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For D/D/1 queueing systems, the P-K formulas do not apply.

Examples

Example 1

Suppose an application passes data transmission requests to the data communications protocol in an end host. The requests arrive according to a Poisson distribution. The data from each request is encapsulated in a single frame for transmission across the outgoing link, and the frame lengths have an exponential distribution. The capacity of the point-to-point link is denoted as C bits/second.

In this example, the customers are frames, and the server is the hardware that transmits frames onto the link.

Because there are Poisson arrivals, the interarrival times are memoryless and exponentially distributed and we start our Kendall's notation for the queueing system as M/*/*. The frames are said to have exponentially distributed lengths. This means the service times will be exponentially distributed (i.e. longer frames take longer to service). Thus, our queueing system is M/M/*. Finally, because we are transmitting onto a single point-to-point link (i.e. a bus), only one transmission can occur at once. Thus, there must be a single server, and we complete our system classification as M/M/1.

The average frame length is 1/μ_l or 1000 bits/frame. Thus, the average service time is 1/μ or 1/μ_lC seconds/frame.

Suppose the utilization of the transmission channel is determined to be 0.6.

What is the average number of frames queued or in transmission?

We can directly reference the P-K formulas for an M/M/1 queueing system, and compute:

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If an average of one second elapses from the time that a data request occurs until transmission of the corresponding frame is complete, what is the link capacity?

The P-K formulas for our M/M/1 queueing system also come in handy here:

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So:

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What is the average fraction of time the channel is busy, and what is the average number of frames in service?

This is a trick question. Both of these are different ways to ask for the utilization of the system. The utilization was given to us as 0.6, thus the answer to both these questions is 0.6.

Can there be exactly ρ frames in service at any instant?

Since we are modeling a real-world, physical system with a single server that works in units of frames, either a single frame is being serviced or no single frame is being serviced. Thus, the answer is no; there must be either 0 or 1 frames in service at any given instant. ρ, on the other hand, denotes the probability that a frame is being serviced at an arbitrary moment in time.

Example 2

Example 3

Example 4

Example 5