Question

Consider two different machines A and B that could be used at a station. Machine A has a mean effective process time te of 1.0 hours and an SCV c = 0.25. Machine B has a mean effective process time of 0.85 hour and an SCV of 4. (a) For an arrival rate of 0.92 job per hour with ca = 1, which machine will have a shorter average cycle time? (b) Now put two identical machines of type A (in parallel) at the station and double the arrival rate. What happens to cycle time? Do the same for machine B. Which type of machine produces shorter average cycle time? (c) With only one machine at a station, let the arrival rate be 0.95 job per hour with c = 1. Recompute the average time spent at the stations for both machine A and machine B.

Answer 1

Machine A is likely to have a shorter average cycle time due to its lower SCV, despite a higher effective process time. When doubling the arrival rate and using two machines in parallel, type A machines maintain a more stable cycle time compared to type B machines due to A's lower variability. Even with an increased arrival rate of 0.95 and c = 1, machine A will typically present shorter cycle times than machine B.

Step-by-step explanation:

To evaluate which machine, A or B, has a shorter average cycle time for an arrival rate of 0.92 jobs per hour with ca = 1, we can use the Pollaczek-Khintchine (P-K) formula to compute the average waiting time in the queue, Wq. The formula is:

Wq = (lambda * ca^2 * te^2) / (2 * (1 - lambda * te))

where lambda is the arrival rate, ca is the coefficient of variation of arrivals, te is the mean effective process time and SCV (the coefficient of variation of service times) is c.

For machine A, Wq will be less than machine B due to the lower SCV, despite its higher effective process time. This indicates a more consistent processing time with less variability, which often translates to a shorter cycle time in stable systems.

(a) Placing two machines of type A in parallel and doubling the arrival rate essentially halves the effective arrival rate to each machine (0.92 jobs per hour / 2 = 0.46 jobs per hour to each machine). The system behaves similar to a single machine with the original arrival rate, thus the cycle time would remain relatively unchanged. Replicating this with machine B will lead to a different outcome due to its high SCV, resulting in a longer cycle time compared to type A machines in a parallel setup.

(b) To recompute the average time spent at the stations for both machine A and machine B with an arrival rate of 0.95 jobs per hour and c = 1, we apply the P-K formula again with the updated arrival rate keeping other factors constant. Machine A will still exhibit a shorter cycle time as compared to Machine B.

Answer 2

The question involves comparing cycle times of two machines using mean effective process time and SCV, adjusting for arrival rates, and considering parallel machine configurations.

Step-by-step explanation:

The student is asking about evaluating process times and the statistical coefficient of variation (SCV) for two different machines to determine which machine has a shorter average cycle time. These calculations are performed using queueing theory, which is an essential concept in operations research and industrial engineering. The question includes mathematical calculations of mean effective process time, SCV, average cycle time, and adjusting for different arrival rates and parallel machines configurations.

For part (a) of the question, we need to compare cycle times for Machine A and Machine B given their mean effective process times (te) and SCV values. The cycle time is essentially the total time a job spends in the system, from arrival to departure. The mean effective process time (te) and SCV are both factors in the calculation of cycle time, with lower values generally indicating better performance in a queueing system.

Part (b) asks what happens to cycle time when the arrival rate is doubled and two identical machines (Machine A or Machine B) are set in parallel. This scenario simulates increasing system capacity to handle a higher workload. Lastly, part (c) requires recalculating the time spent at the station with a new arrival rate for both machines individually.