Welcome to our community

Be a part of something great, join today!

probability of failure

hemanth

New member
May 31, 2012
9
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.
 

chisigma

Well-known member
Feb 13, 2012
1,704
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

$\chi$ $\sigma$
 

hemanth

New member
May 31, 2012
9
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
 

CaptainBlack

Well-known member
Jan 26, 2012
890
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
 

hemanth

New member
May 31, 2012
9
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.
 

CaptainBlack

Well-known member
Jan 26, 2012
890
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
 

hemanth

New member
May 31, 2012
9
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.
 

CaptainBlack

Well-known member
Jan 26, 2012
890
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