D/D/1 Queuing The case of deterministic arrivals and departures with one departure channel (D/D/1 queue) is an excellent starting point in understanding queuing models because of its simplicity. The D/D/1 queue lends itself to an intuitive graphical or mathematical solution that is best illustrated by an example.

EXAMPLE D/D/1 QUEUING WITH CONSTANT ARRIVAL AND DEPARTURE RATES

Vehicles arrive at an entrance to a recreational park. There is a single gate (at which all vehicles must stop), where a park attendant distributes a free brochure. The park opens at 8:00 A.M., at which time vehicles begin to arrive at a rate of 480 veh/h. After 20 minutes, the arrival flow rate declines to 120 veh/h, and it continues at that level for the remainder of the day. If the time required to distribute the brochure is 15 seconds, and assuming D/D/1 queuing, describe the operational characteristics of the queue.

SOLUTION

Begin by putting arrival and departure rates into common units of vehicles per minute:

480 veh/h 8 veh/min for 20 min

60 min/h 120 veh/h

2 veh/min for 20 min 60 min/h 60 s/min

4 veh/min for all 15 s/veh

t

t

t

λ

λ

μ

= = ≤

= = >

= =

Equations for the total number of vehicles that have arrived and departed up to a specified time, t, can now be written. Define t as the number of minutes after the start of the queuing process (in this case the number of minutes after 8:00 A.M.). The total number of vehicle arrivals at time t is equal to

≤8 for 20 mint t

and

( )160 2 20 for 20 m in t t + − >

Similarly, the number of vehicle departures is

4 for all t t

The preceding equations can be illustrated graphically as shown in Fig. 5.7. When the arrival curve is above the departure curve, a queue condition exists. The point at which the arrival curve meets the departure curve is the moment when the queue dissipates (no more queue exists). In this example, the point of queue dissipation can be determined graphically by inspection of Fig. 5.7, or analytically by equating appropriate arrival and departure equations, that is,

( )160 2 20 4 t t+ − =

Solving for t gives t = 60 minutes. Thus the queue that began to form at 8:00 A.M. will dissipate 60 minutes later (9:00 A.M.), at which time 240 vehicles will have arrived and departed (4 veh/min × 60 min).