# What's the average processing time $E(X)$ if $\lambda=\lambda_1=.. = \lambda_n$?

$$n$$ 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

• What is $X$ supposed to be? – Falrach Dec 10 '17 at 13:33
• @Falrach $X$ is total time – 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 $X$:
$$f_X(t) = n(1-e^{-\lambda t})^{n-1} \lambda e^{-\lambda t}$$
$$\Bbb{E}[X] = \int_0^\infty t f_X(t) \text{d}t$$
• Your calculation of $\Bbb{E}[X]$ suggests that $X \sim Exp(\lambda)$. But this is just the distribution of one node. – Falrach Dec 10 '17 at 14:15