Dual Failures with General Densities

Suppose a system contains two independent widgets, denoted by A and 
B, in parallel, only one of which is required for the system to be 
operational, and the system performs a series of missions each 
lasting D hours.  If each widget has a failure rate characterized 
by the density distribution f(t), and the system is run for T hours 
or until the first mission when at least one widget fails (whichever 
comes first), what is the probability that BOTH widgets will fail 
during a single mission?

The joint density function for failure of widget A at time tA and 
widget B at time tB is just the product f(tA)f(tB).  Graphically we 
can represent the space of possible outcomes as the plane of tA 
versus tB, and then integrate the joint density function over the 
region(s) of interest.  For the case we have described, the region 
of interest is the shaded region shown in the figure below.

           

One way of performing the integration is to split up the range into 
three separate spans, over which the boundaries vary linearly, as 
shown below.

            

This leads to the following sum of three double-integrals for the 
total probability

           D     x+D          T-D    x+D           T     T
           /     /             /     /             /     /
  P(T,D) = |f(x) |f(y)dydx  +  |f(x) |f(y)dydx  +  |f(x) |f(y)dydx
           /     /             /     /             /     /
           0     0             D     x-D          T-D    x-D

However, there are more economical ways of evaluating this overall
probability.  We could reduce the number of double integrals from 
three to two by noting the symmetry about the diagonal line, and 
just integrating the region above the line (and then multiplying 
by 2).  An even better approach is to integrate this same "upper"
region diagonally instead of orthogonally, taking the coincidence
closeness as the outer parameter, as illustrated below.

            

This leads to an expression for P(T,D) as just one double integral

                           D T-u
                           / /
              P(T,D)  =  2 | | f(v) f(v+u) dv du             (1)
                           / /
                          0  0

This formulation is particularly nice because it can easily be
modified to include "turnback" credit.  This relates to cases where
the failure of the first widget during a mission causes a turnback
that truncates the mission in a time equal to the time from the
start of the mission, as indicated in the figure below.

            

To capture this turnback credit we simply replace the "0" lower
bound on the inner integral with "u".  In comparison, the original
orthogonal method requires a sum of FOUR double integrals to give
the same result.

For any given value of D the probability P(T,D) converges on a
constant value as T goes to infinity.  This limit represents 
the probability that the two widgets would fail within D hours 
of each other assuming we remove the "hard" limit of T hours 
entirely.  Interestingly, if we take the exponential distribution 
as f(t), then the limit of P(T,D) as T goes to infinity equals 
the cumulative distribution function F(D) of the exponential 
distribution.  In other words, the coincidence threshold for 
two exponentially distributed events (with no truncation) is 
itself exponentially distributed.

To illustrate the use of formula (1), suppose we have a system with
two widgets in parallel, and the failure characteristic of each 
widget has a Weibull distribution with scale parameter a=385 and
shape parameter b=5.55.  Recall that the cumulative Weibull 
distribution has the form
                                
                             -(t/a)^b
             F(t)  =  1  -  e

This is just a generalization of the exponential distribution, and
the parameter "a" simply scales the time variable, whereas "b" affects
the shape of the distribution.  If we take b=1 we have the exponential
distribution (constant failure rate), whereas with b other than 1 we
get either an increasing or decreasing failure rate.  Typically "b"
is greater than 1, signifying the tendancy of the failure rate to
increase with time, due to the "wear out" of parts.

Differentiating F(t) with respect to t, it follows that the Weibull 
density distribution is

                     b  / t \ b-1  -(t/a)^b
            f(t)  =  - ( --- )    e
                     a  \ a /

For the stated values of "a" and "b", a plot of this density function
is shown below.

            

The density of the dual-widget system, where each widget has the
individual failure density f(t), is illustrated in the plot below.

            

To find the probability that both widgets will fail within 12 hours
of each other over a period of T, we apply equation (1), which gives
the cumulative probability curve shown below

            

This would enable us to determine a suitable exposure time for a
dual-widget system to ensure against a complete system failure during
any 12-hour mission.