At a production machine, parts arrive according to a Poisson process at the rate of 0.35 parts per minute. Processing time for parts has an exponential distribution with a mean of 2 minutes. What is the probability that a random part arrival finds that there are already 8 parts in the system (in machine + in the queue)

Question

Prepp Team · Accepted Answer

Understanding the M/M/1 Queueing System

This problem describes a common scenario in operations research known as an M/M/1 queue. The 'M/M/1' stands for:

M: Markovian (or Poisson) arrival process.
M: Markovian (or Exponential) service time distribution.
1: Single server.

We are given the arrival rate and the mean service time, and we need to find the probability of a specific number of parts being in the system.

Identifying Parameters

From the question, we can extract the following information:

Arrival rate ($\lambda$): Parts arrive according to a Poisson process at a rate of 0.35 parts per minute. So, $\lambda = 0.35$ parts/minute.
Mean service time: The average time taken to process a part is 2 minutes.
Service rate ($\mu$): The service rate is the reciprocal of the mean service time. If it takes 2 minutes on average to serve one part, then the machine can serve $1/2$ parts per minute. So, $\mu = \frac{1}{2} = 0.5$ parts/minute.
Number of parts in the system ($n$): We are interested in the probability of having 8 parts in the system (queue + machine). So, $n=8$.

Calculating Traffic Intensity ($\rho$)

The traffic intensity ($\rho$) is a key parameter in queueing theory, representing the utilization of the server. It is calculated as the ratio of the arrival rate to the service rate:

$\rho = \frac{\lambda}{\mu}$

Substituting the values:

$\rho = \frac{0.35}{0.5}$

$\rho = 0.7$

Since $\rho < 1$, the system is stable, and steady-state probabilities exist.

Finding the Probability of n Parts in the System

For an M/M/1 queue in a steady state, the probability ($P_n$) that there are exactly $n$ parts in the system (in the queue plus in service) is given by the formula:

$P_n = \rho^n (1 - \rho)$

We want to find the probability that there are 8 parts in the system, so we set $n=8$.

$P_8 = \rho^8 (1 - \rho)$

Substitute the value of $\rho = 0.7$:

$P_8 = (0.7)^8 (1 - 0.7)$

$P_8 = (0.7)^8 (0.3)$

Performing the Calculation

Now, we calculate $(0.7)^8$:

$(0.7)^8 \approx 0.05764801$

Now, multiply by 0.3:

$P_8 \approx 0.05764801 \times 0.3$

$P_8 \approx 0.017294403$

Conclusion

The probability that a random part arrival finds that there are already 8 parts in the system is approximately 0.017294403.

Comparing this value to the given options:

0.0247
0.0576
0.0173
0.082

The calculated probability of 0.017294403 is closest to 0.0173.

Prepp Team · Answer

0.0247

Prepp Team · Answer

0.0576

Prepp Team · Answer

0.082