About this previous year paper
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
| k-out-of-n | Parallel system |
|---|---|
| At least k units must work | Any one unit must work |
| Partial redundancy | Full redundancy |
(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=R1R2R3...
(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
| Design FMEA | Process FMEA |
|---|---|
| Product design focused | Manufacturing focused |
| Failure due to design | Failure due to process |
| Early stage | Production stage |
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∑nkkΔt
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−λ1t)(1−e−λ2t) 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−λ1t+e−λ2t−e−(λ1+λ2)t