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Production and Operations Management ­MGT613
VU
Lesson 10
The formula for the moving average is:
Ft = w  1A  t -1 + w 2 A  t - 2 + w  3 A  t -3 + ... + w  n A  t -n
n
∑w
=1
wt = weight given to time period "t" occurrence (weights must add to one)
i
i=1
Weighted Moving Average Problem (1) Data
Question: Given the weekly demand and weights, what is the forecast for the 4th period or
Week 4?
Week
Demand
Weights:
1
650
t-1
.5
2
678
t-2
.3
3
720
t-3
.2
4
Weighted Moving Average Problem (1) Solution
Week
Demand Forecast
1
650
2
678
3
720
4
693.4
F4 = 0.5(720)+0.3(678)+0.2(650)=693.4
Note: More weight age would be given to recent most values.
Weighted Moving Average Problem (2) Data
Question: Given the weekly demand information and weights, what is the weighted moving
average forecast of the 5th period or week?
Week
Demand
Weights:
1
820
t-1
0.7
2
775
t-2
0.2
3
680
t-3
0.1
4
655
Weighted Moving Average Problem (2) Solution
Week
Demand Forecast
1
820
2
775
3
680
4
655
5
672
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Production and Operations Management ­MGT613
VU
F5 = (0.1)(755)+(0.2)(680)+(0.7)(655)= 672
Note: More weight age would be given to recent most values.
Exponential Smoothing Model
Ft = Ft-1 + a(At-1 - Ft-1)
Where:
Ft = Forcast vaue for thecomingt timeperiod
l
Ft - 1 = Forecast v luein 1 past timeperiod
a
At - 1 = Actualoccurancein thepast t tim period
e
α = Alphasmoothingconstant
Exponential Smoothing Problem (1) Data
Question: Given the weekly demand data, what are the exponential smoothing forecasts
for periods 2-10 using a=0.10 and a=0.60?
Assume F1=D1
Week
Demand
1
820
2
775
3
680
4
655
5
750
6
802
7
798
8
689
9
775
10
Exponential Smoothing Solution (1)
Week
Demand
0.1
0.6
1
820
820.00
820.00
2
775
820.00
820.00
3
680
815.50
820.00
4
655
801.95
817.30
5
750
787.26
808.09
6
802
783.53
795.59
7
798
785.38
788.35
8
689
786.64
786.57
9
775
776.88
786.61
10
776.69
780.77
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Production and Operations Management ­MGT613
VU
Exponential Smoothing Problem (2) Data
Question: What are the exponential smoothing forecasts for periods 2-5 using Alpha
=0.5? Assume F1=D1
Week
Demand
1
820
2
775
3
680
4
655
5
Exponential Smoothing Problem (2) Solution
Week
Demand
1
820
2
775
3
680
4
655
5
F1=820+(0.5)(820-820)=820
F3=820+(0.5)(775-820)=797.75
Example 3 - Exponential Smoothing
Period
Actual
Alpha = 0.1 Error
Alpha = 0.4 Error
1
42
2
40
42
-2.00
42
-2
3
43
41.8
1.20
41.2
1.8
4
40
41.92
-1.92
41.92
-1.92
5
41
41.73
-0.73
41.15
-0.15
6
39
41.66
-2.66
41.09
-2.09
7
46
41.39
4.61
40.25
5.75
8
44
41.85
2.15
42.55
1.45
9
45
42.07
2.93
43.13
1.87
10
38
42.36
-4.36
43.88
-5.88
11
40
41.92
-1.92
41.53
-1.53
12
41.73
40.92
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Production and Operations Management ­MGT613
VU
Common Nonlinear Trends
Parabolic
Exponential
Growth
· Parabolic Trends
·
Concaved Upwards and Concaved Downwards
The left and right arms are widening as the value increases or the parabola is
·
opening upwards.
It represents the quadratic function
·
Linear Trend Equation
Ft = a + bt
Where:
·  Ft = Forecast for period t
·  t = Specified number of time periods
·  a = Value of Ft at t = 0
·  b = Slope of the line
n ∑ (ty) - ∑ t∑ y
b=
n∑ t 2 - ∑ t 2
y - ∑t
a= ∑
n
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Production and Operations Management ­MGT613
VU
Linear Trend Equation Example
y
2
Week
t
Sales
ty
1
1
150
150
2
4
157
314
3
9
162
486
4
16
166
664
5
25
177
885
Σ t2 = 55 Σ y = 812 Σ  ty
Σ t = 15
=
2499
(Σ t)2
=
225
Linear Trend Calculation
5 (2499) - 15(812)
12495 -12180
b=
=
= 6.3
5(55) - 225
275 -225
812 - 6.3(15)
a=
= 143.
5
y = 143.5 + 6.3t
Associative Forecasting
1. Predictor variables - used to predict values of variable interest
2. Regression - technique for fitting a line to a set of points
3. Least squares line - minimizes sum of squared deviations around the line
Forecast Accuracy
·
Error - difference between actual value and predicted value
·
Mean Absolute Deviation (MAD)
·  Average absolute error
·
Mean Squared Error (MSE)
·  Average of squared error
·
Mean Absolute Percent Error (MAPE)
·  Average absolute percent error
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Production and Operations Management ­MGT613
VU
Simple Linear Regression Formulas for Calculating "a" and "b"
a = y - bx
∑  xy - n( y)(x)
b=
∑  x - n(x)
2
2
Simple Linear Regression Problem Data
Question: Given the data below, what is the simple linear regression model that can be used to
predict sales in future weeks?
Week
Sales
1
150
2
157
3
162
4
166
5
177
Answer: First, using the linear regression formulas, we can compute "a" and "b"
Week Week*Week
Sales Week*Sales
1
1
150
150
2
4
157
314
3
9
162
486
4
16
166
664
5
25
177
885
3
55
162.4
2499
Average
Sum  Average
Sum
∑  xy - n( y)(x) = 2499 - 5(162.4)(3) = 63 = 6.3
b=
∑  x - n(x)
55 - 5(9)
2
2
10
a = y - bx = 162.4 - (6.3)(3) = 143.5
The resulting regression model is:
Yt = 143.5 + 6.3x
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