Training Systems Engineering System Reliability — Series & Parallel
1 / 5

System Reliability — Series & Parallel

24 min Systems Engineering

System Reliability — Series & Parallel

Reliability is the probability that a system performs its intended function under specified conditions for a given time period.

Series System Reliability

All components must work. If components have reliabilities $R_1, R_2, \ldots, R_n$:

$$R_{\text{series}} = R_1 \times R_2 \times \cdots \times R_n = \prod_{i=1}^{n} R_i$$

Parallel (Redundant) System Reliability

At least one component must work:

$$R_{\text{parallel}} = 1 - \prod_{i=1}^{n}(1 - R_i)$$

Exponential Reliability Model

$$R(t) = e^{-\lambda t}$$

where $\lambda$ = failure rate (failures/hr). Mean Time Between Failures: $MTBF = 1/\lambda$.

Example 1

Three components in series have reliabilities 0.95, 0.90, and 0.99. Find system reliability.

$$R = 0.95 \times 0.90 \times 0.99 = 0.846$$

Example 2

Two identical components ($R = 0.90$) in parallel. Find system reliability.

$$R = 1 - (1 - 0.90)^2 = 1 - 0.01 = 0.99$$

Redundancy dramatically improves reliability.

Example 3

A component has $\lambda = 0.001$ failures/hr. Find $R(100)$ and MTBF.

$R(100) = e^{-0.001 \times 100} = e^{-0.1} = 0.905$

$MTBF = 1/0.001 = 1{,}000$ hrs

Practice Problems

1. Five components in series, each with $R = 0.98$. Find system reliability.
2. Three identical components ($R = 0.85$) in parallel. Find system reliability.
3. A series-parallel system: two paths in parallel, each path has two components in series ($R = 0.9$ each). Find system reliability.
4. $\lambda = 0.0005$/hr. Find $R(500)$ and MTBF.
5. If MTBF = 5{,}000 hrs, find the reliability at 1{,}000 hrs.
6. A system needs $R \ge 0.99$ for 100 hrs. Find the maximum $\lambda$.
7. Adding a redundant component ($R = 0.9$) in parallel to an existing one ($R = 0.9$). How much does reliability improve?
8. Availability $= MTBF/(MTBF + MTTR)$. If MTBF $= 500$ hrs, MTTR $= 5$ hrs, find availability.
9. A k-out-of-n system: 2-out-of-3 with $R = 0.95$ each. Find system reliability.
10. If a system has 100 components in series, each at $R = 0.999$, find system $R$.
11. Failure rate of a system in series: $\lambda_{\text{sys}} = \sum \lambda_i$. If 5 components each have $\lambda = 0.002$/hr, find $\lambda_{\text{sys}}$ and MTBF.
12. A standby system has a primary ($R_1 = 0.9$) and backup ($R_2 = 0.9$). Reliability $= R_1 + (1-R_1)R_2$. Find system $R$.
Show Answer Key

1. $R = 0.98^5 = 0.904$

2. $R = 1 - (0.15)^3 = 1 - 0.003375 = 0.9966$

3. Each path: $0.9 \times 0.9 = 0.81$. Parallel: $1 - (0.19)^2 = 0.9639$

4. $R(500) = e^{-0.25} = 0.779$; MTBF $= 2{,}000$ hrs

5. $\lambda = 1/5000 = 0.0002$; $R(1000) = e^{-0.2} = 0.819$

6. $e^{-100\lambda} \ge 0.99$; $\lambda \le -\ln(0.99)/100 = 0.01005/100 = 1.005 \times 10^{-4}$/hr

7. From $0.9$ to $1 - 0.01 = 0.99$; improvement of $0.09$ (10% increase in reliability)

8. $A = 500/505 = 0.9901 = 99.01\%$

9. $R = \binom{3}{2}(0.95)^2(0.05) + \binom{3}{3}(0.95)^3 = 3(0.9025)(0.05) + 0.857 = 0.1354 + 0.8574 = 0.9928$

10. $R = 0.999^{100} = e^{100 \ln 0.999} = e^{-0.1} = 0.905$

11. $\lambda_{\text{sys}} = 0.01$/hr; MTBF $= 100$ hrs

12. $R = 0.9 + 0.1(0.9) = 0.99$

Feedback Systems & Block Diagrams