Question

# Trucks are required to pass through a weighing station so that they can be checked for...

Trucks are required to pass through a weighing station so that they can be checked for weight violations. Trucks arrive at the station at the rate of 35 an hour between 7:00 p.m. and 9:00 p.m. Currently two inspectors are on duty during those hours, each of whom can inspect 25 trucks an hour.

a. How many trucks would you expect to see at the weighing station, including those being inspected?

b.If a truck was just arriving at the station, how long could the driver expect to be at the station, including time waiting in line and being weighed?

c. What is the probability that both inspectors would be busy at the same time?

d. How long, on average, would a truck that is not immediately inspected have to wait?

We have the following information

Arrival rate (lambda) = 35 per hour

Service rate (mu) = 25 per hour

Number of servers (s) = 2

Utilization factor (rho) = lambda/s*mu = 35/50 = 0.7

Using the multiserver queueing model table, we know that at rho = 0.7 and s = 2, the probability of 0 customers in queue (P0) is 0.17647

a)

Number of customers in the system is given by the equation

L = {[lambda*mu*(lambda*mu)^s]/[(s-1)!*(s*mu-lambda)^2)]}*P0 + lambda/mu

Using this formula we get

L = 2.7451

b)

Total time in the system is calculated using the formula

W = L/lambda = 2.7451/35 = 0.078 hours or 4.7 minutes

c)

If there is 0 customers in the system then both of them will be free. The probability is P0 = 0.17647. If there is 1 customer in the system then one of them will be free. The probability of Pn where n<=s is calculated by

Pn = (1/n!)*(lambda/mu)^n*P0

Using this we get

P1 = 0.2471

Probability of 0 or 1 customer in the system is 0.2471+0.1764 = 0.4235. This means the probability that both the servers will be busy is 1-0.4235=0.5765 or 57.65%

d)

The waiting time in the queue is given by

Wq = W – 1/mu = 0.078 – 1/25 = 0.038 or 2.28 minutes

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