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 report. 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?

B) On average, how many reports are waiting to be graded by the auto-grader?

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)?

D) 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 an 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 a report to be graded under this new policy (i.e. the amount of time the reports spend in the system)?

Explanation / Answer

A. From information given, lambda=5/60, mu=5,

P(atleast one report is waiting to be graded)=1-P(no report is waiting to be graded)=1-lambda/mu=1-(5/60)/5=0.9833

B. The average number of reports waiting to be graded by the auto grader is: L=lambda/(mu-lambda)=(5/60)/(5-5/60)=0.0169

C. Amount of time reports spend in the system, W=L/lambda=0.0169/(5/60)=0.2028 minutes

D. The amount of time spent in the system under new policy, W'=L'/lambda'=3/3=0.6667 minutes

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