Question

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?

Answer #1

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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