### expected waiting time probability

Out of these, the cookies that are categorized as necessary are stored on your browser as they are essential for the working of basic functionalities of the website. This type of study could be done for any specific waiting line to find a ideal waiting line system. With probability $q$ the first toss is a tail, so $M = W_H$ where $W_H$ has the geometric $(p)$ distribution. Rename .gz files according to names in separate txt-file. \], 17.4. \end{align} We also use third-party cookies that help us analyze and understand how you use this website. Keywords. \mathbb P(W>t) &= \sum_{n=0}^\infty \mathbb P(W>t\mid L^a=n)\mathbb P(L^a=n)\\ \begin{align} Torsion-free virtually free-by-cyclic groups. Let \(E_k(T)\) denote the expected duration of the game given that the gambler starts with a net gain of \(k\) dollars. Think of what all factors can we be interested in? The number of distinct words in a sentence. All of the calculations below involve conditioning on early moves of a random process. &= \sum_{n=0}^\infty \mathbb P\left(\sum_{k=1}^{L^a+1}W_k>t\mid L^a=n\right)\mathbb P(L^a=n). 1 Expected Waiting Times We consider the following simple game. To assure the correct operating of the store, we could try to adjust the lambda and mu to make sure our process is still stable with the new numbers. a=0 (since, it is initial. Models with G can be interesting, but there are little formulas that have been identified for them. How to react to a students panic attack in an oral exam? Queuing Theory, as the name suggests, is a study of long waiting lines done to predict queue lengths and waiting time. - ovnarian Jan 26, 2012 at 17:22 Define a trial to be a success if those 11 letters are the sequence datascience. As discussed above, queuing theory is a study of long waiting lines done to estimate queue lengths and waiting time. Answer 1. document.getElementById( "ak_js_1" ).setAttribute( "value", ( new Date() ).getTime() ); How to Read and Write With CSV Files in Python:.. To subscribe to this RSS feed, copy and paste this URL into your RSS reader. I think that implies (possibly together with Little's law) that the waiting time is the same as well. x ~ = ~ E(W_H) + E(V) ~ = ~ \frac{1}{p} + p + q(1 + x)
a)If a sale just occurred, what is the expected waiting time until the next sale? where P (X>) is the probability of happening more than x. x is the time arrived. The first waiting line we will dive into is the simplest waiting line. @Nikolas, you are correct but wrong :). That is X U ( 1, 12). Thats \(26^{11}\) lots of 11 draws, which is an overestimate because you will be watching the draws sequentially and not in blocks of 11. With this article, we have now come close to how to look at an operational analytics in real life. With probability $pq$ the first two tosses are HT, and $W_{HH} = 2 + W^{**}$ This means that the passenger has no sense of time nor know when the last train left and could enter the station at any point within the interval of 2 consecutive trains. Calculation: By the formula E(X)=q/p. Jordan's line about intimate parties in The Great Gatsby? What is the expected number of messages waiting in the queue and the expected waiting time in queue? \mathbb P(W>t) = \sum_{n=0}^\infty \sum_{k=0}^n\frac{(\mu t)^k}{k! }\\ This is a Poisson process. Define a trial to be 11 letters picked at random. So An important assumption for the Exponential is that the expected future waiting time is independent of the past waiting time. Beta Densities with Integer Parameters, 18.2. Is Koestler's The Sleepwalkers still well regarded? Planned Maintenance scheduled March 2nd, 2023 at 01:00 AM UTC (March 1st, M/M/1 queue with customers leaving based on number of customers present at arrival. Some analyses have been done on G queues but I prefer to focus on more practical and intuitive models with combinations of M and D. Lets have a look at three well known queues: An example of this is a waiting line in a fast-food drive-through, where everyone stands in the same line, and will be served by one of the multiple servers, as long as arrivals are Poisson and service time is Exponentially distributed. W = \frac L\lambda = \frac1{\mu-\lambda}. $$ (starting at 0 is required in order to get the boundary term to cancel after doing integration by parts). Solution If X U ( a, b) then the probability density function of X is f ( x) = 1 b a, a x b. The response time is the time it takes a client from arriving to leaving. Is email scraping still a thing for spammers. Why does Jesus turn to the Father to forgive in Luke 23:34? This can be written as a probability statement: \(P(X>a)=P(X>a+b \mid X>b)\) &= (1-\rho)\cdot\mathsf 1_{\{t=0\}}+(1-\rho)\cdot\mathsf 1_{\{t=0\}} + \sum_{n=1}^\infty (1-\rho)\rho^n \int_0^t \mu e^{-\mu s}\frac{(\mu s)^{n-1}}{(n-1)! With probability $p$ the first toss is a head, so $M = W_T$ where $W_T$ has the geometric $(q)$ distribution. We will also address few questions which we answered in a simplistic manner in previous articles. I can't find very much information online about this scenario either. This phenomenon is called the waiting-time paradox [ 1, 2 ]. This takes into account the clarification of the the OP in a comment that the correct assumptions to take are that each train is on a fixed timetable independent of the other and of the traveller's arrival time, and that the phases of the two trains are uniformly distributed, $$ p(t) = (1-S(t))' = \frac{1}{10} \left( 1- \frac{t}{15} \right) + \frac{1}{15} \left(1-\frac{t}{10} \right) $$. Your home for data science. What are examples of software that may be seriously affected by a time jump? This means that we have a single server; the service rate distribution is exponential; arrival rate distribution is poisson process; with infinite queue length allowed and anyone allowed in the system; finally its a first come first served model. In general, we take this to beinfinity () as our system accepts any customer who comes in. For example, waiting line models are very important for: Imagine a store with on average two people arriving in the waiting line every minute and two people leaving every minute as well. &= (1-\rho)\cdot\mathsf 1_{\{t=0\}}+(1-\rho)\cdot\mathsf 1_{\{t=0\}} + \sum_{n=1}^\infty (1-\rho)\rho^n \int_0^t \mu e^{-\mu s}\frac{(\mu s)^{n-1}}{(n-1)! We can find $E(N)$ by conditioning on the first toss as we did in the previous example. From $\sum_{n=0}^\infty\pi_n=1$ we see that $\pi_0=1-\rho$ and hence $\pi_n=\rho^n(1-\rho)$. What's the difference between a power rail and a signal line? Why did the Soviets not shoot down US spy satellites during the Cold War? The following is a worked example found in past papers of my university, but haven't been able to figure out to solve it (I have the answer, but do not understand how to get there). Learn more about Stack Overflow the company, and our products. They will, with probability 1, as you can see by overestimating the number of draws they have to make. How did StorageTek STC 4305 use backing HDDs? The given problem is a M/M/c type query with following parameters. Round answer to 4 decimals. HT occurs is less than the expected waiting time before HH occurs. In a 45 minute interval, you have to wait $45 \cdot \frac12 = 22.5$ minutes on average. Service rate, on the other hand, largely depends on how many caller representative are available to service, what is their performance and how optimized is their schedule. How can I change a sentence based upon input to a command? You can check that the function $f(k) = (b-k)(k-a)$ satisfies this recursion, and hence that $E_0(T) = ab$. Typically, you must wait longer than 3 minutes. Here are the possible values it can take: C gives the Number of Servers in the queue. Let {N_1 (t)} and {N_2 (t)} be two independent Poisson processes with rates 1=1 and 2=2, respectively. Let's call it a $p$-coin for short. \], \[
x = \frac{q + 2pq + 2p^2}{1 - q - pq} For example, Amazon has found out that 100 milliseconds increase in waiting time (page loading) costs them 1% of sales (source). \], \[
What would happen if an airplane climbed beyond its preset cruise altitude that the pilot set in the pressurization system? The number of trials till the first success provides the framework for a rich array of examples, because both trial and success can be defined to be much more complex than just tossing a coin and getting heads. For example, your flow asks for the Estimated Wait Time shortly after putting the interaction on a queue and you get a value of 10 minutes. How can I recognize one? Necessary cookies are absolutely essential for the website to function properly. M/M/1//Queuewith Discouraged Arrivals : This is one of the common distribution because the arrival rate goes down if the queue length increases. Therefore, the 'expected waiting time' is 8.5 minutes. Does exponential waiting time for an event imply that the event is Poisson-process? $$ Lets call it a \(p\)-coin for short. $$ In my previous articles, Ive already discussed the basic intuition behind this concept with beginnerand intermediate levelcase studies. MathJax reference. a) Mean = 1/ = 1/5 hour or 12 minutes If there are N decoys to add, choose a random number k in 0..N with a flat probability, and add k younger and (N-k) older decoys with a reasonable probability distribution by date. if we wait one day $X=11$. I think that the expected waiting time (time waiting in queue plus service time) in LIFO is the same as FIFO. On service completion, the next customer $$, $$ It only takes a minute to sign up. So the average wait time is the area from $0$ to $30$ of an array of triangles, divided by $30$. Should the owner be worried about this? $$ 2. Probability simply refers to the likelihood of something occurring. Now that we have discovered everything about the M/M/1 queue, we move on to some more complicated types of queues. Rather than asking what the average number of customers is, we can ask the probability of a given number x of customers in the waiting line. probability probability-theory operations-research queueing-theory Share Cite Follow edited Nov 6, 2019 at 5:59 asked Nov 5, 2019 at 18:15 user720606 However, in case of machine maintenance where we have fixed number of machines which requires maintenance, this is also a fixed positive integer. (1500/2-1000/6)\frac 1 {10} \frac 1 {15}=5-10/9\approx 3.89$$, Assuming each train is on a fixed timetable independent of the other and of the traveller's arrival time, the probability neither train arrives in the first $x$ minutes is $\frac{10-x}{10} \times \frac{15-x}{15}$ for $0 \le x \le 10$, which when integrated gives $\frac{35}9\approx 3.889$ minutes, Alternatively, assuming each train is part of a Poisson process, the joint rate is $\frac{1}{15}+\frac{1}{10}=\frac{1}{6}$ trains a minute, making the expected waiting time $6$ minutes. With the remaining probability $q$ the first toss is a tail, and then. Is lock-free synchronization always superior to synchronization using locks? E(W_{HH}) ~ = ~ \frac{1}{p^2} + \frac{1}{p}
All the examples below involve conditioning on early moves of a random process. Thanks! Can trains not arrive at minute 0 and at minute 60? We can also find the probability of waiting a length of time: There's a 57.72 percent probability of waiting between 5 and 30 minutes to see the next meteor. The most apparent applications of stochastic processes are time series of . Answer: We can find \(E(N)\) by conditioning on the first toss as we did in the previous example. }=1-\sum_{j=0}^{59} e^{-4d}\frac{(4d)^{j}}{j! An example of an Exponential distribution with an average waiting time of 1 minute can be seen here: For analysis of an M/M/1 queue we start with: From those inputs, using predefined formulas for the M/M/1 queue, we can find the KPIs for our waiting line model: It is often important to know whether our waiting line is stable (meaning that it will stay more or less the same size). Connect and share knowledge within a single location that is structured and easy to search. Let's get back to the Waiting Paradox now. \end{align}, \begin{align} M/M/1, the queue that was covered before stands for Markovian arrival / Markovian service / 1 server. Lets understand it using an example. @Aksakal. Browse other questions tagged, Start here for a quick overview of the site, Detailed answers to any questions you might have, Discuss the workings and policies of this site. Was Galileo expecting to see so many stars? Is email scraping still a thing for spammers, How to choose voltage value of capacitors. (Round your answer to two decimal places.) Suppose that the average waiting time for a patient at a physician's office is just over 29 minutes. Your branch can accommodate a maximum of 50 customers. X=0,1,2,. In this article, I will give a detailed overview of waiting line models. Let's call it a $p$-coin for short. In the common, simpler, case where there is only one server, we have the M/D/1 case. We've added a "Necessary cookies only" option to the cookie consent popup. b)What is the probability that the next sale will happen in the next 6 minutes? $$. There is a blue train coming every 15 mins. A mixture is a description of the random variable by conditioning. Step 1: Definition. He is fascinated by the idea of artificial intelligence inspired by human intelligence and enjoys every discussion, theory or even movie related to this idea. In order to do this, we generally change one of the three parameters in the name. The probability that we have sold $60$ computers before day 11 is given by $\Pr(X>60|\lambda t=44)=0.00875$. E_{-a}(T) = 0 = E_{a+b}(T) Does With(NoLock) help with query performance? 5.What is the probability that if Aaron takes the Orange line, he can arrive at the TD garden at . Conditioning and the Multivariate Normal, 9.3.3. I am probably wrong but assuming that each train's starting-time follows a uniform distribution, I would say that when arriving at the station at a random time the expected waiting time for: Suppose that red and blue trains arrive on time according to schedule, with the red schedule beginning $\Delta$ minutes after the blue schedule, for some $0\le\Delta<10$. At what point of what we watch as the MCU movies the branching started? Acceleration without force in rotational motion? Another way is by conditioning on $X$, the number of tosses till the first head. An average service time (observed or hypothesized), defined as 1 / (mu). So if $x = E(W_{HH})$ then We know that $E(X) = 1/p$. Answer 2: Another way is by conditioning on the toss after \(W_H\) where, as before, \(W_H\) is the number of tosses till the first head. Learn more about Stack Overflow the company, and our products. And the expected value is obtained in the usual way: $E[t] = \int_0^{10} t p(t) dt = \int_0^{10} \frac{t}{10} \left( 1- \frac{t}{15} \right) + \frac{t}{15} \left(1-\frac{t}{10} \right) dt = \int_0^{10} \left( \frac{t}{6} - \frac{t^2}{75} \right) dt$. Is there a more recent similar source? This is called Kendall notation. In the supermarket, you have multiple cashiers with each their own waiting line. Could you explain a bit more? The answer is variation around the averages. Finally, $$E[t]=\int_x (15x-x^2/2)\frac 1 {10} \frac 1 {15}dx= Let $E_k(T)$ denote the expected duration of the game given that the gambler starts with a net gain of $\$k$. It has 1 waiting line and 1 server. We know that \(W_H\) has the geometric \((p)\) distribution on \(1, 2, 3, \ldots \). The average wait for an interval of length $15$ is of course $7\frac{1}{2}$ and for an interval of length $45$ it is $22\frac{1}{2}$. as in example? $$, \begin{align} Think about it this way. E(X) = \frac{1}{p} 1. Clearly you need more 7 reps to satisfy both the constraints given in the problem where customers leaving. @Tilefish makes an important comment that everybody ought to pay attention to. &= (1-\rho)\cdot\mathsf 1_{\{t=0\}}+\rho(1-\rho)\int_0^t \mu e^{-\mu(1-\rho)s}\ \mathsf ds\\ If letters are replaced by words, then the expected waiting time until some words appear . x= 1=1.5. Let \(W_H\) be the number of tosses of a \(p\)-coin till the first head appears. a is the initial time. Planned Maintenance scheduled March 2nd, 2023 at 01:00 AM UTC (March 1st, Expected travel time for regularly departing trains. Utilization is called (rho) and it is calculated as: It is possible to compute the average number of customers in the system using the following formula: The variation around the average number of customers is defined as followed: Going even further on the number of customers, we can also put the question the other way around. number" system). For some, complicated, variants of waiting lines, it can be more difficult to find the solution, as it may require a more theoretical mathematical approach. Define a trial to be a "success" if those 11 letters are the sequence. As discussed above, queuing theory is a study oflong waiting lines done to estimate queue lengths and waiting time. They will, with probability 1, as you can see by overestimating the number of draws they have to make. I however do not seem to understand why and how it comes to these numbers. Overlap. Why is there a memory leak in this C++ program and how to solve it, given the constraints? }\\ Waiting Till Both Faces Have Appeared, 9.3.5. That seems to be a waiting line in balance, but then why would there even be a waiting line in the first place? Site design / logo 2023 Stack Exchange Inc; user contributions licensed under CC BY-SA. Moreover, almost nobody acknowledges the fact that they had to make some such an interpretation of the question in order to obtain an answer. There isn't even close to enough time. I am new to queueing theory and will appreciate some help. &= e^{-\mu t}\sum_{k=0}^\infty\frac{(\mu\rho t)^k}{k! A second analysis to do is the computation of the average time that the server will be occupied. +1 At this moment, this is the unique answer that is explicit about its assumptions. Browse other questions tagged, Start here for a quick overview of the site, Detailed answers to any questions you might have, Discuss the workings and policies of this site. We want \(E_0(T)\). x = E(X) + E(Y) = \frac{1}{p} + p + q(1 + x) is there a chinese version of ex. This means that the passenger has no sense of time nor know when the last train left and could enter the station at any point within the interval of 2 consecutive trains. Sometimes Expected number of units in the queue (E (m)) is requested, excluding customers being served, which is a different formula ( arrival rate multiplied by the average waiting time E(m) = E(w) ), and obviously results in a small number. A is the Inter-arrival Time distribution . Could very old employee stock options still be accessible and viable? I think there may be an error in the worked example, but the numbers are fairly clear: You have a process where the shop starts with a stock of $60$, and over $12$ opening days sells at an average rate of $4$ a day, so over $d$ days sells an average of $4d$. One way is by conditioning on the first two tosses. In this article, I will bring you closer to actual operations analytics usingQueuing theory. A coin lands heads with chance $p$. $$ But I am not completely sure. $$ You are expected to tie up with a call centre and tell them the number of servers you require. if we wait one day X = 11. What is the expected waiting time in an $M/M/1$ queue where order The longer the time frame the closer the two will be. This gives To address the issue of long patient wait times, some physicians' offices are using wait-tracking systems to notify patients of expected wait times. In terms of service times, the average service time of the latest customer has the same statistics as any of the waiting customers, so statistically it doesn't matter if the server is treating the latest arrival or any other arrival, so the busy period distribution should be the same. }e^{-\mu t}\rho^n(1-\rho) One day you come into the store and there are no computers available. \end{align}, $$ The probability of having a certain number of customers in the system is. Why was the nose gear of Concorde located so far aft? i.e. Let $X$ be the number of tosses of a $p$-coin till the first head appears. Another name for the domain is queuing theory. \lambda \pi_n = \mu\pi_{n+1},\ n=0,1,\ldots, Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. (a) The probability density function of X is With probability p the first toss is a head, so R = 0. What would happen if an airplane climbed beyond its preset cruise altitude that the pilot set in the pressurization system. Maybe this can help? Here is a quick way to derive $E(X)$ without even using the form of the distribution. $$ $$ $$ When to use waiting line models? (1) Your domain is positive. So The time spent waiting between events is often modeled using the exponential distribution. So the real line is divided in intervals of length $15$ and $45$. Sums of Independent Normal Variables, 22.1. }e^{-\mu t}\rho^k\\ Reversal. &= \sum_{n=0}^\infty \mathbb P(W_q\leqslant t\mid L=n)\mathbb P(L=n)\\ (15x^2/2-x^3/6)|_0^{10}\frac 1 {10} \frac 1 {15}\\= With probability 1, at least one toss has to be made. So when computing the average wait we need to take into acount this factor. Mark all the times where a train arrived on the real line. \], \[
Between $t=0$ and $t=30$ minutes we'll see the following trains and interarrival times: blue train, $\Delta$, red train, $10$, red train, $5-\Delta$, blue train, $\Delta + 5$, red train, $10-\Delta$, blue train. Here are the possible values it can take : B is the Service Time distribution. (2) The formula is. I think that the expected waiting time (time waiting in queue plus service time) in LIFO is the same as FIFO. These cookies do not store any personal information. Find the probability that the second arrival in N_1 (t) occurs before the third arrival in N_2 (t). If you arrive at the station at a random time and go on any train that comes the first, what is the expected waiting time? }e^{-\mu t}(1-\rho)\sum_{n=k}^\infty \rho^n\\ The time between train arrivals is exponential with mean 6 minutes. So we have It uses probabilistic methods to make predictions used in the field of operational research, computer science, telecommunications, traffic engineering etc. Red train arrivals and blue train arrivals are independent. So expected waiting time to $x$-th success is $xE (W_1)$. S. Click here to reply. Should I include the MIT licence of a library which I use from a CDN? After reading this article, you should have an understanding of different waiting line models that are well-known analytically. Stack Exchange network consists of 181 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. The second criterion for an M/M/1 queue is that the duration of service has an Exponential distribution. By Ani Adhikari
Look for example on a 24 hours time-line, 3/4 of it will be 45m intervals and only 1/4 of it will be the shorter 15m intervals. You are setting up this call centre for a specific feature queries of customers which has an influx of around 20 queries in an hour. @whuber I prefer this approach, deriving the PDF from the survival function, because it correctly handles cases where the domain of the random variable does not start at 0. Connect and share knowledge within a single location that is structured and easy to search. Since 15 minutes and 45 minutes intervals are equally likely, you end up in a 15 minute interval in 25% of the time and in a 45 minute interval in 75% of the time. }\ \mathsf ds\\ &= (1-\rho)\cdot\mathsf 1_{\{t=0\}}+\rho(1-\rho)\int_0^t \mu e^{-\mu(1-\rho)s}\ \mathsf ds\\ Hence, it isnt any newly discovered concept. Let $L^a$ be the number of customers in the system immediately before an arrival, and $W_k$ the service time of the $k^{\mathrm{th}}$ customer. Use MathJax to format equations. Browse other questions tagged, Start here for a quick overview of the site, Detailed answers to any questions you might have, Discuss the workings and policies of this site. The average response time can be computed as: The average time spent waiting can be computed as follows: To give a practical example, lets apply the analysis on a small stores waiting line. An educated guess for your "waiting time" is 3 minutes, which is half the time between buses on average. That is, with probability \(q\), \(R = W^*\) where \(W^*\) is an independent copy of \(W_H\). What does a search warrant actually look like? The expected waiting time for a success is therefore = E (t) = 1/ = 10 91 days or 2.74 x 10 88 years Compare this number with the evolutionist claim that our solar system is less than 5 x 10 9 years old. x = \frac{q + 2pq + 2p^2}{1 - q - pq}
If $\tau$ is uniform on $[0,b]$, it's $\frac 2 3 \mu$. An average arrival rate (observed or hypothesized), called (lambda). This is popularly known as the Infinite Monkey Theorem. For example, suppose that an average of 30 customers per hour arrive at a store and the time between arrivals is . The number at the end is the number of servers from 1 to infinity. Your expected waiting time can be even longer than 6 minutes. If $\Delta$ is not constant, but instead a uniformly distributed random variable, we obtain an average average waiting time of This gives a expected waiting time of $\frac14 \cdot 7.5 + \frac34 \cdot 22.5 = 18.75$. $$. But conditioned on them being sold out, the posterior probability of for example being sold out with three days to go is $\frac{\frac14 P_9}{\frac14 P_{11}+ \frac14 P_{10}+ \frac14 P_{9}+ \frac14 P_{8}}$ and similarly for the others. Solution: m = [latex]\frac{1}{12}[/latex] [latex]\mu [/latex] = 12 . Dealing with hard questions during a software developer interview. Let \(N\) be the number of tosses. = \frac{1+p}{p^2}
The best answers are voted up and rise to the top, Not the answer you're looking for? Answer 1: We can find this is several ways. In tosses of a $p$-coin, let $W_{HH}$ be the number of tosses till you see two heads in a row. How many people can we expect to wait for more than x minutes? Also W and Wq are the waiting time in the system and in the queue respectively. Did you like reading this article ? Also, please do not post questions on more than one site you also posted this question on Cross Validated. With probability \(p\) the first toss is a head, so \(M = W_T\) where \(W_T\) has the geometric \((q)\) distribution. To find the distribution of $W_q$, we condition on $L$ and use the law of total probability: For example, if the first block of 11 ends in data and the next block starts with science, you will have seen the sequence datascience and stopped watching, even though both of those blocks would be called failures and the trials would continue. The mean of X is E ( X) = ( a + b) 2 and variance of X is V ( X) = ( b a) 2 12. Does Cast a Spell make you a spellcaster? &= \sum_{n=0}^\infty \mathbb P(W_q\leqslant t\mid L=n)\mathbb P(L=n)\\ $$. If as usual we write $q = 1-p$, the distribution of $X$ is given by. &= (1-\rho)\cdot\mathsf 1_{\{t=0\}} + 1-\rho e^{-\mu(1-\rho)t)}\cdot\mathsf 1_{(0,\infty)}(t). For example, the string could be the complete works of Shakespeare. Dave, can you explain how p(t) = (1- s(t))' ? It only takes a minute to sign up. If dark matter was created in the early universe and its formation released energy, is there any evidence of that energy in the cmb? There is a red train that is coming every 10 mins. Suppose we do not know the order This website uses cookies to improve your experience while you navigate through the website. We have the balance equations Is Koestler's The Sleepwalkers still well regarded? rev2023.3.1.43269. The expected waiting time for a single bus is half the expected waiting time for two buses and the variance for a single bus is half the variance of two buses. Like. rev2023.3.1.43269. The exact definition of what it means for a train to arrive every $15$ or $4$5 minutes with equal probility is a little unclear to me. This calculation confirms that in i.i.d. This means, that the expected time between two arrivals is. Here, N and Nq arethe number of people in the system and in the queue respectively. Your simulator is correct. However, the fact that $E (W_1)=1/p$ is not hard to verify. $$\int_{y

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