advanced24 min studyLesson 402 in Engineering

Series and parallel formulas need a shared mission and independence

Calculate required-path and redundant-path reliability while accounting for switching dependencies, mission duration, and the difference from repairable availability.

On this page

What you’ll understand

  • Derive series and parallel success probabilities from event logic.
  • State independence and mission assumptions before using product formulas.
  • Include a shared required element without hiding it inside redundancy.

A request that needs every dependency has a different success rule from a request that can use either of two alternatives. Series and parallel models turn those rules into simple probability expressions. Their convenience makes them easy to misuse: several highly reliable boxes on a diagram do not establish independent failure, successful switching, or enough spare capacity. Define the mission and its success logic before multiplying component figures.

Required components multiply only under the right assumption

For a series success rule, every required component must succeed during the defined mission. The exact probability is their joint success probability; a product of marginal reliabilities requires mutual independence. Serial dependency does not mean physical cabling must form a straight line. A user request needing an API, authentication service, and database can have series logic even though their network topology is complex. Probabilities must describe the same condition and horizon: mixing annual mission reliability, instantaneous availability, and per-request success creates a number with no coherent interpretation.

For independent exponential no-repair lifetimes, component survival is exp(−λt), and a required series path's rates add. A rate has inverse-time units; survival is dimensionless. This mission-survival model says no component failed before time t and is different from the fraction of time a repairable service is working. A component that fails and is repaired during a day may be available at the day's end while having failed the no-failure mission. State which question the model answers instead of labeling all probabilities uptime.

Rseries(t)=∏iRi(t),Ri(t)=e−λit⇒Rseries(t)=e−t∑iλiR_{\rm series}(t)=\prod_i R_i(t),\qquad R_i(t)=e^{-\lambda_i t}\Rightarrow R_{\rm series}(t)=e^{-t\sum_i\lambda_i}
The product assumes mutually independent component survival events over the same mission. The exponential specialization assumes constant failure hazards and no repair; λ_i is inverse time and t uses the matching time unit.↔ Scroll wide equations horizontally.

Alternative paths work only if an alternative really serves

For an ideal parallel success rule, the system succeeds if at least one required alternative succeeds. Its failure event is the intersection of all alternative failures. Independent alternatives give one minus the product of their failure probabilities. The model assumes that a surviving alternative has the capacity, data, routing, and activation needed to serve. A standby that cannot take load in time is not a successful alternative for a short user deadline. Failover logic can itself be a required component or a conditional activation probability.

Shared dependencies belong outside the independent-alternative product unless a richer conditional model justifies another placement. Two compute replicas using the same failing credential, regional network, or configuration are not automatically independent service paths. A load balancer required by both paths can cap the combined success probability. Redundancy may also introduce complexity and synchronization failure modes. Compare the complete modeled service success rule, not just the component-level improvement achieved by adding a second worker. The next lesson makes common causes explicit rather than hiding them in an optimistic product.

Rparallel=1−∏i(1−Ri),Rwith shared=RsRparallelR_{\rm parallel}=1-\prod_i(1-R_i),\qquad R_{\rm with\ shared}=R_sR_{\rm parallel}
The first expression assumes mutually independent alternative success events and usable surviving capacity. The second additionally assumes the shared required element's success is independent of the alternatives; all probabilities use the same mission definition.↔ Scroll wide equations horizontally.

Use sensitivity to choose what to investigate

A required shared component can dominate an otherwise excellent redundant subsystem. Improving a nonbinding leaf may yield less user benefit than removing a common dependency or making failover credible. Compute alternatives with explicit assumptions, then vary uncertain inputs and inspect whether the decision changes. The arithmetic is most useful as a model comparison and a prompt for evidence collection. Many decimal places do not establish the rare-event accuracy of estimated component figures or prove a future service objective.

The binomial lab shows independent identical alternatives. Reset, explicitly set BOTH p=0.9 and n=2, predict all-success probability 0.81 and at-least-one probability 0.99, then Play or Step. Reset and re-set BOTH p=0.9 and n=3; predict 0.729 and 0.999. Edits retain t; playback only highlights exact probability bars. The same distribution supports different series or parallel questions depending on which counts qualify. It includes no repairs, shared failures, switching delay, capacity saturation, or actual dependency graph.

Add a shared required component explicitly

Two independent alternatives have mission reliabilities 0.9 and 0.8. A shared required gateway succeeds with probability 0.99, independently of both. Assume either surviving alternative can serve the full mission.

  1. If both alternatives are required, their series success probability is 0.9×0.8=0.72 before including the gateway.

  2. If either can serve, both fail with probability 0.1×0.2=0.02, so ideal parallel success is 0.98.

  3. The gateway is still required: total success becomes 0.99×0.98=0.9702. Redundant alternatives do not remove that shared requirement.

The result

The full idealized redundant design succeeds with probability 0.9702, not 0.98 and not a guarantee derived from replica count alone.

Explore the model

Interactive lab

Interactive lab · probability

One distribution, two different success rules

For EACH run Reset, set BOTH p and n, predict, then Play or Step. With p=0.9, compare n=2 and n=3. All-success masses are 0.81 and 0.729; at-least-one masses are 0.99 and 0.999. Exact independent trials contain no common cause, repair, switching, or capacity model.

One distribution, two different success rulesThese are exact binomial probabilities for independent trials with one fixed success probability. Playback tours the distribution; it does not resample outcomes. Current readings: Mean · np: 5; Variance · np(1−p): 2.5; Total probability: 1. All parameters are adjustable with the labeled controls below.03581000.070.150.220.3Successes kP(X = k)k = 0 · P = 9.77e-4
Exact probabilityHighlighted outcomeMean np
Mean · np
5
Variance · np(1−p)
2.5
Total probability
1
t = 0 s

These are exact binomial probabilities for independent trials with one fixed success probability. Playback tours the distribution; it does not resample outcomes. 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 forReliability depends on the required success rule as well as component probabilities and dependence assumptions.

Check your understanding

Which condition is essential before applying 1−(1−R1)(1−R2) to two alternative service paths?

Try it yourself

Two independent no-repair components have constant failure hazards 0.001 and 0.002 per hour and are both required. Calculate survival for 100 hours and explain why this is not repairable uptime fraction.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Add hazards for the required exponential path and keep the exponent dimensionless.

Compare with the solution

The combined hazard is 0.003 per hour, so survival is exp(−0.003×100)=exp(−0.3)≈0.7408. It is the probability of no mission-ending failure over 100 hours; repairs and time spent operational after repairs are absent.

Take it with you

  • Series and parallel models follow explicit joint success rules.
  • Product formulas require compatible horizons and independence assumptions.
  • Shared requirements and usable failover belong in the full service model.

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.