Availability rates and step probabilities are different model inputs
Derive two-state repairable availability, distinguish continuous-time rates from discrete transitions, and inspect transient and degenerate behavior.
On this page
The question for today
What you’ll understand
- Calculate steady and transient availability for an exponential failure-repair model.
- Keep inverse-time transition rates separate from dimensionless step probabilities.
- Explain when a two-state Markov model omits age, queues, or repair phases.
A component can fail and recover many times while a service continues operating. Availability asks whether it is working at a chosen time or what fraction of time it works, rather than whether it survived a mission without failure. A two-state model is a useful starting point if the transition assumptions are explicit. Its small size makes every rate and probability inspectable, but also makes the omitted repair history easy to overlook.
Continuous-time rates describe exponential holding times
Let an up service fail at constant rate λ and a down service recover at constant rate μ. These are nonnegative inverse-time quantities, not probabilities. The continuous-time Markov model assumes exponential state holding times with no memory of elapsed time in the state. With positive rates, mean up duration is 1/λ and mean down duration 1/μ. A repair process with staged diagnosis, a scheduled human handoff, or an aging hazard may not fit that assumption. Adding states or using a non-Markov renewal model can be more appropriate than assigning one convenient average rate.
The stationary up probability is μ/(λ+μ), while transient availability also depends on the starting state. Starting up gives a decaying approach from one to the stationary value. The stationary result is a long-run model property, not a promise about every finite interval or every customer. If both rates are zero there is no unique stationary distribution and the initial state remains fixed. If exactly one is zero, an absorbing state changes the interpretation. Keep these cases visible rather than dividing by zero or announcing that every two-state model necessarily mixes.
Sampling a continuous process is not copying its rates into sliders
A discrete-time chain instead assigns a probability p of moving from up to down in one step and q of moving back. Its stationary up probability q/(p+q), when defined, resembles the continuous formula but uses different inputs. For an exact sampled two-state continuous process at interval Δ, the transition probabilities include the possibility of several transitions inside the interval. Simply setting p=λΔ and q=μΔ is only a small-step approximation under appropriate limits and can even produce invalid probabilities for large Δ.
Discrete chains can also behave in ways that no finite-rate two-state continuous process sampled at a positive finite interval does. With p=q=1, the chain alternates forever from its specified initial state rather than converging to its stationary distribution. With p=q=0 it never moves. A displayed stationary vector alone does not establish convergence or a physical failure rate. State the observation interval, initial state, and whether the model is a sampled process, an approximation, or a discrete operational rule before interpreting its animation as time spent unavailable.
Compare state observations with the model's assumptions
Measure actual up and down intervals, transition counts, repair phases, and missing observations. Monitoring that polls once a minute can miss short outages and repairs, so observed state changes are not automatically the underlying transition events. A stable average can hide correlated incidents or changing repair policy. Fit and validate the model against the question being asked: instantaneous availability, finite-window downtime, and mission survival are different targets. Report parameter uncertainty and test whether exponential holding times and a constant environment are plausible for the chosen data.
The Markov lab is explicitly discrete. Reset, set BOTH p=0.1 and q=0.4, predict down probabilities 0.1 and 0.15 after one and two transitions, then Step twice. The stationary down value is 0.2. Reset and re-set BOTH p=q=1 to inspect alternation. Each Step advances one transition because n=floor(2t); animation seconds are not physical repair hours. Edits retain t and recompute from initial up state. No continuous rates or exponential repair durations are simulated, so do not label the sliders λ and μ.
Work it through
Compute a transient before declaring steady state
A fictional continuous-time service starts up with λ=0.1 failures per hour while up and μ=0.4 recoveries per hour while down. Assume exponential holding times and ask for availability after two hours.
The stationary availability is 0.4/(0.1+0.4)=0.8. Mean up and down durations are ten hours and 2.5 hours respectively.
At t=2 hours, the exponent is −(0.1+0.4)×2=−1, which is dimensionless.
Availability is 0.8+0.2e^(−1)≈0.8736. Starting up leaves the two-hour probability above the stationary value; it is not already exactly 0.8.
The transient availability is about 87.36 percent under this model, while its long-run up fraction is 80 percent.
The living notebook
Explore the model
Interactive lab · markov
Discrete availability with explicit steps
For EACH run Reset, set BOTH p and q, predict, then Step. With (0.1,0.4), two Steps give down probabilities 0.1 and 0.15; stationary down is 0.2. With (1,1), states alternate. n=floor(2t), edits retain t/recompute from initial state 0. This exact discrete chain is not a continuous-time failure/repair simulation.
- P(state 1) at step 0
- 0%
- Stationary P(state 1)
- 60%
- Chain behavior
- Monotone convergence
Exact finite-step probabilities for a two-state chain starting in state 0, not sampled trajectories. Playback advances two transitions per second. When p = q = 0 every distribution is stationary; when p = q = 1 the chain alternates forever. A single zero transition creates an absorbing state. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.
Look forMatching-looking stationary formulas do not make per-step probabilities interchangeable with rates per hour.
Pause & reason
Check your understanding
Your turn
Try it yourself
A discrete chain starts up with p=0.2 and q=0.3 per step. Calculate down probability after one and two steps and its stationary value. Are these numbers failure rates per second?
Take a moment to make a prediction or work it out on paper. Open a hint when you need one.
Give me a hint
Combine the second-step paths up→down and down→down; keep step count separate from physical time.
Compare with the solution
After one step down probability is 0.2. After two it is 0.8×0.2+0.2×0.7=0.30. Stationary down probability is p/(p+q)=0.4. These are probabilities, not rates per second; a physical interpretation needs an explicitly defined observation interval and model.
Keep the idea
Take it with you
- Continuous two-state availability assumes exponential failure and repair holding times.
- Discrete transition probabilities need explicit steps and are not inverse-time rates.
- Transient, stationary, periodic, and absorbing behavior require different interpretations.
Read further
Original sources and deeper explanations to follow at your own pace.
Optional progress, saved in this browser when available. Revisit any lesson at any time.
Lesson not marked complete.