Question

Consider two machines. Machine \(i\) operates for an exponential time with rate \(\lambda_{i}\) and then fails; its repair time is exponential with rate \(\mu_{i}, i=1,2 .\) The machines act independently of each other. Define a four-state continuous-time Markov chain that jointly describes the condition of the two machines. Use the assumed independence to compute the transition probabilities for this chain and then verify that these transition probabilities satisfy the forward and backward equations.

Short answer

To jointly model the condition of two machines, we create a four-state continuous-time Markov chain with states A (both operating), B (Machine 1 operating, Machine 2 failed), C (Machine 1 failed, Machine 2 operating), and D (both failed). The machines act independently, allowing us to compute the transition probabilities. For example, the transition rate from state A to B is given by \(\lambda_2\), while the rate from state B to A is \(\mu_2\). Using the transition rates, we form the forward and backward equations and verify that they hold for all possible transitions. Thus, the transition probabilities satisfy both forward and backward equations.

Step-by-step solution

Define the state space of the Markov chain

First, we define the state space of the continuous-time Markov chain. Since we have two machines and each machine can be in one of two states (operating or failed), there are a total of 4 possible states, which can be defined as follows: 1. State A: Both machines are operating 2. State B: Machine 1 is operating and Machine 2 is failed 3. State C: Machine 1 is failed and Machine 2 is operating 4. State D: Both machines are failed

Define the transition rates between these states

To find the transition rates between the states, we can use the fact that the machines act independently. The transition rates are given by: 1. Transition from state A to B: Machine 1 continues operating, and Machine 2 fails. Therefore, the rate is \(\lambda_{2}\). 2. Transition from state A to C: Machine 1 fails, and Machine 2 continues operating. Therefore, the rate is \(\lambda_{1}\). 3. Transition from state B to A: Machine 1 continues operating, and Machine 2 is repaired. Therefore, the rate is \(\mu_{2}\). 4. Transition from state B to D: Machine 1 fails, and Machine 2 remains failed. Therefore, the rate is \(\lambda_{1}\). 5. Transition from state C to A: Machine 2 continues operating, and Machine 1 is repaired. Therefore, the rate is \(\mu_{1}\). 6. Transition from state C to D: Machine 2 fails, and Machine 1 remains failed. Therefore, the rate is \(\lambda_{2}\). 7. Transition from state D to B: Machine 1 is repaired, and Machine 2 remains failed. Therefore, the rate is \(\mu_{1}\). 8. Transition from state D to C: Machine 2 is repaired, and Machine 1 remains failed. Therefore, the rate is \(\mu_{2}\).

Verify that the transition probabilities satisfy the forward and backward equations

Now we form the forward and backward equations to show that the system is consistent: Forward equations: \[\frac{dp_{AB}(t)}{dt} = -\lambda_{2} p_{AB}(t) + \mu_{2} p_{BA}(t)\] \[\frac{dp_{BA}(t)}{dt} = \lambda_{2} p_{AB}(t) - \mu_{2} p_{BA}(t)\] \[\frac{dp_{AC}(t)}{dt} = -\lambda_{1} p_{AC}(t) + \mu_{1} p_{CA}(t)\] \[\frac{dp_{CA}(t)}{dt} = \lambda_{1} p_{AC}(t) - \mu_{1} p_{CA}(t)\] Backward equations: \[\frac{dp_{BD}(t)}{dt} = -\lambda_{1} p_{BD}(t) + \mu_{1} p_{DB}(t)\] \[\frac{dp_{DB}(t)}{dt} = \lambda_{1} p_{BD}(t) - \mu_{1} p_{DB}(t)\] \[\frac{dp_{CD}(t)}{dt} = -\lambda_{2} p_{CD}(t) + \mu_{2} p_{DC}(t)\] \[\frac{dp_{DC}(t)}{dt} = \lambda_{2} p_{CD}(t) - \mu_{2} p_{DC}(t)\] Since the forward and backward equations hold for all possible transitions, the transition probabilities satisfy both the forward and backward equations.