## 4.5 Example Application of Steady State Queuing Theory

The practical application of the theory outlined above for queues with random arrivals and service times is illustrated by the following example.

At a major sporting venue, patrons arrive in motor vehicles. At one of the entrances to the car parking area surrounding the venue, a single line of vehicles approaches the manually operated entry gate, where a cash payment is required to gain entry. The process of stopping at the payment point, offering payment, receiving change if necessary and departing the payment point averages 12 seconds per vehicle. Vehicles are arriving randomly at an average rate of 216 vehicles per hour. Appropriate design of the entry arrangements requires knowledge of the following values:

- the queue storage length required prior to the payment point, assuming that the length provided must be adequate for at least 95% of the time
- the proportion of vehicles that will have to wait in a queue before entering the payment point
- the average total delay, including time spent in the payment process, to vehicles entering the parking area at this gate.

Value (1), the required queue storage length, is determined by application of Equation 4.5, as the aim to find the smallest value of N for which n, the number of vehicles in the queuing system, satisfies:

Pr (n > N) = \(\rho ^{N+1}\le \) 0.05 |

The average arrival rate of vehicles, r, is 216 veh/h and the average service time of 12 s/veh means that the average service rate, s, is 300 veh/h. Therefore, by Equation 4.1, ρ = r/s = 0.72 and the required value of N is identified as 9, by calculating:

Pr (n > 8) = (0.72)^{9} = 0.052 | |||

and | Pr (n > 9) = (0.72)^{10} = 0.037 |

Given that n includes the vehicle occupying the payment point, space must be provided for eight vehicles to queue before the payment point. If a length of 6 m is allowed for each queued vehicle, a storage length of at least 48 m should be provided.

The following are other results related to queue lengths that may be of interest:

- The probability that no vehicles are present at any given instant is derived from Equation 4.2 as:

\[P_{0}= 1 - \rho = 1 - 0.72 = 0.28 \ or \ 28\% \] |

- The probability of there being exactly 6 vehicles present, including that in service is given by Equation 4.3 as:

\[P_{6} = (1-\rho ) \rho ^{6} = (1 - 0.72) (0.72)^{6} = 0.0390 \ or \ 3.9\% \] |

- The mean number of vehicles present, including that in service, is given by Equation 4.4 as:

\(E(n) = \frac{\rho }{1-\rho } = \frac{0.72}{0.28} = 2.57\) vehicles |

- The mean number of vehicles present, excluding that in service, is given by Equation 4.6 as:

\(E(m) = \frac{\rho ^{2} }{1-\rho } = \frac{0.72^{2} }{0.28} = 1.85\) vehicles |

which is less than E(n) by \(\rho = 0.72\) vehicles, in accordance with Equation 4.7.

- The variance of the mean number of vehicles present, including that in service is given by Equation 4.8 as:

\(\sigma ^{2} (n) = \frac{\rho }{(1-\rho )^{2} } = \frac{0.72}{0.28^{2} } = 9.18\) (vehicles)^{2} |

which implies a standard deviation of 3.03 vehicles in the queue length distribution.

Value (2), the proportion of vehicles that will have to wait in a queue before entering the payment point, is obtained by noting that Equation 4.9 calculates that the probability that any arriving vehicle will **not** have to wait is 1 – ρ. Therefore, the probability that an arriving vehicle **will** have to wait is ρ, that is, 0.72. Hence 72% of vehicles will have to wait before entering the payment point.

Value (3), the average total delay to entering vehicles, including time spent in the payment process, is given by Equation 4.14 as:

\[E(\tau ) = \frac{1}{s - r} \] |

As observed above, the average service rate, s, is 300 veh/h, while the average arrival rate, r, is 216 veh/h. Each of these rates is converted to the units of veh/s by division by 3600. Therefore, the average total delay to entering vehicles is:

\(\frac{1}{s - r} = \frac{3600}{300 - 216} = 42.9\) s/veh |

This total delay includes the time spent in the payment process, which averages 12 s/veh. Therefore, the average waiting time in the queue before entering the payment point (see Equation 4.15) is 42.9 – 12 = 30.9 s/veh.

The following are other results related to delays that may be of interest:

- The probability of a vehicle having to wait more than 20 s before entering the payment point is given by Equation 4.11 as:

Pr (w>\(20) = \rho e^{-(s-r) 20} = 0.72 e^{-(300-216) 20 / 3600} = 0.4515\ or\ 45.15\% \) |

- Over all arriving vehicles, the average waiting time before entering the payment point is also given by Equation 4.12 (consistent with the result obtained above) as:

\(E(w) = \frac{\rho }{s-r} = \frac{3600 .( 0.72)}{300 - 216} = 30.9\) s |

- Over only those arriving vehicles that cannot immediately enter, the average waiting time before entering the payment point is given by Equation 4.13 as:

\(E(w|w\) > \(0) = \frac{1}{s-r} = \frac{3600}{300 - 216} = 42.9\) s |