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Chapter 7: Problem 43

Consider a renewal process having interarrival distribution \(F\) such that $$ F(x)=\frac{1}{2} e^{-x}+\frac{1}{2} e^{-x / 2}, \quad x>0 $$ That is, interarrivals are equally likely to be exponential with mean 1 or exponential with mean 2 . (a) Without any calculations, guess the equilibrium distribution \(F_{e}\) (b) Verify your guess in part (a).

Short Answer

Expert verified
Our guess for the equilibrium distribution, \(F_e(x)\), is a combination of two exponential distributions with the same means 1 and 2: \[ F_e(x) = a e^{-x} + b e^{-x/2}, x > 0 \] After verifying our guess using the renewal equation and solving for coefficients, we find the equilibrium distribution to be: \[ F_e(x) = \frac{1}{2} e^{-x} + \frac{1}{2} e^{-x/2}, \quad x > 0 \]

Step by step solution

01

Understanding Equilibrium Distribution

Equilibrium distribution represents the long-term behavior of the renewal process when the initial transient behavior has disappeared. In essence, it characterizes the steady-state probability distribution over the states.
02

Guess the Equilibrium Distribution

Given the interarrival distribution F(x): \[ F(x) = \frac{1}{2} e^{-x} + \frac{1}{2} e^{-x/2}, x > 0 \] We can see that interarrivals have equal probabilities of having exponential distribution with mean 1 or mean 2. Therefore, we can make an educated guess that the equilibrium distribution F_e should have the same property, i.e., the form of F_e should also be a combination of two exponential distributions with the same means: \[ F_e(x) = a e^{-x} + b e^{-x/2}, x > 0 \] Here, we still need to verify our guess and find the correct coefficients `a` and `b`.
03

Verify the Guess

We know that for a renewal process, the equilibrium distribution can be found using the renewal equation: \[ F_e(x) = F(x) + \int_{0}^{x} F_e(x-u) \,dF(u) \] Let's plug our guess for F_e(x) and F(x) into this equation: \[ a e^{-x} + b e^{-x/2} = \frac{1}{2}e^{-x} + \frac{1}{2}e^{-x/2} + \int_{0}^{x} (a e^{-(x-u)} + b e^{-(x-u)/2}) \, d\left(\frac{1}{2} e^{-u} + \frac{1}{2} e^{-u/2}\right) \] Differentiating both sides with respect to `x`, we get: \[ -a e^{-x} - \frac{1}{2}b e^{-x/2} = (a e^{-x} + b e^{-x/2}) \left(-\frac{1}{2} e^{-x} - \frac{1}{4} e^{-x/2}\right) \] Now, we can compare the coefficients of \(e^{-x}\) and \(e^{-x/2}\) terms on both sides of the equation: For \(e^{-x}\) terms: \[ -a = -\frac{1}{2}a - \frac{1}{4}b \] For \(e^{-x/2}\) terms: \[ -\frac{1}{2}b = -\frac{1}{2}a - \frac{1}{4}b \] Solving the simultaneous equations, we find the values for `a` and `b`: \[ a = \frac{1}{2}, \quad b = \frac{1}{2} \] Now, substituting the values of `a` and `b`, we have the equilibrium distribution: \[ F_e(x) = \frac{1}{2} e^{-x} + \frac{1}{2} e^{-x/2}, \quad x > 0 \] Our guess for the equilibrium distribution was correct, and we have verified it mathematically.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Equilibrium Distribution
The concept of an equilibrium distribution revolves around understanding the long-term, steady behavior of a process after the initial transient dynamics have settled. In cases like the renewal process, once we've waited enough time after 'renewals' (or events), the process tends to stabilize into an equilibrium. This steady state probability distribution does not change over time. So basically, it tells us how likely it is to be in any given state after the process has evolved for a long time.
The equilibrium distribution is often linked with the balance between arrivals and completions of events in systems. To find the equilibrium distribution, we usually need a renewal equation, a powerful tool that helps us formalize this steady state scenario.
Exponential Distribution
The exponential distribution is key in modeling interarrival times for stochastic processes, like the one described in the exercise. It is particularly popular because it's memoryless, which means that the probability of an event occurring in the future doesn't depend on when the last event occurred.
Mathematically, an exponentially distributed random variable has the probability density function (pdf) given by:
  • \( f(x; \lambda) = \lambda e^{-\lambda x} \) for \( x \geq 0 \)
where \( \lambda \) is the rate parameter, representing the number of events per unit time. In our exercise, there are two exponential distributions combined in an interarrival distribution. This combination reflects variability in the process, where arrivals might favor one rate over another at different times.
Interarrival Distribution
An interarrival distribution is pivotal in renewal processes as it captures the time between two consecutive arrivals. This is significant because it often dictates the dynamic of the overall process. In essence, understanding interarrival times helps predict the future course and behavior of a process.
In our example, the interarrival distribution is mixed, composed of an exponential with mean 1 and another with mean 2. This mix indicates that any arrival is equally likely to follow either of these mean arrival times. The distribution takes the form:
  • \( F(x) = \frac{1}{2} e^{-x} + \frac{1}{2} e^{-x/2} \)
which reflects a blend of rapid and slower arrivals, introducing variability or flexibility in process timing.
Renewal Equation
The renewal equation is a centerpiece in renewal theory, used to unravel the equilibrium distribution in such stochastic models. It provides a mathematical method to analyze and solve questions about the long-term behavior of renewal processes.
This equation is represented as:
  • \( F_e(x) = F(x) + \int_{0}^{x} F_e(x-u) \,dF(u) \)
where \( F_e(x) \) is the equilibrium distribution, and \( F(x) \) is the interarrival distribution. In essence, it helps calculate the steady-state probabilities in processes with repeated, potentially overlapping cycles of counts or arrivals. It achieves this by equating total arrivals to initial arrivals plus those arriving due to renewals.

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Most popular questions from this chapter

Consider a single-server bank for which customers arrive in accordance with a Poisson process with rate \(\lambda .\) If a customer will enter the bank only if the server is free when he arrives, and if the service time of a customer has the distribution \(G\), then what proportion of time is the server busy?

For a renewal reward process consider $$ W_{n}=\frac{R_{1}+R_{2}+\cdots+R_{n}}{X_{1}+X_{2}+\cdots+X_{n}} $$ where \(W_{n}\) represents the average reward earned during the first \(n\) cycles. Show that \(W_{n} \rightarrow E[R] / E[X]\) as \(n \rightarrow \infty\)

A system consists of two independent machines that each function for an exponential time with rate \(\lambda .\) There is a single repairperson. If the repairperson is idle when a machine fails, then repair immediately begins on that machine; if the repairperson is busy when a machine fails, then that machine must wait until the other machine has been repaired. All repair times are independent with distribution function \(G\) and, once repaired, a machine is as good as new. What proportion of time is the repairperson idle?

There are three machines, all of which are needed for a system to work. Machine \(i\) functions for an exponential time with rate \(\lambda_{i}\) before it fails, \(i=1,2,3 .\) When a machine fails, the system is shut down and repair begins on the failed machine. The time to fix machine 1 is exponential with rate \(5 ;\) the time to fix machine 2 is uniform on \((0,4) ;\) and the time to fix machine 3 is a gamma random variable with parameters \(n=3\) and \(\lambda=2 .\) Once a failed machine is repaired, it is as good as new and all machines are restarted. (a) What proportion of time is the system working? (b) What proportion of time is machine 1 being repaired? (c) What proportion of time is machine 2 in a state of suspended animation (that is, neither working nor being repaired)?

Consider a renewal process \(\\{N(t), t \geqslant 0\\}\) having a gamma \((r, \lambda)\) interarrival distribution. That is, the interarrival density is $$ f(x)=\frac{\lambda e^{-\lambda x}(\lambda x)^{r-1}}{(r-1) !}, \quad x>0 $$ (a) Show that $$ P[N(t) \geqslant n]=\sum_{i=n r}^{\infty} \frac{e^{-\lambda t}(\lambda t)^{i}}{i !} $$ (b) Show that $$ m(t)=\sum_{i=r}^{\infty}\left[\frac{i}{r}\right] \frac{e^{-\lambda t}(\lambda t)^{i}}{i !} $$ where \([i / r]\) is the largest integer less than or equal to \(i / r\). Hint: Use the relationship between the gamma \((r, \lambda)\) distribution and the sum of \(r\) independent exponentials with rate \(\lambda\) to define \(N(t)\) in terms of a Poisson process with rate \(\lambda\).
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