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solution_hw4

# solution_hw4 - Assignment 4 172-3 Answer a As the basic...

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Unformatted text preview: Assignment 4 172-3. Answer: a) As the basic queueing process said, a queueing system services customers who are generated over time by an input source. These customers enter the system and join queues. The way customers are selected from the queues is known as the queue discipline. The required service is performed for the customers by the service mechanism. Here, we have the customers are the cars, the serves are the lots. And the service being provided when a car is been parked in a lot. The service time is constituted by the arrival time, service time and the leave time. The queue capacity is 0 in this problem. b) L=2nnPn=0X0.2+1x0.3+2X0.3+3x0.2 =1.5 Lq: Znasﬁl _ S)Pl’1 : 0 c) A car spends an average of 45 minutes in a parking space. 17.2-4. a) False. As the definition, the queue is where customers wait before being served. The key words are “before being served”, but not “service is complete”. b) False. There have no Simitation on the capacity'of‘a queue. S'o tlrata queueearrbe called inﬁnite or ﬁnite depends on the maximal numbers of customers. 0) True. The most common is ﬁrst-come-ﬁrst-served (FCFS). 17.3-1. (a) The checkout stand in a grocery store. Customer: buyers waiting checkout Server: checkers (b) A ﬁre station. Customer: building or anything that is in ﬁre Server: ﬁreman and fire engine (c) The toll booth for a bridge. Customer: cars that are passing the bridge Server: counter and toll collector (d) A bicycle repair shop. Customer: broken bicycle Server: maintainer (e) A shipping dock. Customer: merchandise that is needed to be shipped Server: porter (f) A group of semiautomatic machines assigned to one operator. Customer: semiautomatic machines Server: operator (g) The material-handling equipment in a factory area. Customer: materials Server: material-handling eauinment (h) A plumbing shop. Customer: people who need plumbing products Server: the plumbing shop (i) A job shop producing custom orders. Customer: people who want to ﬁnd a job Server: the job shop (j) A secretarial typing pool. Customer: secret information Server: typing pool 17.4-1. Answer: From the question we have 1 = g for n > 0, pi = gand an = 1 for at E 2 a) P{next arrival before 1: 00} = 1 — e‘l/2 = 0.393 1 1 P{next arrival between 1:00 and 2: 00} = 1 -— {9(2) —- (1 — e_('2')) = 0.239 1 P{next arrival after 2: 00} = {(99) = 0.368 13) P{next arrival between 1: 00 and 2: 00|no arrivals between 12: 00 and 1: 00} = 1 1 — 5(2) 2 0.393 (mue-lt c) P{no arrival between 1:00 and 2: 00} = or = Fri/2 = 0.607 At 1e‘“ 1 P{next arrival between 1: 00 and 2:00} = —(——%—— = 59—1/2 = 0.303 ' 1 1 1 P{two or more arrivals between 1: 00 and 2: 00} = 1 —— e—‘Z’ — Ee7 = 0.0902 d) P{none served by 2:00} 2 e'1 = 0.368 P{none served by 1: 10} = e‘lufﬁ) = 0.846 P{none served by 1: 01} = e‘lwﬁﬁ) = 0.983 17.4—6 Answer: a) From aggregation property of Poisson process, the arrival process is still Poisson with mean rate 10 per hour. So, distribution of time between consecutive arrivals is exponential with mean of 6 minutes. b) The probability distribution is the minimum of two exponential random variables. By property 3, it is exponential with mean of 5 minutes. 17.6-2. Answer: A 1 A: 1041: 5,130 = (1—?) =§ That is the proportion of time no one is waiting. ...
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solution_hw4 - Assignment 4 172-3 Answer a As the basic...

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