The time between arrivals of parts in a single machine queuing system is uniformly distributed from...

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The time between arrivals of parts in a single machine queuingsystem is uniformly distributed from 1 to 20 minutes (forsimplicity round off all times to the nearest whole minute.) Thepart's processing time is either 8 minutes or 14 minutes. Considerthe following case of probability mass function for service times:Prob. of processing (8 min.) = .5, Prob. of processing (14 min.) =.5 Simulate the case, you need to estimate average waiting time insystem. Start the system out empty and generate the first arrivaltime, etc. You need to submit random numbers used (Use the thirdblock of the Random number table to generate interarrival time andthe fourth block to generate service time), interarrival time andservice time for each part, discuss how you compute (10%) and (youneed to generate 20 parts):
(1) clock arrival time for each part (10%),
(2) clock time starts to process each part (10%),
(3) clock departure time for each part (10%),
(4) total waiting time in system for each part (20%),
(5) plot average total waiting time in system versus part number(20%),
(6) plot average idle time of the machine versus part number(compute at the end of each processing) (20%)

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The time between arrivals of parts in a single machine queuingsystem is uniformly distributed from 1 to 20 minutes Thereforethe probabilities of a part arriving after 1 or 2 or 3 or so ontill 20 minutes are equalInterarrival timeInterarrival time minutesProbabilityCumulative ProbabilityRandom Nos See Answer

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