# Failure rate of a system at time 't'

by francisg3
Tags: failure, rate, time
 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!
PF Gold
P: 1,951
 Quote by francisg3 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!
Well, Assuming that your first equation reads as such:

$$\lambda = \lambda_0 \left(1-\frac{\mu}{\alpha}\right)$$

You should be able to substitute λ=dμ/dt and get a seperable differential equation in μ. Then you differentiate that equation to get λ.

So you just need to solve:

$$\frac{d\mu}{dt} = \lambda_0 \left(1-\frac{\mu}{\alpha}\right)$$
 P: 32 so i just differentiate with respect to μ?
 P: 32 Failure rate of a system at time 't' so the resulting integration would be: -α ln (μ -α) evaluated at 0 and 't' correct?
 PF Gold P: 1,951 Well, don't EVALUATE it at those two points. Instead, set that equal to t+C.
 P: 610 francis has started two threads for the same question. another thread at http://www.physicsforums.com/showthread.php?t=483125 I am answering the same question at this thread. Beware francis

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