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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

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Measure

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  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. Z = 4 - 3 = 1 = 2 => Z-table 0,5 0,5 p = 2,28E-02 = 2,28% s p 1 2 3 4 5 x 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. 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? Securing deliveries Std=5 days

  5. You can see the Z-value as a safety margin, because it’s the distance the process is from the customer limit. p = 1% Z = 2,32 2,32 = Safety margin Safety margin = 2,32 x 5 5 = 11,6 days Securing deliveries Z =xc - xm sx 99% delivery accuracy => p = 1% Sx = 5 days

  6. You can see the Z-value as a safety margin, because it’s the distance the process is from the customer limit. sx p xm xc Z p = 25% Z = 0,67 0,67 = Safety margin Safety margin = 0,67 x 5 5 = 3,35 days Securing deliveries Z =xc - xm sx late in 5 of 20 occasions => p = 25% Sx = 5 days

  7. Z sx sx p = 25% p = 1% Z xm xc xm xc Safety margin = 11,6 days Safety margin = 3,35 days Is it possible to release the order 11,6 - 3,35 = 8,25 days earlier to assure the delivery? Securing deliveries

  8. 2 5 1 3 6 7 4 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’? ms (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

  9. 2 5 1 3 6 7 4 • Esercizio 1. Soluzione • Qual e’ la probabilita di completare il processo descritto in piu’ di 220 ore note le attuali prestazioni delle singole attivita’? • ms (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 => p = 9.27%

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