Transcript Measure

Measure
Solutions
Normalizing the process
The time for an order acknowledgement is in average 3 days for a
certain product. The predictability is at 0,5 days one standard deviation.
In a customer contract it is required a:
• Order acknowledgement within 4 days
Does this request any immediate improvement activity of the order
acknowledge process?
Commercial in confidence | © Ericsson AB 2010 | 2010-04-13 | Page 2
Normalizing the process
Average, = 3 days.
Standard deviation, s = 0,5 days.
Order acknowledgement , USL = 4 days.
Z = xc - xm
sx
Z = 4 - 3 = 1 = 2 => Z-table
0,5
0,5
p = 2,28E-02
= 2,28%
s
So approx 2% of the order acknowledgement
will be beyond 4 days (consequently 98% will
be with in the 4 days).
Was this according to the contract?
Commercial in confidence | © Ericsson AB 2010 | 2010-04-13 | Page 3
p
1
2
3
x
4
5
Securing deliveries
The product from a particular supplier has
been late in 5 of 20 occasions. You have now
a very important delivery to make. You want
to assure 99% delivery accuracy. How big
safety margin should you apply to your
supply organization?
Std=5 days
Commercial in confidence | © Ericsson AB 2010 | 2010-04-13 | Page 4
Securing deliveries
Distribution Plot
Normal; Mean=0; StDev=5
0,09
0,08
0,07
0,06
Density
You can see the Z-value
as a safety margin,
because it’s the distance
the process is from the
customer limit.
0,05
0,04
0,03
0,02
0,01
0,00
Z = xc - xm
11,6
99% delivery accuracy => p = 1%
Sx = 5 days
sx
p = 1%
0,01
0
X
Z = 2,32
2,32 = Safety margin
5
Commercial in confidence | © Ericsson AB 2010 | 2010-04-13 | Page 5
Safety margin = 2,32 x 5
= 11,6 days
Securing deliveries
You can see the Z-value as a safety
margin, because it’s the distance the
process is from the customer limit.
sx
p
Z
xm
Z = xc - xm
sx
p = 25%
xc
late in 5 of 20 occasions => p = 25%
Sx = 5 days
Z = 0,67
0,67 = Safety margin
5
Commercial in confidence | © Ericsson AB 2010 | 2010-04-13 | Page 6
Safety margin = 0,67 x 5
= 3,35 days
Securing deliveries
sx
sx
xm
p = 25%
p = 1%
Z
xc
Safety margin = 11,6 days
Z
xm
xc
Safety margin = 3,35 days
Is it possible to release the order 11,6 - 3,35 = 8,25 days
earlier to assure the delivery?
Commercial in confidence | © Ericsson AB 2010 | 2010-04-13 | Page 7
Esercizio 1. Risk calculation.
Qual e’ la probabilita di completare il processo descritto in piu’ di 220 ore note
le attuali prestazioni delle singole attivita’?
task 1
task 2
task 3
task 4
task 5
task 6
task 7
m
(hours)
100
50
25
40
10
15
40
s
(hours)
10
10
5
2
3.5
4
4
2
1
5
3
6
4
Commercial in confidence | © Ericsson AB 2010 | 2010-04-13 | Page 8
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›
Esercizio 1. Soluzione
›
Qual e’ la probabilita di completare il processo descritto in piu’ di 220 ore note
le attuali prestazioni delle singole attivita’?
(hours)
Path 1 (task 1, 2, 5, 7)
path 2 (task 1, 3, 6, 7)
path 3 (task 1, 4, 7)
200
180
108
0.025
(hours)
0.020
15.1
12.52
10.95
0.015
0.010
2
5
0.005
0.0927
0.000
200
X
220
Path 2
Normal; Mean=180; StDev=12.52
0.035
0.030
1
Density
0.025
0.020
3
6
0.015
0.010
0.005
›
›
Il cammino critico e’ dato dal path1
0.000
0.04
Path 3
Normal; Mean=180; StDev=10.95
180
X
0.000699
220
0.03
Z = (220-200)/15.1 = 1.32 => p = 9.27%
Density
›
›
›
s
Density
m
›
Path 1
Normal; Mean=200; StDev=15.1
0.030
0.02
4
0.01
0.00
Commercial in confidence | © Ericsson AB 2010 | 2010-04-13 | Page 9
180
X
0.000130
220
7