# Tag Info

## Hot answers tagged queueing

20

This is one of the most instructive and fun kinds of simulation to perform: you create independent agents in the computer, let them interact, keep track of what they do, and study what happens. It is a marvelous way to learn about complex systems, especially (but not limited to) those that cannot be understood with purely mathematical analysis. The best ...

14

The Poisson process involves a "memoryless" waiting time until the arrival of the next customer. Suppose the average time from one customer to the next is $\theta$. A memoryless continuous probability distribution until the next arrival is one in which the probability of waiting an additional minute, or second, or hour, etc., until the next arrival, does ...

13

There are a lot of moving parts in this problem, which makes it ripe for simulation. First off, as Elvis mentioned in the comments, it seems like Stacey should take about 16 appointments, as each one is about half an hour. But you know that as the appointments start to get delayed, things start shifting later and later - so if Stacey is going to only start ...

12

The exponential distribution has the property of being "memoryless," i.e. (using your analogy) the length of your journey so far has no effect on the length of the remaining journey. If the density of distribution decays faster than that of the exponential distribution, then a longer journey will mean a shorter remaining journey; conversely, a density that ...

8

I propose taking, as a point of departure, the concept of a homogeneous Poisson process. This is a point process on the line (often thought of, and referred to, as a "time" line). The realizations are sets of points. Almost surely, any bounded set of real numbers will contain only finitely many points. The fundamental properties enjoyed by this process, ...

7

The answer by bnaul gives the general property you are looking for. Rather than a Weibull distribution though, I'd suggest the Pareto distribution as the best example. The general pdf is $$f(x) = \frac{\alpha x_m}{x^{\alpha - 1}}$$ with support $[x_m, \infty)$ and $\alpha>0$. This has the nice property that conditional on $x>y$, the distribution has ...

5

I am not going to wade through the initial queueing model, so I'll just take your word for it that you want to find the expected value of $e^{vS}$ where $S$ is an exponential random variable. This being the case, what you are looking for here is the moment generating function of an exponential random variable. From what you say in your comments, it appears ...

4

It's a mixture of three distributions, and can be found pretty easily by brute force, if one allows oneself to handwave over some important details (e.g., "is $\lambda < \mu$"). Let $n$ be the number of customers in the system; $n=0$ means no-one is being served or waiting, $n=1$ means one customer is being served but no-one is waiting, etc. Let $p_n$ ...

4

Pretty much any intro to queuing theory or stochastic processes book will cover this, e.g., Ross, Stochastic Processes, or the Kleinrock, Queuing Theory. For an outline of a proof that memoryless arrivals lead to an exponential dist'n: Let G(x) = P(X > x) = 1 - F(x). Now, if the distribution is memoryless, G(s+t) = G(s)G(t) i.e., the probability ...

3

$\newcommand{\E}{\mathbb{E}}$If I'm understanding your question correctly, these are the step-by-step solutions: $$\E[T]=\sum_{i=0}^n p_i \E[T|I=i]=\sum_{i=0}^n p_i α_i$$ and $$\operatorname{Var}[T]=\E[T^2]-\E^2[T]=\sum_{i=0}^n p_i \E[T^2|I=i]-\Big[\sum_{i=0}^n p_i α_i \Big]^2$$ $$=2 \sum_{i=0}^n p_i α_i^2 - \Big[\sum_{i=0}^n p_i α_i\Big]^2$$ Basically, you ...

3

What you are struggling with is part of the derivation of the Takacs integrodifferential equation. The derivation of the expression you are trying to understand starts with: $$P_w(t+\Delta t) = (1-\lambda \Delta t)P_{w+\Delta t}(t) + \dots$$ where the $w$ represents a generic waiting time. This expression says that (part of) the probability that the ...

3

1) Yes, the mean of the service time distribution is just the mean of the Gamma(2,$\lambda)$ distribution. 2) The traffic intensity of the system is the arrival rate / the service rate, in this case: $$\rho = \lambda \mu / 2$$. These questions are a bit simplistic, but perhaps the intent of the first is to get you away from thinking of the mean service ...

3

(This is an expansion on whuber's comment.) Your data are arrival times, which are commonly summarized by the waiting time distribution (i.e. time difference between consecutive arrivals). For example, a simple single-number summary might be the coefficient of variation of the arrival times. For the standard "memoryless" null model, the arrival time PDF ...

3

If $X$ has $k$ distinct values, each appearing with equal probability $1/k$, then it follows discrete uniform distribution. Probability of drawing a specific $x$ in a single draw is $$\Pr(X=x) = \frac{1}{k}$$ probability of drawing value other than $x$ is $$\Pr(X \ne x) = \frac{k-1}{k} = 1-\frac{1}{k}$$ so if you make $n$ draws, probability that in ...

2

Unless I'm misunderstanding things, this is a fairly typical queueing problem. Two software systems that could allow you to simulate this easily are SimPy (Python) and Simulink (Matlab). I'm not sure of an R package for this, though it wouldn't be too hard to create a model for regular arrivals. Bursty arrivals are a little more work to handle, but real ...

2

In this situation, queue length will depend on the number of processors for queued events. Assuming there is one server: Events arrive at $N$ events per hour: so $\lambda =\frac{1}{N}$ Average processing time of $t$: so $\mu = \frac{1}{t}$ the probability, $p$, that the server will be busy is therefore: $\frac{t}{N}$ We can use the properties of the ...

2

Each time you grab a marble you have a 1 out of 100 chance of selecting the correct one. What you are asking is what are the odds that at least one of these will be correct which would be an inclusive "or" statement $$P_{selected} = p \cup p \cup p \cup p \cup p... \cup p$$ However this is not trivial to calculate since you could select your marble ...

2

Note that this is NOT an attempt to fully answer the problem, but to show how to overcome the lack of the Markov property for a special case that may not apply - one that is far too long to put in comments. Unfortunately, as you have realized, this is not a Markov process, but a semi-Markov process. If you a) have integer $k_+$ and $k_-$, and b) are ...

2

The question concerns a queue of infinite capacity in which people are served as soon they arrive, the so-called $M/M/\infty$ queue. All servers work independently. We are interested in keeping track of how many people are in the queue at any time $t.$ Let this be $X(t),$ a random variable that varies with $t.$ In order to appreciate what's going on, it ...

2

I will illustrate how this 'discretization' of a continuous-time Markov process can be done, using a process that is a little simpler than your queue. Continuous-time Markov process. Consider a continuous time Markov process roughly modeling decay of performance of a machine and rate of its repair. States are $S = \{1,2,3\},$ where 1 = poor, 2 = fair, 3 = ...

1

Here's the shove. We can agree that the number of job requests $Y_t$ has probability generating function (pgf) $\Phi_Y(s)$ regardless of $t$, right? Then obtain $\Phi_{q,t+1}(s)=\text{E}[s^{X_{t+1}}]$, the pgf of the number of jobs in the queue at $t+1$ as follows using the system equations you provide, where it is understood that $\Phi_{q,t}(0)=\text{Prob}[... 1 This problem seems ideally suited to Bayesian Decision Theory as it is essentially a gambling problem. It is a case where the differences between Frequentist and Bayesian methods are important. Bayesian methods have the statistical property that they are coherent, whereas Frequentist methods are not coherent. A probability is coherent if a bookie, which ... 1 +1 to @geomatt22 and whuber for their comments and suggestions. Their suggestions work when the information (jumps, first differences, etc.) is not extreme valued. A recent paper by Lin and Tegmark Critical Behavior from Deep Dynamics: A Hidden Dimension in Natural Language (available here ... https://ai2-s2-pdfs.s3.amazonaws.com/5ba0/... 1 Here's the situation for the first arrival time. Let$T_1$be the first arrival time. Sorry for making this capitalized...I like random variables to be capital letters. Pick any$t_1 < t$, where$t$is the end time. Also let$\Lambda(t') = \int_0^{t'}\lambda(s) ds$,$p = \Lambda(t_1) / \Lambda(t)$, and$r = s-1. \begin{align*} P(T_1 \le t_1 | N(t) = n)... 1 Conditioned onX_n = i$, the distribution of$X_{n+1}$is determined by the number of customers that arrive between the$n^{\mathrm{th}}$and$(n+1)^{\mathrm{th}}$departures, and is independent of$\sigma(X_0,\ldots, X_{n-1})$, so the Markov property holds.$\{X_n:n\geqslant 0\}$is not a birth-death chain, however. For example, the probabilities$\mathbb ...

1

Making comment into answer as requested by OP: If the minimum service time is very small compared to the mean, the results may still be somewhat useful, but in many cases a better approximation to service time might be a shifted-exponential. As to when that (or indeed any other) approximation may be useful, that depends on what aspects of the system you'...

1

Let $M=100$. Let's say you make $N$ selections and you are looking for the occurrence of marble $m$. Let $E_n$ be the event that marble $m$ is drawn at draw $n$. Let $F_n=E_n^c$ be the event that marble $m$ is NOT drawn. Note that $P(E_n)=1/M$. This means, $P(F_n)=1-1/M$. Let $F$ be the failure event that marble $m$ is not drawn for any $N$. Success is ...

1

Since the clerk's service time is 90 seconds, this means that he has an average rate of 40 customers per hour. Therefore $\mu = 40$ and so the notes might have made a typo.

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