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Failure rate of a system at time 't'

by francisg3
Tags: failure, rate, time
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francisg3
#1
Mar21-11, 06:59 PM
P: 32
I need to solve the following problem for a school assignment.

Let λ(t) denote the failuer rate of a system at time 't'. The failure rate is simple the number of failures in unit time. For example, if the unit time is one day, then λ is the average of failures per day. Let μ(t) denote the total number of failures from the first release (time t=0) until the current time, 't'. Then we have

(1) λ= dμ/dt

(2) μ = ∫λ(T) where the limits of integration are T=0 (lower) and T=t (upper)

Two models are used for estimating λ and μ. In the forumlae below, λ0 is the failure rate at time t=0, and α and β are constants

λ=λ0(1-μ/α)

λ=λ0e^- β μ



Use (1) or (2) to find λ and μ as functions of time for each model.



.....I just need some direction. Thanks!
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issacnewton
#2
Mar21-11, 09:13 PM
P: 610
for the case 1

[tex] \lambda = \lambda_o (1-\frac{\mu}{\alpha})[/tex]

[tex] \frac{d\mu}{dt}=\lambda_o(1-\frac{\mu}{\alpha})[/tex]

[tex] \frac{d\mu}{\left(1-\frac{\mu}{\alpha}\right)}=\lambda_o \,\, dt [/tex]

now integrate within the given limits
francisg3
#3
Mar21-11, 09:46 PM
P: 32
so the resulting integration would be:

-α ln (μ -α) evaluated at 0 and 't' correct?

issacnewton
#4
Mar21-11, 10:15 PM
P: 610
Failure rate of a system at time 't'

no....check the integration..........remember to integrate both sides

and francis, i see that you have doubled up this thread.......two threads are off
by half an hour. This is NOT a good practice. Somebody will report this to the mods.

Another thread going at

http://www.physicsforums.com/showthread.php?t=483138
francisg3
#5
Mar21-11, 10:26 PM
P: 32
I know I double posted, realized that this was not homework/coursework section but I don't know how to delete a post. Sorry.
kazo0
#6
Feb19-12, 02:57 PM
P: 2
I am also having problems with this question, I integrated:

[itex]\int\frac{dμ}{(1-\frac{μ}{α})}[/itex] = [itex]\int[/itex][itex]\lambda[/itex]0dt

And I got:

-αln(μ-α) = [itex]\lambda[/itex]0t

I'm not sure if this is going in the right direction and what would I have to do after this in order to find μ(t) and λ(t)?

Thanks
issacnewton
#7
Feb19-12, 03:25 PM
P: 610
kazo, use properties of logarithm...

[tex]-\alpha \ln (\mu-\alpha)=\lambda_o t [/tex]

[tex] \ln (\mu-\alpha)=-\frac{\lambda_o t}{\alpha}[/tex]

[tex]\mu-\alpha= \mbox{exp}\left[-\frac{\lambda_o t}{\alpha}\right ] [/tex]

[tex]\mu (t) =\alpha +\mbox{exp}\left[-\frac{\lambda_o t}{\alpha}\right ] [/tex]

and plug this to get [itex]\lambda[/itex] as function of t
Char. Limit
#8
Feb19-12, 06:04 PM
PF Gold
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P: 1,951
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