Exercise 3 (C) At the post office, patrons wait in a single line for the first open window. An average of 100 patrons per hour enter the post office, and each window can serve an average of 45 patrons...


Exercise 3 (C) At the post office, patrons wait in a single line for the first open window. An<br>average of 100 patrons per hour enter the post office, and each window can serve an average of<br>45 patrons per hour. The post office estimates a cost of £0.1 for each minute a patron waits in<br>line and believes that it costs £20 per hour to keep a window. Inter-arrival times and service<br>times are independent and exponentially distributed.<br>(a) How many windows should be kept in order to minimize the total expected hourly cost?<br>What is the minimal total expected hourly cost (round to three decimal places)? [20 marks]<br>

Extracted text: Exercise 3 (C) At the post office, patrons wait in a single line for the first open window. An average of 100 patrons per hour enter the post office, and each window can serve an average of 45 patrons per hour. The post office estimates a cost of £0.1 for each minute a patron waits in line and believes that it costs £20 per hour to keep a window. Inter-arrival times and service times are independent and exponentially distributed. (a) How many windows should be kept in order to minimize the total expected hourly cost? What is the minimal total expected hourly cost (round to three decimal places)? [20 marks]
(b) What is the minimal number of windows that should be kept in order to ensure that at<br>most 5% of all patrons will spend more than 5 minutes in line? [20 marks]<br>(You need to explain how you obtained your answer to get full mark.)<br>Hint: Concerning the distribution of the waiting time W in an M/M/c queueing system,<br>the following information is useful: with IIw being the delay probability,<br>P(W > 0) = IIw,<br>P(W = 0) = 1 – P(W > 0) = 1 – IIw,<br>%3D<br>and<br>P(W > t, W > 0)<br>P(W > t|W > 0) =<br>= e-CH(1-p)t<br>P(W > 0)<br>In other words, with probability 1 – IIw, W equals zero, and with probability IIw, W ~<br>exp(cu(1 — р)).<br>

Extracted text: (b) What is the minimal number of windows that should be kept in order to ensure that at most 5% of all patrons will spend more than 5 minutes in line? [20 marks] (You need to explain how you obtained your answer to get full mark.) Hint: Concerning the distribution of the waiting time W in an M/M/c queueing system, the following information is useful: with IIw being the delay probability, P(W > 0) = IIw, P(W = 0) = 1 – P(W > 0) = 1 – IIw, %3D and P(W > t, W > 0) P(W > t|W > 0) = = e-CH(1-p)t P(W > 0) In other words, with probability 1 – IIw, W equals zero, and with probability IIw, W ~ exp(cu(1 — р)).

Jun 11, 2022
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