I want to calculate reliability at 750 hours

tahirawan11

Involved In Discussions
#1
Hi,

I have the following test data

Nr of hours / test results

2400 / not failed, test stopped
2400 / not failed, test stopped
2400 / not failed, test stopped
2600 / failed

I want to calculate reliability at 750 hours?

/Tahir
 
Last edited:
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Miner

Forum Moderator
Staff member
Admin
#2
You do not have enough failures to perform a traditional reliability analysis. Do you have historical test or field information on a similar design? If you can assume a shape parameter based on past history, you can perform a Weibayes analysis to estimate what you want.
 

tahirawan11

Involved In Discussions
#3
Many thanks for the reply, unfortunately i do not have any past data as i am working on a new design. If i treat the censor data as failure points i guess i can get a conservative estimate of the reliability of the product?
 

Steve Prevette

Deming Disciple
Staff member
Super Moderator
#5
Hi,

I have the following test data

Nr of hours / test results

2400 / not failed, test stopped
2400 / not failed, test stopped
2400 / not failed, test stopped
2600 / failed

I want to calculate reliability at 750 hours?

/Tahir
IF (and this is a BIG IF) you are willing to assume your failures are Exponentially Distributed, you can total up the hours run (2400 times 3 plus 2600) and say you had one failure over that total number of hours. You can then plug that into the exponential distribution and determine the probability of failure after 750 hours.
 

Miner

Forum Moderator
Staff member
Admin
#6
IF (and this is a BIG IF) you are willing to assume your failures are Exponentially Distributed...
Before considering this, what type of failures do you experience? What is your product and more importantly what is the failure mode?

The exponential distribution is only suitable for random failures. These make a very small percentage of the failures that I see, but what really matters is the type of failures that you see.
 

Steve Prevette

Deming Disciple
Staff member
Super Moderator
#7
Certainly worth the caution, thanks, Miner.

I would like to say that "random failures" wouldn't be the criteria for using the exponential, it would be if the failures were independent of each other, independent of how long the component has been on service, and independent of any external factors, then exponential is a good starting point. The basis of the exponential is - if the probability the item will fail in the next minute (hour, day, year) is the same as the previous minute (hour, day, year) then you generally have an exponential.
 

Bev D

Heretical Statistician
Staff member
Super Moderator
#8
Certainly worth the caution, thanks, Miner.

I would like to say that "random failures" wouldn't be the criteria for using the exponential, it would be if the failures were independent of each other, independent of how long the component has been on service, and independent of any external factors, then exponential is a good starting point. The basis of the exponential is - if the probability the item will fail in the next minute (hour, day, year) is the same as the previous minute (hour, day, year) then you generally have an exponential.
Unfortunately, most products have dependent failure rates. The exponential will provide a great 'baseline' to compare actual results to however...
 

tahirawan11

Involved In Discussions
#9
the part is the 'belt transmission system' of a floor cleaning machine. the failure mode observed was 'Belt broken'. We have identified two critical characteristics for this subsystem, which are 'belt tension' and 'perpendicularity to belt plane'. We think the failure was due to 'perpendicularity was out of specs'

/Tahir
 
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