y = 183.015, or 183,015 units
Linear Regression
300 — Sales (thousands of units) s 250 — 200 —
Month 1 2 150 — 3 4 100 — 5
50
Sales (000 units)
Advertising (000 $) 2.5 1.3 1.4 1.0 2.0 + 109.229X
264 116 165 101 209 – 8.136 Y=
a = b = r = r2 = syx =
– 8.136 109.229X 0.98 0 96 0.96 15.61
| | | | 1.0 1.5 2.0 2.5 Advertising (thousands of dollars)
Linear Regression
300 — Sales (thousands of units) s 250 — 200 —
Month 1 2 150 — 3 4 100 — 5
50
Sales (000 units)
Advertising (000 $) 2.5 1.3 1.4 1.0 2.0 + 109.229X
|
264 116 165 101 209 – 8.136 Y=
| | |
a = b = r = r2 = syx =
– 8.136 109.229X 0.98 0 96 0.96 15.61
If 1.0 current stock = 62,500 units, 1.5 2.0 2.5
Advertising (thousands of dollars)
Production = 183,015 – 62,500 = 120,015 units
Summarizing • Exponential smoothing
F t +1 = Ft + ?(At – Ft )
• Trend adjusted ES
• Ft+1= Ft + b t • Ft = ? ?At + (1 – ??)(Ft-1 + bt-1) • b t = ?? (Ft – Ft-1) + (1 – ?)bt-1
• Seasonality (multiplicative)
– Follows using time series decomposition method – If seasonality is additive, simply subtract this from the series and proceed with new series. When making prediction, add back seasonal value for period.
TimeTime-Series Methods
Seasonal Influences
Quarter 1 2 3 4 Total Average Year 1 45 335 520 100 1000 250 Year 2 70 370 590 170 1200 300 Year 3 100 585 830 285 1800 450 Year 4 100 725 1160 215 2200 550
TimeTime-Series Methods
Seasonal Influences
Quarter 1 2 3 4 Total Average Year 1 45 335 520 100 1000 250 Year 2 70 370 590 170 1200 300 Year 3 100 585 830 285 1800 450 Year 4 100 725 1160 215 2200 550
Seasonal Index =
Actual Demand Average Demand
TimeTime-Series Methods
Seasonal Influences
Quarter 1 2 3 4 Total Average Year 1 45 335 520 100 1000 250 Year 2 70 370 590 170 1200 300 Year 3 100 585 830 285 1800 450 Year 4 100 725 1160 215 2200 550
Seasonal Index =
45 250
= 0.18
TimeTime-Series Methods
Seasonal Influences
Quarter 1 2 3 4 Year 1 45/250 = 0.18 335 520 100 1000 250 Year 2 70 370 590 170 1200 300 Year 3 100 585 830 285 1800 450 Year 4 100 725 1160 215 2200 550
Total Average
Seasonal Index =
45 250
= 0.18
TimeTime-Series Methods
Seasonal Influences
Quarter 1 2 3 4 Year 1 Year 2 70/300 = 0.23 370/300 = 1.23 590/300 = 1.97 170/300 = 0.57 Year 3 Year 4 45/250 = 0.18 335/250 = 1.34 520/250 = 2.08 100/250 = 0.40 100/450 = 0.22 100/550 = 0.18 585/450 = 1.30 725/550 = 1.32 830/450 = 1.84 1160/550 = 2.11 285/450 = 0.63 215/550 = 0.39
TimeTime-Series Methods
Seasonal Influences
Quarter 1 2 3 4 Year 1 Year 2 70/300 = 0.23 370/300 = 1.23 590/300 = 1.97 170/300 = 0.57 Year 3 Year 4 45/250 = 0.18 335/250 = 1.34 520/250 = 2.08 100/250 = 0.40 100/450 = 0.22 100/550 = 0.18 585/450 = 1.30 725/550 = 1.32 830/450 = 1.84 1160/550 = 2.11 285/450 = 0.63 215/550 = 0.39
Quarter 1 2 3 4
Average Seasonal Index (0.18 0.23 0.22 0.18)/4 0.20 (0 18 + 0 23 + 0 22 + 0 18)/4 = 0 20
TimeTime-Series Methods
Seasonal Influences
Quarter 1 2 3 4 Year 1 Year 2 70/300 = 0.23 370/300 = 1.23 590/300 = 1.97 170/300 = 0.57 Year 3 Year 4 45/250 = 0.18 335/250 = 1.34 520/250 = 2.08 100/250 = 0.40 100/450 = 0.22 100/550 = 0.18 585/450 = 1.30 725/550 = 1.32 830/450 = 1.84 1160/550 = 2.11 285/450 = 0.63 215/550 = 0.39
Quarter 1 2 3 4
Average Seasonal Index (0.18 0.23 0.22 0.18)/4 0.20 (0 18 + 0 23 + 0 22 + 0 18)/4 = 0 20 (1.34 + 1.23 + 1.30 + 1.32)/4 = 1.30 (2.08 + 1.97 + 1.84 + 2.11)/4 = 2.00 (0.40 + 0.57 + 0.63 + 0.39)/4 = 0.50
TimeTime-Series Methods
Seasonal Influences
Quarter 1 2 3 4 Year 1 Year 2 Year 3 Year 4 45/250 = 0.18 70/300 = 0.23 100/450 = 0.22 Projected Annual Demand = 2600 100/550 = 0.18 335/250 = 1.34 370/300 = 1.23 585/450 = 1.30 725/550 = 1.32 Average 590/300 = 1.97 830/450 2600/4 = 650 520/250 = 2.08 Quarterly Demand = = 1.84 1160/550 = 2.11 100/250 = 0.40 170/300 = 0.57 285/450 = 0.63 215/550 = 0.39
Quarter 1 2 3 4
Average Seasonal Index (0.18 0.23 0.22 0.18)/4 0.20 (0 18 + 0 23 + 0 22 + 0 18)/4 = 0 20 (1.34 + 1.23 + 1.30 + 1.32)/4 = 1.30 (2.08 + 1.97 + 1.84 + 2.11)/4 = 2.00 (0.40 + 0.57 + 0.63 + 0.39)/4 = 0.50
Forecast
TimeTime-Series Methods
Seasonal Influences
Quarter 1 2 3 4 Year 1 Year 2 Year 3 Year 4 45/250 = 0.18 70/300 = 0.23 100/450 = 0.22 Projected Annual Demand = 2600 100/550 = 0.18 335/250 = 1.34 370/300 = 1.23 585/450 = 1.30 725/550 = 1.32 Average 590/300 = 1.97 830/450 2600/4 = 650 520/250 = 2.08 Quarterly Demand = = 1.84 1160/550 = 2.11 100/250 = 0.40 170/300 = 0.57 285/450 = 0.63 215/550 = 0.39
Quarter 1 2 3 4
Average Seasonal Index (0.18 0.23 0.22 0.18)/4 0.20 (0 18 + 0 23 + 0 22 + 0 18)/4 = 0 20 (1.34 + 1.23 + 1.30 + 1.32)/4 = 1.30 (2.08 + 1.97 + 1.84 + 2.11)/4 = 2.00 (0.40 + 0.57 + 0.63 + 0.39)/4 = 0.50
Forecast 650(0.20) 650(0 20) = 130
TimeTime-Series Methods
Seasonal Influences
Quarter 1 2 3 4 Year 1 Year 2 70/300 = 0.23 370/300 = 1.23 590/300 = 1.97 170/300 = 0.57 Year 3 Year 4 45/250 = 0.18 335/250 = 1.34 520/250 = 2.08 100/250 = 0.40 100/450 = 0.22 100/550 = 0.18 585/450 = 1.30 725/550 = 1.32 830/450 = 1.84 1160/550 = 2.11 285/450 = 0.63 215/550 = 0.39
Quarter 1 2 3 4
Average Seasonal Index (0.18 0.23 0.22 0.18)/4 0.20 (0 18 + 0 23 + 0 22 + 0 18)/4 = 0 20 (1.34 + 1.23 + 1.30 + 1.32)/4 = 1.30 (2.08 + 1.97 + 1.84 + 2.11)/4 = 2.00 (0.40 + 0.57 + 0.63 + 0.39)/4 = 0.50
Forecast 650(0.20) 650(0 20) = 650(1.30) = 650(2.00) = 650(0.50) = 130 845 1300 325
Anchor
Steps in forecasting with seasonal data • Find seasonal index • Divide term with appropriate seasonal index and pp p find de-seasonalized series • Use ES or MA or some similar method for forecasting the next term’s figure • Multiply by seasonal index for term • Modify forecast using judgment.
Time Series Decomposition
The oil supply problem (seasonality)
• Mr Ram Kumar is a manager in a major oil company. He is responsible for supplies to the ‘Lensdown area’, and is planning supply arrangements for the next year. The demand for oil (petrol etc.) is seasonal and is less in the winter months with a peak during the summer tourist season. Quarterly demand for the last two years is given below. Can you help Mr Ram Kumar forecast the supplies for the next year? Does he need more data?
Demand in KL Qtr 1 2 3 4 Yr 1 40 350 290 210 Yr 2 60 440 320 280 1100
Total 890
Solution to Ram Kumar’s Oil Supply problem
Basic_Oil_data_Ram_kumar 500 400 300 200 100 0 1 2 3 4 5 6 7 8 Time Period
Demand
DATA DeSeason
Guide to forecast method selection _JBF_1992
Forecast method Simple exponential smoothen Amnt of historical data 5-10 obs Data Pattern Stationary ForeDev. cast time horizon Short short Personnel background
Moderate sophistication Moderate sophistication
Trend Adjust 10 – 15 obs exponential smoothen Trend models
Trend
Short mediu m Short mediu m Short mediu m Short, med., long
short
Trend 10 -20 obs for seasonal > 5 per season
Enough to see
short
Little sophistication
seasonality
two peaks and troughs Causal regression models
Cyclical and seasonal patterns
Short - Little sophistication moderate Long Considerable sophistication
10 obs per Complex independent patterns variable
Demand Forecast Applications
Time Horizon Short Term (0–3 months) Individual products or services Inventory management Final assembly scheduling Workforce scheduling Master production scheduling Time series Causal Judgment Medium Term (3 months– 2 years) Total sales Groups or families of products or services Staff planning Production planning Master production scheduling g Purchasing Distribution Causal Judgment Long Term (more than 2 years) Total sales Application Forecast quantity
Decision area
Facility location Capacity planning Process management g
Forecasting technique
Causal Judgment
Thank You
doc_239183969.pdf