数学代写|运筹学作业代写operational research代考|The Erlang Loss Model

数学代写|运筹学作业代写operational research代考|The Erlang Loss Model

This model is one of the most important models from queueing theory. It can be described as follows. Consider a multiserver model where customer arrivals follow a Poisson process with arrival rate $\lambda$. The service station has no waiting room. There are $c$ identical servers, where $c \geq 1$. An arriving customer who finds all servers busy leaves and has no influence on the system. A customer who arrives when not all servers are busy is immediately served by one of the free servers. The customers’ service times are independent of one another and have a common exponential distribution with expected value $1 / \mu$. An important performance measure for the model is the fraction of customers whose demand is lost. This queueing model is called the Erlang loss model (or the $M / G / c / c$ queue) and was first considered by the Danish engineer A. K. Erlang in the early years of telephony (early 20th century). The Erlang distribution is also named after him. In Erlang’s situation, the service station was a telephone exchange with $c$ telephone lines (the servers), where a new incoming call (the customer) was lost if all lines were busy. Although the Erlang loss model is relatively old, it is still current and even now still has applications to various problems (see also the exercises).
To analyze the Erlang loss model, we define
$$X(t)=\text { number of busy servers at time } t .$$
The stochastic process ${X(t), t \geq 0}$ is a continuous-time Markov chain with state space $I={0,1, \ldots, c}$. The transition rate diagram is given in Figure 8.6.

The transition rate $q_{i, i-1}=i \mu$ requires some explanation. Suppose that at a given time $t$, the process is in state $i$. The probability that the process transitions to state $i-1$ during the coming interval $(t, t+\Delta t)$ is equal to the probability that one of the $i$ busy servers finishes during the interval $(t, t+\Delta t)$. For $\ell=1, \ldots, i$, define $A_{\ell}$ as the event that the $\ell$ th server finishes during the coming interval $(t, t+\Delta t)$. Based on the failure rate representation of the exponential distribution, we have $\mathbb{P}\left(A_{\ell}\right)=\mu \Delta t+o(\Delta t)$ for $\ell=1, \ldots, i$. The probability that two or more servers finish during the interval $(t, t+\Delta t)$ is of order $(\Delta t)^2$ and therefore $o(\Delta t)$. This gives
$$\mathbb{P}\left(A_1 \cup A_2 \cup \ldots \cup A_i\right)=\sum_{\ell=1}^i \mathbb{P}\left(A_{\ell}\right)+o(\Delta t)=i \mu \Delta t+o(\Delta t),$$
that is, $q_{i, i-1}=i \mu$.

数学代写|运筹学作业代写operational research代考|Recursive Calculation of the Equilibrium Probabilities

In both the single-server queue and the Erlang loss model, it is possible to calculate the equilibrium probabilities recursively. This is no coincidence. In general, for a continuous-time Markov chain, the equilibrium distribution can be calculated recursively if the state space $I$ is given by $I={0,1, \ldots, N}$ for some $N \leq \infty$ and if for every $i \in I$, the transition rates $q_{i j}$ satisfy
$$q_{i j}=0 \quad \text { for } j \leq i-2$$
In the continuous-time Markov chain formulation of many queueing systems, these conditions hold. Here is a probabilistic derivation of the recursive relation for the equilibrium probabilities. For every subset $A$ of states with $A \neq I$, the long-run average number of transitions per unit of time from inside $A$ to outside $A$ is equal to the average number of transitions per unit of time from outside $A$ to inside $A$. In other words,
rate out of $A=$ rate into $A$.
Now, choose $A$ well, namely equal to $A={i, i+1, \ldots, N}$ for a fixed $i$ with $i \neq 0$ so that $A \neq I$. The process can only leave the set of states $A$ through state $i$, which happens at a rate of $q_{i, i-1}$. From a state $k \notin A$, the process is pulled into the set of states $A$ at a rate of $\sum_{\ell \geq i} q_{k \ell}$. This therefore leads to the recursive relation
$$\pi_i q_{i, i-1}=\sum_{k=0}^{i-1} \pi_k\left(\sum_{\ell \geq i} q_{k \ell}\right) \quad \text { for } i=1,2, \ldots, N .$$
Because the balance equations uniquely determine the equilibrium probabilities up to a multiplicative constant, we can formulate the following simple algorithm:

Algorithm 8.1.
Step 1. Let $\bar{\pi}0:=1$. Step 2. Successively calculate $\bar{\pi}_1, \bar{\pi}_2, \ldots, \bar{\pi}_N$ from the recursive relation (8.13). Step 3. Normalize by setting $\pi_i:=\bar{\pi}_i /\left(\sum{k=0}^N \bar{\pi}_k\right)$ for $i=0,1, \ldots, N$.

运筹学代考

数学代写|运筹学作业代写operational research代考|The Erlang Loss Model

$$X(t)=\text { number of busy servers at time } t$$

$$\mathbb{P}\left(A_1 \cup A_2 \cup \ldots \cup A_i\right)=\sum_{\ell=1}^i \mathbb{P}\left(A_{\ell}\right)+o(\Delta t)=i \mu \Delta t+o(\Delta t),$$那是， $q_{i, i-1}=i \mu$.

数学代写|运筹学作业代写operational research代考|Recursive Calculation of the Equilibrium Probabilities

$$q_{i j}=0 \quad \text { for } j \leq i-2$$

$$\pi_i q_{i, i-1}=\sum_{k=0}^{i-1} \pi_k\left(\sum_{\ell \geq i} q_{k \ell}\right) \quad \text { for } i=1,2, \ldots, N .$$

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