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

What's the average processing time $E(X)$ in special case $\lambda = \lambda_1 = .... = \lambda_n$ ?

Not sure but I do like this:

In general formula for average processing time is $$E(X) = \int_{0}^{\infty} \lambda x e^{-\lambda x} \, dx= \frac{1}{\lambda}$$

we have $$E(X) = \int_{-\infty}^{+\infty}x \cdot f_X(x) \, dx = \int_{0}^{+\infty} x \lambda e^{-\lambda x} \, dx$$

$$=\lambda \left(\left[x \cdot \left(-\frac{1}{\lambda} e^{-\lambda x}\right)\right]_{0}^{+\infty} - \left[\frac{1}{~\lambda^2} e^{-\lambda x } \cdot 1\right]_{0}^{+\infty}\right)=\lambda\left(0-0-(0-\frac{1}{\lambda^2})\right)= \frac{1}{\lambda}$$

But I think is strange I have as result $E(X) = \frac{1}{\lambda}$

Is good like this or all wrong? :s

  • $\begingroup$ What is $X$ supposed to be? $\endgroup$ – Falrach Dec 10 '17 at 13:33
  • $\begingroup$ @Falrach $X$ is total time $\endgroup$ – eyesima Dec 10 '17 at 13:34

The time when all jobs are done is $X = \max (T_1,\ldots,T_n)$. Compute the probability density of $X$:

$$f_X(t) = n(1-e^{-\lambda t})^{n-1} \lambda e^{-\lambda t}$$

Then compute:

$$\Bbb{E}[X] = \int_0^\infty t f_X(t) \text{d}t$$

  • $\begingroup$ But this isn't what I do already? I don't see mistake maybe you can show? I try again with formula of you and it same I think $\endgroup$ – eyesima Dec 10 '17 at 14:12
  • $\begingroup$ Your calculation of $\Bbb{E}[X]$ suggests that $X \sim Exp(\lambda)$. But this is just the distribution of one node. $\endgroup$ – Falrach Dec 10 '17 at 14:15

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