Question

1. A disk server receives requests from many client machines and requires 8 milliseconds to respond to each request. Let N = the number of additional requests that arrive before the service interval is complete. The probability of exactly k additional requests in the 8-millisecond service interval is P(N = k) = -0.9 (0.9k/k!, for k = 0,1, 2..., . If 2 or more new calls arrive while the service interval is only partially complete, what is the probability that exactly 3 new calls will arrive before the server is ready to respond? Hint: Use rules for conditional probability.

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Answer 1

Answer:

Given that a disk server recevies, requests from many client machines and requires 8 milliseconds to respond to each request.

where N= number of additional requests that arrives before the service interval is complete.

Probablity that exatly 3 new calls arrive before 2 the server is ready to respond:

P(exactly 3 calls given 2 (or) more calls)

$\Rightarrow$P(N=3 / N>=2)

$\Rightarrow$P(N=3 / P(N>=2))

$\Rightarrow$P[N=3 / 1-P(N=0)-P(N=1)]

(Substitute N=3 in the given equation)

$\Rightarrow$$\frac{(e^{-0.9}\times (0.9)^{3})/3!}{1-\frac{e^{-0.9\times (0.9)^{0}}}{0!}-\frac{e^{-0.9}\times (0.9)^{1}}{1!}}$

$\Rightarrow$$\frac{(0.4065\times 0.729)/6}{1-\frac{0.4065\times 1}{1}-\frac{0.4065\times 0.9}{1}}$

$\Rightarrow$$\frac{0.0493}{0.5935-0.36585}$

$\Rightarrow$$\frac{0.0493}{0.22765}$

$\Rightarrow$