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 40 an hour between 7 P.M. and 9 P.M. Currently, two inspectors are on duty during those hours, and each of whom can inspect 25 trucks an hour.
How many trucks would you expect to see at the weighing station, including those being inspected?
If a truck was just arriving at the station, about how many times could the driver be at the station?
What is the probability that both inspectors would be busy at the same time?
How many minutes, on average, would a truck that is not immediately inspected have to wait?
To determine the answers to these questions, we need to apply queuing theory. Queuing theory is a mathematical study of waiting lines or queues. It provides us with tools and formulas to analyze different aspects of a queuing system.
1. How many trucks would you expect to see at the weighing station, including those being inspected?
We can use the M/M/2 queuing model to calculate the average number of trucks in the system. In this model, "M" refers to the exponential arrival rate, and "2" represents the number of service channels (inspectors).
Given:
- Arrival rate (λ) = 40 trucks per hour
- Service rate (μ) = 25 trucks per inspector per hour
First, we need to calculate the effective arrival rate (λ') because it is affected by the number of inspectors:
λ' = λ / (number of inspectors) = 40 / 2 = 20 trucks per hour
Next, we can use the formula for the average number of trucks in the system (L) in an M/M/2 queuing model:
L = (λ' / (μ - λ')) + λ'
Substituting the values, we get:
L = (20 / (25 - 20)) + 20
L = 4 + 20
L = 24 trucks
Therefore, we would expect to see an average of 24 trucks at the weighing station, including those being inspected.
2. If a truck was just arriving at the station, about how many times could the driver be at the station?
To determine this, we need to find the average number of trucks in the system when a truck arrives.
The average number of trucks in the system (L) when a truck arrives is given by:
Ls = L + λ / μ
Substituting the values we calculated earlier:
Ls = 24 + 40 / 25
Ls = 24 + 1.6
Ls = 25.6
Therefore, if a truck has just arrived at the station, on average, the driver could expect to find 25.6 trucks present at the weighing station.
3. What is the probability that both inspectors would be busy at the same time?
The probability that both inspectors would be busy at the same time can be found using the formula for the probability of all servers being busy in an M/M/n queuing model, where "n" is the number of servers (inspectors).
The probability of all servers being busy (P0) in an M/M/n queuing model is given by:
P0 = 1 / (1 + ∑(k=1 to n-1) ((nρ)^k / k! ) + ((nρ)^n / n! ) * (1 / (1 - ρ/n) )
In this case, since we have 2 inspectors (n=2), we can calculate the probability of both inspectors being busy (Pbothbusy) when a truck arrives.
First, we need to calculate the traffic intensity (ρ):
ρ = λ' / μ = 20 / 25 = 0.8
Using the above formula, we get:
Pbothbusy = (1 / (1 + (2 * 0.8) + ((2 * 0.8)^2 / 2 ) * (1 / (1 - 0.8/2) )
Pbothbusy = (1 / (1 + 1.6 + 0.64) * (1 / (1 - 0.4) )
Pbothbusy = (1 / (2.24) * (1 / 0.6 )
Pbothbusy ≈ 0.595
Therefore, the probability that both inspectors would be busy at the same time is approximately 0.595 or 59.5%.
4. How many minutes, on average, would a truck that is not immediately inspected have to wait?
To calculate the average waiting time (W) for a truck, we can use Little’s Law, which states that the average number of customers in a system (L) is equal to the average arrival rate (λ) multiplied by the average waiting time (W).
Thus, we can rearrange the formula to find the average waiting time:
W = L / λ = (λ' / (μ - λ')) / λ
Substituting the values we calculated earlier:
W = (20 / (25 - 20)) / 40
W = (20 / 5) / 40
W = 4 / 40
W = 0.1 hour
Since there are 60 minutes in an hour, the average waiting time for a truck is:
0.1 * 60 = 6 minutes
Therefore, a truck that is not immediately inspected would have to wait, on average, for approximately 6 minutes.