Measure Solutions Normalizing the process The time for

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 p = 2, 28 E-02 = 2, 28% 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 s 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 You can see the Z-value as a safety margin, because it’s the distance the process is from the customer limit. Z = xc - xm 99% delivery accuracy => p = 1% Sx = 5 days sx p = 1% Z = 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 = 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 p = 1% Z xm xc Safety margin = 11, 6 days p = 25% 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 7

› 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 › (hours) › › › Path 1 (task 1, 2, 5, 7) path 2 (task 1, 3, 6, 7) path 3 (task 1, 4, 7) 200 180 108 s (hours) 15. 1 12. 52 10. 95 2 1 5 3 6 › Il cammino critico e’ dato dal path 1 › Z = (220 -200)/15. 1 = 1. 32 => p = 9. 27% 4 Commercial in confidence | © Ericsson AB 2010 | 2010 -04 -13 | Page 9 7