Question

A single repairperson looks after both machines 1 and \(2 .\) Each time it is repaired, machine \(i\) stays up for an exponential time with rate \(\lambda_{i}, i=1,2 .\) When machine \(i\) fails, it requires an exponentially distributed amount of work with rate \(\mu_{i}\) to complete its repair. The repairperson will always service machine 1 when it is down. For instance, if machine 1 fails while 2 is being repaired, then the repairperson will immediately stop work on machine 2 and start on \(1 .\) What proportion of time is machine 2 down?

Short answer

The proportion of time machine 2 is down can be calculated using the formula: \(p_{2\_down} = \frac{\lambda_{2}(\mu_{1} + \lambda_{1})}{(\lambda_{1} + \lambda_{2})(\mu_{1} + \mu_{2} - \lambda_{2})}\). Plug in the given values for λ₁, λ₂, μ₁, and μ₂ to find the proportion of downtime for machine 2.

Step-by-step solution

Identify possible system states

In this problem, there are four possible scenarios in which machines 1 and 2 can be down or up: 1. Both machines are up (State UU). 2. Machine 1 is down and machine 2 is up (State DU). 3. Machine 1 is up and machine 2 is down (State UD). 4. Both machines are down (State DD).

Set up rate equations for each state

The balance of flow into and out of each state can be described by rate equations. In these equations, p_UU, p_DU, p_UD, and p_DD represent the steady-state probabilities of states UU, DU, UD, and DD, respectively. Rate equations: 1. State UU: \( (\lambda_{1} + \lambda_{2}) p_{UU} = \mu_{1} p_{DU} \) 2. State DU: \( \lambda_{1} p_{DU} = \mu_{1} p_{UU} + \mu_{2} p_{DD} \) 3. State UD: \( \lambda_{2} p_{UD} = \mu_{2} p_{UU} \) 4. State DD: \( \lambda_{1} p_{DD} = \mu_{1} p_{UD} + \lambda_{2} p_{DU} \)

Solve the rate equations

Solving these rate equations, we can get the steady-state probabilities of each state: 1. State UU: \(p_{UU} = \frac{\mu_{1} \mu_{2}}{(\lambda_{1} + \lambda_{2})(\mu_{1} + \mu_{2} - \lambda_{2})}\) 2. State DU: \(p_{DU} = \frac{\lambda_{1} \mu_{2}}{(\lambda_{1} + \lambda_{2})(\mu_{1} + \mu_{2} - \lambda_{2})}\) 3. State UD: \(p_{UD} = \frac{\lambda_{2} (\mu_{1} - \lambda_{2})}{(\lambda_{1} + \lambda_{2})(\mu_{1} + \mu_{2} - \lambda_{2})}\) 4. State DD: \(p_{DD} = \frac{\lambda_{1} \lambda_{2}}{(\lambda_{1} + \lambda_{2})(\mu_{1} + \mu_{2} - \lambda_{2})}\)

Calculate the proportion of time machine 2 is down

Machine 2 is down either when both machines are down (State DD) or when machine 1 is up and machine 2 is down (State UD). Hence, the proportion of time machine 2 is down can be given by: Proportion of downtime for machine 2: \(p_{2\_down} = p_{UD} + p_{DD}\) Replace p_UD and p_DD with their values from Step 3: \(p_{2\_down} = \frac{\lambda_{2} (\mu_{1} - \lambda_{2})}{(\lambda_{1} + \lambda_{2})(\mu_{1} + \mu_{2} - \lambda_{2})} + \frac{\lambda_{1} \lambda_{2}}{(\lambda_{1} + \lambda_{2})(\mu_{1} + \mu_{2} - \lambda_{2})}\) Combining the terms: \(p_{2\_down} = \frac{\lambda_{2}(\mu_{1} + \lambda_{1})}{(\lambda_{1} + \lambda_{2})(\mu_{1} + \mu_{2} - \lambda_{2})}\) Thus, the proportion of time machine 2 is down can be calculated by plugging in the given values for λ₁, λ₂, μ₁, and μ₂ into the above formula.