Using a continuous failure probability to get reliability $R(T) = exp(-int_t=0^T h(t) dt)$
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I have a continuous process for which I want to compute the probability that it did not fail until time $T$.
I tried the reliability / survival function:
$$ beginalign
R(T) = exp(-int_t=0^T h(t) dt)
endalign $$
I do not have a failure rate $h(t)$, but a failure probability given by a continuous but not necessarily monotonic function $p_fail(t)$, which gives the probability that the process fails at any time $t$ (with $t in R, t ge 0$).
Here, $p_fail(t)$ does not have a unit, which messes up the whole calculation. Therefore, I think my approach is wrong.
How can I calculate the reliability of the process?
To illustrate my problem:
Consider the simple function $p_1,fail(t [s]) = frac0.013600s cdot t$, for $t$ in seconds. This (in my opinion) is equal to $p_2,fail(t [h]) = frac0.011h cdot t$, for $t$ in hours.
With $T = 1h = 3600s$ the integration gives different results, depending on the unit of time:
beginarray c
hline & int_t=0^T p(t) dt \
hline p_1 & 18.00 \
hline p_2 & 0.005 \
hline
endarray
probability unit-of-measure reliability
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up vote
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down vote
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I have a continuous process for which I want to compute the probability that it did not fail until time $T$.
I tried the reliability / survival function:
$$ beginalign
R(T) = exp(-int_t=0^T h(t) dt)
endalign $$
I do not have a failure rate $h(t)$, but a failure probability given by a continuous but not necessarily monotonic function $p_fail(t)$, which gives the probability that the process fails at any time $t$ (with $t in R, t ge 0$).
Here, $p_fail(t)$ does not have a unit, which messes up the whole calculation. Therefore, I think my approach is wrong.
How can I calculate the reliability of the process?
To illustrate my problem:
Consider the simple function $p_1,fail(t [s]) = frac0.013600s cdot t$, for $t$ in seconds. This (in my opinion) is equal to $p_2,fail(t [h]) = frac0.011h cdot t$, for $t$ in hours.
With $T = 1h = 3600s$ the integration gives different results, depending on the unit of time:
beginarray c
hline & int_t=0^T p(t) dt \
hline p_1 & 18.00 \
hline p_2 & 0.005 \
hline
endarray
probability unit-of-measure reliability
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I have a continuous process for which I want to compute the probability that it did not fail until time $T$.
I tried the reliability / survival function:
$$ beginalign
R(T) = exp(-int_t=0^T h(t) dt)
endalign $$
I do not have a failure rate $h(t)$, but a failure probability given by a continuous but not necessarily monotonic function $p_fail(t)$, which gives the probability that the process fails at any time $t$ (with $t in R, t ge 0$).
Here, $p_fail(t)$ does not have a unit, which messes up the whole calculation. Therefore, I think my approach is wrong.
How can I calculate the reliability of the process?
To illustrate my problem:
Consider the simple function $p_1,fail(t [s]) = frac0.013600s cdot t$, for $t$ in seconds. This (in my opinion) is equal to $p_2,fail(t [h]) = frac0.011h cdot t$, for $t$ in hours.
With $T = 1h = 3600s$ the integration gives different results, depending on the unit of time:
beginarray c
hline & int_t=0^T p(t) dt \
hline p_1 & 18.00 \
hline p_2 & 0.005 \
hline
endarray
probability unit-of-measure reliability
I have a continuous process for which I want to compute the probability that it did not fail until time $T$.
I tried the reliability / survival function:
$$ beginalign
R(T) = exp(-int_t=0^T h(t) dt)
endalign $$
I do not have a failure rate $h(t)$, but a failure probability given by a continuous but not necessarily monotonic function $p_fail(t)$, which gives the probability that the process fails at any time $t$ (with $t in R, t ge 0$).
Here, $p_fail(t)$ does not have a unit, which messes up the whole calculation. Therefore, I think my approach is wrong.
How can I calculate the reliability of the process?
To illustrate my problem:
Consider the simple function $p_1,fail(t [s]) = frac0.013600s cdot t$, for $t$ in seconds. This (in my opinion) is equal to $p_2,fail(t [h]) = frac0.011h cdot t$, for $t$ in hours.
With $T = 1h = 3600s$ the integration gives different results, depending on the unit of time:
beginarray c
hline & int_t=0^T p(t) dt \
hline p_1 & 18.00 \
hline p_2 & 0.005 \
hline
endarray
probability unit-of-measure reliability
edited Aug 30 at 13:22
asked Aug 6 at 8:49
Stanley F.
1196
1196
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