Quality Control Problems: System Reliability And Components

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This assignment encompasses several key topics in quality control and reliability engineering, including system reliability calculations, failure rate computations, Weibull distribution application for machinery failure analysis, binomial probability assessments, and basic statistical analysis in design of experiments. The core tasks involve analyzing system configurations, failure data, probability distributions, and statistical measures derived from experimental data. The following sections present comprehensive solutions to each problem based on fundamental principles of reliability engineering, probability theory, and statistical analysis.

Paper For Above instruction

1. System Reliability of Series and Parallel Configurations

In reliability engineering, understanding the reliability of systems composed of multiple components is crucial. When components are arranged in series, the system fails if any component fails; conversely, in parallel arrangements, the system fails only if all components fail.

a) For a system with five components in series, each with a reliability of 0.96, the overall reliability (R_series) can be calculated by multiplying individual reliabilities:

R_series = (0.96)^5 ≈ 0.8154.

b) For a parallel system where all five components are considered, the reliability (R_parallel) is one minus the probability that all components fail simultaneously:

R_parallel = 1 - (1 - 0.96)^5 ≈ 1 - (0.04)^5 ≈ 1 - 1.024 \times 10^{-7} ≈ 0.9999999.

c) The change in reliability from series to parallel configuration is approximately:

ΔR = R_parallel - R_series ≈ 0.9999999 - 0.8154 ≈ 0.1846, indicating a significant increase in system dependability when components are arranged in parallel.

2. Failure Rate Analysis of Medical Devices

(a) Failure Rate Calculation for the Initial 60-Hour Run

Given 10 items and 3 failures at 25, 27, and 35 hours, the failure rate (λ) can be estimated as the total number of failures divided by the total operational time:

Total failure time = 3 failures \times (average of failure times approximated or sum of failure times)

However, more precisely, failure rate is often modeled as failures per unit time, so:

λ = number of failures / total person-hours

Total person-hours before failures = 10 items \times 60 hours = 600 person-hours

Failures during this period = 3

Failure rate per hour = 3 failures / 600 hours = 0.005 failures/hour.

(b) Failure Rate after 80-Hour Continuation with Replacements

During an extended run till 80 hours, including additional failures (two more devices failing during the next 20 hours), the total operational hours per device account for replacement:

- The initial 60 hours with active failures.

- Additional 20 hours with 2 failures.

Total failures = 3 + 2 = 5 failures over the total operational time: (8 devices \times 80 hours) = 640 person-hours.

Failure rate = 5 failures / 640 hours ≈ 0.0078125 failures/hour.

(c) Mean Life Calculation

The mean life (or mean time to failure) based on failure rate λ is given by:

Mean life (T) = 1 / λ

For the initial failure rate: T_1 = 1 / 0.005 = 200 hours.

For the extended period: T_2 = 1 / 0.0078125 ≈ 128 hours.

This indicates the average device life decreases with increased failure activity over time.

3. Reliability of an Engine Using Weibull Distribution

The Weibull distribution is characterized by its shape parameter (β) and scale parameter (η). The reliability function R(t) for a Weibull distribution is:

R(t) = exp[-(t/η)^β].

Given the mean life during the early failure phase T_m = 6260 hours and β = 1.4, we need to estimate η.

Since the mean (T_m) relates to parameters via:

T_m = η \times \Gamma(1 + 1/β),

where Γ is the gamma function. Calculating Γ(1 + 1/1.4) ≈ 0.902, then:

η = T_m / 0.902 ≈ 6260 / 0.902 ≈ 6944 hours.

Therefore, the reliability at 5000 hours is:

R(5000) = exp[-(5000/6944)^1.4] ≈ exp[-(0.72)^1.4] ≈ exp[-0.533] ≈ 0.586.

4. Binomial Distribution: Probability of Nonconforming Units

Given a batch with 6% nonconformance (p=0.06), sampling 5 units, with the probability formula:

P(d) = (n! / [d! (n – d)!]) \times p^d \times q^{n – d},

where q = 1 – p = 0.94.

Part (a) Calculate P(1) and P(2):

P(1) = (5! / [1! 4!]) \times 0.06^1 \times 0.94^4 ≈ 5 \times 0.06 \times 0.778 \approx 0.233.

P(2) = (5! / [2! 3!]) \times 0.06^2 \times 0.94^3 ≈ 10 \times 0.0036 \times 0.830 ≈ 0.030.

Part (b) Probability of 2 or more nonconforming units:

P(≥2) = 1 – P(0) – P(1)

P(0) = (5! / [0! 5!]) \times 0.06^0 \times 0.94^5 ≈ 1 \times 1 \times 0.735 ≈ 0.735.

P(≥2) ≈ 1 – 0.735 – 0.233 ≈ 0.032.

This indicates a 3.2% chance of observing two or more nonconforming units in the sample.

5. Statistical Analysis of Piston Cylinder Diameters

The diameters are: 15.0, 15.1, 15.8, 15.3, 15.5 cm.

Calculate sum of squares (SS):

First, compute the mean:

Mean = (15.0 + 15.1 + 15.8 + 15.3 + 15.5) / 5 = 76.7 / 5 = 15.34 cm.

Sum of squares:

SS = Σ (x_i – mean)^2

= (15.0–15.34)^2 + (15.1–15.34)^2 + (15.8–15.34)^2 + (15.3–15.34)^2 + (15.5–15.34)^2

= (–0.34)^2 + (–0.24)^2 + (0.46)^2 + (–0.04)^2 + (0.16)^2

= 0.1156 + 0.0576 + 0.2116 + 0.0016 + 0.0256 = 0.412.

Calculate variance (Mean Square):

Variance (s^2) = SS / (n - 1) = 0.412 / 4 ≈ 0.103.

Standard deviation (s):

s = √variance ≈ √0.103 ≈ 0.321 cm.

Conclusion

Reliability calculations for systems in series and parallel configurations reveal significant improvements in system dependability with parallel arrangements. Failure rate analysis of medical devices underscores the importance of maintenance and lifespan estimation, where failure rates directly influence mean life estimations. Weibull distribution fitting enables accurate prediction of engine reliability over extended periods, essential for maintenance scheduling and safety assessments. Binomial probability computations assist quality control by estimating the likelihood of defective units, guiding inspection standards. Finally, statistical analysis of manufacturing data such as piston diameters helps in quality assurance, process control, and identifying variability sources. These integrated approaches form a foundation for effective quality management in engineering and manufacturing environments.

References

• Austin, S. M. (2010). Operational Reliability and Risk Analysis. Wiley.
• Meeker, W. Q., & Escobar, L. A. (1998). Statistical Methods for Reliability Data. Wiley.
• Mann, J. H., & Kelleher, J. (2017). Reliability Engineering and Risk Analysis. Springer.
• Crowder, W., & Treat, J. (2000). Reliability of engineering systems. IEEE Transactions.
• Levi, H. (2018). The Weibull Distribution: Theory and Applications. Journal of Reliability Engineering.
• Montgomery, D. C. (2019). Design and Analysis of Experiments. Wiley.
• Fenton, L., & Emad, S. (2017). Quality Control: Principles and Practice. International Journal of Quality & Reliability Management.
• Kimura, F. (2020). Probabilistic Models in Manufacturing and Quality Control. Quality Engineering Journal.
• Subramanian, N., & Ganesh, K. (2016). Failure data analysis using Weibull distribution. Reliability Engineering & System Safety.
• Ghosh, S., & Mukhopadhyay, S. (2017). Statistical analysis in reliability engineering. Applied Stochastic Models in Business and Industry.