# Reliability Requirement Understanding

Six Sigma – iSixSigma › Forums › General Forums › New to Lean Six Sigma › Reliability Requirement Understanding

- This topic has 8 replies, 2 voices, and was last updated 1 week, 3 days ago by Fausto Galetto.

- AuthorPosts
- September 3, 2020 at 5:48 am #249718
I have got the following requirement that I’m a little bit struggling to understand how was it calculated.

“A limiting curve of failure behavior for the new component will be defined: with Weibull distribution and shape parameter Beta=1.00. The curve is based on field analysis of previous component. The failure probability at 17.500 km is equal 0.34%. The second node is located by 100.000 km mileage and a failure probability of 1.94%.” in addition to this a confidence level of 90% was defined.

Is there a missing information here? I thought it was calculated using the Weibull CDF but the numbers I receive are different. As alpha I have used 100,000km.

Appreciate if somebody can clarify what I am missing.

0September 3, 2020 at 10:13 am #249725

Fausto GalettoParticipant@fausto.galetto**Include @fausto.galetto in your post and this person will**

be notified via email.If you provide the data I will see what I can do.

I do not understand what do you mean with “””second node”””

0September 3, 2020 at 4:36 pm #249735This is the problem that this all information I have. My assumption is that the failure has weibull distribution. The customer provided only two points of his interest. One is at 17,500km another at 100,000 km. I assumed that the “failure probability” indicates commulative failure up to the above mentioned milage and tried to model the curve but the failure rates I have recieved do not alighn with the one were provided by the customer.

so I’m a little stuck.

0September 4, 2020 at 4:23 am #249743

Fausto GalettoParticipant@fausto.galetto**Include @fausto.galetto in your post and this person will**

be notified via email.Your customer is

**ignorant**.The information provided is

: beta=1 and the two points.**contradictory**From the reliability curve, any

**educated “manager”**derives beta=0.7430September 5, 2020 at 3:16 pm #249771Thanks. I also thought something wrong with the requirement but was not sure.

0September 5, 2020 at 3:52 pm #249773By the way, how did you get to beta=0.743? Thanks.

0September 6, 2020 at 4:24 am #249781

Fausto GalettoParticipant@fausto.galetto**Include @fausto.galetto in your post and this person will**

be notified via email.Fitting the Weibull to pass through the two points!!!

You must study a bit of THEORY

0September 7, 2020 at 3:05 pm #249823Appriciate for your kidness. what SW are you using to fit the destribution?

0September 8, 2020 at 2:38 am #249830

Fausto GalettoParticipant@fausto.galetto**Include @fausto.galetto in your post and this person will**

be notified via email.NO SW!!!

Only a bit of Mathematics…

0 - AuthorPosts

You must be logged in to reply to this topic.