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From the provided information, the assignment involves analyzing two queuing system scenarios based on Poisson arrivals and exponential service times, determining key performance metrics such as utilization rate, average waiting times, queue lengths, and probabilities of system states. The detailed calculations and formulas for both cases are outlined below, with clear step-by-step solutions for each part of the problem.

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Analysis of Service System at Rockwell Electronics Corporation

Rockwell Electronics Corporation manages a service crew responsible for repairing machine breakdowns, which occur at an average rate of 3 per day, following a Poisson distribution. The repair crew operates at an average service rate of 8 machines per day, modeled as exponential distribution. Using queuing theory, particularly the M/M/1 model, we analyze various performance metrics. These include the utilization rate, average downtime per machine, average queue length, and system probabilities.

Part a: Utilization Rate

The utilization rate (U) is calculated as the ratio of the arrival rate (λ) to the service rate (μ):

U = λ / μ

Given:

  • λ = 3 machines/day
  • μ = 8 machines/day

U = 3 / 8 = 0.375 or 37.5%

This indicates that the system is busy 37.5% of the time, with ample capacity to handle peak loads without congestion.

Part b: Average Downtime for a Broken Machine

The average downtime (W) includes the waiting time plus the repair time. For an M/M/1 queue, the average waiting time in the system is:

W = 1 / (μ - λ)

Calculating:

W = 1 / (8 - 3) = 1 / 5 = 0.2 days

To convert days into hours, assuming an 8-hour workday:

0.2 days × 8 hours/day = 1.6 hours

Therefore, on average, a broken machine experiences about 1.6 hours of downtime before repair completion.

Part c: Average Number of Machines Waiting to be Serviced (Queue Length)

The average number of machines in the queue (Lq) is given by:

Lq = (λ^2) / (μ × (μ - λ))

Calculations:

  • λ^2 = 3^2 = 9
  • μ(μ - λ) = 8 × (8 - 3) = 8 × 5 = 40

Lq = 9 / 40 = 0.225 machines

This suggests that, on average, less than one machine is waiting for service at any given moment, indicating a relatively unbusy queue.

Part d: Probability that More Than One Machine is in System

The probability that more than k machines are in the system follows the geometric distribution:

P(n > k) = (λ / μ)^{k+1}

Given λ/μ = 3/8 = 0.375

  • More than 1 machine (k=1): P >1 = (0.375)^2 = 0.1406 (~14.06%)
  • More than 2 machines: P >2 = (0.375)^3 ≈ 0.0527 (~5.27%)
  • More than 3 machines: P >3 ≈ 0.0198 (~1.98%)
  • More than 4 machines: P >4 ≈ 0.0074 (~0.74%)

These probabilities indicate decreasing likelihoods of higher system congestion, consistent with the low utilization rate.

Analysis of Car Wash Queue System Using M/M/1 Model

Harry's car wash estimates an arrival rate of 10 cars per hour, with a service rate of one car every 5 minutes, or 12 cars per hour. The system has a single server, and arrivals are Poisson, with service times exponentially distributed.

Part A: Average Number of Cars in Queue (Lq)

The formula for Lq in an M/M/1 system:

Lq = (λ^2) / (μ × (μ - λ))

Calculations:

  • λ = 10 cars/hour
  • μ = 12 cars/hour
  • λ^2 = 100
  • μ(μ - λ) = 12 × (12 - 10) = 12 × 2 = 24

Lq = 100 / 24 ≈ 4.1667 cars

This indicates that, on average, about four cars wait in line before being washed.

Part B: Average Waiting Time in Queue (Wq)

Wq = Lq / λ = 4.1667 / 10 = 0.4167 hours, or about 25 minutes.

This is the mean time a car spends waiting before the washing process begins.

Part C: Average Time a Car Spends in the System (W)

W = 1 / (μ - λ) = 1 / (12 - 10) = 1 / 2 = 0.5 hours, or 30 minutes.

This includes both waiting and service times, confirming that each vehicle spends about half an hour in the system on average.

Part D: Utilization Rate of the Car Wash

ρ = λ / μ = 10 / 12 ≈ 0.8333 or 83.33%

The system operates with high efficiency but still maintains some capacity to handle fluctuations.

Part E: Probability No Cars Are in the System (P0)

For an M/M/1 queue, P0 = 1 - ρ:

P0 = 1 - 0.8333 = 0.1667 or 16.67%

This means that approximately 1 in 6 times, the wash system is idle with no cars waiting or being serviced.

Conclusion

Both queuing systems analyzed exhibit typical characteristics derived from their load intensities. The Rockwell Electronics system's low utilization indicates prompt repairs with minimal backlog, while Harry's car wash, with higher utilization, experiences moderate queue lengths and waiting times but operates efficiently overall. These insights assist in resource planning and process optimization to minimize customer or machine downtime and improve service quality.

References

  • Gross, D., Shortle, J. F., Thompson, J. M., & Harris, C. M. (2008). Fundamentals of Queueing Theory. Wiley.
  • Green, L. (2008). Introduction to Queueing Theory. Springer.
  • Kleinrock, L. (1975). Queueing Systems, Volume 1: Theory. Wiley-Interscience.
  • Buzacott, J. A., & Shanthikumar, J. G. (1993). Stochastic Models of Manufacturing and Service Operations. Prentice Hall.
  • Anderson, D., Sweeney, D., & Williams, T. (2007). An Introduction to Management Science: Quantitative Approaches to Decision Making. South-Western Cengage Learning.
  • Walrand, J. (1988). An Introduction to Queueing Networks. Prentice Hall.
  • Hopp, W. J., & Spearman, M. L. (2008). Factory Physics. McGraw-Hill.
  • Kolen, J. (1983). Stationary Queueing Models. Springer.
  • Stevenson, W. J. (2012). Operations Management. McGraw-Hill Education.
  • Lasdon, L. S. (2000). Optimization of Large Systems. John Wiley & Sons.

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