question #40

After an evening class, a student stops at Steak and Shake. There is a single cashier who takes, on average, 60 seconds to serve a customer. On average, a customer arrives every 80 seconds. Arrivals are random, with CVa=1. Service has Markovian behavior, with CVp=1. This results in a variability factor of 1.

Using the graph below, how long does the student expect to wait before being able to place their order?

quiz40

Question

question #40

After an evening class, a student stops at Steak and Shake. There is a single cashier who takes, on average, 60 seconds to serve a customer. On average, a customer arrives every 80 seconds. Arrivals are random, with CVa=1. Service has Markovian behavior, with CVp=1.  This results in a variability factor of 1.

Using the graph below, how long does the student expect to wait before being able to place their order?

quiz40

BlackTom AI · Accepted Answer

To tackle this question, I’ll break down the scenario and compare each choice before pointing to the given answer.
Option 1: 'Since utilization is less than 100%, the customer does not expect to wait.' While ρ = 0.75 (less than 1) means the system is stable and a queue can form, it does not imply there is no waiting. Even with ρ < 1, an arriving customer can experience a waiting time in the queue depending on the traffic intensity and variability. This option incorrectly assumes zero wait simply because the server isn’t fully utilized.
Option 2: '2 time units' This corresponds to the scenario described in the problem where the arrivals and service processes have CV values of 1 (low variability factors) and thus the system behaves like an M/M/1 queue. With λ = 1/80 per second and μ = 1/60 per second, the utilization is ρ = λ/μ = 0.75. In an M/M/1 system (or a GI/G/1 with CVs=1, i.e., Markovian assumptions), the expected waiting time in queue Wq is a function of ρ and the service/arrival rates. Given the graph provided (which maps Wq against utilization for CVa = CVp = 1), the point corresponding to ρ = 0.75 falls near a Wq of about 2 time units, making this the most consistent choice with the graph and the stated variability. This option aligns with the expected value shown by the curve for unit variability.
Option 3: '1 time unit' While reducing variability and a moderate ρ might yield a smaller waiting time, the calculated and graphical reference for ρ = 0.75 in a 1-1 variability setting does not typically place Wq as low as 1 time unit; the queue still contributes a nonzero wait.
Option 4: '3 or more time units' This would correspond to a higher Wq than the graph indicates for ρ = 0.75 in a CVa = CVp = 1 setting. Since the given graph and the system parameters do not place the wait in the '3 or more' region for these conditions, this option overestimates the waiting time.
In summary, the correct interpretation given the problem’s CV values and the plotted relationship is that the student’s expected wait in the queue is about 2 time units, which matches option 2. The other options rely on either overestimating or underestimating wait time by assuming no wait, too little wait, or too much wait for this utilization and variability profile.

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