Reliability of Machine Components
Mean Time Between Failure  MTB and reliability of machine components and systems.
Reliability characterizes components or system of components by the probability they will perform the desired functions for a given time.
In general 
 more components and/or more complicated systems reduces reliability
 simpler systems with few components increases reliability
Reliability equations:
Reliability
Reliability at a given time:
R = e^{}^{λt} (1)
where
R = reliability. Values between 0  1 where value 1 indicates 100% live components and value 0 indicates 0% live components.
λ = proportional failure rate  a failure rate expressed as a proportion of initial number of live components  N_{o } at time t_{ }
t = time (hours)  Note! other units can be used as long as the use is consistent through the calculations.
The failure rate at time t can be expressed as
λ = N_{F} / (N_{o }t)_{ }
= (N_{o}  N_{s}) / (N_{o }t) (2)
where
N_{F} = N_{o}  N_{s} = number of failing components at time t
N_{s} = number of live surviving components at time t
N_{o} = initial number of live components at time zero
Example  Failure Rate and Reliability
A car manufacturer sells 400000 cars of a certain model in one year. During the the first tree years the owners of 50000 of these cars experience major failures. The failure rate can be calculated as
λ = (50000 cars)/ ((400000 cars) (3 years))
= 0.042 (per year)
The reliability of a model of this car within three years can be calculated as
R = e^{(}^{0.042 1/year) (3 year)}
= 0.88
= 88 %
Unreliability  the Probability for a Device to Fail
The connection between reliability and unreliability:
R + Q = 1 (3)
where
Q = unreliability. Values between 0  1 where value 1 indicates 0% live components and value 0 indicates 100% live components.
(1) and (2) can be used to express unreliability
Q = 1 e^{}^{λt} (4)
Example  Unreliability
The unreliability of the car model in the example above within three years can be calculated as
Q = 1  e^{(}^{0.042 1/year) (3 year)}
= 0.12
= 12 %
Number of Live Components
The number of live surviving components in a system at a given time:
N_{s} = N_{o} e^{}^{λt} (5)
Number of Failure Components
The number of failure dead components in a system at a given time:
N_{s} = N_{o} (1  e^{}^{λt}) (6)
Mean Time Between Failures  MTBF
Mean time between failures  MTBF:
MTBF = 1 / λ (7)
where
MTBF = Mean Time Between Failure (hours)
 MTTF  Mean Time To Failure is an alternative to MTBF
Example  Mean Time Between Failure
From the example above the failure rate is 0.42 per year. MTBF can be calculated as
MTBF = 1 / (0.042 1/year)
= 23.8 years
Mean Time Between Failure (MTBF) can be determined by rating Total Surviving Hours against Number of Failures as
MTBF = t_{s} / N_{F} (8)
where
t_{s} = total surviving hours (hours)
Combining (5) with the formulas for reliability and more
R = e^{}^{t/m} (1b)
Q = 1 e^{}^{t/m} (4b)
N_{s} = N_{o} e^{}^{t/m} (5b)
N_{f} = N_{o} (1  e^{}^{t/m}) (6b)
MTBF vs. MTTF vs. MTTR
 MTBF  Mean Time Between Failures > commonly used to determine average time between failures
 MTTF  Mean Time To Failure > commonly used for replaceable products that cannot be repaired
 MTTR  Mean Time To Repair > commonly used to determine how long it will take to get a failed product running again
MTBF and MTTF are often used interchangeably.
Systems Reliability
There are two basic types of reliability systems  series and parallel  and combinations of them.
 in a series system  all devices in the system must work for the system to work
 in a parallel system  the system works if at least one device in the system works
Reliability of Systems in Series
Reliability of systems in series can be expressed as
R_{s} = R_{1} R_{2} (9)
where
R_{s} = system reliability
R_{1,2} = subsystem reliability
Example  Reliability of Systems in Series
From the example above the reliability of a car over three year is 0.88. If we depends on two cars for a mission  whatever it is  the reliability for our mission will be
R_{s} = 0.88 0.88
= 0.77
= 77 %
Reliability of Systems in Parallel
Reliability of systems in parallel can be expressed as
R_{s} = R_{1} Q_{2} + R_{2} Q_{1} + R_{1} R_{2} (10)
where
Q_{1,2} = (1  R_{1,2}) subsystem unreliability
Example  Reliability of Systems in Parallel
From the example above the reliability of a car over three year is 0.88. If it is enough with one of our two cars for our mission  the reliability for our mission can be calculated as
R_{s} = 0.88 (1 0.88) + 0.88 (1 0.88) + 0.88 0.88
= 0.99
= 99 %
Note!  two systems in parallel with 99 % reliability vs. one system alone with reliability 88 % shows the power of redundancy.
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