Discussion
What is the failure rate function of the series system of two independent components when the failure rate functions of the two components are respectively \(r_{1}(t)\) and \(r_{2}(t)\) ?
Answer: B
The failure rate is \(r_{1}(t) + r_{2}(t)\).
A series system works only while both components work, so its reliability is the product of theirs:
\[ \begin{aligned} R(t) &= R_{1}(t)R_{2}(t) \\ &= e^{-\int_{0}^{t} (r_{1} + r_{2})\,du} \end{aligned} \]
since each \(R_{i}(t) = \exp(-\int_{0}^{t} r_{i})\). Hence the system's failure rate \(-R'/R\) is \(r_{1}(t) + r_{2}(t)\): hazards add in series. The minimum or maximum would ignore one component's risk, and the product has no reliability meaning here.
No comment is present. Be the first to comment.
Loading…
Post your comment
Please log in or register to post a comment.
Examination Categories
- SSC Exams
- Banking Exams
- Teaching Exams
- Civil Services Exams
- Railways Exams
- Engineering Recruitment Exams
- Defence Exams
- State Government Exams
- Police Exams
- Insurance Exams
- Nursing Exams
- Other Government Exams
- NRA CET
- PG Entrance Exams
- Campus Placement
- MBA Entrance Exams
- CUET and UG Entrance
- Judiciary Exams
- Regulatory Body Exams
- Software Development