

MILHDBK338B: Electronic Reliability Design Handbook 
 

6.3.5 Minimization of Effort Algorithm
6.3.5 Minimization of Effort
Algorithm
This algorithm considers minimization of total effort expended
to meet system reliability requirements. It assumes a system comprised of n
subsystems in series. Certain assumptions are made concerning the effort
function. It assumes that the reliability of each subsystem is measured at the
present stage of development, or is estimated, and apportions reliability such
that greater reliability improvement is demanded of the lower reliability
subsystems.
Let R1, R2, . . ., Rn denote
subsystem reliabilities, and the system reliability R would be given
by:
Let R* be the required
reliability of the system, where R* > R. It is then required to increase at
least one of the values of the Ri
to the point that the required reliability
R* will be met. To accomplish such an increase takes a certain effort, which
is to be allocated in some way among the subsystems. The amount of effort
would be some function of the number of tests, amount of engineering manpower
applied to the task, etc.
The algorithm assumes that each
subsystem has associated with it the same effort function,
G(R_{i},_{}), which measures the amount of effort needed to
increase the reliability of the i^{th} subsystem from R_{i} to _{}
The
problem, then, is to determine _{}
is minimized subject to the
condition:
With the preceding
assumptions, it can be shown that the unique solution is:
where the subsystem
reliabilities R_{1},
R_{2}, ...,
R_{n} are ordered in an increasing fashion (assuming such an
ordering is implicit in the notation).
R_{1} £ R_{2}
£ ...
£ R_{n}
and the number K_{o} is determined as:
K_{o} = maximum value of j such that
where R_{n+1} = 1 by
definition.
The number _{} is determined as
It is evident that the
system reliability will then be R*, since the new reliability is:
when the relationship
for _{} is substituted.
Example 8:
As an example, consider a
system that consists of three subsystems (A, B, and C), all of which
must function without failure in
order to achieve system success. The system reliability requirement has been set at 0.70. We have
predicted subsystem reliabilities as R_{A} = 0.90,
R_{B} = 0.80, and R_{C} = 0.85. How should we apportion reliability to the
subsystem in order that the total effort be minimized and that the system reliability requirement be
satisfied? Assume identical effort
functions for the three subsystems.
The resulting minimum effort
apportionment goals are found as follows:
(1) Arrange
subsystem reliability values in ascending order:
R_{1} =
R_{B} = 0.80,
R_{2} = R_{C} =
0.85, R_{3} = R_{A}
= 0.90
(2)
Determine K_{o} , the maximum value of j, such that:
"1">
(3)
When j = 1,
Note that R_{n+1}
was previously defined as 1 (Eq. 6.16).
(4) When j =
2,
"1">
(5) When j =
3,
"1">
(6) Since
R_{1} < r_{1} , R_{2}
< r_{2} , but R_{3} > r_{3}
, then K_{o} = 2 because 2 is the largest subscript
j such that R_{j}
< r_{j} . Thus,
"1">
which means that the effort
is to be allotted so that subsystem B increases in reliability from 0.80
to 0.882, and subsystem C increases
in reliability from 0.85 to 0.882, whereas subsystem A is left alone with a reliability of 0.90. The resulting
reliability of the entire system is, as required, 0.70 = (0.882)^{2}(0.90) . This means
that effort should be expended on subsystems C and B to raise their respective reliabilities to 0.882
with no developmental effort spent on subsystem A. This policy would minimize the total expended
effort required to meet system reliability requirements. The minimization, however, is dependent upon the effort
in meeting the initial assumptions,
which may not be possible.




 
 