- #1
dmatador
- 120
- 1
Consider two machines, both of which have an exponential lifetime with mean 1/λ. There is one repairman who can service machines at an exponential rate of μ.
How does one set up the Kolmogorov backward equations for such a scenario? I am not sure after finding the rates how to work those into the formula, or turn them into the formula, rather.
How does one set up the Kolmogorov backward equations for such a scenario? I am not sure after finding the rates how to work those into the formula, or turn them into the formula, rather.