Calculating Probability of Machine Failure and Repair with Exponential Durations

In summary: The simplest interpretation has the second system service an arrival once if the first system is on when the arrival occurs irrespective of what subsequently happens to the first system's state, and all arrivals in the second system are serviced without having to queue if the first is on.Then the service time is sampled from a distribution with density \(B(t)\) with probability \(\beta/(\alpha+\beta)\) and from the distribution of the sum of a service time from distribution with density \(B(t)\) plus an off time of system 1 with probability \(1-\beta/(\alpha+\beta)\). The density of a sum of RV is the convolution of their individual...
  • #1
hemanth
9
0
i have a machine which runs for exponential duration with mean alpha,and fails.
once it fails, to repair it, it takes an exponential duration with mean beta then again it comes back to service and this continues.
what is the probability that it is in non working(repair) state.
 
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  • #2
hemanth said:
i have a machine which runs for exponential duration with mean alpha,and fails.
once it fails, to repair it, it takes an exponential duration with mean beta then again it comes back to service and this continues.
what is the probability that it is in non working(repair) state.

If we indicate with $T_{ok}$ and $T_{ko}$ the time before a failure and the repairing time [both are random variables...], then their p.d.f. are...

$\displaystyle f_{ok} (x) = \frac{1}{\alpha}\ e^{- \frac{x}{\alpha}} ,\ f_{ko} (x) = \frac{1}{\beta}\ e^{- \frac{x}{\beta}} $ (1)

... and their expected values are

$\displaystyle t_{ok}= \frac{1}{\alpha}\ \int_{0}^{\infty} x\ e^{- \frac{x}{\alpha}}=\alpha,\ t_{ko}=\frac{1}{\beta}\ \int_{0}^{\infty} x\ e^{- \frac{x}{\beta}}=\beta$ (2)

From (2) You can easy find that the probability that the machine is non working is...

$\displaystyle P_{ko}= \frac{t_{ko}}{t_{ok}+t_{ko}}= \frac{\beta}{\alpha+\beta}$ (3)

Kind regards

$\chi$ $\sigma$
 
  • #3
chisigma said:
If we indicate with $T_{ok}$ and $T_{ko}$ the time before a failure and the repairing time [both are random variables...], then their p.d.f. are...

$\displaystyle f_{ok} (x) = \frac{1}{\alpha}\ e^{- \frac{x}{\alpha}} ,\ f_{ko} (x) = \frac{1}{\beta}\ e^{- \frac{x}{\beta}} $ (1)

... and their expected values are

$\displaystyle t_{ok}= \frac{1}{\alpha}\ \int_{0}^{\infty} x\ e^{- \frac{x}{\alpha}}=\alpha,\ t_{ko}=\frac{1}{\beta}\ \int_{0}^{\infty} x\ e^{- \frac{x}{\beta}}=\beta$ (2)

From (2) You can easy find that the probability that the machine is non working is...

$\displaystyle P_{ko}= \frac{t_{ko}}{t_{ok}+t_{ko}}= \frac{\beta}{\alpha+\beta}$ (3)

Kind regards

$\chi$ $\sigma$
thanks a lot $\chi$ $\sigma$,
i was wondering whether this could be done directly, this was obvious but not sure with the proof. was thinking to find the pdf and do the long way.
thanks a lot
 
  • #4
hemanth said:
i have a machine which runs for exponential duration with mean alpha,and fails.
once it fails, to repair it, it takes an exponential duration with mean beta then again it comes back to service and this continues.
what is the probability that it is in non working(repair) state.

We consider a long period of time \(T\) so that the fraction of time spent being repaired is close to the mean. Then if \(n\) is the (mean) number of failures the machine is off-line for a total duration \(n\beta\), and:

\(\displaystyle n=\frac{T-n\beta}{\alpha}\)

so:

\( \displaystyle n=\frac{T}{\alpha+\beta}\).

Hence the fraction of time machine is off-line is:

\( \displaystyle \frac{n\beta}{T}=\frac{T \beta}{(\alpha+\beta)T}=\frac{\beta}{\alpha+\beta}\)

CB
 
  • #5
Thanks a lot CB, i have one more question,

i have a process which is ON for exponential duration with mean alpha,and then it goes OFF.
once it is OFF, it remains OFF for an exponential duration with mean beta then again it comes back to ON and this ON-OFF continues.
This is a kind of alternating renewal process.

associated with this i have another system, where arrivals are generated according to a general distribution A(t) and have a general service time distribution B(t), these arrivals are served whenever the first system(alternating renewal process) is on.

i need to evaluate the distribution of the expanded service time(i.e expanded by the off duration s).
assuming that first ON start and first arrival are synchronized.
thanks a lot once again.
 
  • #6
hemanth said:
Thanks a lot CB, i have one more question,

i have a process which is ON for exponential duration with mean alpha,and then it goes OFF.
once it is OFF, it remains OFF for an exponential duration with mean beta then again it comes back to ON and this ON-OFF continues.
This is a kind of alternating renewal process.

associated with this i have another system, where arrivals are generated according to a general distribution A(t) and have a general service time distribution B(t), these arrivals are served whenever the first system(alternating renewal process) is on.

i need to evaluate the distribution of the expanded service time(i.e expanded by the off duration s).
assuming that first ON start and first arrival are synchronized.
thanks a lot once again.

We need more information, what happens if the first system goes off while the second is servicing an arrival? Is there any queuing? ...

The simplest interpretation has the second system service an arrival once if the first system is on when the arrival occurs irrespective of what subsequently happens to the first system's state, and all arrivals in the second system are serviced without having to queue if the first is on.

Then the service time is sampled from a distribution with density \(B(t)\) with probability \(\beta/(\alpha+\beta)\) and from the distribution of the sum of a service time from distribution with density \(B(t)\) plus an off time of system 1 with probability \(1-\beta/(\alpha+\beta)\). The density of a sum of RV is the convolution of their individual densities.

CB
 
  • #7
When the first system goes off when an arrival is being served, the service is continued from the left point when it again becomes on (preemptive resume). subsequent arrivals when one is being served are queued, and served one after the other during the ON periods of the first system.
 
  • #8
hemanth said:
When the first system goes off when an arrival is being served, the service is continued from the left point when it again becomes on (preemptive resume). subsequent arrivals when one is being served are queued, and served one after the other during the ON periods of the first system.

I would run a simulation to determine an emprical distribution as the system is becoming too complicated for an analytic approach to be worth while (at for me, I would not trust the analysis without simulation support anyway)

CB
 

What is "probability of failure"?

The probability of failure is a measure of the likelihood that a certain event or system will not meet its intended outcome or performance standards. It is often expressed as a percentage or decimal value between 0 and 1.

How is probability of failure calculated?

The calculation of probability of failure depends on the specific context and factors involved. In general, it involves identifying all potential failure modes or events, determining the likelihood of each event occurring, and then combining these probabilities using mathematical techniques such as multiplication, addition, or Bayesian analysis.

What factors influence probability of failure?

The probability of failure can be influenced by a variety of factors, including the complexity of the system or process, the quality of materials or components used, the experience and training of individuals involved, external environmental conditions, and human error. It is important to consider all of these factors when assessing the probability of failure in a particular situation.

How can probability of failure be reduced?

There are several strategies that can be employed to reduce the probability of failure. These include implementing quality control measures, conducting thorough risk assessments and testing, incorporating redundant systems or backup plans, and continuously monitoring and improving processes. Additionally, utilizing experienced and trained personnel and regularly reviewing and updating procedures can also help to decrease the probability of failure.

Why is understanding probability of failure important?

Understanding the probability of failure is crucial in many fields, including engineering, finance, and healthcare. It allows for informed decision making and risk management, helps to identify potential weaknesses or vulnerabilities in systems, and enables the development of effective mitigation strategies. By understanding the probability of failure, scientists can ensure the safety, reliability, and success of their projects and processes.

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