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25000 hr\text{25000 hr}

Step-by-step solution

3 steps

:
is incomplete. To determine the number of functioning/failed units, unreliability, and MTS (Mean Time To System failure or Mean Time To Failure), additional information such as the number of observed failures, the duration of operation, or a specific failure rate for this equipment is required. The problem statement only provides the total number of units (1000) but no data on their operational status or failure events.

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The problem asks to draw a reliability attenuation curve for a constant failure rate.

Given:
Failure rate λ=4×10−5hr−1\lambda = 4 \times 10^{-5} \text{hr}^{-1}
Times T=[5000hr,10000hr,15000hr,20000hr,25000hr,30000hr,35000hr,40000hr,45000hr,50000hr]T = [5000 \text{hr}, 10000 \text{hr}, 15000 \text{hr}, 20000 \text{hr}, 25000 \text{hr}, 30000 \text{hr}, 35000 \text{hr}, 40000 \text{hr}, 45000 \text{hr}, 50000 \text{hr}]

  1. Recall the Mean Time To Failure (MTTF) and reliability values.

    From the previous calculation, the MTTF is 25000 hr\boxed{\text{25000 hr}}.
    The reliability values R(t)=e−λtR(t) = e^{-\lambda t} at the given times are:

    Time (hr)Reliability R(t)50000.8187100000.6703150000.5488200000.4493250000.3679300000.3012350000.2466400000.2019450000.1653500000.1353\begin{array}{|c|c|} \hline \text{Time (hr)} & \text{Reliability } R(t) \\ \hline 5000 & 0.8187 \\ 10000 & 0.6703 \\ 15000 & 0.5488 \\ 20000 & 0.4493 \\ 25000 & 0.3679 \\ 30000 & 0.3012 \\ 35000 & 0.2466 \\ 40000 & 0.2019 \\ 45000 & 0.1653 \\ 50000 & 0.1353 \\ \hline \end{array}
  2. Describe the axes for the graph.

    To plot the reliability attenuation curve:

    • The x-axis represents Time (hr). It should range from 0 to at least 50000 hr.
    • The y-axis represents Reliability R(t)R(t). It should range from 0 to 1.
  3. Describe how to plot the points and the resulting curve.

    • Plot the points from the table, with time on the x-axis and reliability on the y-axis. For example, plot (5000, 0.8187), (10000, 0.6703), and so on.
    • At time t=0t=0, the reliability is R(0)=e0=1R(0) = e^0 = 1. So, the curve starts at the point (0, 1).
    • Connect these plotted points with a smooth curve. The resulting graph will be an exponential decay curve. It will start at 1 on the y-axis and continuously decrease as time increases on the x-axis, approaching 0 but never quite reaching it. The curve will pass through the point (25000 hr,0.3679)(25000 \text{ hr}, 0.3679), which corresponds to the MTTF where reliability is e−1e^{-1}.
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