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]
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]
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
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![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]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F430c6f4f-c306-47e0-b0fa-05492b3caf8f%2F5667689f-245c-454d-a254-795ad2752db1%2F9e79kgk_processed.png&w=3840&q=75)
Transcribed Image 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
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)).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F430c6f4f-c306-47e0-b0fa-05492b3caf8f%2F5667689f-245c-454d-a254-795ad2752db1%2Fte4oc7h_processed.png&w=3840&q=75)
Transcribed Image 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)).
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