Two Pumps Problem

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, t₁ is the variable time to failure of unit 1 and is 0 ≤ t₁ ≤ t, likewise for t₂.

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.

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

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

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Related:

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

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

Exponential Reliability (article)