Let’s say we have two identical pumps share a load in parallel. The failure rate for a pump in this mode of operation is 0.0002 failures per hour. If one pump has to carry the full load alone, that pumps failure rate increases to 0.0009 failures per hour.

What is the reliability of the two pump system over a 168 hour week of operation?

The solution is not the simple two units in parallel as the change from shared load to full load on one unit also changes the expected failure rate. The approach is to consider the time to failure for the shared load units, then if one fails, the remaining time is under a higher load and higher failure rate.

According to Dr. Kececioglu, Reliability Engineering Handbook, volume 2, page 147, in general this can be written as

$$ \large\displaystyle R(t)={{R}_{1}}(t){{R}_{2}}(t)+{{Q}_{1}}({{t}_{1}}){{R}_{2}}({{t}_{1}}){{R}_{{{2}’}}}[{{t}_{1e}},(t-{{t}_{1}})]+{{Q}_{2}}({{t}_{2}}){{R}_{1}}({{t}_{2}}){{R}_{{{1}’}}}[{{t}_{2e}},(t-{{t}_{2}})]$$

where the prime elements are the change in reliability expectation due to the change in load. Also, t_{1} is the variable time to failure of unit 1 and is 0 ≤ t_{1} ≤ t, likewise for t_{2}.

When the failure rates are described by the exponential distribution the above equation becomes

$$ \large\displaystyle R(t)={{e}^{-{{\lambda }_{1}}t}}{{e}^{-{{\lambda }_{2}}t}}+\int_{0}^{t}{{{\lambda }_{1}}}{{e}^{-{{\lambda }_{1}}{{t}_{1}}}}{{e}^{-{{\lambda }_{2}}{{t}_{1}}}}{{e}^{-{{{{\lambda }’}}_{2}}(t-{{t}_{1}})}}d{{t}_{1}}+\int_{0}^{t}{{{\lambda }_{2}}}{{e}^{-{{\lambda }_{2}}{{t}_{2}}}}{{e}^{-\lambda 1{{t}_{2}}}}{{e}^{-{{{{\lambda }’}}_{1}}(t-{{t}_{2}})}}d{{t}_{2}}$$

Which integrates to

$$ \large\displaystyle R(t)={{e}^{-({{\lambda }_{1}}+{{\lambda }_{2}})t}}\left( 1-\frac{{{\lambda }_{1}}}{{{\lambda }_{1}}+{{\lambda }_{2}}-{{\lambda }_{2}}^{\prime }}-\frac{{{\lambda }_{2}}}{{{\lambda }_{1}}+{{\lambda }_{2}}-{{\lambda }_{1}}^{\prime }} \right)+\frac{{{\lambda }_{1}}{{e}^{-{{\lambda }_{2}}^{\prime }t}}}{{{\lambda }_{1}}+{{\lambda }_{2}}-{{\lambda }_{2}}^{\prime }}+\frac{{{\lambda }_{2}}{{e}^{-{{\lambda }_{1}}^{\prime }t}}}{{{\lambda }_{1}}+{{\lambda }_{2}}-{{\lambda }_{1}}^{\prime }}$$

And if the two units are identical then

$$ \large\displaystyle {{\lambda }_{1}}={{\lambda }_{2}}={{\lambda }_{{}}}\text{ and }{{\lambda }_{1}}^{\prime }={{\lambda }_{2}}^{\prime }={\lambda }’$$

and the above reduces to

$$ \large\displaystyle R(t)={{e}^{-2\lambda t}}+\frac{2\lambda }{{\lambda }’-2\lambda }\left( {{e}^{-2\lambda t}}-{{e}^{-{\lambda }’t}} \right)$$

Note that the change in failure rate cannot equal two times the original rate else the formula divides by zero. In this special case, there is a solution, and let’s leave that for an exercise for the interested student. Yet, for the stated problem we can solve for the reliability at time t.

$$ \large\displaystyle R(168)={{e}^{-2\times 0.0002\times 168}}+\frac{2\times 0.0002}{0.0009-2\times 0.0002}\left( {{e}^{-2\times 0.0002\times 168}}-{{e}^{-0.0009\times 168}} \right)$$

$- R(168)=0.935+(0.8)(0.935-0.860)=0.995 -$

Related:

Standby Redundancy, Equal Failure Rates, Imperfect Switching (article)

Standby Redundancy with Equal Failure Rates and Perfect Switching (article)

Exponential Reliability (article)

Mark Powell says

Fred,

This answer is completely spurious since it requires use of the exponential distribution for everything. Neither of these pumps will fail with the exponential model.

Mark Powell

Fred Schenkelberg says

And, recall that this forum is for those preparing for the ASQ CRE exam – which often includes such quickly solvable problems assuming exponential distribution. While you certainly know of my dislike for anything assumed constant, when presented with a problem that has only failure rate (not including a time related function) then the answer is the best we have at the moment.

I agree, in practice, I would look for the appropriate function or relationship that describes the rate of change of failure rate. Maybe I should have used a restaurant example. When both waitresses are working then each have a 0.0002 broken ceramic mugs per hour failure rate, when one is covering for the other during a break (covering all tables), then the failure rate jumps to 0.0009. How many mugs should I plan to buy each week? Same basic problem, yet, maybe only slightly more matched to the failure mechanism.

Thanks for the comments – and for those wanting to know more about my thoughts related to assuming the exponential distribution (and MTBF) see http://nomtbf.com/

cheers,

Fred

Mark Powell says

Fred,

Sorry, I missed the “CRE Prep” part. I try to avoid commenting when I see that.

But this does raise an issue about CRE certification (and just about every other certification process I know of (PMP, CSEP, CSP, etc.). To pass the exam, you have to learn to do a lot of things that you really should not do (and have to “unlearn”) in practice. For most of these certifying institutions, it takes forever if it is even possible to change the established body of knowledge to change the exams to reflect the practices we actually need to use.

I don’t know what the answer is, but I have become extremely wary of anyone who touts their certs as any sort of qualification for a job.

Sorry for that post.

Mark Powell