Question

In the service system there is one waiting place (in addition to the consumer receiving service). The instance process is a Poisson process and the service times are distributed $Exp(\mu)$ and independent. A consumer who completes a service for the first time, is not satisfied with it with probability p and then receives another service. The additional service is divided $Exp(\theta)$. If the consumer is satisfied with the service the first time, or if he ends the service the second time, he leaves the system. We will describe the process using a five-state continuous-time Markov chain: the number of consumers in the system and whether the consumer served (if any) is a first service or a second service. For uniformity, we will mark the states with 0,(1,1), (1,2), (2,1),(2,2). Write a state diagram and transition rates. .1 Write down the steady state equations. .2 Explain why the particular case $p = 0$ Merges with the private case $\theta = \infty$. .3 Solve the steady state equations for the particular case $p = 1, \theta = \mu$.

          In the service system there is one waiting place (in addition to the consumer receiving service). The instance process is a Poisson process and the service times are distributed $Exp(\mu)$ and independent. A consumer who completes a service for the first time, is not satisfied with it with probability p and then receives another service. The additional service is divided $Exp(\theta)$. If the consumer is satisfied with the service the first time, or if he ends the service the second time, he leaves the system.
We will describe the process using a five-state continuous-time Markov chain: the number of consumers in the system and whether the consumer served (if any) is a first service or a second service. For uniformity, we will mark the states with 0,(1,1), (1,2), (2,1),(2,2).
Write a state diagram and transition rates. .1
Write down the steady state equations. .2
Explain why the particular case $p = 0$ Merges with the private case $\theta = \infty$. .3
Solve the steady state equations for the particular case $p = 1, \theta = \mu$.

In the service system there is one waiting place (in addition to the consumer receiving service). The instance process is a Poisson process and the service times are distributed Exp(μ) and independent. A consumer who completes a service for the first time, is not satisfied with it with probability p and then receives another service. The additional service is divided Exp(θ). If the consumer is satisfied with the service the first time, or if he ends the service the second time, he leaves the system.
We will describe the process using a five-state continuous-time Markov chain: the number of consumers in the system and whether the consumer served (if any) is a first service or a second service. For uniformity, we will mark the states with 0,(1,1), (1,2), (2,1),(2,2).
Write a state diagram and transition rates. .1
Write down the steady state equations. .2
Explain why the particular case p = 0 Merges with the private case θ = ∞. .3
Solve the steady state equations for the particular case p = 1, θ = μ.

Added by Sara Q.

Question

Please give Ace some feedback