8.
Assessing Product Reliability
8.3. Reliability Data Collection 8.3.1. How do you plan a reliability assessment test?


How to plan a Bayesian test to confirm a system meets its MTBF objective 
Review Bayesian Basics and
assumptions,
if needed. We start at the point when gamma prior parameters
\(a\) and \(b\)
have already been determined. Assume we have a given MTBF objective, \(M\),
and a desired confidence level of \(100(1\alpha)\).
We want to confirm the system will have an MTBF of at least \(M\)
at the \(100(1\alpha)\)
confidence level. As in the section on
classical (HPP) test plans, we pick a number of
failures, \(r\),
that we can allow on the test. We need a test time \(T\)
such that we can observe up to \(r\)
failures and still
"pass" the test. If the test time is too long (or too short), we can
iterate with a different choice of \(r\).
When the test ends, the posterior gamma distribution will have (worst case  assuming exactly \(r\) failures) new parameters of $$ a' = a+r, \,\,\,\,\, b' = b + T \, ,$$ and passing the test means that the failure rate \(\lambda_{1\alpha}\), the upper \(100(1\alpha)\) percentile for the posterior gamma, has to equal the target failure rate \(1/M\). But this percentile is, by definition, \(G^{1}(1\alpha; \, a', \, b')\), with \(G^{1}\) denoting the inverse of the gamma distribution with parameters \(a', \, b'\). We can find the value of \(T\) that satisfies \(G^{1}(1\alpha; \, a', \, b')=1/M\) by trial and error. However, based on the properties of the gamma distribution, it turns out that we can calculate \(T\) directly by using $$ T = M \cdot G^{1}(1\alpha; \, a', \, 1)  b \, .$$ Special Case: The Prior Has \(a\) = 1 (The "Weak" Prior) 

When the prior is a weak prior with \(a\) = 1, the Bayesian test is always shorter than the classical test 
There is a very simple way to calculate the required Bayesian test time
when the prior is a weak prior
with \(a\)
= 1. Just use the
Test Length Guide Table
to calculate the classical test time. Call this \(T_c\).
The Bayesian test time \(T\) is just \(T_c\)
minus the prior parameter \(b\)
(i.e., \(T = T_c  b\)).
If the \(b\)
parameter was set equal to \((\mbox{ln } 2) \cdot \mbox{MTBF }_{50}\)
(where \(\mbox{MTBF}_{50}\)
is the consensus choice for an "even money" MTBF), then
$$ T = T_c  (\mbox{ln } 2) \cdot \mbox{MTBF }_{50} $$
This shows that when a weak prior is used, the Bayesian test time is always
less than the corresponding classical test time. That is why this prior
is also known as a friendly prior.
Note: In general, Bayesian test times can be shorter, or longer, than the corresponding classical test times, depending on the choice of prior parameters. However, the Bayesian time will always be shorter when the prior parameter \(a\) is less than, or equal to, 1. 

Example 
A new piece of equipment has to meet a MTBF
requirement of 500 hours at 80 % confidence. A group of engineers decide
to use their collective experience to determine a Bayesian gamma prior
using the 50/95 method described in
Section 2.
They think 600 hours is a likely MTBF value and they are very confident
that the MTBF will exceed 250. Following the
example
in Section 2, they determine that the gamma prior parameters are \(a\)
= 2.863 and \(b\)
= 1522.46.
Now they want to determine an appropriate test time so that they can confirm a MTBF of 500 with at least 80 % confidence, provided they have no more than two failures. We obtain a test time of 1756.117 hours using $$ 500 \cdot G^{1}(10.2; \, 2.863+2, \, 1)  1522.46 \, . $$ To compare this result to the classical test time required, use the Test Length Guide Table. The table factor is 4.28, so the test time needed is 500 × 4.28 = 2140 hours for a nonBayesian test. The Bayesian test saves about 384 hours, or an 18 % savings. If the test is run for 1756 hours, with no more than two failures, then an MTBF of at least 500 hours has been confirmed at 80 % confidence. If, instead, the engineers had decided to use a weak prior with an \(\mbox{MTBF}_{50}\) of 600, the required test time would have been $$ 2140  600 \cdot \mbox{ln } 2 = 1725 \mbox { hours} \, . $$ 