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Reliability Block Diagrams
In ReliaLearnR: Learning Modules for Reliability Analysis
library(learnr)
library(DiagrammeR)
Introduction
Welcome to the tutorial on Reliability Block Diagrams (RBDs) and System Reliability! In the RAM module, you learned to calculate reliability metrics for individual components. This tutorial extends those concepts to systems, collections of components whose arrangement determines whether the system succeeds or fails.
RBDs are a graphical tool for modeling how component reliabilities combine to produce system-level reliability. They are widely used in reliability engineering to analyze designs, identify vulnerabilities, and evaluate the benefit of redundancy.
Learning Objectives
By the end of this module, learners will be able to:
- Define reliability block diagrams and relate them to physical systems.
- Compute system reliability for series, parallel, and mixed configurations.
- Interpret k-out-of-n (voting) systems and calculate their reliability.
- Describe the relationship between RBDs and Fault Tree Analysis.
- Apply RBD calculations to a realistic multi-component system.
What is a Reliability Block Diagram?
A Reliability Block Diagram represents each component in a system as a block with a known reliability value. The blocks are connected by lines that show how the components relate to each other functionally:
- A path from left (input) to right (output) through functioning blocks represents a working system.
- The RBD is not a wiring diagram. It shows logical dependencies, not physical connections.
For example, a pump system that requires a motor, a pump, and a valve all to function would be drawn as three blocks in series. A backup generator that can substitute for a primary generator would appear as two blocks in parallel.
RBDs directly connect to the RAM metrics you already know:
- Each block has a reliability $R_i(t) = e^{-\lambda_i t}$ (exponential model) or a Weibull $R_i(t)$ (see the Life Data Analysis module).
- The system reliability is derived from the block reliabilities using the formulas in this tutorial.
Series Systems
In a series system, all components must function for the system to function. This is the most common configuration, think of a chain where every link must hold.
$$R_{sys} = R_1 \times R_2 \times \cdots \times R_n = \prod_{i=1}^{n} R_i$$
Key insight: A series system is always less reliable than its weakest component. Adding more components in series can only reduce system reliability.
Example
A water pumping system requires three components to all be operational: a motor (R = 0.95), a pump (R = 0.90), and a control valve (R = 0.98).
grViz("
digraph series {
rankdir = LR
graph [bgcolor = transparent]
node [shape = rectangle, style = filled, fillcolor = '#AED6F1',
fontname = 'sans-serif', fontsize = 12, margin = '0.2,0.1']
edge [arrowsize = 0.8]
I [shape = point, width = 0.15, fillcolor = black]
A [label = 'Motor\nR = 0.95']
B [label = 'Pump\nR = 0.90']
C [label = 'Valve\nR = 0.98']
O [shape = point, width = 0.15, fillcolor = black]
I -> A -> B -> C -> O
}
")
```r
R_motor <- 0.95
R_pump <- 0.90
R_valve <- 0.98
R_series <- R_motor * R_pump * R_valve
R_series
The system reliability is approximately 83.8%, lower than any individual component.
quiz(caption = "Quiz: Series Systems",
question("Four components in series have reliabilities 0.99, 0.97, 0.95, and 0.93. What is the system reliability?",
answer("0.99", message = "That is only the most reliable component."),
answer("0.846", correct = TRUE,
message = "Correct! R_sys = 0.99 × 0.97 × 0.95 × 0.93 ≈ 0.846."),
answer("0.960", message = "That is the average, not the product."),
answer("1.00", message = "System reliability is always ≤ the weakest component."),
random_answer_order = TRUE,
allow_retry = TRUE
),
question("In a series system, which component has the greatest influence on system reliability?",
answer("The most reliable component",
message = "The weakest component limits the system, not the strongest."),
answer("The least reliable component", correct = TRUE,
message = "Correct! The least reliable component is the bottleneck — improving it gives the largest gain in system reliability."),
answer("The component with the lowest failure rate",
message = "The component with the lowest failure rate is the most reliable one — but in a series system it is the least reliable component that constrains the system."),
answer("All components equally",
message = "In a series system, the least reliable component has a disproportionate effect on system reliability."),
random_answer_order = TRUE,
allow_retry = TRUE
)
)
Calculate series reliability yourself.
# A conveyor belt system has 5 components in series.
# Component reliabilities: 0.98, 0.96, 0.99, 0.94, 0.97
# Calculate the system reliability.
# R_sys <- prod(c(0.98, 0.96, 0.99, 0.94, 0.97))
R_components <- c(0.98, 0.96, 0.99, 0.94, 0.97)
R_series <- prod(R_components)
R_series # ~0.845
Parallel Systems
In a parallel system, only one component needs to function for the system to succeed. The system fails only if all components fail simultaneously. This is called active redundancy.
$$R_{sys} = 1 - \prod_{i=1}^{n}(1 - R_i)$$
Key insight: Adding parallel components always increases system reliability. Redundancy is a powerful tool for critical systems.
Example
A backup power system has a primary generator (R = 0.90) and a standby generator (R = 0.85). Either one alone keeps the system running.
grViz("
digraph parallel {
rankdir = LR
graph [bgcolor = transparent]
node [shape = rectangle, style = filled, fillcolor = '#A9DFBF',
fontname = 'sans-serif', fontsize = 12, margin = '0.2,0.1']
edge [arrowsize = 0.8]
I [shape = point, width = 0.15, fillcolor = black]
A [label = 'Generator 1\nR = 0.90']
B [label = 'Generator 2\nR = 0.85']
O [shape = point, width = 0.15, fillcolor = black]
I -> A -> O
I -> B -> O
}
")
```r
R_primary <- 0.90
R_standby <- 0.85
R_parallel <- 1 - (1 - R_primary) * (1 - R_standby)
R_parallel
The parallel system reliability is 98.5%, much higher than either generator alone.
Use the slider to explore how the number of redundant components affects system reliability.
sliderInput("n_comp", "Number of parallel components:", min = 1, max = 8, value = 2, step = 1)
numericInput("R_ind", "Individual component reliability:", value = 0.90, min = 0.50, max = 0.999, step = 0.01)
plotOutput("parallelPlot")
output$parallelPlot <- renderPlot({
n_vals <- 1:10
Rc <- input$R_ind
R_par <- 1 - (1 - Rc)^n_vals
R_sel <- 1 - (1 - Rc)^input$n_comp
plot(n_vals, R_par, type = "b", col = "steelblue", lwd = 2, pch = 19,
xlab = "Number of redundant components",
ylab = "System Reliability",
main = "Parallel System Reliability vs. Redundancy",
ylim = c(0, 1))
points(input$n_comp, R_sel, col = "red", pch = 19, cex = 2.5)
abline(h = Rc, col = "gray50", lty = 2)
legend("bottomright",
legend = c("System reliability", paste0("n = ", input$n_comp),
paste0("Single component (", Rc, ")")),
col = c("steelblue", "red", "gray50"),
lty = c(1, NA, 2), pch = c(19, 19, NA))
})
quiz(caption = "Quiz: Parallel Systems",
question("Two components each have reliability 0.80. What is the reliability of a parallel system?",
answer("0.80", message = "That is a single component's reliability."),
answer("0.64", message = "0.64 = 0.8 × 0.8 is the series reliability."),
answer("0.96", correct = TRUE,
message = "Correct! R_sys = 1 − (1−0.80)² = 1 − 0.04 = 0.96."),
answer("1.60", message = "Reliability cannot exceed 1."),
random_answer_order = TRUE,
allow_retry = TRUE
),
question("Adding a third identical component (R = 0.80) in parallel to the previous two-component system gives a system reliability of approximately:",
answer("0.992", correct = TRUE,
message = "Correct! R_sys = 1 − (1−0.80)³ = 1 − 0.008 = 0.992."),
answer("0.96", message = "That is the two-component parallel reliability."),
answer("0.80", message = "That is a single component's reliability."),
answer("0.512", message = "0.512 = 0.8³ is three components in series, not parallel."),
random_answer_order = TRUE,
allow_retry = TRUE
)
)
Calculate parallel reliability yourself.
# A critical pump station has 3 pumps in parallel.
# Each pump has reliability 0.88.
# What is the system reliability?
# R_sys <- 1 - prod(1 - c(0.88, 0.88, 0.88))
R_components <- c(0.88, 0.88, 0.88)
R_parallel <- 1 - prod(1 - R_components)
R_parallel # ~0.9983
Mixed Systems
Most real systems combine series and parallel blocks. To analyze a mixed system, decompose it into subsystems and apply series and parallel rules step by step, working from the innermost blocks outward.
Use the selector below to compare how the topology changes across the three core configurations.
selectInput("topology", "Select topology:",
choices = c("Series", "Parallel", "Mixed"), selected = "Mixed")
DiagrammeROutput("topologyDiagram")
output$topologyDiagram <- renderGrViz({
if (input$topology == "Series") {
grViz("digraph { rankdir=LR; graph [bgcolor=transparent]
node [shape=rectangle, style=filled, fillcolor='#AED6F1', fontname='sans-serif', fontsize=12]
I [shape=point, width=0.15, fillcolor=black]
A [label='Component A\nR=0.95']; B [label='Component B\nR=0.90']; C [label='Component C\nR=0.97']
O [shape=point, width=0.15, fillcolor=black]
I->A->B->C->O }")
} else if (input$topology == "Parallel") {
grViz("digraph { rankdir=LR; graph [bgcolor=transparent]
node [shape=rectangle, style=filled, fillcolor='#A9DFBF', fontname='sans-serif', fontsize=12]
I [shape=point, width=0.15, fillcolor=black]
A [label='Component A\nR=0.95']; B [label='Component B\nR=0.90']
O [shape=point, width=0.15, fillcolor=black]
I->A->O; I->B->O }")
} else {
grViz("digraph { rankdir=LR; graph [bgcolor=transparent]
node [shape=rectangle, style=filled, fillcolor='#F9E79F', fontname='sans-serif', fontsize=12]
I [shape=point, width=0.15, fillcolor=black]
S [label='Sensor\nR=0.95']; T [label='Transmitter\nR=0.97']
A1 [label='Actuator 1\nR=0.90']; A2 [label='Actuator 2\nR=0.90']
O [shape=point, width=0.15, fillcolor=black]
I->S->T->A1->O; T->A2->O }")
}
})
Example
A safety system has:
- Subsystem A: a sensor (R = 0.95) and a transmitter (R = 0.97) in series.
- Subsystem B: two redundant actuators (each R = 0.90) in parallel.
- Subsystems A and B are in series (both must work).
# Subsystem A (series)
R_A <- 0.95 * 0.97
R_A
# Subsystem B (parallel)
R_B <- 1 - (1 - 0.90) * (1 - 0.90)
R_B
# Overall system (A and B in series)
R_system <- R_A * R_B
R_system
quiz(caption = "Quiz: Mixed Systems",
question("A system has two subsystems in series. Subsystem 1 has two components in parallel (each R = 0.85); Subsystem 2 is a single component with R = 0.92. What is the system reliability?",
answer("0.85 × 0.92 = 0.782", message = "That ignores the parallel redundancy in Subsystem 1."),
answer("[1 − (1−0.85)²] × 0.92 ≈ 0.899", correct = TRUE,
message = "Correct! R_sub1 = 1 − 0.15² = 0.9775, then R_sys = 0.9775 × 0.92 ≈ 0.899."),
answer("0.85 + 0.92 = 1.77", message = "Reliabilities cannot be added like this."),
answer("0.92", message = "That ignores Subsystem 1 entirely."),
random_answer_order = TRUE,
allow_retry = TRUE
),
question("To analyze a complex mixed RBD, the recommended approach is to:",
answer("Average all component reliabilities",
message = "Averaging does not account for the series/parallel topology."),
answer("Decompose into series and parallel subsystems, then apply the formulas step by step", correct = TRUE,
message = "Correct! Work from the innermost blocks outward, replacing each parallel or series group with its equivalent single-block reliability."),
answer("Use only the reliability of the most critical component",
message = "This would ignore the contributions of all other components."),
random_answer_order = TRUE,
allow_retry = TRUE
)
)
k-out-of-n Systems
A k-out-of-n system succeeds if at least k of its n identical components function. This generalizes series (k = n) and parallel (k = 1):
- k = n: all must work → series.
- k = 1: at least one works → parallel.
- 1 < k < n: voting or load-sharing systems.
The reliability of a k-out-of-n system with identical components (each with reliability p) follows the binomial distribution:
$$R_{k/n} = \sum_{i=k}^{n} \binom{n}{i} p^i (1-p)^{n-i} = 1 - \text{pbinom}(k-1, n, 1-p)$$
Example
A flight control system uses 3 redundant computers. At least 2 of the 3 must agree for the system to function safely (a 2-out-of-3 voter). Each computer has reliability R = 0.99.
grViz("
digraph koon {
rankdir = LR
graph [bgcolor = transparent]
node [shape = rectangle, style = filled, fillcolor = '#D7BDE2',
fontname = 'sans-serif', fontsize = 12, margin = '0.2,0.1']
edge [arrowsize = 0.8]
I [shape = point, width = 0.15, fillcolor = black]
C1 [label = 'Computer 1\nR = 0.99']
C2 [label = 'Computer 2\nR = 0.99']
C3 [label = 'Computer 3\nR = 0.99']
V [label = 'Voter\n(2-of-3)', shape = diamond, fillcolor = '#F1948A', fontsize = 11]
O [shape = point, width = 0.15, fillcolor = black]
I -> C1 -> V
I -> C2 -> V
I -> C3 -> V
V -> O
}
")
n <- 3 # total components
k <- 2 # minimum required
p <- 0.99 # individual reliability
R_voting <- 1 - pbinom(k - 1, n, 1 - p)
R_voting
The 2-out-of-3 system has reliability 0.9997.
quiz(caption = "Quiz: k-out-of-n Systems",
question("A 2-out-of-4 system has 4 identical components each with R = 0.90. What R function computes its reliability?",
answer("1 - pbinom(1, 4, 0.10)", correct = TRUE,
message = "Correct! k=2, n=4, failure probability = 1−0.90 = 0.10. Use pbinom(k−1, n, 1−p) = pbinom(1, 4, 0.10)."),
answer("pbinom(2, 4, 0.90)",
message = "This gives P(X ≤ 2 successes), not the probability that at least 2 succeed."),
answer("1 - pbinom(2, 4, 0.90)",
message = "The third argument should be the failure probability (1−p), not p."),
answer("prod(rep(0.90, 4))",
message = "That formula gives the series reliability (all 4 must work), not 2-out-of-4."),
random_answer_order = TRUE,
allow_retry = TRUE
),
question("How does a 1-out-of-n system relate to a parallel system?",
answer("They are different — a 1-out-of-n system requires only the first component to work",
message = "A 1-out-of-n system requires at least one of any n components to work, which is exactly a parallel configuration."),
answer("They are equivalent — both require only one component to function", correct = TRUE,
message = "Correct! When k = 1, the k-out-of-n formula reduces to the parallel reliability formula."),
answer("A 1-out-of-n system is less reliable than a parallel system",
message = "They are the same configuration."),
random_answer_order = TRUE,
allow_retry = TRUE
)
)
Calculate k-out-of-n reliability in R.
# A 3-out-of-5 redundant sensor array.
# Each sensor has reliability p = 0.95.
# Calculate the system reliability.
n <- 5
k <- 3
p <- 0.95
# R_sys <- 1 - pbinom(k - 1, n, 1 - p)
n <- 5
k <- 3
p <- 0.95
R_sys <- 1 - pbinom(k - 1, n, 1 - p)
R_sys # ~0.9988
System MTTF
For a series system of components with constant failure rates (exponential distribution), the system failure rate is the sum of component failure rates:
$$\lambda_{sys} = \sum_{i=1}^{n} \lambda_i \qquad MTTF_{sys} = \frac{1}{\lambda_{sys}}$$
For $n$ identical parallel components each with failure rate $\lambda$:
$$MTTF_{parallel} = \frac{1}{\lambda}\left(1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n}\right)$$
Example
Two pumps in parallel, each with $\lambda = 0.01$ failures/hour:
lambda <- 0.01
MTTF_single <- 1 / lambda
MTTF_parallel <- (1 / lambda) * (1 + 1/2)
MTTF_single # 100 hours
MTTF_parallel # 150 hours — 50% longer with one spare
quiz(caption = "Quiz: System MTTF",
question("Three components in series have failure rates 0.02, 0.03, and 0.05 failures/hour. What is the system MTTF?",
answer("10 hours", correct = TRUE,
message = "Correct! λ_sys = 0.02 + 0.03 + 0.05 = 0.10, so MTTF = 1/0.10 = 10 hours."),
answer("16.7 hours", message = "That is 1/0.06 — you may have missed one failure rate."),
answer("100 hours", message = "100 hours is the MTTF of a single component with λ = 0.01."),
answer("0.1 hours", message = "0.1 is the system failure rate, not the MTTF."),
random_answer_order = TRUE,
allow_retry = TRUE
),
question("Compared to a single component, the MTTF of two identical components in parallel is:",
answer("Twice as large",
message = "Not quite. For two identical parallel components, MTTF_parallel = (3/2) × MTTF_single — a 50% increase."),
answer("1.5 times as large", correct = TRUE,
message = "Correct! MTTF_parallel = (1/λ)(1 + 1/2) = 1.5 × MTTF_single."),
answer("The same",
message = "Redundancy always increases MTTF."),
answer("Three times as large",
message = "Three times applies to three identical parallel components using the harmonic series formula."),
random_answer_order = TRUE,
allow_retry = TRUE
)
)
Introduction to Fault Tree Analysis
Fault Tree Analysis (FTA) is a top-down approach to reliability analysis that starts with an undesirable system event (the top event) and works backward to identify the combinations of component failures that could cause it.
While an RBD asks "What must work for the system to succeed?", a fault tree asks "What can cause the system to fail?"
Gates
A fault tree uses logic gates to combine failure events:
- AND gate ($\land$): the top event occurs only if all input events occur simultaneously. This corresponds to parallel components in an RBD, all must fail to cause system failure.
- OR gate ($\lor$): the top event occurs if any input event occurs. This corresponds to series components in an RBD, any single failure causes system failure.
RBD — Fault Tree Duality
Every RBD has a corresponding fault tree and vice versa:
| RBD configuration | Fault tree gate for system failure |
|:---:|:---:|
| Series (all must work) | OR gate (any failure causes system failure) |
| Parallel (any can work) | AND gate (all must fail for system failure) |
This duality means the two tools provide complementary views of the same system. RBDs are better for computing reliability; fault trees are better for tracing failure causes and identifying critical combinations.
The fault tree below illustrates a simple two-component system: the top event (system failure) occurs if either component fails — an OR gate, which corresponds to a series RBD.
grViz("
digraph fta {
rankdir = TB
graph [bgcolor = transparent]
node [fontname = 'sans-serif', fontsize = 12, style = filled]
edge [arrowsize = 0.8]
Top [label = 'System\nFailure', shape = rectangle, fillcolor = '#E74C3C', fontcolor = white]
OR [label = 'OR Gate', shape = diamond, fillcolor = '#F39C12']
F1 [label = 'Component A\nFails', shape = ellipse, fillcolor = '#AED6F1']
F2 [label = 'Component B\nFails', shape = ellipse, fillcolor = '#AED6F1']
Top -> OR
OR -> F1
OR -> F2
}
")
For complex fault trees, the R package FaultTree on CRAN provides tools for building and analyzing fault tree models programmatically.
quiz(caption = "Quiz: Fault Tree Analysis",
question("A fault tree has an AND gate combining two component failures. This corresponds to which RBD configuration?",
answer("Series — both components in series",
message = "A series RBD means any single failure causes system failure, which maps to an OR gate in a fault tree."),
answer("Parallel — both components in parallel", correct = TRUE,
message = "Correct! An AND gate means the system fails only if both components fail — exactly the parallel (redundant) RBD configuration."),
answer("k-out-of-n with k = 2",
message = "A 2-out-of-n system is more complex than a simple AND/OR gate."),
random_answer_order = TRUE,
allow_retry = TRUE
),
question("Which statement best distinguishes FTA from RBD?",
answer("FTA models reliability; RBD models failure modes",
message = "It is the other way around — FTA is failure-oriented and RBD is success-oriented."),
answer("RBD models what must succeed for the system to work; FTA traces what can cause the system to fail", correct = TRUE,
message = "Correct! RBDs are success-space models; fault trees are failure-space models. They are logically dual and provide complementary views."),
answer("RBD applies only to series systems; FTA applies to parallel systems",
message = "Both methods apply to any system topology."),
random_answer_order = TRUE,
allow_retry = TRUE
)
)
Case Study: Industrial Cooling System
Let's apply everything to a realistic example. An industrial cooling system has the following architecture:
- Water supply subsystem: two pumps in parallel (each R = 0.92), followed by a filter in series (R = 0.99).
- Control subsystem: a primary controller (R = 0.97) and a backup controller in parallel (R = 0.95).
- The water supply and control subsystems must both work (series at system level).
# Step 1 — Water supply subsystem
R_pump_parallel <- 1 - (1 - 0.92)^2
R_water <- R_pump_parallel * 0.99
R_water
# Step 2 — Control subsystem
R_control <- 1 - (1 - 0.97) * (1 - 0.95)
R_control
# Step 3 — Overall system (both subsystems in series)
R_system <- R_water * R_control
R_system
Now explore what happens if the filter is upgraded.
# Modify the case study: the filter is upgraded to R = 0.999.
# Recalculate the system reliability with the improved filter.
R_pump1 <- 0.92
R_pump2 <- 0.92
R_filter <- 0.999 # upgraded from 0.99
R_ctrl1 <- 0.97
R_ctrl2 <- 0.95
# R_pump_parallel <- 1 - (1 - R_pump1) * (1 - R_pump2)
# R_water <- R_pump_parallel * R_filter
# R_control <- 1 - (1 - R_ctrl1) * (1 - R_ctrl2)
# R_system <- R_water * R_control
R_pump1 <- 0.92
R_pump2 <- 0.92
R_filter <- 0.999
R_ctrl1 <- 0.97
R_ctrl2 <- 0.95
R_pump_parallel <- 1 - (1 - R_pump1) * (1 - R_pump2)
R_water <- R_pump_parallel * R_filter
R_control <- 1 - (1 - R_ctrl1) * (1 - R_ctrl2)
R_system <- R_water * R_control
R_system # ~0.985
Summary
Congratulations on completing the Reliability Block Diagrams and System Reliability tutorial!
Key takeaways:
- Series: $R_{sys} = \prod R_i$ — every component must work; reliability decreases with more components.
- Parallel: $R_{sys} = 1 - \prod(1-R_i)$ — redundancy; only one component needs to work.
- Mixed: decompose into subsystems, apply formulas step by step.
- k-out-of-n:
1 - pbinom(k-1, n, 1-p) for voting / load-sharing systems.
- Series MTTF: $1/\sum\lambda_i$; Parallel MTTF of $n$ identical units: $(1/\lambda)\sum_{i=1}^{n}(1/i)$.
- FTA duality: series RBD ↔ OR gate; parallel RBD ↔ AND gate.
References
-
Abernethy, R.B. (2004) The New Weibull Handbook. Fifth Edition.
-
Aden-Buie G, Schloerke B, Allaire J, Rossell Hayes A (2023). learnr: Interactive Tutorials for R. https://rstudio.github.io/learnr/, https://github.com/rstudio/learnr.
-
Billinton R, Allan R (1992). Reliability Evaluation of Engineering Systems. Second Edition. Plenum Press, New York.
-
Vesely W, Goldberg F, Roberts N, Haasl D (1981). Fault Tree Handbook. US Nuclear Regulatory Commission, NUREG-0492.
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library(learnr) library(DiagrammeR)
Introduction
Welcome to the tutorial on Reliability Block Diagrams (RBDs) and System Reliability! In the RAM module, you learned to calculate reliability metrics for individual components. This tutorial extends those concepts to systems, collections of components whose arrangement determines whether the system succeeds or fails.
RBDs are a graphical tool for modeling how component reliabilities combine to produce system-level reliability. They are widely used in reliability engineering to analyze designs, identify vulnerabilities, and evaluate the benefit of redundancy.
Learning Objectives
By the end of this module, learners will be able to:
- Define reliability block diagrams and relate them to physical systems.
- Compute system reliability for series, parallel, and mixed configurations.
- Interpret k-out-of-n (voting) systems and calculate their reliability.
- Describe the relationship between RBDs and Fault Tree Analysis.
- Apply RBD calculations to a realistic multi-component system.
What is a Reliability Block Diagram?
A Reliability Block Diagram represents each component in a system as a block with a known reliability value. The blocks are connected by lines that show how the components relate to each other functionally:
- A path from left (input) to right (output) through functioning blocks represents a working system.
- The RBD is not a wiring diagram. It shows logical dependencies, not physical connections.
For example, a pump system that requires a motor, a pump, and a valve all to function would be drawn as three blocks in series. A backup generator that can substitute for a primary generator would appear as two blocks in parallel.
RBDs directly connect to the RAM metrics you already know:
- Each block has a reliability $R_i(t) = e^{-\lambda_i t}$ (exponential model) or a Weibull $R_i(t)$ (see the Life Data Analysis module).
- The system reliability is derived from the block reliabilities using the formulas in this tutorial.
Series Systems
In a series system, all components must function for the system to function. This is the most common configuration, think of a chain where every link must hold.
$$R_{sys} = R_1 \times R_2 \times \cdots \times R_n = \prod_{i=1}^{n} R_i$$
Key insight: A series system is always less reliable than its weakest component. Adding more components in series can only reduce system reliability.
Example
A water pumping system requires three components to all be operational: a motor (R = 0.95), a pump (R = 0.90), and a control valve (R = 0.98).
grViz(" digraph series { rankdir = LR graph [bgcolor = transparent] node [shape = rectangle, style = filled, fillcolor = '#AED6F1', fontname = 'sans-serif', fontsize = 12, margin = '0.2,0.1'] edge [arrowsize = 0.8] I [shape = point, width = 0.15, fillcolor = black] A [label = 'Motor\nR = 0.95'] B [label = 'Pump\nR = 0.90'] C [label = 'Valve\nR = 0.98'] O [shape = point, width = 0.15, fillcolor = black] I -> A -> B -> C -> O } ") ```r R_motor <- 0.95 R_pump <- 0.90 R_valve <- 0.98 R_series <- R_motor * R_pump * R_valve R_series
The system reliability is approximately 83.8%, lower than any individual component.
quiz(caption = "Quiz: Series Systems", question("Four components in series have reliabilities 0.99, 0.97, 0.95, and 0.93. What is the system reliability?", answer("0.99", message = "That is only the most reliable component."), answer("0.846", correct = TRUE, message = "Correct! R_sys = 0.99 × 0.97 × 0.95 × 0.93 ≈ 0.846."), answer("0.960", message = "That is the average, not the product."), answer("1.00", message = "System reliability is always ≤ the weakest component."), random_answer_order = TRUE, allow_retry = TRUE ), question("In a series system, which component has the greatest influence on system reliability?", answer("The most reliable component", message = "The weakest component limits the system, not the strongest."), answer("The least reliable component", correct = TRUE, message = "Correct! The least reliable component is the bottleneck — improving it gives the largest gain in system reliability."), answer("The component with the lowest failure rate", message = "The component with the lowest failure rate is the most reliable one — but in a series system it is the least reliable component that constrains the system."), answer("All components equally", message = "In a series system, the least reliable component has a disproportionate effect on system reliability."), random_answer_order = TRUE, allow_retry = TRUE ) )
Calculate series reliability yourself.
# A conveyor belt system has 5 components in series. # Component reliabilities: 0.98, 0.96, 0.99, 0.94, 0.97 # Calculate the system reliability.
# R_sys <- prod(c(0.98, 0.96, 0.99, 0.94, 0.97))
R_components <- c(0.98, 0.96, 0.99, 0.94, 0.97) R_series <- prod(R_components) R_series # ~0.845
Parallel Systems
In a parallel system, only one component needs to function for the system to succeed. The system fails only if all components fail simultaneously. This is called active redundancy.
$$R_{sys} = 1 - \prod_{i=1}^{n}(1 - R_i)$$
Key insight: Adding parallel components always increases system reliability. Redundancy is a powerful tool for critical systems.
Example
A backup power system has a primary generator (R = 0.90) and a standby generator (R = 0.85). Either one alone keeps the system running.
grViz(" digraph parallel { rankdir = LR graph [bgcolor = transparent] node [shape = rectangle, style = filled, fillcolor = '#A9DFBF', fontname = 'sans-serif', fontsize = 12, margin = '0.2,0.1'] edge [arrowsize = 0.8] I [shape = point, width = 0.15, fillcolor = black] A [label = 'Generator 1\nR = 0.90'] B [label = 'Generator 2\nR = 0.85'] O [shape = point, width = 0.15, fillcolor = black] I -> A -> O I -> B -> O } ") ```r R_primary <- 0.90 R_standby <- 0.85 R_parallel <- 1 - (1 - R_primary) * (1 - R_standby) R_parallel
The parallel system reliability is 98.5%, much higher than either generator alone.
Use the slider to explore how the number of redundant components affects system reliability.
sliderInput("n_comp", "Number of parallel components:", min = 1, max = 8, value = 2, step = 1) numericInput("R_ind", "Individual component reliability:", value = 0.90, min = 0.50, max = 0.999, step = 0.01) plotOutput("parallelPlot")
output$parallelPlot <- renderPlot({ n_vals <- 1:10 Rc <- input$R_ind R_par <- 1 - (1 - Rc)^n_vals R_sel <- 1 - (1 - Rc)^input$n_comp plot(n_vals, R_par, type = "b", col = "steelblue", lwd = 2, pch = 19, xlab = "Number of redundant components", ylab = "System Reliability", main = "Parallel System Reliability vs. Redundancy", ylim = c(0, 1)) points(input$n_comp, R_sel, col = "red", pch = 19, cex = 2.5) abline(h = Rc, col = "gray50", lty = 2) legend("bottomright", legend = c("System reliability", paste0("n = ", input$n_comp), paste0("Single component (", Rc, ")")), col = c("steelblue", "red", "gray50"), lty = c(1, NA, 2), pch = c(19, 19, NA)) })
quiz(caption = "Quiz: Parallel Systems", question("Two components each have reliability 0.80. What is the reliability of a parallel system?", answer("0.80", message = "That is a single component's reliability."), answer("0.64", message = "0.64 = 0.8 × 0.8 is the series reliability."), answer("0.96", correct = TRUE, message = "Correct! R_sys = 1 − (1−0.80)² = 1 − 0.04 = 0.96."), answer("1.60", message = "Reliability cannot exceed 1."), random_answer_order = TRUE, allow_retry = TRUE ), question("Adding a third identical component (R = 0.80) in parallel to the previous two-component system gives a system reliability of approximately:", answer("0.992", correct = TRUE, message = "Correct! R_sys = 1 − (1−0.80)³ = 1 − 0.008 = 0.992."), answer("0.96", message = "That is the two-component parallel reliability."), answer("0.80", message = "That is a single component's reliability."), answer("0.512", message = "0.512 = 0.8³ is three components in series, not parallel."), random_answer_order = TRUE, allow_retry = TRUE ) )
Calculate parallel reliability yourself.
# A critical pump station has 3 pumps in parallel. # Each pump has reliability 0.88. # What is the system reliability?
# R_sys <- 1 - prod(1 - c(0.88, 0.88, 0.88))
R_components <- c(0.88, 0.88, 0.88) R_parallel <- 1 - prod(1 - R_components) R_parallel # ~0.9983
Mixed Systems
Most real systems combine series and parallel blocks. To analyze a mixed system, decompose it into subsystems and apply series and parallel rules step by step, working from the innermost blocks outward.
Use the selector below to compare how the topology changes across the three core configurations.
selectInput("topology", "Select topology:", choices = c("Series", "Parallel", "Mixed"), selected = "Mixed") DiagrammeROutput("topologyDiagram")
output$topologyDiagram <- renderGrViz({ if (input$topology == "Series") { grViz("digraph { rankdir=LR; graph [bgcolor=transparent] node [shape=rectangle, style=filled, fillcolor='#AED6F1', fontname='sans-serif', fontsize=12] I [shape=point, width=0.15, fillcolor=black] A [label='Component A\nR=0.95']; B [label='Component B\nR=0.90']; C [label='Component C\nR=0.97'] O [shape=point, width=0.15, fillcolor=black] I->A->B->C->O }") } else if (input$topology == "Parallel") { grViz("digraph { rankdir=LR; graph [bgcolor=transparent] node [shape=rectangle, style=filled, fillcolor='#A9DFBF', fontname='sans-serif', fontsize=12] I [shape=point, width=0.15, fillcolor=black] A [label='Component A\nR=0.95']; B [label='Component B\nR=0.90'] O [shape=point, width=0.15, fillcolor=black] I->A->O; I->B->O }") } else { grViz("digraph { rankdir=LR; graph [bgcolor=transparent] node [shape=rectangle, style=filled, fillcolor='#F9E79F', fontname='sans-serif', fontsize=12] I [shape=point, width=0.15, fillcolor=black] S [label='Sensor\nR=0.95']; T [label='Transmitter\nR=0.97'] A1 [label='Actuator 1\nR=0.90']; A2 [label='Actuator 2\nR=0.90'] O [shape=point, width=0.15, fillcolor=black] I->S->T->A1->O; T->A2->O }") } })
Example
A safety system has:
- Subsystem A: a sensor (R = 0.95) and a transmitter (R = 0.97) in series.
- Subsystem B: two redundant actuators (each R = 0.90) in parallel.
- Subsystems A and B are in series (both must work).
# Subsystem A (series) R_A <- 0.95 * 0.97 R_A # Subsystem B (parallel) R_B <- 1 - (1 - 0.90) * (1 - 0.90) R_B # Overall system (A and B in series) R_system <- R_A * R_B R_system
quiz(caption = "Quiz: Mixed Systems", question("A system has two subsystems in series. Subsystem 1 has two components in parallel (each R = 0.85); Subsystem 2 is a single component with R = 0.92. What is the system reliability?", answer("0.85 × 0.92 = 0.782", message = "That ignores the parallel redundancy in Subsystem 1."), answer("[1 − (1−0.85)²] × 0.92 ≈ 0.899", correct = TRUE, message = "Correct! R_sub1 = 1 − 0.15² = 0.9775, then R_sys = 0.9775 × 0.92 ≈ 0.899."), answer("0.85 + 0.92 = 1.77", message = "Reliabilities cannot be added like this."), answer("0.92", message = "That ignores Subsystem 1 entirely."), random_answer_order = TRUE, allow_retry = TRUE ), question("To analyze a complex mixed RBD, the recommended approach is to:", answer("Average all component reliabilities", message = "Averaging does not account for the series/parallel topology."), answer("Decompose into series and parallel subsystems, then apply the formulas step by step", correct = TRUE, message = "Correct! Work from the innermost blocks outward, replacing each parallel or series group with its equivalent single-block reliability."), answer("Use only the reliability of the most critical component", message = "This would ignore the contributions of all other components."), random_answer_order = TRUE, allow_retry = TRUE ) )
k-out-of-n Systems
A k-out-of-n system succeeds if at least k of its n identical components function. This generalizes series (k = n) and parallel (k = 1):
- k = n: all must work → series.
- k = 1: at least one works → parallel.
- 1 < k < n: voting or load-sharing systems.
The reliability of a k-out-of-n system with identical components (each with reliability p) follows the binomial distribution:
$$R_{k/n} = \sum_{i=k}^{n} \binom{n}{i} p^i (1-p)^{n-i} = 1 - \text{pbinom}(k-1, n, 1-p)$$
Example
A flight control system uses 3 redundant computers. At least 2 of the 3 must agree for the system to function safely (a 2-out-of-3 voter). Each computer has reliability R = 0.99.
grViz(" digraph koon { rankdir = LR graph [bgcolor = transparent] node [shape = rectangle, style = filled, fillcolor = '#D7BDE2', fontname = 'sans-serif', fontsize = 12, margin = '0.2,0.1'] edge [arrowsize = 0.8] I [shape = point, width = 0.15, fillcolor = black] C1 [label = 'Computer 1\nR = 0.99'] C2 [label = 'Computer 2\nR = 0.99'] C3 [label = 'Computer 3\nR = 0.99'] V [label = 'Voter\n(2-of-3)', shape = diamond, fillcolor = '#F1948A', fontsize = 11] O [shape = point, width = 0.15, fillcolor = black] I -> C1 -> V I -> C2 -> V I -> C3 -> V V -> O } ")
n <- 3 # total components k <- 2 # minimum required p <- 0.99 # individual reliability R_voting <- 1 - pbinom(k - 1, n, 1 - p) R_voting
The 2-out-of-3 system has reliability 0.9997.
quiz(caption = "Quiz: k-out-of-n Systems", question("A 2-out-of-4 system has 4 identical components each with R = 0.90. What R function computes its reliability?", answer("1 - pbinom(1, 4, 0.10)", correct = TRUE, message = "Correct! k=2, n=4, failure probability = 1−0.90 = 0.10. Use pbinom(k−1, n, 1−p) = pbinom(1, 4, 0.10)."), answer("pbinom(2, 4, 0.90)", message = "This gives P(X ≤ 2 successes), not the probability that at least 2 succeed."), answer("1 - pbinom(2, 4, 0.90)", message = "The third argument should be the failure probability (1−p), not p."), answer("prod(rep(0.90, 4))", message = "That formula gives the series reliability (all 4 must work), not 2-out-of-4."), random_answer_order = TRUE, allow_retry = TRUE ), question("How does a 1-out-of-n system relate to a parallel system?", answer("They are different — a 1-out-of-n system requires only the first component to work", message = "A 1-out-of-n system requires at least one of any n components to work, which is exactly a parallel configuration."), answer("They are equivalent — both require only one component to function", correct = TRUE, message = "Correct! When k = 1, the k-out-of-n formula reduces to the parallel reliability formula."), answer("A 1-out-of-n system is less reliable than a parallel system", message = "They are the same configuration."), random_answer_order = TRUE, allow_retry = TRUE ) )
Calculate k-out-of-n reliability in R.
# A 3-out-of-5 redundant sensor array. # Each sensor has reliability p = 0.95. # Calculate the system reliability. n <- 5 k <- 3 p <- 0.95
# R_sys <- 1 - pbinom(k - 1, n, 1 - p)
n <- 5 k <- 3 p <- 0.95 R_sys <- 1 - pbinom(k - 1, n, 1 - p) R_sys # ~0.9988
System MTTF
For a series system of components with constant failure rates (exponential distribution), the system failure rate is the sum of component failure rates:
$$\lambda_{sys} = \sum_{i=1}^{n} \lambda_i \qquad MTTF_{sys} = \frac{1}{\lambda_{sys}}$$
For $n$ identical parallel components each with failure rate $\lambda$:
$$MTTF_{parallel} = \frac{1}{\lambda}\left(1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n}\right)$$
Example
Two pumps in parallel, each with $\lambda = 0.01$ failures/hour:
lambda <- 0.01 MTTF_single <- 1 / lambda MTTF_parallel <- (1 / lambda) * (1 + 1/2) MTTF_single # 100 hours MTTF_parallel # 150 hours — 50% longer with one spare
quiz(caption = "Quiz: System MTTF", question("Three components in series have failure rates 0.02, 0.03, and 0.05 failures/hour. What is the system MTTF?", answer("10 hours", correct = TRUE, message = "Correct! λ_sys = 0.02 + 0.03 + 0.05 = 0.10, so MTTF = 1/0.10 = 10 hours."), answer("16.7 hours", message = "That is 1/0.06 — you may have missed one failure rate."), answer("100 hours", message = "100 hours is the MTTF of a single component with λ = 0.01."), answer("0.1 hours", message = "0.1 is the system failure rate, not the MTTF."), random_answer_order = TRUE, allow_retry = TRUE ), question("Compared to a single component, the MTTF of two identical components in parallel is:", answer("Twice as large", message = "Not quite. For two identical parallel components, MTTF_parallel = (3/2) × MTTF_single — a 50% increase."), answer("1.5 times as large", correct = TRUE, message = "Correct! MTTF_parallel = (1/λ)(1 + 1/2) = 1.5 × MTTF_single."), answer("The same", message = "Redundancy always increases MTTF."), answer("Three times as large", message = "Three times applies to three identical parallel components using the harmonic series formula."), random_answer_order = TRUE, allow_retry = TRUE ) )
Introduction to Fault Tree Analysis
Fault Tree Analysis (FTA) is a top-down approach to reliability analysis that starts with an undesirable system event (the top event) and works backward to identify the combinations of component failures that could cause it.
While an RBD asks "What must work for the system to succeed?", a fault tree asks "What can cause the system to fail?"
Gates
A fault tree uses logic gates to combine failure events:
- AND gate ($\land$): the top event occurs only if all input events occur simultaneously. This corresponds to parallel components in an RBD, all must fail to cause system failure.
- OR gate ($\lor$): the top event occurs if any input event occurs. This corresponds to series components in an RBD, any single failure causes system failure.
RBD — Fault Tree Duality
Every RBD has a corresponding fault tree and vice versa:
| RBD configuration | Fault tree gate for system failure | |:---:|:---:| | Series (all must work) | OR gate (any failure causes system failure) | | Parallel (any can work) | AND gate (all must fail for system failure) |
This duality means the two tools provide complementary views of the same system. RBDs are better for computing reliability; fault trees are better for tracing failure causes and identifying critical combinations.
The fault tree below illustrates a simple two-component system: the top event (system failure) occurs if either component fails — an OR gate, which corresponds to a series RBD.
grViz(" digraph fta { rankdir = TB graph [bgcolor = transparent] node [fontname = 'sans-serif', fontsize = 12, style = filled] edge [arrowsize = 0.8] Top [label = 'System\nFailure', shape = rectangle, fillcolor = '#E74C3C', fontcolor = white] OR [label = 'OR Gate', shape = diamond, fillcolor = '#F39C12'] F1 [label = 'Component A\nFails', shape = ellipse, fillcolor = '#AED6F1'] F2 [label = 'Component B\nFails', shape = ellipse, fillcolor = '#AED6F1'] Top -> OR OR -> F1 OR -> F2 } ")
For complex fault trees, the R package FaultTree on CRAN provides tools for building and analyzing fault tree models programmatically.
quiz(caption = "Quiz: Fault Tree Analysis", question("A fault tree has an AND gate combining two component failures. This corresponds to which RBD configuration?", answer("Series — both components in series", message = "A series RBD means any single failure causes system failure, which maps to an OR gate in a fault tree."), answer("Parallel — both components in parallel", correct = TRUE, message = "Correct! An AND gate means the system fails only if both components fail — exactly the parallel (redundant) RBD configuration."), answer("k-out-of-n with k = 2", message = "A 2-out-of-n system is more complex than a simple AND/OR gate."), random_answer_order = TRUE, allow_retry = TRUE ), question("Which statement best distinguishes FTA from RBD?", answer("FTA models reliability; RBD models failure modes", message = "It is the other way around — FTA is failure-oriented and RBD is success-oriented."), answer("RBD models what must succeed for the system to work; FTA traces what can cause the system to fail", correct = TRUE, message = "Correct! RBDs are success-space models; fault trees are failure-space models. They are logically dual and provide complementary views."), answer("RBD applies only to series systems; FTA applies to parallel systems", message = "Both methods apply to any system topology."), random_answer_order = TRUE, allow_retry = TRUE ) )
Case Study: Industrial Cooling System
Let's apply everything to a realistic example. An industrial cooling system has the following architecture:
- Water supply subsystem: two pumps in parallel (each R = 0.92), followed by a filter in series (R = 0.99).
- Control subsystem: a primary controller (R = 0.97) and a backup controller in parallel (R = 0.95).
- The water supply and control subsystems must both work (series at system level).
# Step 1 — Water supply subsystem R_pump_parallel <- 1 - (1 - 0.92)^2 R_water <- R_pump_parallel * 0.99 R_water # Step 2 — Control subsystem R_control <- 1 - (1 - 0.97) * (1 - 0.95) R_control # Step 3 — Overall system (both subsystems in series) R_system <- R_water * R_control R_system
Now explore what happens if the filter is upgraded.
# Modify the case study: the filter is upgraded to R = 0.999. # Recalculate the system reliability with the improved filter. R_pump1 <- 0.92 R_pump2 <- 0.92 R_filter <- 0.999 # upgraded from 0.99 R_ctrl1 <- 0.97 R_ctrl2 <- 0.95
# R_pump_parallel <- 1 - (1 - R_pump1) * (1 - R_pump2) # R_water <- R_pump_parallel * R_filter # R_control <- 1 - (1 - R_ctrl1) * (1 - R_ctrl2) # R_system <- R_water * R_control
R_pump1 <- 0.92 R_pump2 <- 0.92 R_filter <- 0.999 R_ctrl1 <- 0.97 R_ctrl2 <- 0.95 R_pump_parallel <- 1 - (1 - R_pump1) * (1 - R_pump2) R_water <- R_pump_parallel * R_filter R_control <- 1 - (1 - R_ctrl1) * (1 - R_ctrl2) R_system <- R_water * R_control R_system # ~0.985
Summary
Congratulations on completing the Reliability Block Diagrams and System Reliability tutorial!
Key takeaways:
- Series: $R_{sys} = \prod R_i$ — every component must work; reliability decreases with more components.
- Parallel: $R_{sys} = 1 - \prod(1-R_i)$ — redundancy; only one component needs to work.
- Mixed: decompose into subsystems, apply formulas step by step.
- k-out-of-n:
1 - pbinom(k-1, n, 1-p)for voting / load-sharing systems. - Series MTTF: $1/\sum\lambda_i$; Parallel MTTF of $n$ identical units: $(1/\lambda)\sum_{i=1}^{n}(1/i)$.
- FTA duality: series RBD ↔ OR gate; parallel RBD ↔ AND gate.
References
-
Abernethy, R.B. (2004) The New Weibull Handbook. Fifth Edition.
-
Aden-Buie G, Schloerke B, Allaire J, Rossell Hayes A (2023). learnr: Interactive Tutorials for R. https://rstudio.github.io/learnr/, https://github.com/rstudio/learnr.
-
Billinton R, Allan R (1992). Reliability Evaluation of Engineering Systems. Second Edition. Plenum Press, New York.
-
Vesely W, Goldberg F, Roberts N, Haasl D (1981). Fault Tree Handbook. US Nuclear Regulatory Commission, NUREG-0492.
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