10.8.1.1 Fixed Sample
Size Plans
This test plan is based on having the system perform a fixed
number of cycles R. The result is R pairs of timestofailure and down times
(X_{1}, Y_{1}), ......., (X_{R}, Y_{R}).
Let A^{R} = the observed
availability of the test
where:
and
A^{R} = the maximum likelihood estimate of A
Let,
and
The procedure to be followed is:
If r_{o}
ZR > C
reject H_{o}
_{ }r_{o}
ZR £ C accept H_{o} 
(10.106)

where C will be derived in the
following paragraphs.
Assume that the uptimes, X_{i}, are
gamma distributed with parameters (m, q) and the down times, Y_{i}, are
gamma distributed with parameters (n, f) with n f = 1.
Then it can be shown that
rZ_{R}
is Fdistributed with parameters (2nR,
2mR)
The critical value, C, and number
of up/down cycles, R, are determined so that the significance test satisfies
the consumer and producer risk requirements, a
and b, i.e.,

P( r_{o}
ZR > CúA_{o}, R)
£ a 
(10.107) 

P( r_{o}
ZR £
CúA_{1} , R)
£ b 
(10.108) 
which is equivalent to:

C ³ F_{a}(2nR,
2mR) 
(10.109) 
Here F_{a}(u_{1}, u_{2}) denotes the upper a
percentile of the Fdistribution with
parameters u_{1} and u_{2}.
This system of inequalities has two
unknowns and is solved numerically by finding the smallest integer R
satisfying
F_{a}(2nR,2mR) • F_{b}(2mR, 2nR)
£ D
where D is the discrimination ratio,
The value of R obtained in this way is used to calculate the
critical value, C:

C = F_{a}(2nR,
2mR) 
(10.113) 
The OC function is
where F(u_{1} , u_{2} ; x) is the c.d.f. of the Fdistribution with
parameters u_{1} and u_{2}.
The expected test duration is:
The variance of the total test duration is:
For large sample size, R > 20, the distribution of T is
approximately normal.
Example 19: Exponential Distribution
Let the timetofailure and downtime distributions be
exponentially distributed. Therefore, n = m = 1. Let A_{o} = 0.9 and
A_{1} = 0.8 and a = b = 0.2. Calculate the parameters of the test.
Therefore,
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Find the smallest integer R satisfying
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From a Table of Percentiles
of the Fdistribution we find
F_{0.2}(16,16) = 1.536
and F_{0.2}(18,18) = 1.497
Therefore,
R = 9 satisfies the inequality
Therefore,
C = 1.497
The OC function is
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