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Reliability & Queuing Theory

From failure rate analysis and life testing to Markovian queuing models and inventory optimization — the mathematics of system performance and operations research.

Reliability Concepts Life Testing System Reliability Elementary Queues Markovian Models Inventory Models

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Reliability Concepts

Fundamental reliability measures: reliability function, hazard rate, failure rate classes, and mean time to failure.

🔧 Reliability & Hazard Functions

Definition — Reliability Function For a non-negative random variable $T$ representing lifetime, the reliability function (survival function) is: \[R(t) = P(T > t) = 1 - F(t) = \int_t^\infty f(u)\,du\] where $F(t)$ is the CDF and $f(t)$ is the PDF. Properties: $R(0) = 1$, $R(\infty) = 0$, $R$ is non-increasing.

Definition — Hazard Function (Failure Rate) The hazard function is the instantaneous failure rate given survival to time $t$: \[h(t) = \frac{f(t)}{R(t)} = -\frac{d}{dt}\ln R(t)\] This gives the fundamental relationship: \[R(t) = \exp\!\left(-\int_0^t h(u)\,du\right) = e^{-H(t)}\] where $H(t) = \int_0^t h(u)\,du$ is the cumulative hazard function.

Definition — Failure Rate Classes

IFR (Increasing Failure Rate): $h(t)$ is non-decreasing — aging components.
DFR (Decreasing Failure Rate): $h(t)$ is non-increasing — burn-in improvement.
Bathtub curve: DFR in early life (infant mortality), constant in useful life, IFR in wear-out phase.
Constant failure rate: $h(t) = \lambda$ implies $T \sim \text{Exp}(\lambda)$ — the memoryless property.

Theorem — Mean Time to Failure & Mean Residual Life The MTTF is: \[\text{MTTF} = E[T] = \int_0^\infty R(t)\,dt\] The mean residual life (MRL) at age $t$ is: \[m(t) = E[T - t \mid T > t] = \frac{1}{R(t)}\int_t^\infty R(u)\,du\] For the exponential distribution, $m(t) = 1/\lambda$ for all $t$ (memoryless).

Example

A component has reliability $R(t) = e^{-0.002t^2}$ for $t \ge 0$. Find the hazard function and classify the failure rate.

$h(t) = -\frac{d}{dt}\ln R(t) = -\frac{d}{dt}(-0.002t^2) = 0.004t$. Since $h(t) = 0.004t$ is strictly increasing for $t > 0$, this is an IFR distribution (Rayleigh-type). MTTF $= \int_0^\infty e^{-0.002t^2}dt = \frac{1}{2}\sqrt{\pi/0.002} \approx 22.36$.

Life Testing & Censoring

Censoring schemes, Kaplan-Meier estimation, and parametric life models.

📉 Censoring & Estimation

Definition — Types of Censoring

Type I censoring: The test ends at a pre-specified time $T_0$. Units surviving beyond $T_0$ are censored. The number of failures $r$ is random.
Type II censoring: The test ends when a pre-specified number $r$ of failures occurs. The test duration $T_{(r)}$ is random.
Random censoring: Each unit $i$ has a potential censoring time $C_i$ independent of $T_i$. We observe $\min(T_i, C_i)$ and an indicator $\delta_i = I(T_i \le C_i)$.

Theorem — Kaplan-Meier (Product-Limit) Estimator Let $t_1 < t_2 < \cdots < t_k$ be the distinct observed failure times. Let $d_j$ = number of failures at $t_j$ and $n_j$ = number at risk just before $t_j$. The Kaplan-Meier estimator of $R(t)$ is: \[\hat{R}(t) = \prod_{j: t_j \le t}\left(1 - \frac{d_j}{n_j}\right)\] This is a non-parametric, consistent estimator. Its variance is estimated by Greenwood's formula: \[\hat{V}[\hat{R}(t)] = [\hat{R}(t)]^2 \sum_{j: t_j \le t}\frac{d_j}{n_j(n_j - d_j)}\]

Definition — Exponential Life Model Under Censoring For $T_i \sim \text{Exp}(\lambda)$ with Type II censoring at $r$ failures, the MLE of $\lambda$ is: \[\hat{\lambda} = \frac{r}{\sum_{i=1}^r T_{(i)} + (n-r)T_{(r)}}\] The denominator is the total time on test (TTT). The quantity $2\lambda \cdot \text{TTT}$ has a $\chi^2_{2r}$ distribution, enabling exact confidence intervals.

Example

In a Type II censored test with $n = 10$ items, 4 failures occur at times 100, 200, 350, 500. Estimate the mean life assuming an exponential model.

TTT $= 100 + 200 + 350 + 500 + (10-4) \times 500 = 1150 + 3000 = 4150$. $\hat{\lambda} = 4/4150 = 0.000964$. Mean life $= 1/\hat{\lambda} = 4150/4 = 1037.5$.

System Reliability

Series, parallel, and k-out-of-n systems; coherent systems and structure functions.

🔗 System Configurations

Definition — Series & Parallel Systems For $n$ independent components with reliabilities $R_1, R_2, \ldots, R_n$:

Series system (fails if any component fails): $R_s = \prod_{i=1}^n R_i$
Parallel system (fails only if all fail): $R_p = 1 - \prod_{i=1}^n (1 - R_i)$

For identical components with reliability $p$: $R_s = p^n$, $R_p = 1 - (1-p)^n$.

Definition — k-out-of-n System A k-out-of-n:G system works if at least $k$ of $n$ components function. For i.i.d. components with reliability $p$: \[R_{k/n} = \sum_{j=k}^n \binom{n}{j}p^j(1-p)^{n-j}\] Special cases: $n$-out-of-$n$ = series; $1$-out-of-$n$ = parallel.

Theorem — Coherent Systems & Structure Functions A structure function $\phi(\mathbf{x})$ maps component state vector $\mathbf{x} = (x_1, \ldots, x_n) \in \{0,1\}^n$ to system state $\{0,1\}$. A system is coherent if:

$\phi$ is non-decreasing in each $x_i$ (improving a component cannot harm the system)
Each component is relevant (there exists $\mathbf{x}$ such that $\phi(1_i, \mathbf{x}) \neq \phi(0_i, \mathbf{x})$)

Every coherent system can be expressed via minimal path sets $P_j$ and minimal cut sets $K_j$: \[\phi(\mathbf{x}) = \max_j \prod_{i \in P_j} x_i = \min_j \left[1 - \prod_{i \in K_j}(1-x_i)\right]\]

Definition — Standby Redundancy In standby redundancy, a backup component activates only when the primary fails. For two independent components with lifetimes $T_1 \sim \text{Exp}(\lambda)$ and $T_2 \sim \text{Exp}(\lambda)$ with perfect switching: \[R_{standby}(t) = P(T_1 + T_2 > t) = e^{-\lambda t}(1 + \lambda t)\] Standby redundancy is more reliable than active (parallel) redundancy when components are exponential.

Example

A bridge system has 5 components: 1-3 in series in the upper path, 2-4 in series in the lower path, and component 5 connecting nodes between 1&2 and 3&4. Each has reliability $p = 0.9$. Find the system reliability.

The minimal path sets are $\{1,3\}$, $\{2,4\}$, $\{1,5,4\}$, $\{2,5,3\}$. Using inclusion-exclusion on the minimal paths (or equivalently computing from the structure function): \[R = 2p^2 + 2p^3 - 5p^4 + 2p^5\] At $p = 0.9$: $R = 2(0.81) + 2(0.729) - 5(0.6561) + 2(0.59049) = 1.62 + 1.458 - 3.2805 + 1.18098 = 0.97848$.

Elementary Queuing Models

Birth-death processes, Kendall's notation, and the foundations of queuing theory.

🚦 Queuing Foundations

Definition — Kendall's Notation A queuing system is described as $A/B/c/K/N/D$ where:

$A$ = inter-arrival time distribution (M = Markovian/exponential, D = deterministic, G = general)
$B$ = service time distribution
$c$ = number of servers
$K$ = system capacity (default: $\infty$)
$N$ = calling population size (default: $\infty$)
$D$ = service discipline (default: FCFS)

Definition — Birth-Death Process A continuous-time Markov chain $\{N(t)\}$ on $\{0, 1, 2, \ldots\}$ with transition rates:

Birth rates: $\lambda_n = $ rate of transition from state $n$ to $n+1$
Death rates: $\mu_n = $ rate of transition from state $n$ to $n-1$

The steady-state probabilities satisfy the balance equations: \[\lambda_n p_n = \mu_{n+1} p_{n+1}, \quad n = 0, 1, 2, \ldots\] giving $p_n = p_0 \prod_{i=0}^{n-1}\frac{\lambda_i}{\mu_{i+1}}$ with $\sum p_n = 1$.

Theorem — Little's Law For any stable queuing system in steady state: \[L = \lambda W, \quad L_q = \lambda W_q, \quad W = W_q + \frac{1}{\mu}\] where $L$ = expected number in system, $L_q$ = expected number in queue, $W$ = expected time in system, $W_q$ = expected waiting time in queue, $\lambda$ = effective arrival rate, $\mu$ = service rate. This holds regardless of distributions or discipline.

Example

Customers arrive at a rate of 20 per hour and spend an average of 15 minutes in the system. Find $L$, $L_q$, and $W_q$ if the service rate is 30 per hour.

$W = 15/60 = 0.25$ hours. By Little's law: $L = \lambda W = 20 \times 0.25 = 5$. $W_q = W - 1/\mu = 0.25 - 1/30 = 0.25 - 0.0333 = 0.2167$ hours $\approx 13$ minutes. $L_q = \lambda W_q = 20 \times 0.2167 = 4.33$.

Markovian Queuing Models

M/M/1, M/M/c, M/M/1/K, and M/G/1 queues with performance measures.

📊 M/M/1 and Extensions

Theorem — M/M/1 Queue Poisson arrivals (rate $\lambda$), exponential service (rate $\mu$), single server. For $\rho = \lambda/\mu < 1$ (stability): \[p_n = (1-\rho)\rho^n, \quad L = \frac{\rho}{1-\rho}, \quad L_q = \frac{\rho^2}{1-\rho}\] \[W = \frac{1}{\mu - \lambda}, \quad W_q = \frac{\rho}{\mu - \lambda} = \frac{\lambda}{\mu(\mu - \lambda)}\] The probability that the system is busy: $P(\text{busy}) = \rho$.

Theorem — M/M/1/K Queue (Finite Capacity) With system capacity $K$, steady-state exists for all $\rho$: \[p_n = \begin{cases}\frac{(1-\rho)\rho^n}{1-\rho^{K+1}} & \rho \neq 1 \\ \frac{1}{K+1} & \rho = 1\end{cases}, \quad n = 0, 1, \ldots, K\] The effective arrival rate is $\lambda_{\text{eff}} = \lambda(1 - p_K)$ since arrivals when the system is full are lost.

Theorem — M/M/c Queue (Multiple Servers) With $c$ servers, $\rho = \lambda/(c\mu) < 1$: \[p_0 = \left[\sum_{n=0}^{c-1}\frac{(c\rho)^n}{n!} + \frac{(c\rho)^c}{c!(1-\rho)}\right]^{-1}\] The Erlang-C formula gives the probability of waiting: \[C(c, c\rho) = P(\text{wait}) = \frac{(c\rho)^c}{c!(1-\rho)} \cdot p_0\] \[W_q = \frac{C(c,c\rho)}{c\mu(1-\rho)}, \quad L_q = \lambda W_q\]

Theorem — M/G/1: Pollaczek-Khinchine Formula For M/G/1 (general service distribution with mean $1/\mu$ and variance $\sigma_s^2$): \[L_q = \frac{\rho^2(1 + \mu^2\sigma_s^2)}{2(1-\rho)} = \frac{\lambda^2 E[S^2]}{2(1-\rho)}\] where $E[S^2] = \sigma_s^2 + 1/\mu^2$ is the second moment of service time. This reduces to the M/M/1 result when $\sigma_s^2 = 1/\mu^2$.

Example

An M/M/1 queue has arrival rate $\lambda = 4$/hr and service rate $\mu = 6$/hr. Find the probability that there are more than 3 customers in the system.

$\rho = 4/6 = 2/3$. $P(N > 3) = \rho^4 = (2/3)^4 = 16/81 \approx 0.1975$.
This follows because $P(N \ge n) = \rho^n$ in the M/M/1 queue, so $P(N > 3) = P(N \ge 4) = \rho^4$.

Inventory Models

Deterministic and stochastic inventory models: EOQ, newsvendor, and reorder point policies.

📦 Deterministic Inventory Models

Theorem — Economic Order Quantity (Wilson Formula) For constant demand rate $D$, ordering cost $K$ per order, and holding cost $h$ per unit per time: \[Q^* = \sqrt{\frac{2KD}{h}}, \quad TC(Q^*) = \sqrt{2KDh}\] The optimal order interval is $T^* = Q^*/D$ and the number of orders per unit time is $D/Q^*$. At the optimum, ordering cost equals holding cost.

Definition — EOQ with Shortages When backorders are allowed with shortage cost $p$ per unit per time: \[Q^* = \sqrt{\frac{2KD}{h}} \cdot \sqrt{\frac{h+p}{p}}, \quad S^* = Q^* \cdot \frac{h}{h+p}\] where $S^*$ is the maximum shortage level. As $p \to \infty$, this reduces to the basic EOQ.

🎲 Stochastic Inventory Models

Theorem — Newsvendor Problem A single-period model with random demand $D$, unit cost $c$, selling price $p$, and salvage value $v$ ($v < c < p$). The optimal order quantity $Q^*$ satisfies: \[F(Q^*) = \frac{p - c}{p - v} = \frac{c_u}{c_u + c_o}\] where $c_u = p - c$ is the underage cost and $c_o = c - v$ is the overage cost. This is the critical ratio solution.

Definition — (s, S) Policy & Reorder Point In continuous-review inventory with stochastic demand:

Reorder point: $s = \mu_L + z_\alpha \sigma_L$ where $\mu_L$ = mean demand during lead time, $\sigma_L$ = std dev of demand during lead time, $z_\alpha$ = safety factor.
Safety stock: $SS = z_\alpha \sigma_L$, providing protection against demand uncertainty during lead time.
(s, S) policy: When inventory drops to or below $s$, order up to level $S$. This is optimal under certain cost structures (fixed + variable ordering cost).

Example

A newsvendor buys newspapers at $0.50 and sells at $1.00. Unsold papers are worthless. Daily demand is $N(100, 20^2)$. Find the optimal order quantity.

Critical ratio: $\frac{p - c}{p - v} = \frac{1.00 - 0.50}{1.00 - 0} = 0.5$. $F(Q^*) = 0.5$, so $Q^* = \mu = 100$ (the median of the normal distribution). Since the normal is symmetric, the newsvendor orders exactly the mean demand when the critical ratio equals 0.5.

Key Takeaways

The hazard function uniquely determines the lifetime distribution via $R(t) = e^{-H(t)}$; constant hazard characterizes the exponential.
The Kaplan-Meier estimator handles censored data non-parametrically; Greenwood's formula gives its variance.
Series systems multiply reliabilities; parallel systems multiply unreliabilities — coherent systems generalize both via minimal path/cut sets.
Little's law $L = \lambda W$ is distribution-free and applies to any stable queue.
The M/M/1 queue has geometric state probabilities; the Pollaczek-Khinchine formula extends results to general service distributions.
The EOQ balances ordering and holding costs; the newsvendor critical ratio balances underage and overage costs.

Practice Problems

Problem 1

A system has Weibull lifetime with $R(t) = e^{-(t/100)^{2.5}}$. Find the hazard function, MTTF, and classify the failure rate.

Show Solution ▼

$h(t) = \frac{2.5}{100}\left(\frac{t}{100}\right)^{1.5} = \frac{2.5 t^{1.5}}{100^{2.5}}$. Since the exponent $\beta = 2.5 > 1$, this is IFR. MTTF $= 100 \cdot \Gamma(1 + 1/2.5) = 100 \cdot \Gamma(1.4) = 100 \times 0.8873 = 88.73$.

Problem 2

Five items are tested; failures at 50, 120, 200 and two items censored at 150, 180. Compute the Kaplan-Meier estimate $\hat{R}(160)$.

Show Solution ▼

At $t = 50$: $n_1 = 5, d_1 = 1$, $\hat{R} = 4/5$. At $t = 120$: $n_2 = 4, d_2 = 1$, $\hat{R} = (4/5)(3/4) = 3/5$. Censored at 150, 180 (after 120, before next failure at 200). At $t = 160$: $\hat{R}(160) = 3/5 = 0.6$.

Problem 3

A 2-out-of-3:G system has component reliabilities $R_1 = 0.9, R_2 = 0.8, R_3 = 0.7$. Compute the system reliability.

Show Solution ▼

$R_{sys} = R_1 R_2 + R_1 R_3 + R_2 R_3 - 2R_1 R_2 R_3$ $= 0.72 + 0.63 + 0.56 - 2(0.504) = 1.91 - 1.008 = 0.902$. This uses inclusion-exclusion on the event that at least 2 of 3 components work.

Problem 4

An M/M/2 queue has $\lambda = 6$/hr and $\mu = 4$/hr per server. Find $p_0$ and the expected number in queue $L_q$.

Show Solution ▼

$\rho = \lambda/(c\mu) = 6/8 = 0.75$. $c\rho = 1.5$. $p_0 = \left[1 + 1.5 + \frac{1.5^2}{2(1-0.75)}\right]^{-1} = \left[1 + 1.5 + 4.5\right]^{-1} = 1/7 \approx 0.143$. $C(2, 1.5) = \frac{1.5^2}{2 \times 0.25} \times p_0 = 4.5/7 \approx 0.643$. $L_q = \frac{C(2,1.5) \times \rho}{1-\rho} = \frac{0.643 \times 0.75}{0.25} = 1.929$.

Problem 5

Annual demand is 10,000 units, ordering cost $100 per order, holding cost $2 per unit per year. Find the EOQ, total cost, and optimal number of orders per year.

Show Solution ▼

$Q^* = \sqrt{2 \times 100 \times 10000 / 2} = \sqrt{1000000} = 1000$ units. $TC = \sqrt{2 \times 100 \times 10000 \times 2} = \sqrt{4000000} = 2000$ per year. Orders per year $= D/Q^* = 10000/1000 = 10$. Order interval $= 365/10 = 36.5$ days.

Self-Assessment Quiz

1. The memoryless property $P(T > s+t \mid T > s) = P(T > t)$ characterizes which distribution?

A Weibull

B Exponential

C Normal

D Lognormal

2. In Type II censoring with $n$ items, the test ends when:

A A pre-specified time is reached

B A pre-specified number of failures occurs

C All items have failed

D Items are randomly withdrawn

3. For an M/M/1 queue with $\rho = 0.8$, the expected number in the system is:

A 3.2

B 4

C 0.8

D 5

4. Little's Law $L = \lambda W$ requires:

A Poisson arrivals

B Single server and FCFS discipline

C Only that the system is in steady state

D Exponential service times

5. In the EOQ model, at the optimal order quantity:

A Annual ordering cost equals annual holding cost

B Total cost is zero

C Holding cost dominates ordering cost

D Average inventory equals demand