Transcript Chapter 10

Demand Management and
Forecasting
1.
2.
3.
4.
Understand the role of forecasting as a basis for supply
chain planning.
Compare the differences between independent and
dependent demand.
Identify the basic components of independent demand:
average, trend, seasonal, and random variation.
Show how to make a time series forecast using moving
averages and exponential smoothing.


The purpose of demand management is to
coordinate and control all sources of demand
Two basic sources of demand
◦ Dependent demand: the demand for a product or service
caused by the demand for other products or services
◦ Independent demand: the demand for a product or
service that cannot be derived directly from that of other
products
LO 2
Not much a firm can do about dependent
demand

◦
It is demand that must be met
There is a lot a firm can do about
independent demand

1. Take an active role to influence demand
2. Take a passive role and respond to demand
LO 1
Time series analysis is based on the idea
that data relating to past demand can be
used to predict future demand
Components of demand


◦
◦
◦
◦
◦
LO 3
Average demand for a period of time
Trend
Seasonal element
Cyclical elements
Random variation


The simple moving average model assumes an
average is a good estimator of future behavior
The formula for the simple moving average is:
A t-1 + A t-2 + A t-3 +...+A t- n
Ft =
n
Ft = Forecast for the coming period
N = Number of periods to be averaged
A t-1 = Actual occurrence in the past period for up to
“n” periods
If N=1 we have the “naïve forecast” – the forecast
equals the current period’s actual
LO 4
Week
1
2
3
4
5
6
7
8
9
10
11
12
LO 4
Demand
650
678
720
785
859
920
850
758
892
920
789
844
A t-1 + A t-2 + A t-3 +...+A t- n
Ft =
n
Question: What are the 3week and 6-week
moving average
forecasts for demand?
Assume you only have 3
weeks and 6 weeks of
actual demand data for
the respective forecasts
8
Calculating the moving averages gives
us: Week Demand 3-Week 6-Week
1
2
3
4
5
6
7
8
9
10
11
12
650 F4=(650+678+720)/3
678
=682.67
720
F7=(650+678+720
+785+859+920)/6
785
682.67
859
727.67
=768.67
920
788.00
850
854.67
768.67
758
876.33
802.00
892
842.67
815.33
920
833.33
844.00
789
856.67
866.50
844
867.00
854.83
LO 4
©The McGraw-Hill Companies, Inc., 2004
Demand
Plotting the moving averages and comparing
them shows how the lines smooth out to reveal
the overall upward trend in this example
950
900
850
800
750
700
650
600
550
500
Demand
3-Week
6-Week
1
2
3
4
5
6
7
Week
LO 4
8
9
10 11 12
Note how the 3Week is
smoother than
the Demand,
and 6-Week is
even smoother
Week
1
2
3
4
5
6
7
LO 4
Demand
820
775
680
655
620
600
575
Question: What is the
3 week moving
average forecast for
this data?
Assume you only
have 3 weeks and 5
weeks of actual
demand data for the
respective forecasts
Week
1
2
3
4
5
6
7
LO 4
Demand
820
775
680
655
620
600
575
3-Week
5-Week
F4=(820+775+680)/3
=758.33
758.33
703.33
651.67
625.00
F6=(820+775+680
+655+620)/5
=710.00
710.00
666.00
Ft = Ft-1 + a(At-1 - Ft-1)
Where:
Ft  Forcast value for thecomingt timeperiod
Ft - 1  Forecast value in 1 past timeperiod
At - 1  Actualoccurancein thepast t time period
a  Alphasmoothingconstant


LO 4
Premise: The most recent observations might
have the highest predictive value
Therefore, we should give more weight to the
more recent time periods when forecasting
Week
1
2
3
4
5
6
7
8
9
10
LO 4
Demand
820
775
680
655
750
802
798
689
775
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
Answer: The respective alphas columns denote the forecast values.
Note that you can only forecast one time period into the future.
Week
1
2
3
4
5
6
7
8
9
10
LO 4
Demand
820
775
680
655
750
802
798
689
775
0.1
820.00
820.00
815.50
801.95
787.26
783.53
785.38
786.64
776.88
776.69
0.6
820.00
820.00
793.00
725.20
683.08
723.23
770.49
787.00
728.20
756.28
Note how that the smaller alpha results in a
smoother line in this example
850
800
d 750
n 700
a
m650
e 600
D
550
500
Demand
0.1
0.6
1
2
3
4
5
6
Week
LO 4
7
8
9
10
Week
1
2
3
4
5
LO 4
Question: What are the
Demand
exponential smoothing
820
forecasts for periods
775 2-5 using a =0.5?
680
655
Assume F1=D1
F1=820+(0.5)(820-820)=820
Week
1
2
3
4
5
LO 4
Demand
820
775
680
655
F3=820+(0.5)(775-820)=797.75
0.5
820.00
820.00
797.50
738.75
696.88
n
A
MAD =


LO 4
t
t=1
- Ft
1 MAD  0.8 standard deviation
1 standard deviation  1.25 MAD
n
The ideal MAD is zero which would
mean there is no forecasting error
The larger the MAD, the less the
accurate the resulting model
Question: What is the MAD value given the forecast values
in the table below?
Month
1
2
3
4
5
LO 4
Sales Forecast
220
n/a
250
255
210
205
300
320
325
315
Month
1
2
3
4
5
Sales
220
250
210
300
325
Forecast Abs Error
n/a
255
5
205
5
320
20
315
10
40
n
A
MAD =
LO 4
t
t=1
n
- Ft
40
=
= 10
4
Note that by itself, the MAD
only lets us know the mean
error in a set of forecasts

Mean Absolute Percentage Error (MAPE)
is another measure often used to
evaluate forecasting accuracy
n
MAPE  100

i 1
actual i  forecast i
actual i
n
A MAPE of under 8% is acceptable for
most applications
LO 4
Time
1
2
3
4
5
6
7
8
9
10
ACTUAL FORECAST ERROR ABS ERROR APE
820
820.00 ------775
820.00
-45.00
45.00
5.81
680
815.50 -135.50
135.50 19.93
655
801.95 -146.95
146.95 22.44
750
787.26
-37.26
37.26
4.97
802
783.53
18.47
18.47
2.30
798
785.38
12.62
12.62
1.58
689
786.64
-97.64
97.64 14.17
775
776.88
-1.88
1.88
0.24
776.69
61.91
8.93
MAD MAPE