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A toll booth on a turnpike is open from 8:00 AM until midnight. Vehicles start arriving at 7:45 AM at a uniform rate of 6 veh/min. At 8:15 AM the rate drops to 2 veh/min. If vehicles are processed at a uniform rate of 6 veh/min, determine when the queue will end, the total delay, the maximum queue length, and the longest vehicle delay.

2 Answers

2 votes

Answer:

a) The queue will end at 8:15 + 22.5 mins = 8:37:30 AM

b) Total time delay = 52 mins 30 sec

c) maximum queue length = 90 vehicles

d) longest vehicle delay = 15 mins

Step-by-step explanation:

At 7:45 AM, uniform rate = 6 veh/min

At 8: 15 AM, uniform rate = 2 veh/min

c) The maximum queue length will be between 8:00 AM to 8:15 AM I.e. 15 mins

The number of vehicles = uniform rate * duration

Number of vehicles = 15 * 6 = 90 vehicles

Maximum queue length = 90 vehicles

a) From 8:15 and above:

Incoming rate = 2 veh/min

The vehicles are processed at a uniform rate of 6 veh/min,

i.e. outgoing rate = 6 veh/min

Dissipation rate = Outgoing rate - incoming rate = 6 - 2 = 4 veh/min

Therefore, 90 vehicles will dissipate at 90/4 = 22.5 min

The queue will end at 8:15 + 22.5 mins = 8:37:30 AM

b) The total delay will be between the time the queue begins and the time it dissipates

The time queue begins = 7:45 AM

The time queue dissipates = 8:37:30

Total delay = 8:37:30 AM - 7:45 AM

Total time delay = 52 mins 30 sec

d) Longest vehicle delay

The longest delay will be experienced by the 90th vehicle between 8:00 AM and 8:15 AM

The longest vehicle delay = 15 mins

User Timothy Clotworthy
by
8.3k points
3 votes

Answer:

The vehicles start arriving at 7:45 AM at a rate of 6 veh/min.

then, until the 8:00 AM that the vehicles start being processed, the amount of vehicles accumulated is 6veh/min*15min = 6*15 vehicles = 90 vehicles.

At 8:00 am, when the vehicles start to being processed, we have a queue of 90 vehicles, and at this point, each minute 6 vehicles arrive and also 6 vehicles are processed, so the queue does not change.

At 8:15 AM, the rate at which the vehicles arrive is equal to 2 veh/min.

then, from this hour we have that the number of vehicles in the queue can be described by the equation:

V (t) = 90 + 2*t - 6*t

where t is the number of minutes that passed since the 8:15 AM

this means: we have an initial number of 90 vehicles, and every minute 2 new arrive and 6 are processed.

The queue will end when we have that V(t) = 0

this is:

90 - 4t = 0

t = 90/4 = 22.5 mins.

This means that the queue will end at:

8:15 am + 22.5mins = 8:37.5 Am

You also ask for the longest delay, I guess this is how much a vehicle must wait until it is processed.

knowing that 6 vehicles are processed per minute, the car in the last place (car number 90) must wait 15 minutes, because of 6*15 = 90

so the longest delay is 15 mins.

and this is the same for the first car that arrives at 7:45 AM, this car also needs to wait for 15 minutes.

User Datasert
by
8.5k points
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