Consider the following queuing system with four servers. Suppose all four server

Question

Consider the following queuing system with four servers. Suppose all four servers are busy and three customers arrive (Customers 5, 6, and 7). The service time of all four servers follows an exponential distribution with a mean of 1 minute. Assume the inter arrival times of customers to the queuing system are exponentially distributed with rate of 2 customers per minute. What is the probability that the next customer arrives after 4 minutes? If no customer arrives during the next 10 minutes, what is the probability that the next arrival will not happen during the following 3 minutes? What is the probability that two customers arrive during the next 6 minutes? Suppose customer-1, customer-2, customer-3, and customer-4 started to be served 5, 4, 2, and 1 minutes ago, respectively. What is the probability that customer-1 will be the first of the four customers (i e customer-1, customer-2, customer-3, and customer-4) to complete the service? Why? What is the probability that Customer-5 will be the last of the five customers (customers 1, 2, 3, 4, and 5) to complete the service if first come first serve policy is used? Why

Explanation / Answer

d> The exponential distribution has loss of memory property. Hence for how many minutes they have been served or when they started, that won't count here. Rather we have to consider as if all 4 have started at the same time. Hence the probability that customer-1 will be the 1st of the 4 customers to complete the service= 3!/4! = 1/4

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