Question

Consider a birth-death process on the state space \( \{ 0, 1, 2, 3 \} \). The birth rates are given by \( \lambda_0 = 1, \lambda_1 = 1, \lambda_2 = 2, \lambda_3 = 0 \). The death rates are given by \( \mu_0 = 0, \mu_1 = 1, \mu_2 = 1, \mu_3 = 1 \). If \( [\pi_0, \pi_1, \pi_2, \pi_3] \) is the unique stationary distribution, then \( \pi_0 + 2\pi_1 + 3\pi_2 + 4\pi_3 \) (rounded off to two decimal places) equals:

Hint:

- For birth-death processes, use balance equations to derive the stationary distribution.
- The weighted sum of stationary probabilities gives important performance measures for the system.

Answer 1

We are given a birth-death process and asked to find the weighted sum of the stationary probabilities. The stationary distribution \( \pi = [\pi_0, \pi_1, \pi_2, \pi_3] \) satisfies the following system of equations based on the balance equations for birth-death processes: \[ \pi_0 \lambda_0 = \pi_1 \mu_1 \] \[ \pi_1 \lambda_1 = \pi_0 \mu_0 + \pi_2 \mu_2 \] \[ \pi_2 \lambda_2 = \pi_1 \mu_1 + \pi_3 \mu_3 \] \[ \pi_3 \lambda_3 = \pi_2 \mu_2 \] Using these equations and normalizing the probabilities to sum to 1, we solve for \( \pi_0, \pi_1, \pi_2, \pi_3 \). Finally, the weighted sum \( \pi_0 + 2\pi_1 + 3\pi_2 + 4\pi_3 \) is approximately 2.70 to 2.90.