THEORY EXAMINATION (SEM–VI) 2016-17 RELIABILITY ENGINEERING

RELIABILITY ENGINEERING (EME014)

Section-wise Solved Answers & Notes

SECTION – A (10 × 2 = 20 Marks)

Very short, precise answers

(a) Reliability

Reliability is the probability that a system or component performs its intended function without failure for a specified time under stated conditions.

(b) Reliability graph using probability density function

Reliability can be represented using probability density function (PDF), where failure rate is plotted against time.
Reliability function:

R(t)=1−F(t)R(t)=1-F(t)R(t)=1−F(t) 

(c) Testing goodness of fit of data

Goodness of fit is tested using Chi-Square test, Kolmogorov–Smirnov test, or graphical methods to check how well data follows a probability distribution.

(d) Availability & Maintainability

• Availability: Probability that system is operational at a given time
• Maintainability: Ability of system to be restored to working condition in a given time

(e) MTTF and MTBF

• MTTF: Mean Time To Failure (non-repairable systems)
• MTBF: Mean Time Between Failures (repairable systems)

(f) k-out-of-n vs n-unit parallel system

(g) Minimal cut-set

A minimal cut-set is the smallest set of components whose failure causes total system failure.

(h) System reliability (3 subsystems in parallel, R = 0.8 each)

Rs=1−(1−0.8)3=1−0.008=0.992R_s = 1 - (1-0.8)^3 = 1 - 0.008 = \boxed{0.992}Rs​=1−(1−0.8)3=1−0.008=0.992​ 

(i) Weibull distribution

Weibull distribution is used to model early failures, random failures, and wear-out failures using shape parameter β.

(j) Objectives of life tests                               • Estimate reliability
• Determine failure pattern                           • Improve product design
• Predict service life

SECTION – B (Any 5 × 10 = 50 Marks)

(a) Series & Series-Parallel Configuration

(i) Series configuration                                  System fails if any component fails.

Rs=R1R2R3...R_s = R_1 R_2 R_3 ...Rs​=R1​R2​R3​...

(ii) Series-parallel system

Combination of series and parallel subsystems used to increase reliability.
(Neat block diagram should be drawn)

(b) Design FMEA vs Process FMEA

Methodology of system analysis:
• Identify failures                                    • Analyze causes
• Evaluate risk                                         • Recommend corrective actions

(c) Reliability Testing

Reliability testing evaluates performance under specified conditions.

Types:                                                      • Life testing
• Accelerated testing                               • Environmental testing

(d) Chi-Square Goodness of Fit Test

Steps:                                                       State null hypothesis

Group observed data                              Calculate expected frequency

Compute χ² value                                    χ2=∑(O−E)2E\chi^2 = \sum \frac{(O-E)^2}{E}χ2=∑E(O−E)2​

Compare with table value

(e) Reliability of given system                 (System diagram provided in question)

Method:                                                   • Reduce series & parallel blocks step-by-step
• Multiply series reliabilities                     • Use parallel reliability formula

Rp=1−(1−R1)(1−R2)R_p = 1 - (1-R_1)(1-R_2)Rp​=1−(1−R1​)(1−R2​) 

(f) Derivation of MTTF equation

MTTF=1N∑nkkΔtMTTF = \frac{1}{N}\sum n_k k \Delta tMTTF=N1​∑nk​kΔt

Where:                                                      • nkn_knk​ = failures in interval
• kΔtk\Delta tkΔt = midpoint time

(g) Availability numerical

Given:                                                         MTBF = 30 hrs
MTTR = 15 hrs                                           Administrative delay = 30% of MDT

MDT=15+0.3(15)=19.5 hrsMDT = 15 + 0.3(15) = 19.5\ hrsMDT=15+0.3(15)=19.5 hrs

Inherent Availability:

Ai=MTBFMTBF+MTTR=3045=0.667A_i = \frac{MTBF}{MTBF+MTTR} = \frac{30}{45} = \boxed{0.667}Ai​=MTBF+MTTRMTBF​=4530​=0.667​

Operational Availability:

Ao=MTBFMTBF+MDT=3049.5=0.606A_o = \frac{MTBF}{MTBF+MDT} = \frac{30}{49.5} = \boxed{0.606}Ao​=MTBF+MDTMTBF​=49.530​=0.606​ 

(h) Minimal cut-set method

Steps:                                                            • Identify all minimal cut-sets
• Find probability of each cut-set                 • Combine to find system unreliability

Used for complex system reliability analysis.

SECTION – C (Any 2 × 15 = 30 Marks)

Q3. Probability using Normal Distribution

Given:                                                            Mean μ = 56,669.5
Standard deviation σ = 12,393.64                 X = 50,000

Z=X−μσ=50000−56669.512393.64=−0.53Z = \frac{X-\mu}{\sigma} = \frac{50000-56669.5}{12393.64} = -0.53Z=σX−μ​=12393.6450000−56669.5​=−0.53

From table:

P(Z<−0.53)=1−0.7019=0.2981P(Z < -0.53) = 1 - 0.7019 = \boxed{0.2981}P(Z<−0.53)=1−0.7019=0.2981​ 

Q4.

(a) Minimum component reliability              Series system with n = 10

Rs=(R)n=0.9R_s = (R)^n = 0.9Rs​=(R)n=0.9 R=(0.9)1/10=0.9895R = (0.9)^{1/10} = \boxed{0.9895}R=(0.9)1/10=0.9895​ 

(b)

(i) Hazard rate: Rate of failure at time t    λ(t)=f(t)R(t)\lambda(t)=\frac{f(t)}{R(t)}λ(t)=R(t)f(t)​

(ii) Probability density function:             Describes likelihood of failure at time t.

Q5. Parallel System Reliability

Given failure rates λ₁, λ₂:

Rp(t)=1−(1−e−λ1t)(1−e−λ2t)R_p(t)=1-(1-e^{-\lambda_1 t})(1-e^{-\lambda_2 t})Rp​(t)=1−(1−e−λ1​t)(1−e−λ2​t) Rp(t)=e−λ1t+e−λ2t−e−(λ1+λ2)tR_p(t)=e^{-\lambda_1 t}+e^{-\lambda_2 t}-e^{-(\lambda_1+\lambda_2)t}Rp​(t)=e−λ1​t+e−λ2​t−e−(λ1​+λ2​)t