2

Consider a manufacturing process with batches of raw materials coming in. Suppose that the interarrival times of batches are i.i.d. exponential random variables with rate λ and their processing times are i.i.d exponential random variables with rate µ. Due to financial constraints, there is only one machine which can only process one batch of raw materials at any given instance. We assume that once a batch gets processed by the machine, it leaves the system immediately.

Let X(t) denote the number of batches in the system at time t and let $T_j := inf\{t ≥ 0 : X(t) = j\}$be the hitting time to have j batches in the system,I want to derive an expression for $E_i[T_j]$ only using $(λ, µ)$

so firstly ,I can derive an expression for $E_i[T_j]$ in terms of $E_{i+1}[T_j ]$, and $E_{i-1}[T_j ]$ (and $λ$, $µ$).

but i can't solve this characteristic equation of this cubic recurrence relation.

please help me to derive an expression for $E_i[T_j]$ only using $(λ, µ)$

sakula
  • 21
  • I am confused here. Why don't you just say $t_i = E[T_i]$. You want the mean hitting time to $j$ starting at $i$? – underdisplayname Jul 26 '19 at 00:08
  • Your question needs another absorbing state other than $j$ to be able to be calculated. For example, you can assume that the queue has a maximum capacity of $N$. Then you can use gambler's ruin similar method to calculate mean hitting time. – underdisplayname Jul 26 '19 at 00:23
  • This is a discrete time queue?? – Nap D. Lover Jul 26 '19 at 16:21
  • (Usually queues are studied in continuous time, and $q_{i i+1}=\lambda$ is the rate of arrivals and the diagonal terms are given by $q_{ii}=-(\lambda +\mu)$ because rows in the generator matrix $Q$ must sum to zero) – Nap D. Lover Jul 26 '19 at 16:23
  • sorry, I REedit the question ,is it clear now? – sakula Jul 27 '19 at 02:06

1 Answers1

1

I'm going to make two assumptions:

(1)$j>i$

(2) $\mu \neq \lambda$

Let's define $\tau_i =$ time to go to $i+1$, starting from $i$.

$$ E[\tau_i] = \frac{1}{\lambda+ \mu} + \frac{\mu}{\mu + \lambda} \cdot(E[\tau_{i-1}]+ E[\tau_{i}])$$

$$ E[\tau_i] = \frac{1}{\lambda} + \frac{\mu}{\lambda} \cdot(E[\tau_{i-1}])$$

Now, let's get the general function for $E[\tau_i]$

$E[\tau_0]= \frac{1}{\lambda}$

$E[\tau_1]= \frac{1}{\lambda} + \frac{1}{\lambda}\cdot\frac{\mu}{\lambda} $

$E[\tau_2]= \frac{1}{\lambda} + \frac{1}{\lambda}\cdot(\frac{\mu}{\lambda} + (\frac{\mu}{\lambda})^2)$

$\vdots$

$E[\tau_i]= \frac{1}{\lambda} + \frac{1}{\lambda}\cdot(\frac{\mu}{\lambda} + (\frac{\mu}{\lambda})^2 + \cdots + (\frac{\mu}{\lambda})^i) = \frac{1-(\frac{\mu}{\lambda})^{i+1}}{\lambda -\mu}$

Therefore, $$E_i[T_j] = \sum_{k=i}^{j-1}\tau_k = \frac{j-i}{\lambda - \mu} - \sum_{k=i}^{j-1}\frac{1-(\frac{\mu}{\lambda})^{i+1}}{\lambda -\mu} = \frac{j-i}{\lambda - \mu} - \frac{(\frac{\mu}{\lambda})^{i+1}}{\lambda -\mu} \cdot \frac{1-(\frac{\mu}{\lambda})^{j-i}}{1-\frac{\mu}{\lambda}}$$

underdisplayname
  • 1,260