## Queue Time

Read this research article about a cross-docking problem, which proposes a nonstationary queuing model to speed up logistics timeframes. As you read, try to think about your next online order and how your order is fulfilled, packaged, transported, sorted, and delivered. Which part of this process had the longest queue time?

### Theoretical Model of the Nonstationary Queue

Figure 2 shows the arrival process of supply trucks at the terminal for a scheduled cross-docking operation, where operators are required to send their inbound vehicles respecting a predefined time window (diagram of Figure 2 is based on Odoni and de Neufville’s article). It is observed that, in situations of this type, with sufficiently strict operating rules, the stochastic queuing process may not reach the steady-state equilibrium mentioned above for the classical queuing models. It is then necessary to apply a type of modeling that does not depend on the stationary characteristics of the process. Newell has developed a rather unique and robust modeling solution for queuing problems with transient and nonstationary states. For this purpose, a continuous representation is used, in which the arrival process is represented as if its elements were a fluid in movement.

Figure 2 Arrival rate of the incoming feeder trucks at the cross-docking terminal, fitted with the third-degree mathematical curve.

It is considered a hypothetical case of a supermarket chain (real data is firm proprietary) that owns a cross-docking terminal, which operates the unloading of supply vehicles during a time window of $T = 8$ hours per day, with a total of $G = 85,000$ cartons per day. In the application, each vehicle has, on average, 463.8 boxes. The integral of $\lambda$  in the interval $0 \leq t \leq T$ should equal the total demand observed in the time window, obtaining the final expression:

$\lambda(t)=\left(\frac{12 G}{\tau^{4}}\right)\left[(T-t)^{3}-2 T(T-t)^{2}+T^{2}(T-t)\right], \quad(0 \leq t \leq T)$

##### 4.1. Queue Forming Process

Let $D(t)$ be the cumulative function of the unloading capacity of the $n$ receiving doors which operate simultaneously and in parallel at the terminal dock. Each door has the capacity to download, on average, $\mu$ boxes per hour. The total capacity average of the discharge doors is therefore equal to $\mu n$. It has been assumed that the average discharge capacity is constant over time; that is, $D(t)$ varies linearly as a function of time:

$D(t) = a + \mu nt$

It was assumed that there is no queue at time $t = 0$, that is, $Q(0) = 0$, where $Q(t)$ expresses the queue size at time $t$. This no-queue situation tends to remain for some time $t_0$, when $\lambda (t)$ reaches a growth rate equal to $\mu n$ and where the curve $A(t) = \int_0^t \lambda (t)$ separates from the line $D(t)$, as shown in Figure 3.

Figure 3 Phase between instant zero and instant $t_0$, when there is practically no queue.

It is therefore necessary to determine $t_0$ such that

$\lambda (t_0) = \mu n$

That is, the instant $t_0$ is one in which the arrival rate of the products becomes equal to the total discharge capacity of the dock, including all the doors; that is,

$\left(\frac{12 G}{T^{4}}\right)\left[(T-t_0)^{3}-2 T(T-t_0)^{2}+T^{2}(T-t_0)\right] = \mu n$

In this application, the instant $t_0$ is determined by minimizing $f(t)$ given by the absolute value of the difference $\lambda(t)$ and $\mu n$, according to relation

$f(t) = \min \{abs[\lambda(t) - \mu n] \}\to t_0$

and applying a golden search numerical method to solve.

In Figure 3 it is noted that $D(t_0) = A(t_0)$, since both are coincident in that position. Thus, the equation of the line $D(t)$ can be expressed specifically for $t = t_0$:

\begin{aligned}A\left(t_{0}\right) &=a+\mu n t_{0} \\D(t) &=A\left(t_{0}\right)-\mu n t_{0}+\mu n t\end{aligned}

Figure 3 shows that the supply vehicle service process ends at time $t = T$, when the last truck starts to be unloaded. But in most cases this is not what actually happens. In fact, $t_Q$ is the point at which $D(t)$ equals the value of $G$, that is, the moment when the service supply $D(t)$ fully satisfies the daily demand $G$. Then the value of $t_Q$ is calculated through of the expression (7), making $D(T) = G$:

$t_Q = \frac{[G - A (t_0) + \mu nt_0]}{\mu n}$

##### 4.2. Queue Extension

Theoretically, it can be considered that, for nonstationary queuing situations of this type, terminal operation could be performed with any value of $n$ equal or greater than unity. The correct action to take is to establish a maximum daily operation time of the dock and to size the number of unloading doors, so that the daily operation time of the terminal does not exceed a certain limit.

The extension $t$ of the queuing process varies between instants $t_0$ and $t_Q$; for $0 \leq t \leq t_0$, the queue is null. We have the following formulations:

(a) $0 \leq t \leq t_{0}$,

$Q(t)=0.$

(b) $t_{o} \leq t \leq T$,

$Q(t)=A(t)-D(t).$

(c) $T < t \leq t_{Q}$,

$Q(t)=G-D(t)$

Three parameters are of interest to the analysis: the average queue, considering all the time of discharging operation, the average queue considering only the congestion phase, and the maximum value of the queue in the three above cases.

##### 4.3. Queue Time

The article by Little and Graves shows the validity of

$Q = \lambda \ast W,$

where $Q$ is the average number of elements in the queue, $W$ is the average waiting time of an element in the queue, and $\lambda$ is the average arrival rate per unit time. Expression, widely used in practice, was given the name “little law” because of its almost universal character.

Expression is used in this work to estimate the average waiting time in the queue, given by $W = Q/\lambda$.

##### 4.4. Random Variation of Queue Extension

If we divide the time window $0 \leq t \leq T$ into infinitesimal intervals $dt$, in each of them the arrivals can be considered as obeying a Poisson distribution, with mean $\lambda(t)$. Because it is a Poisson distribution, the variance is equal to average $\lambda(t)$. On the other hand, the discharge time of a vehicle is governed by a log-normal distribution in our application. The attendance rate is running at a rate equal to $n \mu$, with variance equal to $CV \times n \mu$.

The variance of the difference between the arrival process and the service process is ($A[t]$) and the discharge of the trucks ($D[t]$), accumulating at time $t$. Thus, the queue variance, which is cumulatively generated over time, is given by

$var Q (t) \cong t [\lambda (t) + CV \ast n \mu],$

with the standard deviation

$\sigma_Q(t) = \{t[\lambda (t) + CV \ast n \mu]\}^{1/2}$

A computer program written in Delphi/Pascal, Berlin Version 10.1, by Embarcadero Technology, Inc., was developed to determine the main queue elements indicated in this section. The decision variable of the problem is $n$, the number of discharging doors at the cross-docking terminal, varying in the interval $10 \leq n \leq 29$.

##### 4.5. Distribution of Vehicle Discharge Time

It was assumed that discharge times are governed by log-normal distribution, with mean $\overline{S}$ 43,8 min and standard deviation $\sigma_s$ 12,8 min. The probability density function $f(s)$ is given by

$f(S)=\frac{1}{S v \sqrt{2 \pi}} \exp \left[\frac{-(\ln S-m)^{2}}{2 v^{2}}\right],$

where $m$ and $v$ are two auxiliary parameters given by

\begin{aligned}&m=\ln \left(\frac{\bar{S}^{2}}{\sqrt{\sigma_{S}+\bar{S}^{2}}}\right) \\&v=\sqrt{\ln \left(\frac{\sigma_{S}}{S^{2}}+1\right)}\end{aligned}

The discharge rate at any door of the receiving dock is the inverse of the discharge time

$\mu = \frac{u \ast 60}{\overline{S}}$

With $\overline{S}$ expressed in minutes and $\overline{\mu}$ expressed in boxes unloaded per hour and where $u$ is the average load of a truck (cartons), we obtain $\overline{\mu} = 635.3$ boxes discharged per hour.

The time the receiving dock is effectively in use when unloading vehicles is

$DH = n \times \tau$

where $\tau$ is the daily use time of the dock (hours).

The time $\tau$ is equal to $t_Q$ plus the mean discharge time $\overline{S}$; that is,

$\tau = t_Q + \overline{S}$

In, the term $\overline{S}$ refers to the additional time which is related to the trucks that arrive at the terminal at the last allowed moment.

##### 4.7. Average Time of Stay in the System

Let $y$ be the fraction of the arriving vehicles that do not enter the queue. On the other hand, $\overline{WT}_{fila}$ is the average time in the queue, calculated within the time interval when congestion (i.e., queue) occurs. When queuing occurs, the average time in the system is equal to the sum of the time in the queue and the unloading time. Conversely, when there is no queue, the time in the system is equal to the unloading time. Thus, the average time in the system, $\overline{WT}_{syst}$, is given by

$\overline{W T}_{\text {syst }}=(1-\gamma)\left(\overline{W T}_{\text {fila }}+\bar{S}\right)+\gamma \bar{S}$

##### 4.8. Number of Receiving Doors in the Analysis

To determine the upper end of $n$, which was called $n_Q$ in the program developed in Pascal, it is worth stressing that the line $D(t)$ does cross the curve $A(t)$ for $t_0 \geq (2/3)T$. In such a condition, the line representing $D(t)$ is tangential to the curve $A(t)$. From that point on, it becomes uneconomic to add more discharging doors to the dock, since the benefit would be nil.

On the other hand, for small values of $n$, the value of $t_Q$ might increase too much, above acceptable operating conditions. With these aspects in mind, the following range of variation for the decision variable $n$ was

$10 \leq n \leq 29$

##### 4.9. Results of the Application of the Theoretical Model of Nonstationary Queues

As mentioned, a program was developed in Pascal, within a Delphi XE7 platform, in order to apply the model. Table 2 shows the main results obtained with the application of the model, for $n$ ranging from 10 to 29. The results are referenced to the “box” units (cartons). These values are then divided by $\mu$ 463,8 cartons per vehicle to turn them into “vehicles units.”

Table 2 Results of the theoretical model application.

 Number of positions Average queue(vehicles) Average wait time (h) Average system time (h) Total average usage time of docks (h) Average dock occupancy rate (%) 10 39.9 3.20 3.75 15.43 86.7 11 36.0 2.66 3.22 14.29 85.1 12 32.4 2.23 2.79 13.35 83.5 13 28.9 1.87 2.43 12.56 81.9 14 25.7 1.57 2.14 11.89 80.3 15 22.7 1.31 1.89 11.32 78.8 16 19.9 1.10 1.69 10.83 77.2 17 17.2 0.91 1.51 10.41 75.6 18 14.8 0.75 1.36 10.04 74.1 19 12.6 0.62 1.24 9.71 72.5 20 10.5 0.50 1.13 9.42 71.0 21 8.6 0.40 1.04 9.17 69.5 22 6.9 0.31 0.97 8.94 68.0 23 5.4 0.23 0.90 8.74 66.5 24 3.9 0.17 0.85 8.73 65.1 25 2.7 0.12 0.81 8.73 63.6 26 1.7 0.07 0.78 8.73 62.2 27 0.9 0.04 0.75 8.73 60.8 28 0.4 0.02 0.74 8.73 59.4 29 0.1 0.00 0.73 8.73 58.0

Analyzing Table 2 and Figure 4, the configuration with $n = 23$ receiving positions seems satisfactory, with a mean queue of 5.4 vehicles and a mean waiting time of 0.23 hours. We show that an approximate and nonstationary model of the queue, based on a continuous representation of the central variable, can provide sufficient results for a preliminary analysis of the problem. Through a simulation model, the analysis can be deepened, leading to more accurate values of the results.

Figure 4 Queue waiting time based on dock utilization rate (%).