10.4.3.1 Model A  Based Upon
Probability of Failure During Previous Mission and Probability of Repair
Before Next Mission Demand
In this model, the assumption is made that if no failures
needing repair occurred in the preceding mission, the system is immediately
ready to be used again; and, if such failures did occur, the system will be
ready for the next mission only if its maintenance time is shorter than the
time by which the demand for its use arises. The operational readiness POR may
then be expressed as:

P_{OR} =
R(t) + Q(t) • P(t_{m}
< t_{d}) 
(10.72) 
where:
R(t) 
= 
probability of no failures in the preceding
mission

Q(t) 
= 
probability of one or more failures in
the preceding mission

t 
= 
mission duration

P(t_{m} < t_{d}) 
= 
probability that if failures occur, the
system maintenance time, t_{m},
is shorter than the time, t_{d}, at which the
next demand or call for mission engagement
arrives 
The calculations of R(t) and Q(t) = 1  R(t) are comparatively
simple using standard reliability equations; however, all possible types of
failures that need fixing upon return in order to restore in full the system
reliability and combat capability must be considered, including any failures
in redundant configurations.
As for P(t_{m} < t_{d}), one needs to know the probability distributions of
the system maintenance time and of call arrivals. Denoting by f(tm) the
probability density function of maintenance time and by
g(t_{d}), the probability density
function of time to the arrival of the next call, counted from the instant the
system returned from the preceding mission in a state requiring repair, the
probability that the system will be restored to its full operational
capability before the next call arrives is:
The integral in the square brackets on the right side of the
equation is the probability that the call arrives at t_{d} after a
variable time t_{m}. When this is
multiplied by the density function f(t_{m}) of the duration of maintenance times and integrated
over all possible values of t_{m},
we get P(t_{m}< t_{d}).
Now assume that maintenance time
t_{m} and time to next call arrival t_{d} are
exponentially distributed, with M_{1} being the
mean time to maintain the system and M_{2} the mean time to next
call arrival. The probability density functions are thus:

f(tm) = [exp(t_{m}/M_{1})]/M_{1} 
(10.74) 

f(td) = [exp(t_{d}/M_{2} )]/M_{2} 
(10.75) 
We then obtain


(10.76) 
In this exponential case, system operational readiness
becomes

POR = R(t) + Q(t) • [
M_{2} /(M_{1} +M_{2})
] 
(10.77) 
As a numerical example let us look at a system with a
probability of R = 0.8 of returning from a mission of t = 1 hr duration
without requiring repair and therefore had a probability of Q = 0.2 that it
will require repair. If system mean maintenance time is
M_{1} = 1 hr and the mean time to next call arrival is
M_{2} = 2 hr., the operational readiness of the system
becomes
P = 0.8 + 0.2 (2/3) =
0.933
Comparing this result with the conventional steadystate
availability concept and assuming that the system has a mean maintenance time
of M_{1} = 1 hr and a mean time to failure of
M_{2} = 5 hr (roughly corresponding to the exponential case
of R = 0.8 for a onehour mission), we obtain a system availability
of:
A = M_{2}/(M_{1} + M_{2}) = 5/6 = 0.833
which is a result quite different from POR =
0.933.