15.10 Queuing Theory & Simple Queuing Models

Why Queues Form

A queue forms whenever demand for a service arrives faster than it can be served — but crucially, queues also form when average capacity exceeds average demand, purely because of variability in arrivals and service times. This is the central and counter-intuitive insight of queuing theory, and it explains why a workshop with nominally adequate capacity still has jobs waiting.

In a coal operation, queuing problems are everywhere: dumpers waiting at a shovel, wagons waiting at a loading silo, machines waiting for a fitter, and vehicles waiting at a weighbridge.

Kendall's Notation

A queuing system is described as

$$A/B/c\,:\,(K/N/D)$$

Common symbols for $A$ and $B$:

So M/M/1 means Poisson arrivals, exponential service, one server, infinite capacity and population, first come first served. M/M/c is the multi-server equivalent.

Queue disciplines: FCFS (first come first served), LCFS, SIRO (service in random order), and priority schemes.

The link between Poisson and exponential

If arrivals occur as a Poisson process with mean rate $\lambda$, then the time between successive arrivals is exponentially distributed with mean $1/\lambda$. The two statements are equivalent descriptions of the same process, and the exponential distribution's memoryless property — the time already waited gives no information about the time still to wait — is what makes the mathematics tractable.

The M/M/1 Model

Let $\lambda$ be the mean arrival rate and $\mu$ the mean service rate of the single server. Define the traffic intensity or utilisation

$$\rho = \frac{\lambda}{\mu}$$

The system is stable only if $\rho < 1$. If $\rho \geq 1$, arrivals outpace service and the queue grows without limit.

Standard results

Useful identities: $L_s = L_q + \rho$ and $W_s = W_q + 1/\mu$. The difference between "in system" and "in queue" is exactly the one job being served.

Little's Law

$$\boxed{L = \lambda\,W}$$

The average number in a system equals the arrival rate multiplied by the average time spent in it. Little's law is remarkably general — it holds for any queuing system regardless of the arrival or service distributions, the number of servers, or the discipline. It applies equally to $L_s$ with $W_s$ and to $L_q$ with $W_q$.

Its practical use is that measuring any two of the three quantities gives the third. If a workshop has on average 12 jobs in progress and receives 3 jobs per day, the average throughput time is 4 days — no further data needed.

The Congestion Curve

The behaviour of $L_q$ as $\rho$ rises is the most important qualitative result in the topic:

Queue length rises hyperbolically, not linearly. Going from 80% to 90% utilisation adds ten percentage points of utilisation but more than doubles the queue; going from 90% to 95% doubles it again.

The managerial implication is direct and often resisted: a facility planned for 95% or higher utilisation of a variable process will congest. Buffer capacity is not waste; it is what buys acceptable waiting times. This is the quantitative justification for keeping standby capacity in a maintenance workshop or an extra dumper in a shovel fleet.

Worked Example: Workshop Repairs

Breakdowns arrive at a mine workshop at an average of 4 per day, Poisson distributed. A single fitting bay repairs them at an average of 6 per day, exponentially distributed.

$$\rho = \frac{4}{6} = 0.667$$ $$L_s = \frac{4}{6-4} = 2 \text{ machines in the system}$$ $$L_q = \frac{16}{6\times2} = 1.33 \text{ machines waiting}$$ $$W_s = \frac{1}{6-4} = 0.5 \text{ days} = 12 \text{ hours}$$ $$W_q = \frac{4}{6\times 2} = 0.333 \text{ days} = 8 \text{ hours}$$

Check by Little's law: $L_s = \lambda W_s = 4\times0.5 = 2$. Consistent.

The probability the bay is idle is $P_0 = 1 - 0.667 = 0.333$, so the fitter is idle a third of the time — and yet machines still wait 8 hours on average. That combination of visible idleness and visible waiting is precisely what variability produces, and it is why intuitive capacity planning by average load alone fails.

The M/M/c Multi-Server Model

With $c$ identical servers sharing one queue, utilisation becomes

$$\rho = \frac{\lambda}{c\,\mu}$$

and stability again requires $\rho < 1$. The probability of an empty system is

$$P_0 = \left[\sum_{n=0}^{c-1}\frac{(\lambda/\mu)^n}{n!} + \frac{(\lambda/\mu)^c}{c!\,(1-\rho)}\right]^{-1}$$

and the mean queue length is

$$L_q = P_0\,\frac{(\lambda/\mu)^c\,\rho}{c!\,(1-\rho)^2}$$

with $W_q = L_q/\lambda$, $W_s = W_q + 1/\mu$ and $L_s = \lambda W_s$.

One queue or several?

A practically important result: a single queue feeding $c$ servers always outperforms $c$ separate queues each feeding one server, even though the total capacity is identical. The single queue never leaves a server idle while someone waits elsewhere. This is why banks, airports and well-run workshops use a single feeding line rather than one line per counter.

Other Models in Brief

The M/D/1 result deserves emphasis because of what it implies for practice: simply reducing variability in service times, with no increase in capacity at all, halves the queue. Standardising repair procedures, pre-kitting spares and using standard work packages are therefore direct queue-reduction measures, not merely tidiness.

Economic Analysis

The purpose of the analysis is normally to choose a service capacity. The total cost per unit time is

$$C_{\text{total}} = C_s\,c + C_w\,L_s$$

where $C_s$ is the cost of providing one server and $C_w$ the cost of one unit waiting. Service cost rises linearly with $c$ while waiting cost falls hyperbolically, so the total has a clear minimum — evaluated by computing the total for $c = 1, 2, 3\ldots$ and selecting the lowest.

In a mine, $C_w$ can be very large: a haul truck waiting is not merely an idle asset but lost production from the whole shovel-truck circuit. That high waiting cost is what justifies apparently generous workshop and standby capacity, and quantifying it is exactly what this model is for.