Failure Rates, MTBFs, and All That

Suppose we're given a batch of 1000 widgets, and each functioning 
widget has a probability of 0.1 of failing on any given day, 
regardless of how many days it has already been functioning.  This
suggests that about 100 widgets are likely to fail on the first day,
leaving us with 900 functioning widgets.  On the second day we would
again expect to lose about 0.1 of our functioning widgets, which
represents 90 widgets, leaving us with 810.  On the third day we
would expect about 81 widgets to fail, and so on.  Clearly this is
an exponential decay, where each day we lose 0.1 of the remaining
functional units.  In a situation like this we can say that widgets
have a constant failure rate (in this case, 0.1), which results in
an exponential failure distribution.  The "density function" for a
continuous exponential distribution has the form

                 f(t)  =  L exp(-L t)                     (1)

where L is the rate.  For example, the density function for our
widgets is (0.1)exp(-t/10), which is plotted below:



Notice that by assuming the probability of failure for a functioning
widget on any given day is independent of how long it has already been
functioning we are assuming that widgets don't "wear-out" (nor do they
improve) over time.  This characteristic is sometimes called "lack of
memory", and it's fairly accurate for many kinds of electronic devices
with essentially random failure modes.  However, in each application
it's important to evaluate whether the devices in question really do
have constant failure rates.  If they don't, then use of the exponential
distribution may be misleading.

Anyway, assuming our widgets have an exponential failure density as
defined by (1), the probability that a given widget will fail between 
t=t0 and t=t1 is just the integral of f(t) over that interval.  Thus, 
we have

        Pr{t0 < tfail < t1}  =  exp(-L t0) - exp(-L t1)

Of course, if t0 equals 0 the first term is simply 1, and we have
the cumulative failure distribution

                F(t)  =   1 - exp(-L t)               (2)

which is the probability that a functioning widget will fail at any 
time during the next t units of time.  By the way, for ANY failure
distribution (not just the exponential distribution), the "rate" at
any time t is defined as

                     dF(t)/dt           f(t)
       Rate(t)  =   ----------   =   ----------       (3)
                     1 - F(t)         1 - F(t)

In other words, the "failure rate" is defined as the rate of change 
of the cumulative failure probability divided by the probability
that the unit will not already be failed at time t.  Notice that
for the exponential distribution we have f(t)=Lexp(-L t) and
F(t) = 1 - exp(-L t), so the rate is simply the constant L.

It might also be worth mentioning that the function exp(x) has the
power series representation exp(x) = 1 + x + x^2/2! + ..., and so
if the product Lt is much smaller than 1 we have approximately
exp(x) ~ 1 + x, which when substituted into (2) gives a rough
approximation for the cumultative failure probability F(t) ~ Lt.

Now, we might ask what is the *mean time* to fail for a device with
an arbitrary failure density f(t)?  We just need to take the weighted
average of all time values from zero to infinity, weighted according 
to the density.  Thus the mean time to fail is

                     inf
                    INT   t f(t) dt
                      0
         MTTF  =   -----------------               (4)
                      inf
                     INT  f(t) dt
                       0

Of course, the denominator will ordinarily be 1, because the device
has a cumulative probability of 1 of failing some time from 0 to
infinity.  Thus it is a characteristic of probability density
functions that the integrals from 0 to infinity are 1.  As a result,
the mean time to fail can usually be expressed as

                        inf
              MTTF  =  INT   t f(t) dt              (5)
                        0

If we substitute the exponential density f(t) = L e(-L t) into this
equation and evaluate the integral, we get  MTTF  =  1/L.  Thus the
mean time to fail for an exponential system is the inverse of the
rate.

Now let's try something a lttle more interesting.  Suppose we 
manufacture a batch of dual-redundant widgets, hoping to improve
their reliability in service.  A dual-widget is said to be failed 
only when *both* sub-widgets have failed.  What is the failure 
density for a dual-widget?  Well, this can be derived in several
different ways, but one simple way is to realize that the probability 
of both sub-widgets being failed by time t is 

 [1 - exp(-L t)][1 - exp(-L t)]  =  1 - 2exp(-L t) + exp(-2L t)

so this is the cumulative failure distribution F(t) for dual-widgets.
From this we can immediately infer the density distribution f(t),
which is simply the derivative of F(t) (recalling that F(t) is the 
integral of f(t)), so we have

         f(t)  =  2L exp(-L t)  -  2L exp(-2L t)

Notice that this is NOT a pure exponential distribution anymore 
(unlike the simplex-widgets).  A plot of this density is shown below:



Remember that the failure density for the simplex widgets is a maximum
at t=0, whereas it is ZERO for a dual-widget.  It then rises to a 
maximum and falls off.  What is the mean time for a dual-widget to 
fail?  Well, as always, we get that by evaluating equation (5) above, 
but now we use our new dual-widget density function.  Evaluating
the integral gives MTTF = (3/2)(1/L).

It sometimes strikes people as counter-intuitive that the early
failure probability of a dual-redundant system is so low, and yet
the MTTF is only increased by a factor of 3/2, but it's obvious
from the plot of the dual-widget density f(t) that although it it 
does extremely well for the early time period, it eventually rises
above the simplex widget density.  This stands to reason, because 
we're very unlikely to have them BOTH compoents of a dual-widget
fail at an early point, but on the other hand they each component
still has an individual MTTF of 1/L, so they it isn't likely that 
EITHER of them will survive far past their mean life.

By the way, by this same approach we can determine that a triplex-
widget would have a mean time to failure of (11/6)(1/L), and a 
quad-widget would have an MTTF of of (25/12)(1/L).  Notice that
the leading factors are

         1   =  1
        3/2  =  1 + 1/2
       11/6  =  1 + 1/2 + 1/3
       25/12 =  1 + 1/2 + 1/3 + 1/4

and so on.  For more on this, see Infinite Parallel Redundancy.