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12.6 Finite Population Model (M / M / 1 with Finite Source)

When there is a limited population of potential customers for a service facility, we need to consider a different queuing model. This model would be used, for example, if you were considering equipment repairs in a factory that has five machines, if you were in charge of maintenance for a fleet of 10 commuter airplanes, or if you ran a hospital ward that has 20 beds. The limited population model permits any number of repair people (servers) to be considered.

The reason this model differs from the three earlier queuing models is that there is now a dependent relationship between the length of the queue and the arrival rate. To illustrate the extreme situation, if your factory had five machines and all were broken and awaiting repair, the arrival rate would drop to zero. In general, as the waiting line becomes longer in the limited population model, the arrival rate of customers or machines drops lower.

In this section, we describe a finite calling population model that has the following assumptions:

There is only one server.
The population of units seeking service is finite.⁴
4Although there is no definite number that we can use to divide finite from infinite populations, the general rule of thumb is this: if the number in the queue is a significant proportion of the calling population, use a finite queuing model. Finite Queuing Tables, by L. G. Peck and R. N. Hazelwood (New York: John Wiley & Sons, Inc., 1958), eliminates much of the mathematics involved in computing the operating characteristics for such a model.
Arrivals follow a Poisson distribution, and service times are exponentially distributed.
Customers are served on a first-come, first-served basis.

Equations for the Finite Population Model

Using

λ = mean arrival rate, μ = mean service rate, and N = size of the population,

$λ = mean arrival rate, μ = mean service rate, and N = size of the population,$

the operating characteristics for the finite population model with a single channel or server on duty are as follows:

Probability that the system is empty:

$P_{0} = \frac{1}{\sum_{n = 0}^{N} \frac{N!}{(N - n)!} {(\frac{λ}{μ})}^{n}}$ $P_{0} = \frac{1}{\sum_{n = 0}^{N} \frac{N!}{(N - n)!} {(\frac{λ}{μ})}^{n}}$ (12-23)
Average length of the queue:

$L_{q} = N - (\frac{λ + μ}{λ}) (1 - P_{0})$ $L_{q} = N - (\frac{λ + μ}{λ}) (1 - P_{0})$ (12-24)
Average number of customers (units) in the system:

$L = L_{q} + (1 - P_{0})$ $L = L_{q} + (1 - P_{0})$ (12-25)
Average waiting time in the queue:

$W_{q} = \frac{L_{q}}{(N - L) λ}$ $W_{q} = \frac{L_{q}}{(N - L) λ}$ (12-26)
Average time in the system:

$W = W_{q} + \frac{1}{μ}$ $W = W_{q} + \frac{1}{μ}$ (12-27)
Probability of n units in the system:

$P_{n} = \frac{N!}{(N - n)!} {(\frac{λ}{μ})}^{n} P_{0} for n = 0, 1, \dots, N$ $P_{n} = \frac{N!}{(N - n)!} {(\frac{λ}{μ})}^{n} P_{0} for n = 0, 1, \dots, N$ (12-28)

Department of Commerce Example

Past records indicate that each of the five high-speed “page” printers at the U.S. Department of Commerce, in Washington, D.C., needs repair after about 20 hours of use. Breakdowns have been determined to be Poisson distributed. The one technician on duty can service a printer in an average of 2 hours, following an exponential distribution.

To compute the system’s operation characteristics, we first note that the mean arrival rate is $λ =^{1} /_{20} = 0.05$ $λ =^{1} /_{20} = 0.05$ printer/hour. The mean service rate is $μ =^{1} /_{2} = 0.50$ $μ =^{1} /_{2} = 0.50$ printer/hour. Then

$P_{0} = \frac{1}{\sum_{n = 0}^{5} \frac{5!}{(5 - n)!} {(\frac{0.05}{0.5})}^{n}} = 0.564$ $P_{0} = \frac{1}{\sum_{n = 0}^{5} \frac{5!}{(5 - n)!} {(\frac{0.05}{0.5})}^{n}} = 0.564$ (we leave these calculations for you to confirm)

Program 12.4 Excel QM Solution for Finite Population Model for Department of Commerce Example

Figure 12.4 Full Alternative Text
$\begin{array}{l} L_{q} & = & 5 - (\frac{0.05 + 0.5}{0.05}) (1 - P_{0}) = 5 - (11) (1 - 0.564) = 5 - 4.8 \\ = & 0.2 printer \end{array}$ $\begin{array}{l} L_{q} & = & 5 - (\frac{0.05 + 0.5}{0.05}) (1 - P_{0}) = 5 - (11) (1 - 0.564) = 5 - 4.8 \\ = & 0.2 printer \end{array}$
$L = 0.2 + (1 - 0.564) = 0.64 printer$ $L = 0.2 + (1 - 0.564) = 0.64 printer$
$W_{q} = \frac{0.2}{(5 - 0.64) (0.05)} = \frac{0.2}{0.22} = 0.91 hour$ $W_{q} = \frac{0.2}{(5 - 0.64) (0.05)} = \frac{0.2}{0.22} = 0.91 hour$
$W = 0.91 + \frac{1}{0.50} = 2.91 hours$ $W = 0.91 + \frac{1}{0.50} = 2.91 hours$

If printer downtime costs $120 per hour and the technician is paid $25 per hour, we can also compute the total cost per hour:

\begin{array}{l} Total hourly cost & = & (Average number of printers down)(Cost per downtime hour) \\ + Cost per technician hour \\ = & (0 .64)($120) + $25 = $76 .80 + $25 .00 = $101 .80 \end{array}

$\begin{array}{l} Total hourly cost & = & (Average number of printers down)(Cost per downtime hour) \\ + Cost per technician hour \\ = & (0 .64)($120) + $25 = $76 .80 + $25 .00 = $101 .80 \end{array}$

Solving The Department Of Commerce Finite Population Model With Excel QM

To use Excel QM for this problem, from the Excel QM menu, select Waiting Lines - Limited Population Model (M/M/s). When the spreadsheet appears, enter the arrival rate (8), service rate (12), number of servers, and population size. Once these are entered, the solution shown in Program 12.4 will be displayed. Additional output is also available.

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Table of Contents for 12.6 Finite Population Model (M / M / 1 with Finite Source)

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Table of Contents for
12.6 Finite Population Model (M / M / 1 with Finite Source)