---
title: "Exponential reliability and system reliability"
subject: "Machine Design II"
chapter: 4
key: "semester_7/machine_design_ii:reliability-exponential-series-parallel"
source: https://www.bctengineer.com/q/semester_7/machine_design_ii/reliability-exponential-series-parallel
site: BCT Engineering
---

# Exponential reliability and system reliability

Machine Design II · Chapter 4: Risk and Reliability of Design

- Asked in: year not given, 4+4 marks
- Video: [System Reliability Calculation | Physical Significance of Calculating System Reliability Probability](https://www.youtube.com/watch?v=J4P1WljAoJI) — Engineering Presentations
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_7/machine_design_ii/reliability-exponential-series-parallel
- Whole chapter: https://www.bctengineer.com/md/semester_7/machine_design_ii/chapter-4

## Question

(a) A component follows the exponential failure law with a constant failure rate of $2.5\times10^{-4}$ per hour. Find its reliability for 400 h, its MTBF, the operating time for which reliability is 0.95, and the expected number of failures among 500 such components in 1000 h. (b) A machine has component A (R = 0.95) in series with a parallel pair B and C (each R = 0.90) and then in series with component D (R = 0.98). Draw the reliability block diagram and find the system reliability.

## Short answer

R(400 h) = e^{-0.1} = 0.905, MTBF = 4000 h, time for R=0.95 is 205 h, about 111 failures among 500 components in 1000 h. System reliability with parallel B-C: R_s = 0.95 × [1-(0.1)^2] × 0.98 = 0.922.

## Answer

#### (a) Exponential law
$R(t) = e^{-\lambda t}$, MTBF $= 1/\lambda$, with $\lambda = 2.5\times10^{-4}\ \text{h}^{-1}$.

**Reliability at 400 h**

$$
R(400) = e^{-2.5\times10^{-4}\times 400} = e^{-0.1} = 0.9048
$$

**MTBF**

$$
\text{MTBF} = \frac{1}{2.5\times10^{-4}} = 4000\ \text{h}
$$

**Time for $R = 0.95$**

$$
t = -\frac{\ln 0.95}{\lambda} = \frac{0.05129}{2.5\times10^{-4}} = 205.2\ \text{h}
$$

**Failures among 500 components in 1000 h**

$$
R(1000) = e^{-0.25} = 0.7788,\quad \text{failures} = 500(1-0.7788) = 110.6 \approx 111
$$

#### (b) System reliability
```text
        +--- B (0.90) ---+
 --A----|                |----D----
 (0.95) +--- C (0.90) ---+   (0.98)
```

Parallel pair: $R_{BC} = 1 - (1-0.90)(1-0.90) = 1 - 0.01 = 0.99$

Series system: $R_s = R_A \times R_{BC} \times R_D$

$$
R_s = 0.95 \times 0.99 \times 0.98 = 0.9217
$$

Note that without the redundancy in B and C the reliability would be $0.95\times0.90\times0.98 = 0.838$, so the parallel pair improves the system by about 8 percentage points. The weakest series element (A) now limits reliability.

**Answer:** (a) $R(400)=0.905$, MTBF $=4000$ h, $t_{0.95}=205$ h, about 111 failures; (b) $R_s = 0.922$.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. With a constant failure rate of 2.5e-4 per hour, what is R(400 h)?
   A. 0.905, from e^(-0.1)
   B. 0.100
   C. 0.990
   D. 0.368
   Answer: A. R(t) = e^(-lambda t) = e^(-0.1) = 0.9048. (Key point: Exponential reliability)

2. How is the time for R = 0.95 found, and what is it?
   A. t = lambda x ln 0.95, giving a negative time
   B. t = 1/lambda = 4000 h
   C. t = -ln(0.95)/lambda = 205 h
   D. t = 0.95/lambda = 3800 h
   Answer: C. From R = e^(-lambda t), t = -ln R / lambda = 0.05129 / 2.5e-4 = 205.2 h. (Key point: Time for given reliability)

3. What is the reliability of B and C (each 0.90) in parallel?
   A. 0.81
   B. 0.99
   C. 0.90
   D. 1.80
   Answer: B. R = 1 - (1 - 0.9)(1 - 0.9) = 0.99. Multiplying the reliabilities (0.81) is the series result. (Key point: Parallel pair)

4. What is the reliability of A (0.95) in series with the B-C pair and D (0.98)?
   A. 0.838
   B. 0.950
   C. 0.980
   D. 0.922
   Answer: D. Rs = 0.95 x 0.99 x 0.98 = 0.9217. Without the redundancy it would be 0.838. (Key point: System reliability)

## Related in Machine Design II

- [Reliability from stress-strength interference](https://www.bctengineer.com/question.md?key=semester_7%2Fmachine_design_ii%3Astress-strength-interference-reliability) — Machine Design II, chapter 4
- [FMEA, FTA and design for reliability](https://www.bctengineer.com/question.md?key=semester_7%2Fmachine_design_ii%3Ahazard-analysis-fmea-fta) — Machine Design II, chapter 4
- [Number of plates in a multiplate clutch](https://www.bctengineer.com/question.md?key=semester_7%2Fmachine_design_ii%3Amultiplate-clutch-power-numerical) — Machine Design II, chapter 7

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