Measure

1
Measure

• Solutions

2
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?

3
Normalizing the process
Average, 3 days. Standard deviation, s 0,5
days. Order acknowledgement , USL 4 days.
Z xc - xm sx
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?
4
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?

Std5 days
5
Securing deliveries

• You can see the Z-value as a safety margin,
  because its the distance the process is from the
  customer limit.

Z xc - xm sx
99 delivery accuracy gt p 1 Sx 5 days
6
Securing deliveries

• You can see the Z-value as a safety margin,
  because its the distance the process is from the
  customer limit.

Z xc - xm sx
late in 5 of 20 occasions gt p 25 Sx 5 days
7
Securing deliveries
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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? m s
(hours) (hours) task 1 100
10 task 2 50 10 task 3
25 5 task 4 40
2 task 5 10 3.5 task 6
15 4 task 7 40 4
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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?
• m s
• (hours) (hours)
• Path 1 (task 1, 2, 5, 7) 200 15.1
• path 2 (task 1, 3, 6, 7) 180 12.52
• path 3 (task 1, 4, 7) 108 10.95

• Il cammino critico e dato dal path1
• Z (220-200)/15.1 1.32 gt p 9.27

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