Question

The following two birth and death processes (cf. Section 4, Chapter 4 ) can lor viewed as models for queueing with balking. (a) First consider a birth and death process with parameters $$ \begin{array}{ll} \lambda_{n}=\lambda q^{n}, \quad 00 \quad(n=0,1,2, \ldots) \\ \mu_{n}=\mu, & \mu>0 \\ \mu_{0}=0 \end{array} $$ (b) Let the parameters be $$ \begin{aligned} &\lambda_{n}=\frac{\lambda}{n+1}, \quad \mu_{n}=\mu \quad(n=1,2, \ldots) \\ &\mu_{0}=0 \end{aligned} $$ Determine the stationary distribution in each case.

Short answer

The stationary distributions for the two birth and death processes are: (a) Process A: $$ \pi_n = \left(\frac{\lambda}{\mu}\right)^n q^{(n-1)(n)/2} \pi_{0} \quad (n=1,2,\ldots), $$ with $$ \pi_0 \left(1 + \sum_{n=1}^{\infty} \left(\frac{\lambda}{\mu}\right)^n q^{n(n-1)/2}\right) = 1. $$ (b) Process B: $$ \pi_n = \frac{\lambda^n}{\mu^n (n+1)!} \pi_{0} \quad (n=1,2,\ldots), $$ with $$ \pi_0 \left(1 + \sum_{n=1}^{\infty} \frac{\lambda^n}{\mu^n (n+1)!}\right) = 1. $$

Step-by-step solution

Balance equations for process (a)

For n=0,1,2,... the balance equations are: $$ \begin{cases} \lambda_{0} \pi_{0} = \mu_{1} \pi_{1} \\ \lambda_{n} \pi_{n} = \mu_{n+1} \pi_{n+1} \quad (n=1,2,\ldots) \end{cases} $$ Step 2: Substitute the given parameters for process (a) into the balance equations

Substituting parameters for process (a)

Using the given \(\lambda_n = \lambda q^n\) and \(\mu_n = \mu\) for n = 0, 1, 2, ... in the balance equations, we have: $$ \begin{cases} \lambda \pi_{0} = \mu \pi_{1} \\ \lambda q^n \pi_{n} = \mu \pi_{n+1} \quad (n=1,2,\ldots) \end{cases} $$ Step 3: Solve the system of equations to find the stationary probabilities for process (a)

Solving for stationary probabilities in process (a)

From the first equation, we get: $$ \pi_{1} = \frac{\lambda}{\mu} \pi_{0}. $$ From the second equation, we get: $$ \pi_{n+1} = \frac{\lambda q^n}{\mu} \pi_n \quad (n=1,2,\ldots), $$ which can be further simplified to: $$ \pi_{n} = \frac{\lambda q^{n-1}}{\mu} \pi_{n-1} \quad (n=2,3,\ldots). $$ Using these equations iteratively, we have: $$ \pi_{n} = \frac{\lambda}{\mu} q^{n-1} \frac{\lambda}{\mu} q^{n-2} \cdots \frac{\lambda}{\mu} \pi_{0} = \left(\frac{\lambda}{\mu}\right)^n q^{(n-1)(n)/2} \pi_{0} \quad (n=1,2,\ldots). $$ Since the sum of probabilities should be equal to 1, we have: $$ \pi_0 \left(1 + \sum_{n=1}^{\infty} \left(\frac{\lambda}{\mu}\right)^n q^{n(n-1)/2}\right) = 1. $$ Now, we have the stationary distribution for process (a). #Process (b)# Step 1: Write the balance equations for process (b)

Balance equations for process (b)

Similarly, for process (b), the balance equations are: $$ \begin{cases} \lambda_{0} \pi_{0} = \mu_{1} \pi_{1} \\ \lambda_{n} \pi_{n} = \mu_{n+1} \pi_{n+1} \quad (n=1,2,\ldots) \end{cases} $$ Step 2: Substitute the given parameters for process (b) into the balance equations

Substituting parameters for process (b)

Using the given \(\lambda_n = \frac{\lambda}{n+1}\) and \(\mu_n = \mu\) for n = 1, 2, ... in the balance equations, we have: $$ \begin{cases} \lambda \pi_{0} = \mu \pi_{1} \\ \frac{\lambda}{n+1} \pi_{n} = \mu \pi_{n+1} \quad (n=1,2,\ldots) \end{cases} $$ Step 3: Solve the system of equations to find the stationary probabilities for process (b)

Solving for stationary probabilities in process (b)

From the first equation, we get: $$ \pi_{1} = \frac{\lambda}{\mu} \pi_{0}. $$ From the second equation, we get: $$ \pi_{n+1} = \frac{\lambda}{\mu(n+1)} \pi_n \quad (n=1,2,\ldots). $$ Using these equations iteratively, we have: $$ \pi_{n} = \frac{\lambda}{\mu(n+1)} \frac{\lambda}{\mu n} \cdots \frac{\lambda}{\mu 2} \pi_{0} = \frac{\lambda^n}{\mu^n (n+1)!} \pi_{0} \quad (n=1,2,\ldots). $$ Since the sum of probabilities should be equal to 1, we have: $$ \pi_0 \left(1 + \sum_{n=1}^{\infty} \frac{\lambda^n}{\mu^n (n+1)!}\right) = 1. $$ Now, we have the stationary distribution for process (b).