10.7.2 Failure and Repair Rate Allocations For Parallel Redundant Systems
10.7.2 Failure and Repair Rate
Allocations For Parallel Redundant Systems
A system comprising several stages of redundant subsystems whose
l/m ratio is less than 0.1 can
be treated as if the stages were statistically independent. The system
steadystate availability A_{s} is:
A_{s}
= A_{1} • A_{2} • A_{3} •
... • A_{n}
where:
A_{i}
= the availability of State
I
The procedure for allocating the failure and repair rates and
the availability is as follows.
This is equivalent to
treating each stage as if it had a repairman assigned to it. It is also
equivalent to saying that a single repairman is assigned to the system but
that the probability of a second failure occurring while the first is being
repaired is very small. If the stages are not statistically independent, the
system availability must be computed by the state matrix approach. In either
case, the system requirement can be obtained with a range of failure and
repair rates. Tradeoff procedures must be used to determine the best set of
these parameters.
It will be recalled (from Eq.
(10.52)) that the steadystate measure of availability for a stage where
at least m out of n equipments must be available for the stage to be available
can be expressed by the binomial expansion
and, where m = 1, i.e., only one equipment of n need be
available at any one time, Eq. (10.97) simplifies to:

A_{s} = 1  (1  A)^{n1} 
(10.98) 
If Eq. (10.97) can be expressed in terms of the operability
ratio m/l, the initial allocation may be made. Eq. (10.97) can
be expressed in terms of the operability ratio as:
Now if a value of A_{s} is specified and
we know the system configuration (at least how many equipments out of
nequipments must be available within each stage), we can solve for the
operability ratios m/l.
For example, consider Table 10.71,
in which the system availability requirement of 0.992 has been allocated to
each of 4 series subsystems (stages) as indicated in column (2). In turn, in
order to achieve the given stage availability, it has been determined that
parallel redundant subsystems are required for each stage (column (3)) in
which at least one of the redundant subsystems per stage must be available for
the system availability requirement to be met.
TABLE 10.71: PRELIMINARY
SYSTEM AND SUBSYSTEM
RELIABILITY SPECIFICATIONS
(1) 
(2) 
(3) 
(4) 
(5) 
Stage

Stage Availability

Number of Subsystems (n)

Number of Subsystems
Required (m) 
Operability Ratio

1 2 3 4 
0.9984 0.9976 0.9984 0.9976 
4 5 4 5 
1 1 1 1 
4.0 2.5 4.0 2.5 
The final column (5) indicates the calculated m/l
(operability ratio) required of each
subsystem in the redundant configuration of each stage in order to achieve the
allocated stage availability. Column (5) results are obtained by the use of
Eqs. (10.98) or (10.99). For example, for Stage 1, m = 1, n = 4. Therefore,
since m = 1, we may use Eq. (10.98).
"1">
This represents an upper bound of
the ratio. All
solutions for which the ratio £
.25 are acceptable.
Obviously, there are a multitude of combinations that would
satisfy this equation as shown in Figure 10.71. Until more information
becomes available concerning the cost of various failure rates and repair
rates of the particular equipments involved, this initial specification allows
preliminary equipment design to start with an availability goal that is
consistent with the system’s requirements. To facilitate calculations of
operability ratio, solutions to Eq. (10.99) for n from two through five (Ref. [25]) are
given in Figures 10.72a through 10.72d. The abscissa of the graphs is
expressed in terms of unavailability since the plot allows for greater
linearity, and, thus, ease of reading. Let us solve an example problem
utilizing the graphs.
"1">
FIGURE 10.71: PERMISSIBLE
EQUIPMENT FAILURE
AND REPAIR RATES FOR l/m
= .25
FIGURE 10.72:
UNAVAILABILITY CURVES
(click image to enlarge)
Example 18:
A system consists of five identical, statistically independent
subsystems connected in a parallel redundant configuration. A system
availability of 0.999 is required. Four out of five subsystems must be
operating for the system availability requirement to be met. What is the
required l/m ratio? The procedure for finding this ratio is as
follows.
Procedure 
(1) 

State the system availability requirement,
A_{s} (e.g., A_{s} =
0.999)

(2) 

Compute the system unavailability,
U_{s}, by subtracting As from 1 (e.g.,
U_{s}
= 10.999 =
0.0010)

(3) 

Enter Figure 10.7.22d using m = 2 and
U_{s} = 0.0010, and read the required ratio (e.g.,
l/m =
.01)  