1 Reliability, Availability, and Maintainability (RAM)

1.1 Introduction

Reliability, Availability, and Maintainability (RAM) are the foundational metrics of reliability engineering. They describe how well a system performs its intended function, how often it is usable, and how quickly it can be restored when it fails. This chapter introduces these concepts and builds toward calculating probability of failure, $B_n$ life, and system-level reliability.

1.2 Learning Objectives

By the end of this chapter, you will be able to:

Define key reliability metrics: reliability, availability, and failure rate.
Describe the significance of MTTR, MTTF, and MTBF.
Calculate probability of failure using reliability data.
Interpret $B_n$ (or $L_n$) life values.
Explain the bathtub curve and identify which failure region a component occupies given its $\beta$ value.
Compute system reliability for series and parallel configurations.

1.3 What is RAM?

Reliability: The ability of an item to perform its intended function without failure over a specified period.

Availability: The proportion of time an item is in a functioning condition.

Maintainability: The ease and speed with which an item can be restored to operational status after a failure.

These three concepts are interrelated: a highly reliable item will be available more often, and an item that is easy to maintain can be restored to service quickly, further improving availability.

1.4 Reliability

Reliability is the probability that an item will not fail under defined conditions for a specified period:

\[\text{Reliability} = \left(1 - \frac{\text{Failed Time}}{\text{Total Time}}\right) \times 100\]

Example: A motor ran for 5 years total and was failed for 10 of those days.

\[\text{Reliability} = \left(1 - \frac{10}{5 \times 365}\right) \times 100 = 94.5\%\]

Unreliability is the complement:

\[\text{Unreliability} = 1 - \text{Reliability} = \frac{\text{Failed Time}}{\text{Total Time}} \times 100\]

The ReliaLearnR package includes helper functions for these calculations:

library(ReliaLearnR)

# Motor: 5 years total, 10 days failed
rel(outageTime = 10, totalTime = 5 * 365)

[1] 0.9945205

Review

A machine ran for 3 years total and was failed for 5 of those days. What are the reliability and unreliability?

Answer

Reliability = $(1 - 5/(3 \times 365)) \times 100 = 99.5\%$. Unreliability = $0.5\%$.

1.5 Availability

Availability accounts for both failed time and scheduled maintenance — any time the item is unavailable for service:

\[\text{Availability} = \left(1 - \frac{\text{Unavailable Time}}{\text{Total Time}}\right) \times 100\]

Example: A motor ran for 5 years, was failed for 10 days, and had 15 days of scheduled maintenance.

\[\text{Availability} = \left(1 - \frac{10 + 15}{5 \times 365}\right) \times 100 = 93.2\%\]

# Motor: 5 years, 10 days failed + 15 days maintenance = 25 unavailable
avail(unavailTime = 25, totalTime = 5 * 365)

[1] 0.9863014

Note: standby time (available but not in use) does not count as unavailable time.

Review

What is the difference between reliability and availability?

Answer

Availability includes scheduled maintenance downtime; reliability does not. Availability $\leq$ Reliability (all failed time reduces both, but only availability is also reduced by planned maintenance).

1.6 Mean Time to Repair (MTTR)

MTTR measures maintainability — the average time required to repair a failed item:

\[\text{MTTR} = \frac{\sum_{i=1}^{n} \text{RepairTime}_i}{\text{RepairCount}}\]

Example: 5 failures with repair times of 5, 10, 15, 8, and 12 days.

\[\text{MTTR} = \frac{5 + 10 + 15 + 8 + 12}{5} = 10 \text{ days}\]

1.7 Mean Time to Failure (MTTF)

MTTF is a reliability measure for non-repairable items. Once the item fails, it is replaced rather than repaired.

\[\text{MTTF} = \frac{\text{Total Time}}{\text{FailureCount}}\]

Example: 100 motors run for 5 years total; 5 failures occur.

\[\text{MTTF} = \frac{5 \times 100}{5} = 100 \text{ years}\]

mttf(failures = 5, totalTime = 5 * 100)

[1] 100

1.8 Mean Time Between Failures (MTBF)

MTBF is the reliability measure for repairable systems, where the item is restored to service after each failure.

\[\text{MTBF} = \frac{\text{Total Time}}{\text{FailureCount}}\]

The formula is identical to MTTF, but the interpretation differs: MTBF is the average time between successive failures on the same system.

Example: A motor fails 5 times during 10,000 total operating hours.

\[\text{MTBF} = \frac{10{,}000}{5} = 2{,}000 \text{ hours}\]

mtbf(failures = 5, totalTime = 10000)

[1] 2000

Try It

A fleet of 50 generators runs for 8 years. During that time, 20 failures occur. Calculate the MTBF in years.

fleet_size <- 50
years      <- 8
failures   <- 20
# Calculate MTBF = (fleet_size * years) / failures

Solution

fleet_size <- 50
years      <- 8
failures   <- 20
MTBF <- (fleet_size * years) / failures
MTBF  # 20 years

[1] 20

Review

What is the difference between MTTF and MTBF?

Answer

MTTF is for non-repairable items (replaced after failure); MTBF is for repairable systems (restored to service after failure). Both use the same formula, but MTBF represents the average time between successive failures of the same system.

1.9 Failure Rate

The failure rate $\lambda$ is the inverse of MTBF:

\[\lambda = \frac{1}{\text{MTBF}} = \frac{\text{FailureCount}}{\text{Total Time}}\]

A key assumption of MTBF is a constant failure rate throughout the item’s life. While this is not always true in practice, it is a reasonable approximation for many systems during their useful life period.

Example: 100 motors run for 10 years; 20 failures occur.

\[\lambda = \frac{20}{10 \times 100} = 0.02 \text{ failures per year}\]

fr(failures = 20, totalTime = 10 * 100)

[1] 0.02

1.10 Probability of Failure

Under the constant failure rate assumption, failure times follow an exponential distribution. The cumulative probability of failure prior to time $t$ is:

\[F(t) = 1 - e^{-\lambda t}\]

And the cumulative probability of survival (reliability function) is:

\[R(t) = 1 - F(t) = e^{-\lambda t}\]

Example: A motor has a failure rate of 0.1 failures/year. Probability of surviving 10 years:

\[R(10) = e^{-10 \times 0.1} = e^{-1} \approx 36.8\%\]

lambda <- 0.1
t      <- 10
R <- exp(-lambda * t)
R

[1] 0.3678794

The Reliability Curve

The shape of $R(t) = e^{-\lambda t}$ depends entirely on $\lambda$. Below are curves for three different failure rates:

t <- seq(0, 30, by = 0.1)

plot(t, exp(-0.05 * t), type = "l", col = "blue", lwd = 2,
     xlab = "Time", ylab = "Reliability R(t)",
     main = "Exponential Reliability for Different Failure Rates",
     ylim = c(0, 1))
lines(t, exp(-0.1 * t),  col = "red",    lwd = 2)
lines(t, exp(-0.3 * t),  col = "darkgreen", lwd = 2)
abline(h = exp(-1), col = "gray50", lty = 2)
legend("topright",
       legend = c("λ = 0.05 (MTBF = 20)", "λ = 0.10 (MTBF = 10)", "λ = 0.30 (MTBF = 3.3)",
                  "R = 36.8% at t = MTBF"),
       col = c("blue", "red", "darkgreen", "gray50"),
       lty = c(1, 1, 1, 2), lwd = 2)

Note that at $t = \text{MTBF} = 1/\lambda$, reliability is always $e^{-1} \approx 36.8\%$ regardless of $\lambda$. This is a universal property of the exponential distribution.

Try It

A pump has a failure rate of 0.05 failures/year. Calculate the probability of survival at $t = 10$ years.

lambda <- 0.05
t      <- 10
# R(t) = exp(-lambda * t)

Solution

lambda <- 0.05
t      <- 10
R <- exp(-lambda * t)
R  # ~0.607 (60.7% survival)

[1] 0.6065307

1.11 $B_n$ or $L_n$ Life

The $B_n$ life is the time at which $n$% of a population is expected to have failed. Some industries use $L_n$ instead, but the concept is identical.

To find $B_n$, set $F(t) = n/100$ and solve for $t$:

\[B_n = -\frac{\ln(1 - n/100)}{\lambda}\]

Example: Motor with $\lambda = 0.2$ failures/year. B10 life:

\[B_{10} = -\frac{\ln(1 - 0.1)}{0.2} = 0.526 \text{ years}\]

lambda <- 0.2
B10 <- -log(1 - 0.10) / lambda
B10  # 0.526 years

[1] 0.5268026

Try It

A component has a failure rate of 0.05 failures/year. Calculate the B10 life.

lambda <- 0.05
# B10 = -log(1 - 0.10) / lambda

Solution

lambda <- 0.05
B10 <- -log(1 - 0.10) / lambda
B10  # ~2.1 years

[1] 2.10721

Review

What is the relationship between failure rate and MTBF?

Answer

Failure rate is the inverse of MTBF: $\lambda = 1/\text{MTBF}$, and $\text{MTBF} = 1/\lambda$.

1.12 From Exponential to Weibull

The exponential model assumes a constant failure rate. This is appropriate for electronic components that fail at random, but many mechanical components have failure rates that change over time.

The Weibull distribution (Abernethy 2004) generalizes the exponential by adding a shape parameter $\beta$:

\[R(t) = e^{-(t/\eta)^\beta}\]

$\beta$	Failure rate	Typical cause
$< 1$	Decreasing	Infant mortality
$= 1$	Constant	Random failures — exponential applies
$> 1$	Increasing	Wear-out (fatigue, corrosion, aging)

When $\beta = 1$, the Weibull reduces to the exponential with $\lambda = 1/\eta$.

1.13 The Bathtub Curve

The failure rate of a population of items over its lifetime traces a characteristic shape known as the bathtub curve — named for its distinctive U-profile. Rather than drawing it as a stylized cartoon, we can build it analytically by compositing the three Weibull failure modes:

\[h(t) = \underbrace{h_{\text{infant}}(t)}_{\beta < 1} \;+\; \underbrace{h_{\text{random}}(t)}_{\beta = 1} \;+\; \underbrace{h_{\text{wear-out}}(t)}_{\beta > 1}\]

where the Weibull hazard rate is $h(t) = (\beta/\eta)(t/\eta)^{\beta-1}$. Each term represents a physically distinct failure mechanism — the bathtub shape is not assumed; it emerges from their sum.

t <- seq(0.01, 15, by = 0.01)

# Three Weibull hazard rate components
h_infant  <- function(t, b = 0.4, e = 2)  (b/e) * (t/e)^(b-1)   # infant mortality
h_random  <- function(t, b = 1.0, e = 5)  (b/e) * (t/e)^(b-1)   # random failures
h_wearout <- function(t, b = 3.5, e = 10) (b/e) * (t/e)^(b-1)   # wear-out

h_total <- h_infant(t) + h_random(t) + h_wearout(t)

plot(t, h_total, type = "l", col = "black", lwd = 3,
     xlab = "Time", ylab = "Failure Rate h(t)",
     main = "The Bathtub Curve: Three Composited Failure Modes",
     ylim = c(0, max(h_total) * 1.1))
lines(t, h_infant(t),  col = "blue",      lwd = 2, lty = 2)
lines(t, h_random(t),  col = "red",       lwd = 2, lty = 2)
lines(t, h_wearout(t), col = "darkgreen", lwd = 2, lty = 2)

abline(v = c(2.5, 9), col = "gray60", lty = 3)
text(1.25,  max(h_total) * 0.85, "Infant\nMortality",    col = "blue",      cex = 0.85)
text(5.75,  max(h_total) * 0.55, "Useful Life\n(Random)", col = "red",       cex = 0.85)
text(12.25, max(h_total) * 0.85, "Wear-out",              col = "darkgreen", cex = 0.85)

legend("top", inset = 0.02,
       legend = c("Total h(t)", "Infant (β = 0.4)", "Random (β = 1)", "Wear-out (β = 3.5)"),
       col = c("black", "blue", "red", "darkgreen"), lwd = c(3, 2, 2, 2), lty = c(1, 2, 2, 2))

Each region maps to a distinct engineering strategy:

Infant mortality ($\beta < 1$): manufacturing defects, assembly errors, material flaws. Mitigated by burn-in testing — running units briefly before shipment to screen out defective items.
Useful life ($\beta \approx 1$): random external shocks, load spikes, operator error. The constant failure rate means failures are unrelated to age; mitigated by redundancy.
Wear-out ($\beta > 1$): fatigue, corrosion, material aging. Mitigated by preventive replacement scheduled before the B₅ or B₁₀ life is reached.

Review

A pump has $\beta = 2.8$. Which region of the bathtub curve does it occupy, and what maintenance strategy is most appropriate?

Answer

$\beta > 1$ places it in the wear-out region. The failure rate is increasing with age, so a time-based preventive replacement strategy — replacing the pump before it reaches its B₅ or B₁₀ life — is most appropriate.

The bathtub curve is a population-level concept: individual units fail at random times, but the aggregate failure rate of a large fleet traces this shape. Fitting a separate Weibull to each failure mode is covered in Chapter 3.

The WeibullR package (Silkworth and Symynck 2022) provides functions for Weibull fitting in R.

library(WeibullR)

failures <- c(500, 820, 1100, 1350, 1590)
fit <- MLEw2p(failures, show = TRUE)

A $\beta$ close to 1 indicates random failures consistent with the exponential model. Higher values indicate wear-out behavior. The Life Data Analysis chapter covers Weibull analysis in depth.

1.14 System Reliability

Real systems combine many components. System reliability depends on how those components are arranged.

Series Systems

In a series configuration, every component must function for the system to function:

\[R_{\text{sys}} = R_1 \times R_2 \times \cdots \times R_n = \prod_{i=1}^{n} R_i\]

A series system is always less reliable than its weakest component.

R_components <- c(0.90, 0.90, 0.90)
R_series <- prod(R_components)
R_series  # 72.9%

[1] 0.729

Parallel Systems

In a parallel configuration, only one component needs to function — redundancy. The system fails only when all components fail:

\[R_{\text{sys}} = 1 - (1 - R_1)(1 - R_2) \cdots (1 - R_n)\]

R_parallel <- 1 - prod(1 - R_components)
R_parallel  # 99.9%

[1] 0.999

The effect of redundancy increases dramatically with the number of parallel components:

n_vals <- 1:8
R_c    <- 0.90
R_par  <- 1 - (1 - R_c)^n_vals

plot(n_vals, R_par, type = "b", col = "steelblue", lwd = 2, pch = 19,
     xlab = "Number of parallel components",
     ylab = "System Reliability",
     main = "Effect of Redundancy on System Reliability (R_component = 0.90)",
     ylim = c(0, 1))
abline(h = R_c, col = "gray50", lty = 2)
legend("bottomright",
       legend = c("System reliability", "Single component (0.90)"),
       col = c("steelblue", "gray50"), lty = c(1, 2), pch = c(19, NA), lwd = 2)

Try It

Four components each have reliability R = 0.85. Calculate the series and parallel system reliability.

R_comp <- 0.85
# Series:   R_sys = prod(R_components)
# Parallel: R_sys = 1 - prod(1 - R_components)

Solution

R_comp <- rep(0.85, 4)
R_series   <- prod(R_comp)
R_parallel <- 1 - prod(1 - R_comp)
R_series    # ~0.522 (52.2%)

[1] 0.5220062

R_parallel  # ~0.9995 (99.95%)

[1] 0.9994937

Review

Three components each have 80% reliability. What is the series system reliability?

Answer

$R_{\text{sys}} = 0.8 \times 0.8 \times 0.8 = 0.512 = 51.2\%$

1.15 Summary

Key takeaways:

$R(t) = e^{-\lambda t}$ — exponential reliability under a constant failure rate.
At $t = \text{MTBF} = 1/\lambda$, reliability is always 36.8%.
MTTF is for non-repairable items; MTBF is for repairable systems.
Series systems: $R_{\text{sys}} = \prod R_i$ — reliability decreases with more components.
Parallel systems: $R_{\text{sys}} = 1 - \prod(1 - R_i)$ — redundancy increases reliability.
When $\beta \neq 1$, use the Weibull distribution (see Chapter 3).

Abernethy, Robert B. 2004. The New Weibull Handbook. 5th ed. R.B. Abernethy.

Silkworth, David, and Jurgen Symynck. 2022. WeibullR: Weibull Analysis for Reliability Engineering. https://doi.org/10.32614/CRAN.package.WeibullR.

# Reliability, Availability, and Maintainability (RAM) {#sec-ram} ## Introduction **Reliability, Availability, and Maintainability (RAM)** are the foundational metrics of reliability engineering. They describe how well a system performs its intended function, how often it is usable, and how quickly it can be restored when it fails. This chapter introduces these concepts and builds toward calculating probability of failure, $B_n$ life, and system-level reliability. ## Learning Objectives By the end of this chapter, you will be able to: - Define key reliability metrics: reliability, availability, and failure rate. - Describe the significance of MTTR, MTTF, and MTBF. - Calculate probability of failure using reliability data. - Interpret $B_n$ (or $L_n$) life values. - Explain the bathtub curve and identify which failure region a component occupies given its $\beta$ value. - Compute system reliability for series and parallel configurations. ## What is RAM? **Reliability**: The ability of an item to perform its intended function without failure over a specified period. **Availability**: The proportion of time an item is in a functioning condition. **Maintainability**: The ease and speed with which an item can be restored to operational status after a failure. These three concepts are interrelated: a highly reliable item will be available more often, and an item that is easy to maintain can be restored to service quickly, further improving availability. ## Reliability Reliability is the probability that an item will not fail under defined conditions for a specified period: $$\text{Reliability} = \left(1 - \frac{\text{Failed Time}}{\text{Total Time}}\right) \times 100$$ **Example**: A motor ran for 5 years total and was failed for 10 of those days. $$\text{Reliability} = \left(1 - \frac{10}{5 \times 365}\right) \times 100 = 94.5\%$$ Unreliability is the complement: $$\text{Unreliability} = 1 - \text{Reliability} = \frac{\text{Failed Time}}{\text{Total Time}} \times 100$$ The `ReliaLearnR` package includes helper functions for these calculations: ```{r} library(ReliaLearnR) # Motor: 5 years total, 10 days failed rel(outageTime = 10, totalTime = 5 * 365) ``` ::: {.callout-tip} ## Review A machine ran for 3 years total and was failed for 5 of those days. What are the reliability and unreliability? <details><summary>Answer</summary> Reliability = $(1 - 5/(3 \times 365)) \times 100 = 99.5\%$. Unreliability = $0.5\%$. </details> ::: ## Availability Availability accounts for both failed time **and** scheduled maintenance — any time the item is unavailable for service: $$\text{Availability} = \left(1 - \frac{\text{Unavailable Time}}{\text{Total Time}}\right) \times 100$$ **Example**: A motor ran for 5 years, was failed for 10 days, and had 15 days of scheduled maintenance. $$\text{Availability} = \left(1 - \frac{10 + 15}{5 \times 365}\right) \times 100 = 93.2\%$$ ```{r} # Motor: 5 years, 10 days failed + 15 days maintenance = 25 unavailable avail(unavailTime = 25, totalTime = 5 * 365) ``` Note: standby time (available but not in use) does not count as unavailable time. ::: {.callout-tip} ## Review What is the difference between reliability and availability? <details><summary>Answer</summary> Availability includes scheduled maintenance downtime; reliability does not. Availability $\leq$ Reliability (all failed time reduces both, but only availability is also reduced by planned maintenance). </details> ::: ## Mean Time to Repair (MTTR) **MTTR** measures maintainability — the average time required to repair a failed item: $$\text{MTTR} = \frac{\sum_{i=1}^{n} \text{RepairTime}_i}{\text{RepairCount}}$$ **Example**: 5 failures with repair times of 5, 10, 15, 8, and 12 days. $$\text{MTTR} = \frac{5 + 10 + 15 + 8 + 12}{5} = 10 \text{ days}$$ ## Mean Time to Failure (MTTF) **MTTF** is a reliability measure for *non-repairable* items. Once the item fails, it is replaced rather than repaired. $$\text{MTTF} = \frac{\text{Total Time}}{\text{FailureCount}}$$ **Example**: 100 motors run for 5 years total; 5 failures occur. $$\text{MTTF} = \frac{5 \times 100}{5} = 100 \text{ years}$$ ```{r} mttf(failures = 5, totalTime = 5 * 100) ``` ## Mean Time Between Failures (MTBF) **MTBF** is the reliability measure for *repairable systems*, where the item is restored to service after each failure. $$\text{MTBF} = \frac{\text{Total Time}}{\text{FailureCount}}$$ The formula is identical to MTTF, but the interpretation differs: MTBF is the average time *between* successive failures on the same system. **Example**: A motor fails 5 times during 10,000 total operating hours. $$\text{MTBF} = \frac{10{,}000}{5} = 2{,}000 \text{ hours}$$ ```{r} mtbf(failures = 5, totalTime = 10000) ``` ::: {.callout-note} ## Try It A fleet of 50 generators runs for 8 years. During that time, 20 failures occur. Calculate the MTBF in years. ```{r} fleet_size <- 50 years <- 8 failures <- 20 # Calculate MTBF = (fleet_size * years) / failures ``` <details><summary>Solution</summary> ```{r} fleet_size <- 50 years <- 8 failures <- 20 MTBF <- (fleet_size * years) / failures MTBF # 20 years ``` </details> ::: ::: {.callout-tip} ## Review What is the difference between MTTF and MTBF? <details><summary>Answer</summary> MTTF is for **non-repairable** items (replaced after failure); MTBF is for **repairable** systems (restored to service after failure). Both use the same formula, but MTBF represents the average time *between* successive failures of the same system. </details> ::: ## Failure Rate The **failure rate** $\lambda$ is the inverse of MTBF: $$\lambda = \frac{1}{\text{MTBF}} = \frac{\text{FailureCount}}{\text{Total Time}}$$ A key assumption of MTBF is a **constant** failure rate throughout the item's life. While this is not always true in practice, it is a reasonable approximation for many systems during their useful life period. **Example**: 100 motors run for 10 years; 20 failures occur. $$\lambda = \frac{20}{10 \times 100} = 0.02 \text{ failures per year}$$ ```{r} fr(failures = 20, totalTime = 10 * 100) ``` ## Probability of Failure Under the constant failure rate assumption, failure times follow an **exponential distribution**. The cumulative probability of failure prior to time $t$ is: $$F(t) = 1 - e^{-\lambda t}$$ And the cumulative probability of **survival** (reliability function) is: $$R(t) = 1 - F(t) = e^{-\lambda t}$$ **Example**: A motor has a failure rate of 0.1 failures/year. Probability of surviving 10 years: $$R(10) = e^{-10 \times 0.1} = e^{-1} \approx 36.8\%$$ ```{r} lambda <- 0.1 t <- 10 R <- exp(-lambda * t) R ``` ### The Reliability Curve The shape of $R(t) = e^{-\lambda t}$ depends entirely on $\lambda$. Below are curves for three different failure rates: ```{r} t <- seq(0, 30, by = 0.1) plot(t, exp(-0.05 * t), type = "l", col = "blue", lwd = 2, xlab = "Time", ylab = "Reliability R(t)", main = "Exponential Reliability for Different Failure Rates", ylim = c(0, 1)) lines(t, exp(-0.1 * t), col = "red", lwd = 2) lines(t, exp(-0.3 * t), col = "darkgreen", lwd = 2) abline(h = exp(-1), col = "gray50", lty = 2) legend("topright", legend = c("λ = 0.05 (MTBF = 20)", "λ = 0.10 (MTBF = 10)", "λ = 0.30 (MTBF = 3.3)", "R = 36.8% at t = MTBF"), col = c("blue", "red", "darkgreen", "gray50"), lty = c(1, 1, 1, 2), lwd = 2) ``` Note that at $t = \text{MTBF} = 1/\lambda$, reliability is always $e^{-1} \approx 36.8\%$ regardless of $\lambda$. This is a universal property of the exponential distribution. ::: {.callout-note} ## Try It A pump has a failure rate of 0.05 failures/year. Calculate the probability of survival at $t = 10$ years. ```{r} lambda <- 0.05 t <- 10 # R(t) = exp(-lambda * t) ``` <details><summary>Solution</summary> ```{r} lambda <- 0.05 t <- 10 R <- exp(-lambda * t) R # ~0.607 (60.7% survival) ``` </details> ::: ## $B_n$ or $L_n$ Life The **$B_n$ life** is the time at which $n$% of a population is expected to have failed. Some industries use $L_n$ instead, but the concept is identical. To find $B_n$, set $F(t) = n/100$ and solve for $t$: $$B_n = -\frac{\ln(1 - n/100)}{\lambda}$$ **Example**: Motor with $\lambda = 0.2$ failures/year. B10 life: $$B_{10} = -\frac{\ln(1 - 0.1)}{0.2} = 0.526 \text{ years}$$ ```{r} lambda <- 0.2 B10 <- -log(1 - 0.10) / lambda B10 # 0.526 years ``` ::: {.callout-note} ## Try It A component has a failure rate of 0.05 failures/year. Calculate the B10 life. ```{r} lambda <- 0.05 # B10 = -log(1 - 0.10) / lambda ``` <details><summary>Solution</summary> ```{r} lambda <- 0.05 B10 <- -log(1 - 0.10) / lambda B10 # ~2.1 years ``` </details> ::: ::: {.callout-tip} ## Review What is the relationship between failure rate and MTBF? <details><summary>Answer</summary> Failure rate is the **inverse** of MTBF: $\lambda = 1/\text{MTBF}$, and $\text{MTBF} = 1/\lambda$. </details> ::: ## From Exponential to Weibull The exponential model assumes a **constant** failure rate. This is appropriate for electronic components that fail at random, but many mechanical components have failure rates that change over time. The **Weibull distribution** [@Weibull] generalizes the exponential by adding a shape parameter $\beta$: $$R(t) = e^{-(t/\eta)^\beta}$$ | $\beta$ | Failure rate | Typical cause | |:---:|---|---| | $< 1$ | Decreasing | Infant mortality | | $= 1$ | Constant | Random failures — exponential applies | | $> 1$ | Increasing | Wear-out (fatigue, corrosion, aging) | When $\beta = 1$, the Weibull reduces to the exponential with $\lambda = 1/\eta$. ## The Bathtub Curve The failure rate of a population of items over its lifetime traces a characteristic shape known as the **bathtub curve** — named for its distinctive U-profile. Rather than drawing it as a stylized cartoon, we can build it analytically by compositing the three Weibull failure modes: $$h(t) = \underbrace{h_{\text{infant}}(t)}_{\beta < 1} \;+\; \underbrace{h_{\text{random}}(t)}_{\beta = 1} \;+\; \underbrace{h_{\text{wear-out}}(t)}_{\beta > 1}$$ where the Weibull hazard rate is $h(t) = (\beta/\eta)(t/\eta)^{\beta-1}$. Each term represents a physically distinct failure mechanism — the bathtub shape is not assumed; it *emerges* from their sum. ```{r} t <- seq(0.01, 15, by = 0.01) # Three Weibull hazard rate components h_infant <- function(t, b = 0.4, e = 2) (b/e) * (t/e)^(b-1) # infant mortality h_random <- function(t, b = 1.0, e = 5) (b/e) * (t/e)^(b-1) # random failures h_wearout <- function(t, b = 3.5, e = 10) (b/e) * (t/e)^(b-1) # wear-out h_total <- h_infant(t) + h_random(t) + h_wearout(t) plot(t, h_total, type = "l", col = "black", lwd = 3, xlab = "Time", ylab = "Failure Rate h(t)", main = "The Bathtub Curve: Three Composited Failure Modes", ylim = c(0, max(h_total) * 1.1)) lines(t, h_infant(t), col = "blue", lwd = 2, lty = 2) lines(t, h_random(t), col = "red", lwd = 2, lty = 2) lines(t, h_wearout(t), col = "darkgreen", lwd = 2, lty = 2) abline(v = c(2.5, 9), col = "gray60", lty = 3) text(1.25, max(h_total) * 0.85, "Infant\nMortality", col = "blue", cex = 0.85) text(5.75, max(h_total) * 0.55, "Useful Life\n(Random)", col = "red", cex = 0.85) text(12.25, max(h_total) * 0.85, "Wear-out", col = "darkgreen", cex = 0.85) legend("top", inset = 0.02, legend = c("Total h(t)", "Infant (β = 0.4)", "Random (β = 1)", "Wear-out (β = 3.5)"), col = c("black", "blue", "red", "darkgreen"), lwd = c(3, 2, 2, 2), lty = c(1, 2, 2, 2)) ``` Each region maps to a distinct engineering strategy: - **Infant mortality** ($\beta < 1$): manufacturing defects, assembly errors, material flaws. Mitigated by *burn-in testing* — running units briefly before shipment to screen out defective items. - **Useful life** ($\beta \approx 1$): random external shocks, load spikes, operator error. The constant failure rate means failures are unrelated to age; mitigated by *redundancy*. - **Wear-out** ($\beta > 1$): fatigue, corrosion, material aging. Mitigated by *preventive replacement* scheduled before the B₅ or B₁₀ life is reached. ::: {.callout-tip} ## Review A pump has $\beta = 2.8$. Which region of the bathtub curve does it occupy, and what maintenance strategy is most appropriate? <details><summary>Answer</summary> $\beta > 1$ places it in the **wear-out** region. The failure rate is increasing with age, so a time-based **preventive replacement** strategy — replacing the pump before it reaches its B₅ or B₁₀ life — is most appropriate. </details> ::: The bathtub curve is a population-level concept: individual units fail at random times, but the aggregate failure rate of a large fleet traces this shape. Fitting a separate Weibull to each failure mode is covered in @sec-lda. The `WeibullR` package [@WeibullR] provides functions for Weibull fitting in R. ```{r} library(WeibullR) failures <- c(500, 820, 1100, 1350, 1590) fit <- MLEw2p(failures, show = TRUE) ``` A $\beta$ close to 1 indicates random failures consistent with the exponential model. Higher values indicate wear-out behavior. The Life Data Analysis chapter covers Weibull analysis in depth. ## System Reliability Real systems combine many components. System reliability depends on how those components are arranged. ### Series Systems In a **series** configuration, every component must function for the system to function: $$R_{\text{sys}} = R_1 \times R_2 \times \cdots \times R_n = \prod_{i=1}^{n} R_i$$ A series system is always *less* reliable than its weakest component. ```{r} R_components <- c(0.90, 0.90, 0.90) R_series <- prod(R_components) R_series # 72.9% ``` ### Parallel Systems In a **parallel** configuration, only one component needs to function — redundancy. The system fails only when *all* components fail: $$R_{\text{sys}} = 1 - (1 - R_1)(1 - R_2) \cdots (1 - R_n)$$ ```{r} R_parallel <- 1 - prod(1 - R_components) R_parallel # 99.9% ``` The effect of redundancy increases dramatically with the number of parallel components: ```{r} n_vals <- 1:8 R_c <- 0.90 R_par <- 1 - (1 - R_c)^n_vals plot(n_vals, R_par, type = "b", col = "steelblue", lwd = 2, pch = 19, xlab = "Number of parallel components", ylab = "System Reliability", main = "Effect of Redundancy on System Reliability (R_component = 0.90)", ylim = c(0, 1)) abline(h = R_c, col = "gray50", lty = 2) legend("bottomright", legend = c("System reliability", "Single component (0.90)"), col = c("steelblue", "gray50"), lty = c(1, 2), pch = c(19, NA), lwd = 2) ``` ::: {.callout-note} ## Try It Four components each have reliability R = 0.85. Calculate the series and parallel system reliability. ```{r} R_comp <- 0.85 # Series: R_sys = prod(R_components) # Parallel: R_sys = 1 - prod(1 - R_components) ``` <details><summary>Solution</summary> ```{r} R_comp <- rep(0.85, 4) R_series <- prod(R_comp) R_parallel <- 1 - prod(1 - R_comp) R_series # ~0.522 (52.2%) R_parallel # ~0.9995 (99.95%) ``` </details> ::: ::: {.callout-tip} ## Review Three components each have 80% reliability. What is the series system reliability? <details><summary>Answer</summary> $R_{\text{sys}} = 0.8 \times 0.8 \times 0.8 = 0.512 = 51.2\%$ </details> ::: ## Summary **Key takeaways:** - $R(t) = e^{-\lambda t}$ — exponential reliability under a constant failure rate. - At $t = \text{MTBF} = 1/\lambda$, reliability is always 36.8%. - MTTF is for non-repairable items; MTBF is for repairable systems. - Series systems: $R_{\text{sys}} = \prod R_i$ — reliability decreases with more components. - Parallel systems: $R_{\text{sys}} = 1 - \prod(1 - R_i)$ — redundancy increases reliability. - When $\beta \neq 1$, use the Weibull distribution (see @sec-lda).

1.1 Introduction

1.2 Learning Objectives

1.3 What is RAM?

1.4 Reliability

1.5 Availability

1.6 Mean Time to Repair (MTTR)

1.7 Mean Time to Failure (MTTF)

1.8 Mean Time Between Failures (MTBF)

1.9 Failure Rate

1.10 Probability of Failure

The Reliability Curve

1.11 \(B_n\) or \(L_n\) Life

1.12 From Exponential to Weibull

1.13 The Bathtub Curve

1.14 System Reliability

Series Systems

Parallel Systems

1.15 Summary