Subjects reliability engineering

Standby Vs Parallel

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Standby Vs Parallel


1. **State the problem:** We have a standby system with one basic component and two standby components, each with an exponential time-to-failure distribution. The failure rate $\lambda$ for each component is 0.006 per hour, and the operation period is 500 hours. We need to find the reliability of this standby system and its mean time to failure (MTTF). Then, compare these results with a parallel system of the same components. 2. **Reliability of a single component:** For an exponential distribution, reliability at time $t$ is given by: $$ R(t) = e^{-\lambda t} $$ Given $\lambda = 0.006$ and $t = 500$ hours, $$ R(500) = e^{-0.006 \times 500} = e^{-3} \approx 0.0498 $$ 3. **Reliability of the standby system:** In a standby system with one active and two standby components, the system fails only if all three components fail. The reliability of the standby system is the probability that at least one component survives. For exponential components in cold standby, the reliability is: $$ R_{standby}(t) = 1 - (1 - R(t))^3 $$ Substitute $R(t) = e^{-3}$: $$ R_{standby}(500) = 1 - (1 - 0.0498)^3 = 1 - (0.9502)^3 = 1 - 0.857 = 0.143 $$ 4. **Mean Time To Failure (MTTF) for a single component:** For exponential distribution, $$ MTTF = \frac{1}{\lambda} = \frac{1}{0.006} = 166.67 \text{ hours} $$ 5. **MTTF for the standby system:** Since the system fails only when all three components fail sequentially, the MTTF is the sum of the MTTFs of all three components: $$ MTTF_{standby} = 3 \times 166.67 = 500 \text{ hours} $$ 6. **Reliability of the parallel system:** In a parallel system, the system works if at least one component works. The reliability is: $$ R_{parallel}(t) = 1 - (1 - R(t))^3 = 0.143 $$ This is the same as the standby system reliability at 500 hours. 7. **MTTF for the parallel system:** For independent components in parallel with exponential failure rates, the system failure rate is: $$ \lambda_{parallel} = \lambda / 3 = 0.006 / 3 = 0.002 $$ So, $$ MTTF_{parallel} = \frac{1}{\lambda_{parallel}} = \frac{1}{0.002} = 500 \text{ hours} $$ **Summary:** - Reliability at 500 hours for both standby and parallel systems is approximately 0.143. - MTTF for both systems is 500 hours. Thus, for this configuration and parameters, the standby and parallel systems have the same reliability and mean time to failure at 500 hours.