Your instructor is tired of grading the reports, so he decides to put his comput

Question

Your instructor is tired of grading the reports, so he decides to put his computing knowledge (not true but we’ll go with it) to good use by creating an auto- grader for the reports. The auto-grader is designed to grade one report at a time, and the time it takes to grade a report is exponentially distributed with mean 5 minutes. Students submit reports at a Poisson rate of 5 per hour. Please assume this is a birth and death CTMC where i in Pi represents the number of reports in the system to be graded.

A) What is the probability that there is at least 1 report waiting to be graded by the auto-grader (i.e. one report being graded and one report in queue to be graded)? Please write the exact    expression to find this value and show all work.

B) On average, how many reports are waiting to be graded by the auto-grader? (Find the exact expression to find this value and show all work)

C) On average, how long does it take for a report to be graded by the auto-grader (i.e. that amount of time reports spend in the system)? Please write the exact expression to find this value and show all work.

D) Your professor is worried that it is taking too long for the auto-grader to grade the reports, and he decides to institute the following policy: assume that a maximum of 2 reports are allowed to be waiting to be graded by the auto-grader (i.e. a maximum of 3 reports can be in the automated system at any given time, two reports waiting to be graded, and one report being graded). If additional reports are submitted while the system is full, the additional reports are immediately graded by your professor in exactly 3 minutes. If a report is submitted while the professor is grading a report and the auto-grader system is still full, that newly submitted report will be graded by the professor with the report he is already grading and finished when the first report is finished grading. On average, how long does it take for an report to be graded under this new policy (i.e. the amount of time the reports spend in the system)?

Explanation / Answer

Solution

These problems are solved treating these as M/M/1 Queue system i.e., Poisson-arrivals-exponential service-single channel queues.

Terminology

Arrivals follow Poisson pattern with average rate , service time follows Exponential Distribution with average service rate of µ

Let n = number of customers in the system and

m = number of customers in the queue.

[Trivially, n = m + number of customers under service.] (1 - ) + (1 - )

Let (/µ) =

The steady-state probability of n customers in the system is given by Pn = n(1 - ) ………(1)

The steady-state probability of no customers in the system is given by P0 = (1 - ) ………(2)

Average queue length = E(m) = (2)/{ µ(µ - )} …………………………………………..(3)

Average number of customers in the system = E(n) = ()/(µ - )…………………………..(4)

Average waiting time = E(w) = ()/{ µ(µ - )} ……………………………………………..(5)

Average time spent in the system = E(v) = {1/(µ - )}……………………………………..(6)

Given,

= 5/hr and µ = 12/hr [5 minutes per report => 60/5 = 12 reports per hour.]

= 5/12

Part (A)

Probability that there is at least 1 report waiting to be graded by the auto-grader

P(n 2) = 1 – (P0 + P1)

= 1- {(1 - ) + (1 - )} [vide (1) and (2)]

= 1- (1 - )(1 + )

= 1- (1 – 2)

= 2

= (5/12)2

= 25/144

= 0.1701 ANSWER

Part (B)

On average, number of reports waiting to be graded by the auto-grader = E(m)

= (2)/{ µ(µ - )} [vide (3)]

= (25)/(12 x 7)

= 0.2976 ANSWER

Part (C)

On average, time it takes for a report to be graded by the auto-grader = E(v)

= {1/(µ - )} [vide (6)]

= 1/7

= 0.1429 hr = 8.57 minutes ANSWER

Part (D)

Average time spent in the system = {1/(µ - )} = 8.57, if n 3

= (8.57 + 3), if n = 4

= (8.57 + 6), if n = 5.

So, Average time spent in the system = {8.57 x P(n 3)} + (11.57 x P4) + (14.57 x P5).

= {8.57 x (P0 + P1 + P2 + P3)} + (11.57 x P4) + (14.57 x P5)

Now, (P0 + P1) = 0.1701 from Part (A); P2 = (25/144)(7/12) = 0.1013; P3 = 0.0422; P4 = 0.0176; P5 = 0.0073.

So, Average time spent in the system = (8.57 x 0.3136) + (11.57 x 0.0176) + (14.57 x 0.0073)

= 2.6876 + 0.2036 + 0.1077

= 2.9989 ANSWER

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