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A group of $m$ customers frequents a single-server station in the following manner. When a customer arrives, he or she either enters service if the server is free o...

Question

A group of $m$ customers frequents a single-server station in the following manner. When a customer arrives, he or she either enters service if the server is free or joins the queue otherwise. Upon completing service the customer departs the system, but then returns after an exponential time with rate $heta .$ All service times are exponentially distributed with rate $mu$.(a) Define states and set up the balance equations.In terms of the solution of the balance equations, find(b) the average rat

A group of $m$ customers frequents a single-server station in the following manner. When a customer arrives, he or she either enters service if the server is free or joins the queue otherwise. Upon completing service the customer departs the system, but then returns after an exponential time with rate $ heta .$ All service times are exponentially distributed with rate $mu$. (a) Define states and set up the balance equations. In terms of the solution of the balance equations, find (b) the average rate at which customers enter the station. (c) the average time that a customer spends in the station per visit.



Answers

In the design of a $\textbf{rapid transit system}$, it is necessary to balance the average speed of a train against the distance between station stops. The more stops there are, the slower the train's average speed. To get an idea of this problem, calculate the time it takes a train to make a 15.0-km trip in two situations: ($a$) the stations at which the trains must stop are 3.0 km apart (a total of 6 stations, including those at the ends); and ($b$) the stations are 5.0 km apart (4 stations total). Assume that at each station the train accelerates at a rate of 1.1 m/s$^2$ until it reaches 95 km/h then stays at this speed until its brakes are applied for arrival at the next station, at which time it decelerates at $-$2.0 m/s$^2$. Assume it stops at each intermediate station for 22 s.

Affects you too. Get your first term. The court to toe times. Second term I'm just due to a second term. Would you go to two ones, too? Times there's term, All one minus X clear and super fight. It gets to one minus thanks, Square. Thanks to purse. Thanks. So here to round going warn Mexico too. 11 hires 2.1 over two tonnes. Point night screen. They couldn't find 1296 You're too 1 60 co two. 26 times 2.6. Well, we're to pass the times 0.4 square is he cooked it full point 875

We're told the rate at which customers arrive at the counter to be served can be modeled by the function effort. Easy Digital Plus six co sign T over pi or tea between zero and 60 and T is in minutes through the nearest hole number. We want to determine how many customers arrive over a 60 minute period. All right, so to find customers, what we're going to do is integrate. So over our time period of our rate function, which in this case is F A T. So we'll go ahead and plug that ends. We have 12 plus six co sign t over pot. DT integrating 12. We should get 12 t integrating co sign. Well, that should give us sign of tea over pie. But then we need to divide by what we have on the inside where it's Constance would be won over one pie, which would be the same thing. It's multiplying by pi, and then we evaluate this from 0 to 60. Almost forgot my team right here. All right, so plug it in 60 would get 12 times 60 which is 7 20 and then we'd have six high sign of 60 over pie and then minus Well, we plug in zero. 12 t zero and then sign of zero is also zero. So we subtract off zero so that we just need to estimate what this is. So six i times sign of 60 over pi would be about 4.6 at that seven. 20 so this would be approximately 724 customers, which would be closest to answer choice me.

In this question. The scenario is that it is lunchtime at a given restaurant and customers are arriving at the drive thru in a random manner, and we were told that they follow a Poisson process with a rate of 0.8 customers per minute, so we can say the rate is 0.8 customers per minute. Now, when something follows a Poisson process, there's two things that we can say about it. The first is that the number of arrivals that occur in non overlapping time intervals are independent, and the second is that the number of arrivals in a given time interval follows a Poisson distribution with mean lambda T. So if we're given a certain time interval, the mean of the Poisson distribution is equal to Lamberti now. For part A were asked, what is the expected number of customers in one hour and what is the corresponding standard deviation? So if we define X as the number of customers that arrive in one hour, we know that X follows a possible distribution with mean equal to Lambda Times T. And so they mean for possible distribution is simply Lambda Times T, which is 0.8 times 60 minutes in one hour because this rate is 0.8 per minute and so that we expect that there will be 48 arrivals on average in one hour now. Another property of the possible distribution is that the variance on the number of arrivals is equal to the expected number of arrivals, which tells us that's the standard deviation on the number of arrivals is equal to the square root of 48. How and this is equal to 6.93 now for Part B. Were given information that the drive thru workers cannot handle more than 10 customers in any five minute span and were asked to find the probability that too many customers arrive for the workers to handle between the time of 12:15 p.m. And 12:20 p.m. So we're talking about a time of five minutes because it's a poison process. It doesn't matter if we're talking about the time interval from 12 05 to 12 10 or 12, 10 to 12 15 or 12 15 to 12 20. All that matters is theory length of time that passes, so if we want to find the probability that the drive thru cannot handle the number of customers that occur in that five minutes. We're looking for the probability that the number of customers who arrive in that five minutes is greater than 10 because we know that they can handle up to 10 customers in five minutes. And this is equal to one, minus the probability that the number of customer arrivals is that most 10. So this is equal to one minus. The summation from X is equal to zero to 10 of E to the minus lander times T, which is equal to minus four. I got that from tee times lambda, which is 0.8 times Lambda Times t to the exponents X over x factorial. You know, just to explain this, remember, the probability of getting X arrivals is equal to eat to the negative and the tea on celebrity to the exponents X over x factorial. So here we're finding the some of the probabilities of getting zero through 10 arrivals in the next five minutes and then subtracting that from one. And so if you calculate this, it comes out to about 0.0 28 So that is the probability that too many customers come between the time of 12:15 p.m. And 12:20 p.m. For the drive through staff to handle and moving on to Part C. We're told that a customer has just arrived and were asked to calculate the probability that another customer will arrive in the next 30 seconds. So one way to think of this question is that for a Poisson process, the inter arrival times are distributed according to an exponential distribution with rate lambda. So if we say that we define a as the inter arrival time, it's the time between subsequent arrivals. It follows an exponential distribution with parameter lambda, so we want the probability that a is less than or equal to 30 seconds. This is the cumulative distribution for the exponential distribution, and we know that our rate is 0.8 per minute. So to keep this consistent, I should probably put this in terms of minutes, so that would be half a minute and remember, for an exponential distribution, a cumulative function is one minus e to the negative Lambda Key. So if we feel these numbers in, we get one minus e to the negative 0.8 times half, and that should come out to zero point 33 with your calculator. So the probability that the next customer arrives in the next 30 seconds is about 0.33 now for Part D, starting at noon. We want to determine the expected arrival time of the 1/100 customer as well as the standard deviation of that time. So recall that the inter arrival times of customers air exponentially distributed, and we want to find the average time for the 1/100 customer to arrive. So if we define the time that it takes for the 1/100 customer to arrive, it is thesis, um, of all of the inter arrival times for the 1st, 2nd, third customer and so on up to the 1/100 customer where all of the teas are distributed according to the exponential distribution with rate parameter lambda. Then why some 100 is distributed as a gamma with parameter and equals 100 and Lambda equals 0.8. Now the expected time for why 7 100? It's simply end times one over Lambda and remember one over Lambda would be the expected enter arrival time between any successive arrivals and since Lambda equals 0.8 would simply be 1.25 minutes. That is the expected time between success of any two successive arrivals. So for the 1/100 arrival, all we're doing is multiplying that by 100. So we end up with 125. So the expected time for the 1/100 arrival is 125 minutes. And the standard deviation for the 1/100 arrival is given by the square root of end times one over Lambda squared and this comes out to 12.5 seconds. So the standard deviation on the expected time for the 1/100 arrival is 12.5 seconds s. Sorry. Actually, that's minutes. So our units are minutes. So the standard deviation has the same units as the expected Valley, which are minutes

This question pertains to non homogeneous place on processes and normally a person process for a normal person. Process has a constant rate of arrival for all time, but for a non homogeneous possum process, the rate of arrival can change with time. So in this question we have a repair facility that's open eight hours each day, and the rate of arrivals is dependent on how much of the time of the day has elapsed. So for the first hour, we can see that the arrival rate is T. And between hours one and seven, the arrival rate is one between our seven and eight. The arrival rate is eight minus T. Now, in part A were asked what the probability is that no customers arrive in both the first and last hours of the day and that four customers arrive in the middle of six hours, so the first thing we could do is find the mean are the expected number of arrivals for each of these periods. So let's call the first one new sub 0 to 1, and this is equal to the integral of the arrival rate. With respect to time from 0 to 1 so is equal to the integral of tea from 0 to 1 just equal to half. So repeat that for the time period from 1 to 7. So these three numbers are our expected number of arrivals during these time periods. So for part A is looking for the probability that the number that arrived by time one is zero and the number that arrived between time, one and time seven is for and the number that arrived between time eight and times seven is zero. So we can we can characterized the number of arrivals in each of these different time periods as they put some random variable with parameter one half six and one half, respectively. So this is equal to eat the negative one half times one half the exponents zero for zero factorial. And this comes out to 0.0 492 now for part B were asked to calculate the probability that the same number of customers arrive in each of these three different time periods. So the first hour, the middle six hours and the last hour So we're looking for the probability at the number in the first hour is equal to the number by the seventh hour minus number by the first hour is equal to the number by the eighth hour, minus the number by the seventh hour. Now the probability that they have the same number of arrivals in each of these different time periods is equal to following. Let's say let's say they all have the same number arrivals, probability that they all have X arrivals would be written like this, except we just want the total probability that they have the same number of arrivals, regardless of how maney arrivals. That is so what we want to dio is some these overall possible values of X and then this could be simplified as follows and this comes out to zero point 00 26


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