Exponential distribution and Poisson process A server handles critical requests

Question

Exponential distribution and Poisson process A server handles critical requests that come according to a Poisson process, on average once every hour. Suppose that at midnight the server will undergo maintenance that will last for half an hour, and requests that arrive during this period will be lost i. What is the probability that no requests will be lost? ii. What is the expected number of requests that will be lost? iii. Suppose that for two hours after the maintenance the server will be in a "warm-up" state and will not be able to process more than 2 requests total, and the rest will be lost. What is the probability that no requests will be lost from midnight till 2:30am?

Explanation / Answer

(i) Here = 1per hour

so expected number of requests to be arrived in half an hour = 0.5

so Pr(X = 0) = POISSON (X = 0 ; 0.5) = e-0.5 0.50 / 0! = 0.6065

(ii) Expected number of requests that will be lost = 1 * 0.5 = 0.5 requests

(iii) Here we have to find the probability that no requests will be lost from midnight till 2:30 pm.

here Pr(No requests lost between 12 to 12:30) * Pr(no request lost in between 12:30 to 2:30)

Pr(No requests lost between 12 to 12:30) = 0.6065

Expected number of requests that would be lost in between 12:30 to 2:30 = 2 * 1 = 2 requests

so Pr(NO requests will be lost) = Pr(X < = 2) = POISSON (X < = 2; 2)

here POISSON (X < = 2; 2) = Pr( X = 0) + Pr(X = 1) + Pr( X =2)

= e-2 20 / 0! + e-2 21/ 1! + e-2 22 / 2!

= 0.1353 + 0.2707 + 0.2707 = 0.6767

Pr( no request will be lost) = 0.6065 * 0.6767 = 0.4104

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