MTBF of n redundant of N equipments
MTBF of n redundant of N equipments
(OP)
Hi,
I am looking for a closed formulae or method for getting the composed MTBF of m redundant equipments of N total.
I.e. 8 pumps, I need 6 working to do the job.
MTBF of 1 pump is X, What is the MTBF_6_of_8_pumps_working?
I got this formula from "introduction to reliability engineering" of E.E.Lewis, John Wiley & Sons but just for 1 of N redundant equipments:
MTBF= Sum_from_1_to_N[ (-1)^(n-1) * C_N_n / (n*Lambda) ]
where C_N_n is de binominal coeficient, Lambda the failure rate, and n the summing variable.
in the same book there is a discussion about m/N redundancy and it gets to a probability of failure (p.o.f.) method stimation,
i.e. for the pumps:
if 1 pump has p.o.f. of 1% so with this method I get 99.994% of probability of at least 6 pumps running
Where I am lost is in how to change from this p.o.f. values to MTBF (if it is possible using i.e. the MTBF of 1 pump)
Thanks for reading!
I am looking for a closed formulae or method for getting the composed MTBF of m redundant equipments of N total.
I.e. 8 pumps, I need 6 working to do the job.
MTBF of 1 pump is X, What is the MTBF_6_of_8_pumps_working?
I got this formula from "introduction to reliability engineering" of E.E.Lewis, John Wiley & Sons but just for 1 of N redundant equipments:
MTBF= Sum_from_1_to_N[ (-1)^(n-1) * C_N_n / (n*Lambda) ]
where C_N_n is de binominal coeficient, Lambda the failure rate, and n the summing variable.
in the same book there is a discussion about m/N redundancy and it gets to a probability of failure (p.o.f.) method stimation,
i.e. for the pumps:
if 1 pump has p.o.f. of 1% so with this method I get 99.994% of probability of at least 6 pumps running
Where I am lost is in how to change from this p.o.f. values to MTBF (if it is possible using i.e. the MTBF of 1 pump)
Thanks for reading!





RE: MTBF of n redundant of N equipments
MTBF and POF are not necessarily directly related. MTBF is a mathematical construct based on the exponential failure rate, and the failure rate does result in probability of failure, but requires knowledge of the time period over which that failure can occur. The POF of an exponential distribution decreases over time.
TTFN
FAQ731-376: Eng-Tips.com Forum Policies
RE: MTBF of n redundant of N equipments
MTBF=1000h (lambda=0.001 failures/hour took as constant)
N=8 pumps total
m=6 pums at least to keep system running
First state: 8 pumps so MTBF_8= 1000/8= 125h for first failure (mean)
Second state: 7 pumps so MTBF_7= 1000/7= 142,85h for second failure (mean)
Third state: 6 pumps so MTBF_6= 1000/6= 166,66h for thir failure (mean)
With the third failure I have the real system fault so:
MTBF_at_least_6_of_8 = 125 + 142.85 + 166.66 = 434.5h (lambda=0.0023 failures/hour)
Regards!
RE: MTBF of n redundant of N equipments
TTFN
FAQ731-376: Eng-Tips.com Forum Policies