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$n$ independent jobs are distributed by parallel computing on $n$ free nodes, where the processing time $T_i$ of job $i$ is exponentially distributed, $T_i \sim Exp(\lambda_i)$.

Determine the distribution function of the entire processing time $Y = \min\left\{T_1,..,T_n\right\}$, where the processing is finished, as soon as one job is completely processed. What's the average processing time $E(Y)$?

I check on internet, the formula for exponential distribution function is $$F(x)=\int_{0}^{x}f_{\lambda}(t) \, dt=\left\{\begin{matrix} \begin{align*}1-e^{-\lambda x} \text{ for } x\geq 0,\\ 0 \text{ for } x<0. \end{align*} \end{matrix}\right.$$

I think because say "as soon as one job..", we set $x=1$. That why we need use the first condition $1-e^{-\lambda x}$.

$$F(x) = \int_{0}^{1}\left(1-e^{-\lambda x}\right) \, dx= \left[\frac{e^{-\lambda x}}{\lambda}+x\right]_{0}^{1}$$

But now is problem, what put for $\lambda$? It should be $n$ because in total we have so many jobs right?

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You should notice that there is something strange in your solution since they ask for a function but what you have obtained is a number.

The fact that they say "as soon as one job is completely processed" is just, in my opinion, another way to say that you are considering a minimum. Now the minimum of indipendent exponential distributed random variables is just another exponential distributed random variable of parameter equal to the sum of the others parameters, in formula: $$\lambda= \sum_{i=1}^{n}\lambda_i$$

so the distribution function is just:

$$F_y(t)=P(Y \le t)=\int_{0}^{t}\lambda e^{-\lambda x}dx,t\ge0.$$

I think you can compute $E(Y)$ on your own now, just apply the definition.

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  • $\begingroup$ Ohh thank you very much!! For $E(Y)$ it mean expected value. Formula for expected value of exponential distribution is $$E(Y) = \int_{0}^{\infty}\lambda xe^{-\lambda x} \, dx = \frac{1}{\lambda}$$ Because it we have $T_i \sim Exp(\lambda_i)$ above, we insert $\lambda_i$ in the formula and have $$E(Y) = \frac{1}{\lambda_i}$$ Is it good like this? $\endgroup$
    – eyesima
    Dec 9 '17 at 14:42
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    $\begingroup$ The formula is correct, but you are doing it wrong. $E(Y)=\frac{1}{\lambda}$, where $Y\sim exp(\lambda)$ and $\lambda$ is defined as i wrote above, $\frac{1}{\lambda_i}=E(T_i) \neq E(Y)$. Once you have the distribution of $Y$ just forget about $T_i$ in order to compute the expected value. You get it now? $\endgroup$
    – chak
    Dec 9 '17 at 17:30
  • $\begingroup$ $$E(Y) = \frac{1}{\frac{1}{\lambda_i}}=\lambda_i$$? $\endgroup$
    – eyesima
    Dec 9 '17 at 17:47
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    $\begingroup$ Suppose that you have $A \sim exp(\beta)$, what is $E(A)$? Now as i wrote $Y \sim exp(\lambda)$, where $\lambda=\sum \lambda_i$, so $E(Y)=?$ $\endgroup$
    – chak
    Dec 9 '17 at 17:59
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    $\begingroup$ yes, that's right. $\endgroup$
    – chak
    Dec 9 '17 at 18:43

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