D/D/1 QUEUING WITH TIME-INVARIANT ARRIVALS AND DEPARTURES Vehicles begin arriving at a closed entrance to a parking lot at a rate of six vehicles per minute for the first 15 minutes and then continue to arrive at two vehicles per minute for times after this initial 15-minute period. The parking-lot entrance opens when the vehicle queue reaches 30 vehicles and the parking-lot attendant services one vehicle every 15 seconds. Assuming D/D/1 queuing, what is the total delay?

SOLUTION Note: Open boxes in equations “ ” are to be completed by the reader

To begin, a few key points for the points for the D/D/1 queuing diagram need to be determined. First, the number of minutes into the process when parking-lot entrance opens can be calculated as,

number of queued vehicles when open 5 min.

vehicle arrival rate opent = = =

Next, at 15 min into the process the arrival rate changes from 6 veh/min to 2 veh/min. So after the first 15 min total vehicle arrivals are,

arrivals15 = 15 × = 90 veh.

188 Chapter 5 Fundamentals of Traffic Flow and Queuing Theory

It is known that the parking-lot attendant services vehicles at a rate of one every 15 sec or, equivalently, 4 veh/min starting 15 min into the process (when parking service begins). So the number of departures after 15 min will be,

departures15 = 10 × = veh, which means that the queue at 15 min into the process is,

queue15 = arrivals15 departures15 = = veh. So the equation for the time after 15 min to queue clearance (t>15) can be determined from,

queue15 + (arrival rate after 15 min)(t>15) = (departure rate after 15 min)(t>15) or,

( ) ( ) 15 15 queue

departure rate after 15 min arrival rate after 15 min

25 min.

t> = −= = −

So the queue will clear 15 min + 25 min or 40 min after the start of the process. For the total delay, we need the area between arrival and departure curves. At the 15th minute, 90 veh have arrived so the area under the arrival curve is a triangle with a height of 90 veh and a base of 15 min. Also at the 15th minute, 40 veh will have departed so the area under the departure curve is a triangle with a height of 40 veh and but the base is now 10 min (15 5) since departures do not start until 5 min after the first vehicle arrival. The subtraction of these two triangles will give the area between arrival and departure curves up to 15 min.

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