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Chapter 8: Problem 39

In a two-class priority queueing model suppose that a cost of \(C_{i}\) per unit time is incurred for each type \(i\) customer that waits in queue, \(i=1,2 .\) Show that type 1 customers should be given priority over type 2 (as opposed to the reverse) if $$ \frac{E\left[S_{1}\right]}{C_{1}}<\frac{E\left[S_{2}\right]}{C_{2}} $$

Short Answer

Expert verified
In a two-class priority queueing model, type 1 customers should be given priority over type 2 customers if \(\frac{E[S_{1}]}{C_{1}} < \frac{E[S_{2}]}{C_{2}}\), where \(E[S_i]\) is the expected waiting time and \(C_i\) is the cost per unit waiting time for each customer type. This is because prioritizing type 1 customers leads to a favorable change in waiting times and an overall reduction in waiting time cost when the inequality holds.

Step by step solution

01

In the given inequality, \(E[S_i]\) represents the expected waiting time for type \(i\) customers in the queue, where \(i=1,2\). Therefore, the ratio \(\frac{E\left[S_{i}\right]}{C_{i}}\) represents the cost per unit time (waiting time) for type \(i\) customers in the queue. By comparing the cost per unit waiting time for both types of customers, the inequality states that type 1 customers should be given priority over type 2 if the cost per unit waiting time for type 1 customers is less than the cost per unit waiting time for type 2 customers. #Step 2: Analyze the benefit of prioritizing type 1 customers#

By prioritizing a type of customer, we can reduce the expected waiting time for that type of customer. The idea behind the priority is to reduce the overall cost for both types of customers' waiting time without sacrificing the service to any type. Let's assume we prioritize type 1 customers over type 2. This will result in a reduction in the expected waiting time for type 1 customers, denoted by \(\Delta S_1\), while possibly increasing the waiting time for type 2 customers, denoted by \(\Delta S_2\). #Step 3: Determine the change in cost per unit waiting time#
02

When we prioritize type 1 customers, we need to calculate the overall change in customer waiting time cost. We can represent the change in cost as: $$ \Delta C = C_1 \Delta S_1 + C_2 \Delta S_2 $$ Since we are prioritizing type 1 customers, \(\Delta S_1\) will be negative (their waiting time will decrease) and \(\Delta S_2\) will be positive (type 2 customers waiting time may increase). #Step 4: Identify the condition for favorable prioritization#

For the overall cost to be reduced, we need the change in cost \(\Delta C\) to be negative, which means: $$ C_1 \Delta S_1 + C_2 \Delta S_2 < 0 $$ We can rearrange the inequality as: $$ \frac{\Delta S_1}{\Delta S_2} < - \frac{C_2}{C_1} $$ The ratio \(\frac{\Delta S_1}{\Delta S_2}\) represents the change in waiting times when prioritizing type 1 customers. Since \(\frac{E\left[S_{1}\right]}{C_{1}}<\frac{E\left[S_{2}\right]}{C_{2}}\) holds, the change in waiting times will be favorable, and the overall waiting time cost will be reduced. Thus, this proves that type 1 customers should be given priority over type 2 customers when the given inequality holds.

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Most popular questions from this chapter

Poisson \((\lambda)\) arrivals join a queue in front of two parallel servers \(A\) and \(B\), having exponential service rates \(\mu_{A}\) and \(\mu_{B}\). When the system is cmpty, arrivals go into server \(A\) with probability \(\alpha\) and into \(B\) with probability \(1-\alpha\). Otherwise, the head of the queue takes the first free server. (a) Define states and set up the balance equations. Do not solve. (b) In terms of the probabilities in part (a), what is the average number in the system? Average number of servers idle? (c) In terms of the probabilities in part (a), what is the probability that an arbitrary arrival will get serviced in \(A\) ?

Consider a closed queueing network consisting of two customers moving among two servers, and suppose that after each service completion the customer is equally likely to go to either server-that is, \(P_{1,2}=P_{2,1}=\frac{1}{2}\). Let \(\mu_{i}\) denote the exponential service rate at server \(i, i=1,2\). (a) Determine the average number of customers at each server. (b) Determine the service completion rate for each server.

A group of \(m\) customers frequents a single-server station in the following manner. When a customer arrives, he or she either enters service if the server is free or joins the queue otherwise. Upon completing service the customer departs the system, but then returns after an exponential time with rate \(\theta .\) All service times are exponentially distributed with rate \(\mu\). (a) Define states and set up the balance equations. In terms of the solution of the balance equations, find (b) the average rate at which customers enter the station. (c) the average time that a customer spends in the station per visit.

Consider a single-server exponential system in which ordinary customers arrive at a rate \(\lambda\) and have service rate \(\mu .\) In addition, there is a special customer who has a service rate \(\mu_{1}\). Whenever this special customer arrives, it goes directly into service (if anyone else is in service, then this person is bumped back into queue). When the special customer is not being serviced, the customer spends an exponential amount of time (with mean \(1 / \theta\) ) out of the system. (a) What is the average arrival rate of the special customer? (b) Define an appropriate state space and set up balance equations. (c) Find the probability that an ordinary customer is bumped \(n\) time.

Compare the M/G/1 system for first-come, first-served queue discipline with one of last-come, first-served (for instance, in which units for service are taken from the top of a stack). Would you think that the queue size, waiting time, and busy-period distribution differ? What about their means? What if the queue discipline was always to choose at random among those waiting? Intuitively which discipline would result in the smallest variance in the waiting time distribution?
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