Mining Economics

Queueing Theory (M/M/1)

The single-server queue — utilization, average number in the system, and average waiting time for shovel–truck and service problems.

PART 1

Topic Breakdown & Traps

Truck-and-shovel and workshop-service problems are modelled as an M/M/1 queue: Poisson arrivals at rate

λ

and exponential service at rate

μ

with one server. The utilization

ρ = λ / μ

must be below 1 for a stable queue. From it follow the average number in the system

L_{s}

and the average waiting time, which grow sharply as

ρ

approaches 1.

Utilization:

ρ = \frac{λ}{μ}

(must be

< 1

).

Average number in system:

L_{s} = \frac{λ}{μ - λ} = \frac{ρ}{1 - ρ}

.

Average number in queue:

L_{q} = \frac{ρ ^{2}}{1 - ρ}

.

Average waiting time in queue:

W_{q} = \frac{λ}{μ ( μ - λ )}

; in system

W_{s} = \frac{1}{μ - λ}

⚠
** $ρ = λ / μ$ , not $μ / λ$ .** Utilization must be below 1; the inverse exceeds 1 and is meaningless here.
⚠
** $L_{s}$ counts the one in service plus those waiting**, so $L_{s} = λ / (μ - λ)$ , larger than $L_{q}$ .
⚠
**Keep $λ$ and $μ$ in the same time unit.** Mixing per-hour and per-minute corrupts every result.

PART 2

Q1BASIC1 Mark · MCQ

Trucks arrive at a crusher at

λ = 6

per hour and are served at

μ = 10

per hour. The server utilization is:

Q2MEDIUM2 Marks · NAT

For

λ = 6

/hr and

μ = 10

/hr (M/M/1), the average number of trucks in the system is ______. (Round off to two decimal places.)

Q3HARD2 Marks · NAT

For

λ = 6

/hr and

μ = 10

/hr, the average waiting time in the queue is ______ hours. (Round off to two decimal places.)