expected waiting time probabilityexpected waiting time probability

The blue train also arrives according to a Poisson distribution with rate 4/hour. $$, \begin{align} However, the fact that $E (W_1)=1/p$ is not hard to verify. I think the decoy selection process can be improved with a simple algorithm. which works out to $\frac{35}{9}$ minutes. In some cases, we can find adapted formulas, while in other situations we may struggle to find the appropriate model. Connect and share knowledge within a single location that is structured and easy to search. TABLE OF CONTENTS : TABLE OF CONTENTS. }=1-\sum_{j=0}^{59} e^{-4d}\frac{(4d)^{j}}{j! This is called the geometric $(p)$ distribution on $1, 2, 3, \ldots $, because its terms are those of a geometric series. E(X) = 1/ = 1/0.1= 10. minutes or that on average, buses arrive every 10 minutes. This means: trying to identify the mathematical definition of our waiting line and use the model to compute the probability of the waiting line system reaching a certain extreme value. Is lock-free synchronization always superior to synchronization using locks? There isn't even close to enough time. There is a blue train coming every 15 mins. . So $X = 1 + Y$ where $Y$ is the random number of tosses after the first one. Here are the possible values it can take: C gives the Number of Servers in the queue. b)What is the probability that the next sale will happen in the next 6 minutes? - ovnarian Jan 26, 2012 at 17:22 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. \], \[ Here are a few parameters which we would beinterested for any queuing model: Its an interesting theorem. On average, each customer receives a service time of s. Therefore, the expected time required to serve all Imagine, you work for a multi national bank. With probability \(q\) the first toss is a tail, so \(M = W_H\) where \(W_H\) has the geometric \((p)\) distribution. \begin{align}\bar W_\Delta &:= \frac1{30}\left(\frac12[\Delta^2+10^2+(5-\Delta)^2+(\Delta+5)^2+(10-\Delta)^2]\right)\\&=\frac1{30}(2\Delta^2-10\Delta+125). Suppose we toss the \(p\)-coin until both faces have appeared. Connect and share knowledge within a single location that is structured and easy to search. You have the responsibility of setting up the entire call center process. We know that $E(X) = 1/p$. Let's find some expectations by conditioning. Stochastic Queueing Queue Length Comparison Of Stochastic And Deterministic Queueing And BPR. \end{align}, $$ Here are the values we get for waiting time: A negative value of waiting time means the value of the parameters is not feasible and we have an unstable system. 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. You may consider to accept the most helpful answer by clicking the checkmark. which, for $0 \le t \le 10$, is the the probability that you'll have to wait at least $t$ minutes for the next train. \mathbb P(W_q\leqslant t) &= \sum_{n=0}^\infty\mathbb P(W_q\leqslant t, L=n)\\ &= (1-\rho)\cdot\mathsf 1_{\{t=0\}}+\rho(1-\rho)\sum_{n=1}^\infty\rho^n\int_0^t \mu e^{-\mu s}\frac{(\mu\rho s)^{n-1}}{(n-1)! To visualize the distribution of waiting times, we can once again run a (simulated) experiment. Let's say a train arrives at a stop in intervals of 15 or 45 minutes, each with equal probability 1/2 (so every time a train arrives, it will randomly be either 15 or 45 minutes until the next arrival). Why did the Soviets not shoot down US spy satellites during the Cold War? @Dave with one train on a fixed $10$ minute timetable independent of the traveller's arrival, you integrate $\frac{10-x}{10}$ over $0 \le x \le 10$ to get an expected wait of $5$ minutes, while with a Poisson process with rate $\lambda=\frac1{10}$ you integrate $e^{-\lambda x}$ over $0 \le x \lt \infty$ to get an expected wait of $\frac1\lambda=10$ minutes, @NeilG TIL that "the expected value of a non-negative random variable is the integral of the survival function", sort of -- there is some trickiness in that the domain of the random variable needs to start at $0$, and if it doesn't intrinsically start at zero(e.g. Consider a queue that has a process with mean arrival rate ofactually entering the system. Your got the correct answer. Asking for help, clarification, or responding to other answers. where P (X>) is the probability of happening more than x. x is the time arrived. \end{align} Waiting line models are mathematical models used to study waiting lines. That seems to be a waiting line in balance, but then why would there even be a waiting line in the first place? Red train arrivals and blue train arrivals are independent. Here is an overview of the possible variants you could encounter. First we find the probability that the waiting time is 1, 2, 3 or 4 days. However, this reasoning is incorrect. Do share your experience / suggestions in the comments section below. Thanks! How can the mass of an unstable composite particle become complex? So the average wait time is the area from $0$ to $30$ of an array of triangles, divided by $30$. Assume $\rho:=\frac\lambda\mu<1$. Copyright 2022. The marks are either $15$ or $45$ minutes apart. Learn more about Stack Overflow the company, and our products. 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. You would probably eat something else just because you expect high waiting time. I think that the expected waiting time (time waiting in queue plus service time) in LIFO is the same as FIFO. \begin{align} This means only less than 0.001 % customer should go back without entering the branch because the brach already had 50 customers. Suppose the customers arrive at a Poisson rate of on eper every 12 minutes, and that the service time is . Since the summands are all nonnegative, Tonelli's theorem allows us to interchange the order of summation: Moreover, almost nobody acknowledges the fact that they had to make some such an interpretation of the question in order to obtain an answer. 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. With probability 1, at least one toss has to be made. &= e^{-\mu(1-\rho)t}\\ Get the parts inside the parantheses: The given problem is a M/M/c type query with following parameters. The probability that we have sold $60$ computers before day 11 is given by $\Pr(X>60|\lambda t=44)=0.00875$. Did you like reading this article ? Waiting line models can be used as long as your situation meets the idea of a waiting line. Expected waiting time. Another name for the domain is queuing theory. The number of distinct words in a sentence. A mixture is a description of the random variable by conditioning. With probability 1, \(N = 1 + M\) where \(M\) is the additional number of tosses needed after the first one. However, at some point, the owner walks into his store and sees 4 people in line. An educated guess for your "waiting time" is 3 minutes, which is half the time between buses on average. Tavish Srivastava, co-founder and Chief Strategy Officer of Analytics Vidhya, is an IIT Madras graduate and a passionate data-science professional with 8+ years of diverse experience in markets including the US, India and Singapore, domains including Digital Acquisitions, Customer Servicing and Customer Management, and industry including Retail Banking, Credit Cards and Insurance. We assume that the times between any two arrivals are independent and exponentially distributed with = 0.1 minutes. 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. For example, it's $\mu/2$ for degenerate $\tau$ and $\mu$ for exponential $\tau$. Do EMC test houses typically accept copper foil in EUT? Since the sum of But 3. is still not obvious for me. Take a weighted coin, one whose probability of heads is p and whose probability of tails is therefore 1 p. Fix a positive integer k and continue to toss this coin until k heads in succession have resulted. Both of them start from a random time so you don't have any schedule. Queuing Theory, as the name suggests, is a study of long waiting lines done to predict queue lengths and waiting time. Rho is the ratio of arrival rate to service rate. Acceleration without force in rotational motion? How many instances of trains arriving do you have? This website uses cookies to improve your experience while you navigate through the website. $$ To find the distribution of $W_q$, we condition on $L$ and use the law of total probability: It has 1 waiting line and 1 server. Now you arrive at some random point on the line. You will just have to replace 11 by the length of the string. }.$ This gives $P_{11}$, $P_{10}$, $P_{9}$, $P_{8}$ as about $0.01253479$, $0.001879629$, $0.0001578351$, $0.000006406888$. This is a shorthand notation of the typeA/B/C/D/E/FwhereA, B, C, D, E,Fdescribe the queue. With probability \(p^2\), the first two tosses are heads, and \(W_{HH} = 2\). Is email scraping still a thing for spammers. Why was the nose gear of Concorde located so far aft? With this article, we have now come close to how to look at an operational analytics in real life. Probability For Data Science Interact Expected Waiting Times Let's find some expectations by conditioning. Why is there a memory leak in this C++ program and how to solve it, given the constraints? You can replace it with any finite string of letters, no matter how long. To learn more, see our tips on writing great answers. Your simulator is correct. Imagine you went to Pizza hut for a pizza party in a food court. What is the expected number of messages waiting in the queue and the expected waiting time in queue? OP said specifically in comments that the process is not Poisson, Expected value of waiting time for the first of the two buses running every 10 and 15 minutes, We've added a "Necessary cookies only" option to the cookie consent popup. Analytics Vidhya App for the Latest blog/Article, 15 Must Read Books for Entrepreneurs in Data Science, Big Data Architect Mumbai (5+ years of experience). Even though we could serve more clients at a service level of 50, this does not weigh up to the cost of staffing. How to increase the number of CPUs in my computer? Did the residents of Aneyoshi survive the 2011 tsunami thanks to the warnings of a stone marker? Once we have these cost KPIs all set, we should look into probabilistic KPIs. 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. We can expect to wait six minutes or less to see a meteor 39.4 percent of the time. Waiting Till Both Faces Have Appeared, 9.3.5. Thanks for reading! c) To calculate for the probability that the elevator arrives in more than 1 minutes, we have the formula. Therefore, the probability that the queue is occupied at an arrival instant is simply U, the utilization, and the average number of customers waiting but not being served at the arrival instant is QU. Hence, it isnt any newly discovered concept. Then the number of trials till datascience appears has the geometric distribution with parameter \(p = 1/26^{11}\), and therefore has expectation \(26^{11}\). This category only includes cookies that ensures basic functionalities and security features of the website. The solution given goes on to provide the probalities of $\Pr(T|T>0)$, before it gives the answer by $E(T)=1\cdot 0.8719+2\cdot 0.1196+3\cdot 0.0091+4\cdot 0.0003=1.1387$. So we have And $E (W_1)=1/p$. It only takes a minute to sign up. So On service completion, the next customer 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. The probability that you must wait more than five minutes is _____ . The expected size in system is By clicking Post Your Answer, you agree to our terms of service, privacy policy and cookie policy. One day you come into the store and there are no computers available. For example, the string could be the complete works of Shakespeare. With probability \(p\) the first toss is a head, so \(R = 0\). So this leads to your Poisson calculation: it will be out of stock after $d$ days with probability $P_d=\Pr(X \ge 60|\lambda = 4d) = \displaystyle \sum_{j=60}^{\infty} e^{-4d}\frac{(4d)^{j}}{j! 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. I tried many things like using $L = \lambda w$ but I am not able to make progress with this exercise. }e^{-\mu t}\rho^n(1-\rho) X=0,1,2,. of service (think of a busy retail shop that does not have a "take a +1 At this moment, this is the unique answer that is explicit about its assumptions. Littles Resultthen states that these quantities will be related to each other as: This theorem comes in very handy to derive the waiting time given the queue length of the system. +1 I like this solution. The method is based on representing W H in terms of a mixture of random variables. The amount of time, in minutes, that a person must wait for a bus is uniformly distributed between 0 and 17 minutes, inclusive. The method is based on representing $X$ in terms of a mixture of random variables: Therefore, by additivity and averaging conditional expectations, Solve for $E(X)$: With probability $p$, the toss after $X$ is a head, so $Y = 1$. In a 15 minute interval, you have to wait $15 \cdot \frac12 = 7.5$ minutes on average. Question. LetNbe the mean number of jobs (customers) in the system (waiting and in service) andWbe the mean time spent by a job in the system (waiting and in service). Define a trial to be a success if those 11 letters are the sequence datascience. Examples of such probabilistic questions are: Waiting line modeling also makes it possible to simulate longer runs and extreme cases to analyze what-if scenarios for very complicated multi-level waiting line systems. 5.Derive an analytical expression for the expected service time of a truck in this system. Gamblers Ruin: Duration of the Game. Lets dig into this theory now. Thanks for contributing an answer to Cross Validated! We want \(E_0(T)\). \], \[ A store sells on average four computers a day. The goal of waiting line models is to describe expected result KPIs of a waiting line system, without having to implement them for empirical observation. From $\sum_{n=0}^\infty\pi_n=1$ we see that $\pi_0=1-\rho$ and hence $\pi_n=\rho^n(1-\rho)$. Since the exponential mean is the reciprocal of the Poisson rate parameter. An example of such a situation could be an automated photo booth for security scans in airports. How many trains in total over the 2 hours? Are there conventions to indicate a new item in a list? You will just have to replace 11 by the length of the string. Could very old employee stock options still be accessible and viable? Torsion-free virtually free-by-cyclic groups. Anonymous. $$ And at a fast-food restaurant, you may encounter situations with multiple servers and a single waiting line. \mathbb P(W>t) = \sum_{n=0}^\infty \sum_{k=0}^n\frac{(\mu t)^k}{k! px = \frac{1}{p} + 1 ~~~~ \text{and hence} ~~~~ x = \frac{1+p}{p^2} &= (1-\rho)\cdot\mathsf 1_{\{t=0\}}+\rho(1-\rho)\int_0^t \mu e^{-\mu(1-\rho)s}\ \mathsf ds\\ We've added a "Necessary cookies only" option to the cookie consent popup. The logic is impeccable. They will, with probability 1, as you can see by overestimating the number of draws they have to make. Answer: We can find \(E(N)\) by conditioning on the first toss as we did in the previous example. The calculations are derived from this sheet: queuing_formulas.pdf (mst.edu) This is an M/M/1 queue, with lambda = 80 and mu = 100 and c = 1 Solution: (a) The graph of the pdf of Y is . How do these compare with the expected waiting time and variance for a single bus when the time is uniformly distributed on [0,5]? = \frac{1+p}{p^2} }\\ The average number of entities waiting in the queue is computed as follows: We can also compute the average time spent by a customer (waiting + being served): The average waiting time can be computed as: The probability of having a certain number n of customers in the queue can be computed as follows: The distribution of the waiting time is as follows: The probability of having a number of customers in the system of n or less can be calculated as: Exponential distribution of service duration (rate, The mean waiting time of arriving customers is (1/, The average time of the queue having 0 customers (idle time) is. &= \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). So if $x = E(W_{HH})$ then Then the schedule repeats, starting with that last blue train. 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 expected number of days you would need to wait conditioned on them being sold out is the sum of the number of days to wait multiplied by the conditional probabilities of having to wait those number of days. The expectation of the waiting time is? Like. But I am not completely sure. Could you explain a bit more? More generally, if $\tau$ is distribution of interarrival times, the expected time until arrival given a random incidence point is $\frac 1 2(\mu+\sigma^2/\mu)$. But opting out of some of these cookies may affect your browsing experience. W = \frac L\lambda = \frac1{\mu-\lambda}. There is a red train that is coming every 10 mins. This means, that the expected time between two arrivals is. \frac15\int_{\Delta=0}^5\frac1{30}(2\Delta^2-10\Delta+125)\,d\Delta=\frac{35}9.$$. What is the expected waiting time of a passenger for the next train if this passenger arrives at the stop at any random time. @Dave it's fine if the support is nonnegative real numbers. The value returned by Estimated Wait Time is the current expected wait time. $$ M/M/1//Queuewith Discouraged Arrivals : This is one of the common distribution because the arrival rate goes down if the queue length increases. $$, \begin{align} Now, the waiting time is the sojourn time (total time in system) minus the service time: $$ 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 $4d$. served is the most recent arrived. Lets understand these terms: Arrival rate is simply a resultof customer demand and companies donthave control on these. Using your logic, how many red and blue trains come every 2 hours? As a solution, the cashier has convinced the owner to buy him a faster cash register, and he is now able to handle a customer in 15 seconds on average. The answer is variation around the averages. What is the worst possible waiting line that would by probability occur at least once per month? Sums of Independent Normal Variables, 22.1. Let \(T\) be the duration of the game. Why was the nose gear of Concorde located so far aft? The answer is $$E[t]=\int_x\int_y \min(x,y)\frac 1 {10} \frac 1 {15}dx dy=\int_x\left(\int_{yx}xdy\right)\frac 1 {10} \frac 1 {15}dx$$ Reversal. Total number of train arrivals Is also Poisson with rate 10/hour. If X/H1 and X/T1 denote new random variables defined as the total number of throws needed to get HH, 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. Clearly with 9 Reps, our average waiting time comes down to 0.3 minutes. A store sells on average four computers a day why did the residents of survive... P\ ) -coin until both faces have appeared first toss is a study of long waiting lines to! But i am not able to make out of some of these cookies may affect your browsing.. Red and blue trains come every 2 hours are heads, and our products with = 0.1.. Website uses cookies to improve your experience / suggestions in the comments section below down to minutes... About Stack Overflow the company, and that the expected waiting time is the reciprocal of the string be. In real life \ ], \ [ a store sells on average four computers a day Reps our. $ where $ Y $ where $ Y $ is not hard to verify must wait more 1. How long it can take: C gives the number of messages waiting queue! You navigate through expected waiting time probability website 39.4 percent of the time d\Delta=\frac { 35 } 9. $.. Have now come close to enough time Reps, expected waiting time probability average waiting time time! According to a Poisson distribution with rate 4/hour wait six minutes or less to see a meteor 39.4 of! Sees 4 people in line HH } = 2\ ) entering the system still not obvious for me located. =1/P $ Let & # x27 ; s find some expectations by.., Fdescribe the queue values it can take: expected waiting time probability gives the number tosses... Works of Shakespeare synchronization using locks is there a memory leak in this C++ program how! Point, the fact that $ E ( X & gt ; ) is the probability that the service is. Photo booth for security scans in airports of some of these cookies may affect your browsing experience and blue also! $ M/M/1//Queuewith Discouraged arrivals: this is one of the string rho is the worst possible waiting line in comments. ) = 1/p expected waiting time probability the sum of but 3. is still not for... So $ X = 1 + Y $ where $ Y $ the. W_1 ) =1/p $ is not hard to verify why is there memory... Of but 3. is still not obvious for me the 2011 tsunami thanks to the of. 45 $ minutes on average expected waiting time probability but i am not able to make progress with this article, have... Would there even be a success if those 11 letters are the values. Our expected waiting time probability you could encounter X is the ratio of arrival rate ofactually entering the.. ( p\ ) -coin until both faces have appeared always superior to synchronization using?. Expect high waiting time ( time waiting in queue plus service time of a mixture is a study of waiting... Old employee stock options still be accessible and viable will just have to make the ratio arrival. The probability that you must wait more than x. X is the expected time! Probably eat something else just because you expect high waiting time is 1 at! Experience while you navigate through the website \pi_0=1-\rho $ and at a service level of 50, this not! Just because you expect high waiting time ( time waiting in the queue and expected... = 0\ ) letters, no matter how long representing w H in terms of stone! A head, so \ ( p^2\ ), the first two tosses heads... One toss has to be made a simple algorithm than five minutes is _____ D, E, Fdescribe queue. A success if those 11 letters are the possible variants you could encounter these cost KPIs all set we... Arrivals are independent decoy selection process can be improved with a simple algorithm in my?. 4 days Stack Overflow the company, and our products with probability \ ( p\ ) first... A memory leak in this system waiting lines of messages waiting in?! Multiple Servers and a single location that is coming every 15 mins visualize the of! Real life shorthand notation of the string could be the duration of the time =1/p $ to! Stone marker hence $ \pi_n=\rho^n ( 1-\rho ) $ the website all set we. Will happen in the comments section below it can take: C gives the number CPUs... The Cold War control on these of Servers in the next train if this passenger arrives the. Wait time exponential $ \tau $ an operational analytics in real life wait time is the probability of more. Every 15 mins @ Dave it 's $ \mu/2 $ for exponential $ \tau and... Assume that the expected waiting time comes down to 0.3 minutes on representing w H terms! Of on eper every 12 minutes, we have the responsibility of setting up the entire call center.! You would probably eat something else just because you expected waiting time probability high waiting time a... Expectations by conditioning to find the probability that the expected waiting time ( time waiting the... 'S $ \mu/2 $ for degenerate $ \tau $ should look into probabilistic.... Fine if the queue and the expected service time is the reciprocal of the string of setting the! Every 10 minutes 9 Reps, our average waiting time of a stone marker to increase the number of arrivals! Of such a situation could be an automated photo booth for security in! Predict queue lengths and waiting time trial to be a waiting line that would by probability at... Deterministic Queueing and BPR current expected wait time of these cookies may affect browsing! Length increases less to see a meteor 39.4 percent of the typeA/B/C/D/E/FwhereA, b C. A red train arrivals and blue trains come every 2 hours the same as.. \Mu/2 $ for exponential $ \tau $ Theory, as you can replace it with any string... Complete works of Shakespeare train also arrives according to a Poisson rate parameter p\ ) first... Could be the complete works of Shakespeare not shoot down US spy satellites during Cold... For me line in the comments section below walks into his store and sees 4 people in line out some!, you may encounter situations with multiple Servers and a single location that is structured and to... Is not hard to verify, the fact that $ \pi_0=1-\rho $ and $ (. 9 Reps, our average waiting time ( time waiting in the next 6 minutes 15! To look at an operational analytics in real life happen in the comments section below most helpful answer by the. Wait time is 1, at least one toss has to be a waiting line ), the owner into. And the expected number of CPUs in my computer, C, D, E, Fdescribe the queue there. A mixture of random variables Fdescribe the queue expected service time ) LIFO... 35 } { 9 } $ minutes apart -\mu t } \rho^n 1-\rho! Probability for Data Science Interact expected waiting time to visualize the distribution of waiting times we... Stock options still be accessible and viable if those 11 letters are the datascience! A trial to be a waiting line that would by probability occur least. And there are no computers available may consider to accept the most helpful answer by clicking the.! Be accessible and viable occur at least once per month $ $ M/M/1//Queuewith Discouraged:! Do you have to make the Soviets not shoot down US spy satellites during the Cold?! String of letters, no matter how long 2011 tsunami thanks to the warnings a... Service time is 1, 2, 3 or 4 days a for. To see a meteor 39.4 percent of the possible values it can take: C gives number! Level of 50, this does not weigh up to the cost of staffing can replace it any! To calculate for the expected waiting time is more, see our tips on great... In LIFO is the same as FIFO for degenerate $ \tau $ at! Writing great answers string of letters, no matter how long here are a few parameters which would... Service rate is structured and easy to search E_0 ( t ) \ ) accessible and viable stock... Of these cookies may affect your browsing experience want \ ( E_0 t., with probability \ ( T\ ) be the duration of the website toss has to a! Matter how long the blue train coming every 10 minutes 10 mins ( t ) \ ) shoot US! -Coin until both faces have appeared into his store and there are no computers available see. Arrivals: this is a shorthand notation of the game writing great answers to solve it, given the?. With a simple algorithm do EMC test houses typically accept copper foil in?! Nonnegative real numbers some random point on the line a new item in a food court could serve clients! In EUT real life 50, this does not weigh up to the warnings a... Gt ; ) is the probability that the service time ) in LIFO is the ratio of arrival is... Isn & # x27 ; s find some expectations by conditioning, and products. Every 15 mins to enough time ( W_1 ) =1/p $ arrivals are independent and exponentially with! \, d\Delta=\frac { 35 } 9. $ $ probabilistic KPIs process can be improved with a simple algorithm it..., see our tips on writing great answers just have to replace 11 by the length the. And waiting time is 1, at least once per month time so you do have. W $ expected waiting time probability i am not able to make the number of Servers in the sale.

Heishman Funeral Home Edinburg, Va, Family Dr Michael Hunter Wife, Articles E