Q11 A system has MTBF of 200 hrs. Calculate the 100 hr. reliability of a system
ID: 1717347 • Letter: Q
Question
Q11
A system has MTBF of 200 hrs. Calculate the 100 hr. reliability of a system in which one such unit is operative and two identical units are standing by.
Two components each with an MTBF of 1000 hours are connected in (1) active (2) standby redundancy. Determine the overall system MTBF.
The failure rate of a device is constant equal to 0.06x10-3 per hr. How many standby devices are required to achieve a reliability of more than 0.985 for an operating period of 10,000 hrs? What is the MTTF of the resulting system?
Explanation / Answer
Reliablity = e^-T/M
T = time = 100 hr
M = MTBF = 200 hrs
R = e^-100 /200
Reliabilty = 2.21
c = 0.06 x 10 ^-3 /hr
Total reliabilty R = 0.985
time T = 10000 hrs
R = e ^-CT
0.985 x n = e^-0.06 x 10^-3 x 10000
n = no of stand by devices = 3
MTTF= 1 /C
MTTF = 1/ 0.06 x 10^-3
= 16666.66 hrs
overall system of MTBF = 1000/2
= 500 hrs
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