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We have two demonstrations (2007+) for this session:
This example shows a simple way, using names, to evaluate the formulas we’ve seen. By assigning values to the names ArvRate and SvcRate, which represent λ and μ respectively, we can create readable and understandable models of service systems. Perhaps you’re wondering why we don’t just use the names Lambda and Mu. We could use Lambda — that would be OK. But Mu presents problems, because it could be confused with the label for column MU of the worksheet. That’s why, in this course, we usually use ArvRate for λ and SvcRate for μ.
For this illustration we suppose that customers arrive at a Poissondistributed rate of 20 per hour. Service times average 2.6 minutes, and are exponentially distributed. How long is the average waiting line?
This is a straightforward application of the formulas of the slide “Performance Measures for Single Server Systems” of the class notes (Office 2007+), or from the reading on Service Systems. We’re looking for the average length of all lines, even when there is no line. Looking at the readings, we find that the quantity we want is given by
L_{q} = λ^{2}/(μ (μ  λ))
The formula is:
Excluding empty lines, what is the average line length?
Now the formula we want is:
L_{a} = μ/(μ  λ)
The formula is:
This example plots the shape of the Poisson distribution. It’s intended to give you a feel for what the arrival distribution looks like. The plot shows the probability of n arrivals on or before time t. The four series shown depict the distributions of the arrival of the n^{th} customer by time t, for n = 0 to 3.
The plot for n = 0 shows the probability of zero arrivals as a function of time. In other words, for t=5 say, it answers the question “What’s the probability that we have no arrivals?” For high λ, this plot moves in towards 0 — more probability is concentrated toward 0. This happens because of the first factor of the Poisson distribution, (λt)^{k}, which for k = 0 becomes (λt)^{0}. At t = 0, this is 0^{0} = 1. The second factor, e^{λt}, then prevails.
For other values of n, the two factors compete. Since the first is a simple power, and the second is an exponential, the second always wins, eventually. Thus the shape of the distribution for n > 0 is humped, with the maximum occurring at greater t as n increases.
Last Modified: Wednesday, 27Apr2016 04:15:26 EDT
Modeling service systems in general is extraordinarily complex, but as we’ve seen, if we make reasonable approximations, we can gain powerful tools that are very easy to apply. In the case of service systems, we assumed that the system was at equilibrium or close to it. Analogously, we can make simplifying assumptions for many other complex questions. Examples are process control, resource scheduling, resource allocation, cost allocation, vehicle routing, and many more.