Concept explainers
a)
Interpretation: utilization of employees.
Concept introduction:The M/M/1 queue assumes poison arrivals, exponential service times, and a single server serving customers in a FCFS fashion. Poisson arrivals are a reasonably good assumption for unscheduled systems. Further if there is a mix of many different types of jobs the exponential distribution can be realistic for service times. Otherwise it tends to be too variable of a distribution.
b)
Interpretation:the average time a customer spends at the drive-through.
Concept introduction:The M/M/1 queue assumes poison arrivals, exponential service times, and a single server serving customers in a FCFS fashion. Poisson arrivals are a reasonably good assumption for unscheduled systems. Further if there is a mix of many different types of jobs the exponential distribution can be realistic for service times. Otherwise it tends to be too variable of a distribution.
c)
Interpretation:cars on average are in the drive-through lane (including those in service)
Concept introduction:The M/M/1 queue assumes poison arrivals, exponential service times, and a single server serving customers in a FCFS fashion. Poisson arrivals are a reasonably good assumption for unscheduled systems. Further if there is a mix of many different types of jobs the exponential distribution can be realistic for service times. Otherwise it tends to be too variable of a distribution.
d)
Interpretation:suggestion for the drive-through to improve customer satisfaction.
Concept introduction:The M/M/1 queue assumes poison arrivals, exponential service times, and a single server serving customers in a FCFS fashion. Poisson arrivals are a reasonably good assumption for unscheduled systems. Further if there is a mix of many different types of jobs the exponential distribution can be realistic for service times. Otherwise it tends to be too variable of a distribution.
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Check out a sample textbook solutionChapter 7 Solutions
Production and Operations Analysis, Seventh Edition
- In a queueing system, customers arrive once every 4 minutes (standard deviation = 7) and services take 3 minutes (standard deviation = 5.5). (Do not round intermediate calculations. Round your answer to three decimal places.) What is the average time a customer will spend in the queue (in minutes)? minutesarrow_forwardA server operates with 75 percent utilization. The average processing time istwo minutes and the standard deviation of processing time is one minute. The coefficient ofvariation of the arrival process is one. What is the average time in the queue for customers?arrow_forward
- Practical Management ScienceOperations ManagementISBN:9781337406659Author:WINSTON, Wayne L.Publisher:Cengage,