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In an \(\mathrm{M} / \mathrm{G} / 1\) queue, (a) what proportion of departures leave behind 0 work? (b) what is the average work in the system as seen by a departure?

Short Answer

Expert verified
In an M/G/1 queue, (a) the proportion of departures leaving behind 0 work is \(P0 = 1 - \frac{1}{\mu} \cdot \frac{\lambda}{\mu}\), and (b) the average work in the system as seen by a departure is \(E(W) = \frac{\rho^2}{1-\rho}\cdot \frac{1}{\mu}\), where \(\rho = \frac{\lambda}{\mu}\), \(\lambda\) is the mean arrival rate, and \(\mu\) is the mean service rate.

Step by step solution

01

Part (a): Proportion of Departures Leaving Behind 0 Work

In an M/G/1 queue, the proportion of departures leaving behind 0 work is equal to the proportion of the system being empty when a general arrival occurs. Let P0 be the probability that the system is empty when a general arrival occurs. We can write this probability as: \[P0 = 1 - E(S)\rho\] where \(E(S)\) is the expected service time and \(\rho\) is the utilization factor. We are given that this is an M/G/1 queue, so we can use the fact that \(E(S)=1/\mu\), where \(\mu\) is the mean service rate. From the definition of the utilization factor, we have: \[\rho = \frac{\lambda}{\mu}\] where \(\lambda\) is the mean arrival rate. Now, we can compute \(P0\): \[P0 = 1 - \frac{1}{\mu} \cdot \frac{\lambda}{\mu}\] Once we have the values of \(\lambda\) and \(\mu\), we can plug them into the above formula to find the proportion of departures leaving behind 0 work.
02

Part (b): Average Work in the System as Seen by a Departure

The average work in the system as seen by a departure in an M/G/1 queue can be found using Little's Law and the fact that the average work seen by a general arrival is equal to the average work seen by a departure: \[E(W) = E(Q) \cdot E(S)\] where \(E(W)\) is the average work in the system as seen by a departure, \(E(Q)\) is the average number of customers in the queue, and \(E(S)\) is the expected service time. We know that \(E(S) = 1/\mu\). To find the average number of customers in the queue, we can use Little's Law: \[E(Q) = \frac{\rho^2}{1-\rho}\] Now, we can find the average work in the system: \[E(W) = \frac{\rho^2}{1-\rho}\cdot \frac{1}{\mu}\] Once we have the values of \(\lambda\) and \(\mu\), we can plug them into the above formula to find the average work in the system as seen by a departure.

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Most popular questions from this chapter

For the \(M / G / 1\) queue, let \(X_{n}\) denote the number in the system left behind by the nth departure. (a) If $$ X_{n+1}=\left\\{\begin{array}{ll} X_{n}-1+Y_{n}, & \text { if } X_{n} \geqslant 1 \\ Y_{n}, & \text { if } X_{n}=0 \end{array}\right. $$ what does \(Y_{n}\) represent? (b) Rewrite the preceding as $$ X_{n+1}=X_{n}-1+Y_{n}+\delta_{n} $$ where $$ \delta_{n}=\left\\{\begin{array}{ll} 1, & \text { if } X_{n}=0 \\ 0, & \text { if } X_{n} \geqslant 1 \end{array}\right. $$ Take expectations and let \(n \rightarrow \infty\) in Equation ( \(8.64\) ) to obtain $$ E\left[\delta_{\infty}\right]=1-\lambda E[S] $$ (c) Square both sides of Equation \((8.64)\), take expectations, and then let \(n \rightarrow \infty\) to obtain $$ E\left[X_{\infty}\right]=\frac{\lambda^{2} E\left[S^{2}\right]}{2(1-\lambda E[S])}+\lambda E[S] $$ (d) Argue that \(E\left[X_{\infty}\right]\), the average number as seen by a departure, is equal to \(L\).

Customers arrive at a single-server station in accordance with a Poisson process with rate \(\lambda\). All arrivals that find the server free immediately enter service. All service times are exponentially distributed with rate \(\mu\). An arrival that finds the server busy will leave the system and roam around "in orbit" for an exponential time with rate \(\theta\) at which time it will then return. If the server is busy when an orbiting customer returns, then that customer returns to orbit for another exponential time with rate \(\theta\) before returning again. An arrival that finds the server busy and \(N\) other customers in orbit will depart and not return. That is, \(N\) is the maximum number of customers in orbit. (a) Define states. (b) Give the balance equations. In terms of the solution of the balance equations, find (c) the proportion of all customers that are eventually served; (d) the average time that a served customer spends waiting in orbit.

Consider a model in which the interarrival times have an arbitrary distribution \(F\), and there are \(k\) servers each having service distribution \(G .\) What condition on \(F\) and \(G\) do you think would be necessary for there to exist limiting probabilities?

Consider an \(M / G / 1\) system in which the first customer in a busy period has the service distribution \(G_{1}\) and all others have distribution \(G_{2}\). Let \(C\) denote the number of customers in a busy period, and let \(S\) denote the service time of a customer chosen at random. Argue that (a) \(a_{0}=P_{0}=1-\lambda E[S]\) (b) \(E[S]=a_{0} E\left[S_{1}\right]+\left(1-a_{0}\right) E\left[S_{2}\right]\) where \(S_{i}\) has distribution \(G_{i}\). (c) Use (a) and (b) to show that \(E[B]\), the expected length of a busy period, is given by $$ E[B]=\frac{E\left[S_{1}\right]}{1-\lambda E\left[S_{2}\right]} $$ (d) Find \(E[C]\).

For the \(M / M / 1\) queue, compute (a) the expected number of arrivals during a service period and (b) the probability that no customers arrive during a service period. Hint: "Condition."

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