Standby Configuration Example: Difference between revisions

From ReliaWiki
Jump to navigation Jump to search
(Created page with '<noinclude>{{Banner BlockSim Examples}} ''This example appears in the [[Time-Dependent_System_Reliability_(Analytical)#Reliability_of_Standby_Systems_with_a_Switching_Device|Syst…')
 
No edit summary
 
(3 intermediate revisions by 2 users not shown)
Line 1: Line 1:
<noinclude>{{Banner BlockSim Examples}}
<noinclude>{{Banner BlockSim Examples}}
''This example appears in the [[Time-Dependent_System_Reliability_(Analytical)#Reliability_of_Standby_Systems_with_a_Switching_Device|System Analysis Reference book]]''.
''This example appears in the [https://help.reliasoft.com/reference/system_analysis System analysis reference]''.


</noinclude>
</noinclude>
Consider a car with four new tires and a full-size spare.  Assume the following failure characteristics:
Consider a car with four new tires and a full-size spare.  Assume the following failure characteristics:


:*The tires follow a Weibull distribution with a <math>\beta=4\,\!</math> and an <math>\eta = 40,000\,\!</math> miles while on the car due to wear.
:*The tires follow a Weibull distribution with a <math>\beta=4\,\!</math> and an <math>\eta = 40,000\,\!</math> miles while on the car due to wear.


:*The tires also have a probability of failing due to puncture or other causes.  For this, assume a constant rate for this occurrence with a probability of 1 every 50,000 miles.
:*The tires also have a probability of failing due to puncture or other causes.  For this, assume a constant rate for this occurrence with a probability of 1 every 50,000 miles.


:*When not on the car (i.e., is a spare), a tire's probability of failing also has a Weibull distribution with a <math>\beta = 2\,\!</math> and <math>\eta = 120,000\,\!</math> miles.
:*When not on the car (i.e., is a spare), a tire's probability of failing also has a Weibull distribution with a <math>\beta = 2\,\!</math> and <math>\eta = 120,000\,\!</math> miles.


Assume a mission of 1,000 miles.  If a tire fails during this trip, it will be replaced with the spare.  However, the spare will not be repaired during the trip.  In other words, the trip will continue with the spare on the car and, if the spare fails, the system will fail.  Determine the probability of system failure.
Assume a mission of 1,000 miles.  If a tire fails during this trip, it will be replaced with the spare.  However, the spare will not be repaired during the trip.  In other words, the trip will continue with the spare on the car and, if the spare fails, the system will fail.  Determine the probability of system failure.
Line 17: Line 17:
The active failure distribution for the tires are:
The active failure distribution for the tires are:


:*Due to wearout, Weibull <math>\beta =4</math> and <math>\eta =40,000\,\!</math> miles.
:*Due to wearout, Weibull <math>\beta =4\,\!</math> and <math>\eta =40,000\,\!</math> miles.


:*Due to random puncture, exponential <math>\mu =50,000.\,\!</math>
:*Due to random puncture, exponential <math>\mu =50,000.\,\!</math>


:*The quiescent failure distribution is a Weibull distribution with <math>\beta =2\,\!</math> and <math>\eta =120,000\,\!</math> miles.
:*The quiescent failure distribution is a Weibull distribution with <math>\beta =2\,\!</math> and <math>\eta =120,000\,\!</math> miles.


The block diagram for each tire has two blocks in series, one block representing the wearout mode and the other the random puncture mode, as shown next:
The block diagram for each tire has two blocks in series, one block representing the wearout mode and the other the random puncture mode, as shown next:




[[Image:small5.png|center|250px|]]
[[Image:small5.png|center|350px|link=]]




Line 32: Line 32:




[[Image:BStirecontainer.png|center|250px|]]
[[Image:BStirecontainer.png|center|350px|link=]]


   
   
For the standby Wear block, set the active failure and the quiescent distributions, but for the Puncture block, set only the active puncture distribution (because the tire cannot fail due to puncture while stored).  Using BlockSim, the probability of system failure is found to be 0.003 or 0.3%.
For the standby Wear block, set the active failure and the quiescent distributions, but for the Puncture block, set only the active puncture distribution (because the tire cannot fail due to puncture while stored).  Using BlockSim, the probability of system failure is found to be 0.003 or 0.3%.

Latest revision as of 20:58, 18 September 2023

BlockSim Examples Banner.png


New format available! This reference is now available in a new format that offers faster page load, improved display for calculations and images and more targeted search.

As of January 2024, this Reliawiki page will not continue to be updated. Please update all links and bookmarks to the latest references at BlockSim examples and BlockSim reference examples.




This example appears in the System analysis reference.


Consider a car with four new tires and a full-size spare. Assume the following failure characteristics:

  • The tires follow a Weibull distribution with a [math]\displaystyle{ \beta=4\,\! }[/math] and an [math]\displaystyle{ \eta = 40,000\,\! }[/math] miles while on the car due to wear.
  • The tires also have a probability of failing due to puncture or other causes. For this, assume a constant rate for this occurrence with a probability of 1 every 50,000 miles.
  • When not on the car (i.e., is a spare), a tire's probability of failing also has a Weibull distribution with a [math]\displaystyle{ \beta = 2\,\! }[/math] and [math]\displaystyle{ \eta = 120,000\,\! }[/math] miles.

Assume a mission of 1,000 miles. If a tire fails during this trip, it will be replaced with the spare. However, the spare will not be repaired during the trip. In other words, the trip will continue with the spare on the car and, if the spare fails, the system will fail. Determine the probability of system failure.

Solution

The active failure distribution for the tires are:

  • Due to wearout, Weibull [math]\displaystyle{ \beta =4\,\! }[/math] and [math]\displaystyle{ \eta =40,000\,\! }[/math] miles.
  • Due to random puncture, exponential [math]\displaystyle{ \mu =50,000.\,\! }[/math]
  • The quiescent failure distribution is a Weibull distribution with [math]\displaystyle{ \beta =2\,\! }[/math] and [math]\displaystyle{ \eta =120,000\,\! }[/math] miles.

The block diagram for each tire has two blocks in series, one block representing the wearout mode and the other the random puncture mode, as shown next:


Small5.png


There are five tires, four active and one standby (represented in the diagram by a standby container with a 4-out-of-5 requirement), as shown next:


BStirecontainer.png


For the standby Wear block, set the active failure and the quiescent distributions, but for the Puncture block, set only the active puncture distribution (because the tire cannot fail due to puncture while stored). Using BlockSim, the probability of system failure is found to be 0.003 or 0.3%.