It is given that :
i). Average arrival rate = 8 vehicle/min
ii). Average departure rate = 10 vehicle/min
We have to assume : D/D/1 queuing
Now let us consider departure starts after 10 minutes of the vehicles arrival,
i.e. [tex]$\mu (t) = 10(t-10)$[/tex]
Therefore,
The time required to clear the queue is given as :
[tex]$\lambda (t) = \mu (t)$[/tex]
[tex]$8t = 10(t-10)$[/tex]
[tex]$ t = 50$[/tex] minutes
Therefore, time required to clear the queue is 50 minutes.
Now we plot a graph that shows the timing and the number of vehicles.
The total number of vehicles delay is :
[tex]$D_1 = A_1-A_2$[/tex]
So from the graph we get,
The area of triangle AOB = [tex]A_1[/tex]
Area of triangle ACB = [tex]A_2[/tex]
Therefore, we get :
[tex]$D_2 = \frac{1}{2} (50 \times 100) - \frac{1}{2} (40 \times 400)$[/tex]
The total number of the vehicles arriving :
[tex]$=8t$[/tex]
[tex]$= 8 \times 50$[/tex]
[tex]= 400[/tex] vehicles
Now the average delay per vehicle is :
[tex]$=\frac{2000}{400}$[/tex]
[tex]=5[/tex] minutes
So the average delay per vehicle,
[tex]$D_a = 5$[/tex] minute
The longest queue is :
[tex]$Q_{max} (t) = \lambda (t) - \mu(t)$[/tex]
[tex]$Q_{max} (10) = (8 \times) - 10(10-10)$[/tex]
[tex]$Q_{max} = 80$[/tex] vehicles
Now the arrival time of the [tex]100^{th}[/tex] vehicle = 12.5 min
The departure time of the [tex]100^{th}[/tex] vehicle = 20 min
Therefore, the wait time for the vehicle is = 20 min - 12.5 min
= 7.5 min
So, [tex]w_t = 7.5[/tex] minute
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