Crow Extended Model for Repairable Systems Analysis Example: Difference between revisions

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The failures and fixes of two repairable systems in the field are recorded. Both systems started operating from time 0. System 1 ends at time = 504 and system 2 ends at time = 541. All the BD modes are fixed at the end of the test. A fixed effectiveness factor equal to 0.6 is used. Answer the following questions:
The failures and fixes of two repairable systems in the field are recorded. Both systems started operating from time 0. System 1 ends at time = 504 and system 2 ends at time = 541. All the BD modes are fixed at the end of the test. A fixed effectiveness factor equal to 0.6 is used. Answer the following questions:


:1) Estimate the parameters of the Crow Extended model.
#Estimate the parameters of the Crow Extended model.
:2) Calculate the projected MTBF after the delayed fixes.
#Calculate the projected MTBF after the delayed fixes.
:3) If no fixes were performed for the future failures, what would be the expected number of failures at time 1,000?
#If no fixes were performed for the future failures, what would be the expected number of failures at time 1,000?




'''Solution'''
'''Solution'''
:1) The next figure shows the estimated Crow Extended parameters.
<ol>
<li>The next figure shows the estimated Crow Extended parameters.


[[Image:rga13.14.png|thumb|center|450px|Crow Extended model for repairable systems.]]
[[Image:rga13.14.png|thumb|center|450px|Crow Extended model for repairable systems.]]


:2) The next figure shows the projected MTBF at time = 541 (i.e., the age of the oldest system).
</li>
<li>The next figure shows the projected MTBF at time = 541 (i.e., the age of the oldest system).


[[Image:rga13.15.png|thumb|center|450px|MTBFs from Crow Extended model.]]
[[Image:rga13.15.png|thumb|center|450px|MTBFs from Crow Extended model.]]


:3) The next figure shows the expected number of failures at time = 1,000.
</li>
<li>The next figure shows the expected number of failures at time = 1,000.


[[Image:rga13.16.png|thumb|center|450px|Cumulative number of failures at time = 1,000.]]
[[Image:rga13.16.png|thumb|center|450px|Cumulative number of failures at time = 1,000.]]
</li>
</ol>

Revision as of 20:36, 6 February 2014

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This example appears in the Reliability Growth and Repairable System Analysis Reference book.


The failures and fixes of two repairable systems in the field are recorded. Both systems started operating from time 0. System 1 ends at time = 504 and system 2 ends at time = 541. All the BD modes are fixed at the end of the test. A fixed effectiveness factor equal to 0.6 is used. Answer the following questions:

  1. Estimate the parameters of the Crow Extended model.
  2. Calculate the projected MTBF after the delayed fixes.
  3. If no fixes were performed for the future failures, what would be the expected number of failures at time 1,000?


Solution

  1. The next figure shows the estimated Crow Extended parameters.
    Crow Extended model for repairable systems.
  2. The next figure shows the projected MTBF at time = 541 (i.e., the age of the oldest system).
    MTBFs from Crow Extended model.
  3. The next figure shows the expected number of failures at time = 1,000.
    Cumulative number of failures at time = 1,000.
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