{"id":238045,"date":"2018-02-20T23:50:30","date_gmt":"2018-02-20T23:50:30","guid":{"rendered":"https:\/\/writemyessayfree.com\/i-semester-2015-16-department-of-mathematics"},"modified":"2018-10-24T09:13:29","modified_gmt":"2018-10-24T09:13:29","slug":"i-semester-2015-16-department-of-mathematics","status":"publish","type":"post","link":"https:\/\/www.benedictsol.com\/blogs\/i-semester-2015-16-department-of-mathematics\/","title":{"rendered":"I Semester 2015-16 Department of Mathematics"},"content":{"rendered":"<p>I Semester 2015-16<br \/>\nDepartment of Mathematics<br \/>\nMTL 106 (Introduction to Probability and Stochastic Processes)<br \/>\nTutorial Sheet No. 8<br \/>\n1. Consider a M\/M\/1 queuing model. Find the waiting time distribution for any customer in this model.<br \/>\nDeduce the mean waiting time from the above distribution.<br \/>\n2. Patients visit a doctor in accordance with a Poisson process at the rate of 8 per hour, and the time doctor<br \/>\ntakes to examine any patient is exponential with mean 6 minutes. All arriving patients attended by the<br \/>\ndoctor.<br \/>\n(a) Find the probability that the patient has to wait on arrival.<br \/>\n(b) Find the expected total time spent (including the service time) by any patient who visits the doctor.<br \/>\n3. Consider a multiplexer that collects traffic formed by messages arriving according to exponentially distributed<br \/>\ninterarrival times. The multiplexer is formed by a buffer and a transmission line. Assume that, the transmission<br \/>\ntime of a message is exponentially distributed with the mean value 10 ms. From measurements on<br \/>\nthe state of the buffer, we know that the idle buffer probability is 0.8.<br \/>\n(a) The underlying queueing model is &#8230;&#8230;&#8230;.<br \/>\n(b) The mean delay (waiting time) for the message is &#8230;&#8230;..<br \/>\n4. Consider that two identical M\/M\/1 queuing systems with the same rates  ,   are in operation side by side<br \/>\n(with separate queues) in a premises. Find the distribution of the total number, N, in the two systems taken<br \/>\ntogether in long-run.<br \/>\n5. Prove that mean time spent in an M=M=1 system having arrival rate   and service rate 2  is less than the<br \/>\nmean time spent in an M=M=2 system with arrival rate   and each service rate  .<br \/>\n6. Ms. H. R. Cutt runs a one-person, unisex hair salon. She does not make appointments, but runs the salon<br \/>\non a first-come, first-served basis. She finds that she is extremely busy on Saturday mornings, so she is<br \/>\nconsidering hiring a part-time assistant and even possibly moving to a larger building. Having obtained a<br \/>\nmaster\u2019s degree in operations research (OR) prior to embarking upon her career, she elects to analyze the<br \/>\nsituation carefully before making a decision. She thus keeps careful records for a succession of Saturday<br \/>\nmornings and finds that customers seem to arrive according to a Poisson process with a mean arrival rate of<br \/>\n5\/hr. Because of her excellent reputation, customers were always willing to wait. The data further showed<br \/>\nthat customer processing time (aggregated female and male) was exponentially distributed with an average<br \/>\nof 10 min. Cutt interested to calculate the following measures:<br \/>\n(a) What is the average number of customers in the shop?<br \/>\n(b) What is the average number of customers waiting for a haircut?<br \/>\n(c) What is the percentage of time an arrival can walk right in without having to wait at all?<br \/>\n(d) If waiting room has only four seats at present, what is the probability that a customer, will not able<br \/>\nto find a seat and have to stand.<br \/>\n7. City Hospital\u2019s eye clinic offers free vision tests everyWednesday evening. There are three ophthalmologists on<br \/>\nduty. A test takes, on the average, 20 min, and the actual time is found to be approximately exponentially<br \/>\ndistributed around this average. Clients arrive according to a Poisson process with a mean of 6\/hr, and<br \/>\npatients are taken on a first-come, first-served basis. The hospital planners are interested in knowing:<br \/>\n(a) What is the average number of people waiting?<br \/>\n(b) What is the average amount of time a patient spends at the clinic?<br \/>\n8. Consider New Delhi International Airport. Suppose that, it has three runway. Airplanes have been found to<br \/>\narrive at the rate of 20 per hour. It is estimated that each landing takes 3 minutes. Assume that a Poisson<br \/>\nprocess for arrivals and an exponential distribution for landing times. Without loss of generality, assume<br \/>\nthat the system is modeled as a birth and death process. What is the steady state probability that the no<br \/>\nwaiting time to land? What is the expected number of airplanes waiting to land? Find the expected waiting<br \/>\ntime to land?<br \/>\n1<br \/>\nI Semester 2015-16<br \/>\n9. Consider a telephone switching system consisting of n trunks with an infinite caller population. We assume<br \/>\nthat an arrival call is lost if all trunks are busy.<br \/>\n(a) Find the expression for  n the steady state probability that n trunks are busy.<br \/>\n(b) This above system must design the number of trunks, n, in order to guarantee a blocking probability<br \/>\nlower than or equal to 2%. The following data are available: The arrival stream is Poisson with rate 6<br \/>\nper minute. Each call has a duration modeled by an exponentially distributed variable with mean value<br \/>\nof 3 minutes. The value of n is &#8230;&#8230;.<br \/>\n10. A service center consists of two servers, each working at an exponential rate of two services per hour. If<br \/>\ncustomers arrive at a Poisson rate of three per hour, then, assuming a system capacity of at most three<br \/>\ncustomers.<br \/>\n(a) What fraction of potential customers enter the system?<br \/>\n(b) What would the value of part (a) be if there was only a single server, and his rate was twice as fast(that<br \/>\nis   = 4) ?<br \/>\n11. Consider an automobile emission inspection station with three inspection stalls, each with room for only one<br \/>\ncar. It is reasonable to assume that cars wait in such a way that when a stall becomes vacant, the car at the<br \/>\nhead of the line pulls up to it. The station can accommodate at most four cars waiting (seven in the station)<br \/>\nat one time. The arrival pattern is Poisson with a mean of one car every minute during the peak periods.<br \/>\nThe service time is exponential with mean 6 min. The chief inspector wishes to know the average number<br \/>\nin the system during peak periods, the average time spent (including service), and the expected number per<br \/>\nhour that cannot enter the station because of full capacity.<br \/>\n12. In a parking lot with N spaces the incoming traffic is according to Poisson process with rate  , but only as<br \/>\nlong as empty spaces are available. The occupancy times have an exponential distribution with mean 1= .<br \/>\nLet X(t) be the number of occupied parking spaces at time t.<br \/>\n(a) Determine rate matrix ? and the forward Kolmogorov equations for the Markov process fX(t); t   0g.<br \/>\n(b) Determine the limiting equilibrium probability distribution of the process.<br \/>\n13. A toll bridge with 10 booths at the entrance can be modeled as a 10 server Markovian queueing system with<br \/>\ninfinite capacity. Assume that the vehicle arrival follows a Poisson process with parameter 8 per minute and<br \/>\nthe service times are independent exponential distributed random variables with mean 1 minute.<br \/>\n(a) Draw the state transition diagram for a birth and death process for the system.<br \/>\n(b) Find the limiting state probabilities.<br \/>\n(c) If 2 more booths are installed, i.e, total 12 booths, what is the maximum arrival rate such that the<br \/>\nlimiting state probabilities exist?<br \/>\n14. Consider the central library of IIT Delhi where there are 4 terminals. These terminals can be used to obtain<br \/>\ninformation about the available literature in the library. If all terminals are occupied when someone wants<br \/>\ninformation, then that person will not wait but leave immediately (to look for the required information<br \/>\nsomewhere else). A user session on a terminal takes exponential distributed time with average 2.5 minutes.<br \/>\nSince the number of potential users is large, it is reasonable to assume that users arrive according to a Poisson<br \/>\nprocess. On average 25 users arrive per hour.<br \/>\n(a) Determine the steady state probability that i terminals are occupied, i = 0; 1; 2; 3; 4.<br \/>\n(b) What is the mean time spent in the system by any person?<br \/>\n(c) How many terminals are required such that at most 5% of the arriving users find all terminals occupied?<br \/>\n2<\/p>\n","protected":false},"excerpt":{"rendered":"<p>I Semester 2015-16 Department of Mathematics MTL 106 (Introduction to Probability and Stochastic Processes) Tutorial Sheet No. 8 1. Consider a M\/M\/1 queuing model. Find the waiting time distribution for any customer in this model. Deduce the mean waiting time <a href=\"https:\/\/www.benedictsol.com\/blogs\/i-semester-2015-16-department-of-mathematics\/\" class=\"read-more\">Read More &#8230;<\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[],"tags":[],"class_list":["post-238045","post","type-post","status-publish","format-standard","hentry"],"_links":{"self":[{"href":"https:\/\/www.benedictsol.com\/blogs\/wp-json\/wp\/v2\/posts\/238045","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.benedictsol.com\/blogs\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.benedictsol.com\/blogs\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.benedictsol.com\/blogs\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.benedictsol.com\/blogs\/wp-json\/wp\/v2\/comments?post=238045"}],"version-history":[{"count":0,"href":"https:\/\/www.benedictsol.com\/blogs\/wp-json\/wp\/v2\/posts\/238045\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.benedictsol.com\/blogs\/wp-json\/wp\/v2\/media?parent=238045"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.benedictsol.com\/blogs\/wp-json\/wp\/v2\/categories?post=238045"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.benedictsol.com\/blogs\/wp-json\/wp\/v2\/tags?post=238045"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}