CQE Reliability Exam Problem - System Reliability Requirement

R

RACHJAIS

#1
Hi...
I have my CQE exam on this coming saturday. I know i am asking a lot of last minute questions but any help would be appreciated.
Q- The reliability of a component is 0.8. How many of these components are required in parallel if the system reliability requirement is 0.99?

Thanks Much,
 

mdurivage

Quite Involved in Discussions
#3
Re: CQE Reliability Exam Problem

The quick way is trial and error.

1-(the unreliability) ^ # of items
1-(.2)^3=.992
1-(.2)^4=.9984
1-(.2)^5= .99968
1-(.2)^6=.999936
1-(.2)^7=.9999872
 
Last edited:

mdurivage

Quite Involved in Discussions
#4
Re: CQE Reliability Exam Problem

Your approach is correct except that you must use unreliabiliy of the item in the calculation and raised to the number of items not multiplied.




I tried but i could not find the solution.... the correct answer is 3...
I was calculating it this way: 0.99 = (1-0.8)*n
n= i get 4.95 this way... however, the correct answer is 3....
Can someone help me? the notes are not being very helpful
Thanks,
 
Last edited:

Top Bottom