The pdf for the Eyring relationship and the exponential distribution is given next.
The pdf of the 1-parameter exponential distribution is given by:
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It can be easily shown that the mean life for the 1-parameter exponential distribution, presented in detail in the Life Distributions chapter, is given by:
(3)

thus,
(4)

The Eyring-exponential model pdf can then be obtained by setting m = L(V) in Eqn. (1),

and substituting for m in Eqn. (4),
(5)
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Eyring Exponential Statistical Properties Summary
Mean or MTTF
The mean,
, or mean time to failure (MTTF) for the Eyring-exponential relationship is given by:

Median
The median,
for the Eyring-exponential relationship is given by:

Mode
The mode,
for the Eyring-exponential relationship is
= 0.
Standard Deviation
The standard deviation,
, for the Eyring-exponential relationship is given by:

Eyring-Exponential Reliability Function
The Eyring-exponential reliability function is given by:
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This function is the complement of the Eyring-exponential cumulative distribution function or,

and,

Conditional Reliability
The conditional reliability function for the Eyring-exponential relationship is given by:
Reliable Life
For the Eyring-exponential relationship, the reliable life, or the mission duration for a desired reliability goal
is given by:

or,

Parameter Estimation
Maximum Likelihood Estimation Method
The complete exponential log-likelihood function of the Eyring model is composed of two summation portions,

where:
·
is the number of groups of exact times-to-failure data points.
·
is the number of times-to-failure in the
time-to-failure data group.
·
is the stress level of the
group.
· A is the Eyring parameter (unknown, the first of two parameters to be estimated).
· B is the second Eyring parameter (unknown, the second of two parameters to be estimated).
·
is the exact failure time of the
group.
· S is the number of groups of suspension data points.
·
is the number of suspensions in the
group of suspension data points.
·
is the running time of the
suspension data group.
The solution (parameter estimates) will be found by solving for the parameters
and
so that
= 0 and
= 0 where:
See Also:
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