# Industrial Engineering Questions and Answers – Reliability System – Series, Parallel and Combination

This set of Industrial Engineering Multiple Choice Questions & Answers (MCQs) focuses on “Reliability System – Series, Parallel and Combination”.

1. Find the reliability of the system having fours units in series with failure rates of 0.001, 0.002, 0.003, and 0.004 for the total operation of 100 hours.
a) 0.366
b) 0.663
c) 0.899
d) 0.998

Explanation: Given, a system with fours unit in series each has failure rates as 0.001, 0.002, 0.003, and 0.004. Therefore, the reliability of each unit is calculated as below:
R = e-λt; λ = Failure rate
R1 = e-0.001 × 100 = 0.904
R2 = e-0.002 × 100 = 0.818
R3 = e-0.003 × 100 = 0.740
R4 = e-0.004 × 100 = 0.670
As all the units are in series, Reliability of the system = 0.904 × 0.818 × 0.740 × 0.670 = 0.366

2. What is the reliability of the system if it has four units parallel to each other with failure rates of 0.001, 0.002, 0.003, and 0.004 for the total operation of 100 hours?
a) 0.366
b) 0.663
c) 0.899
d) 0.998

Explanation: Given, a system with fours unit in parallel to each other has failure rates as 0.001, 0.002, 0.003, and 0.004. Therefore, the reliability of each unit is calculated as below:
R = e-λt; λ = Failure rate
R1 = e-0.001 × 100 = 0.904
R2 = e-0.002 × 100 = 0.818
R3 = e-0.003 × 100 = 0.740
R4 = e-0.004 × 100 = 0.670
As all the units are in parallel, Reliability of the system = 1 – $$\Pi_{i=1}^n$$(1 – Ri) = 1 – [(1 – 0.904) × (1 – 0.818) × (1 – 0.740) × (1 – 0.670)] = 0.9985

3. Calculate the system reliability for the figure shown below.

a) 0.245
b) 0.542
c) 0.452
d) 0.254

Explanation: Given,
R1 = 0.82;R2 = 0.82; R3 = 0.82; R4 = 0.82;
System Reliability = $$\Pi_{i=1}^n$$(Ri) = 0.82 × 0.82 × 0.82 × 0.82 = 0.452

4. Calculate the system reliability for the figure shown below.

a) 0.898
b) 0.892
c) 0.998
d) 0.899

Explanation: Given, units in parallel to each other
R1 = 0.82; R2 = 0.82; R3 = 0.82; R4 = 0.82;
As all the units are in parallel, Reliability of the system = 1 – $$\Pi_{i=1}^n$$(1 – Ri) = 1 – [(1 – 0.82) × (1 – 0.82) × (1 – 0.82) × (1 – 0.82)] = 0.998

5. Calculate the system reliability for the figure shown below.

Consider P as probability of failure.
a) 0.245
b) 0.542
c) 0.452
d) 0.001

Explanation: Given, Probability of failures as 0.82 for all the four units which are in series in the given system.
We know that, R + P = 1 i.e., the sum of reliability and probability of failure is always one.
Hence, the reliability of all the units is,
R1 = 1 – P1 = 1 – 0.82 = 0.18 R2 = 0.18; R3 = 0.18; R4 = 0.18;
∴ Reliability of the system = $$\Pi_{i=1}^n$$(Ri) = 0.18 × 0.18 × 0.18 × 0.18 = 0.001

6. Calculate the system reliability for the figure shown below.

a) 0.245
b) 0.547
c) 0.452
d) 0.254

Explanation: Given, Probability of failures as 0.82 for all the four units which are in series in the given system.
We know that,
R + P = 1 i.e., the sum of reliability and probability of failure is always one.
Hence, the reliability of all the units is,
R1 = 1 – P1 = 1 – 0.82 = 0.18 R2 = 0.18; R3 = 0.18; R4 = 0.18;
As all the units are in parallel, Reliability of the system = 1 – $$\Pi_{i=1}^n$$(1 – Ri) = 1 – [(0.82) × (0.82) × (0.82) × (0.82)] = 0.547

7. Which of the following equation is/are not correct for a system with all the units connected in series?
a) RS = 1 – $$\Pi_{i=1}^n$$(1 – Ri)
b) RS = $$\Pi_{i=1}^n$$(Ri)
c) RS = 1 – $$\sum_{i=1}^n$$(1 – Ri)
d) RS = $$\sum_{i=1}^n$$(Ri)

Explanation: An example of a system with all the units connected in series is shown below.

For a system with components/units connected in series to each other, system reliability can be calculated using the formula RS = $$\Pi_{i=1}^n$$.

8. Which of the following equation is/are not correct for a system with all the units connected in parallel to each other?
a) RS = 1 – $$\Pi_{i=1}^n$$(1 – Ri)
b) RS = $$\Pi_{i=1}^n$$(Ri)
c) RS = 1 – $$\sum_{i=1}^n$$(1 – Ri)
d) RS = $$\sum_{i=1}^n$$(Ri)

Explanation: An example of a system with all the units connected in series is shown below.

For a system with components/units connected in parallel to each other, system reliability can be calculated using the formula RS = 1 – ($$\Pi_{i=1}^n$$(1 – Ri))

9. Determine the system reliability for the hybrid system shown below.

a) 0.849
b) 0.489
c) 0.949
d) 0.994

Explanation: Given, a unit with reliability R1 in series with a sub-system consists of units with reliability R2, R3, R4 and R5 parallel to each other.
∴ System reliability = RS = ($$\Pi_{i=1}^1$$(Ri)) × (1 – ($$\Pi_{i=2}^5$$(1-Ri))) = 0.85 × (1 – ((1 – 0.91) × (1 – 0.96) × (1 – 0.89) × (1 – 0.98))) = 0.849

10. Determine the system reliability for the hybrid system shown below.

a) 0.849
b) 0.489
c) 0.949
d) 0.994

Explanation: Given, a sub-system consists of units with reliabilities R1, R2, R3 and R4 in parallel to each other in series with a unit with reliability R5.
∴ System reliability = RS = (1 – ($$\Pi_{i=1}^4$$(1-Ri))) × ($$\Pi_{i=5}^5$$(Ri))= 0.85 × (1 – ((1 – 0.91) × (1 – 0.96) × (1 – 0.89) × (1 – 0.98))) = 0.849

11. Determine the system reliability for the hybrid system shown below.

a) 0.721
b) 0.271
c) 0.127
d) 0.217

Explanation: Given, a unit with reliability R1 in series with a sub-system consists of units with reliability R2, R3, R4 and R5 parallel to each other which is in series with a unit of reliability R6.
∴ System reliability = RS = ($$\Pi_{i=1}^1$$(Ri)) × (1 – ($$\Pi_{i=2}^5$$(1 – Ri))) × ($$\Pi_{i=1}^1$$(Ri)) = 0.85 × (1 – ((1 – 0.91) × (1 – 0.96) × (1 – 0.89) × (1 – 0.98))) × 0.85 = 0.721

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