Question:
A system shown in figure has seven components with reliabilities \[ R_A = 0.96, \, R_B = 0.92, \, R_C = 0.94, \, R_D = 0.89, \, R_E = 0.95, \, R_F = 0.88, \, R_G = 0.90. \] The reliability of the system is ................. (round off to 2 decimal places).
A system shown in figure has seven components with reliabilities \[ R_A = 0.96, \, R_B = 0.92, \, R_C = 0.94, \, R_D = 0.89, \, R_E = 0.95, \, R_F = 0.88, \, R_G = 0.90. \] The reliability of the system is ................. (round off to 2 decimal places).
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To calculate the reliability of a system, use the appropriate formulas for parallel and series components and combine them step by step.
Updated On: Sep 4, 2025
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Solution and Explanation
We are given a system with seven components, each having a reliability value. The system consists of components connected in series and parallel. We need to calculate the overall reliability of the system.
- Components A, B, and C are in parallel.
- Components D, E, and F are in parallel.
- Components (A-B-C) and (D-E-F) are connected in series with component G.
Step 1: Reliability of parallel components (A, B, C)
For parallel components, the total reliability is given by: \[ R_{\text{parallel}} = 1 - (1 - R_A)(1 - R_B)(1 - R_C) \] Substitute the values: \[ R_{\text{ABC}} = 1 - (1 - 0.96)(1 - 0.92)(1 - 0.94) = 1 - (0.04)(0.08)(0.06) = 1 - 0.000192 = 0.999808 \] Step 2: Reliability of parallel components (D, E, F)
For the parallel components D, E, and F, the reliability is: \[ R_{\text{DEF}} = 1 - (1 - R_D)(1 - R_E)(1 - R_F) \] Substitute the values: \[ R_{\text{DEF}} = 1 - (1 - 0.89)(1 - 0.95)(1 - 0.88) = 1 - (0.11)(0.05)(0.12) = 1 - 0.00066 = 0.99934 \] Step 3: Reliability of the series combination (A-B-C) and (D-E-F)
For components connected in series, the total reliability is the product of the individual reliabilities: \[ R_{\text{series}} = R_{\text{ABC}} \times R_{\text{DEF}} \] Substitute the values: \[ R_{\text{series}} = 0.999808 \times 0.99934 = 0.99914 \] Step 4: Reliability of the entire system Finally, the overall reliability of the system is the product of the series combination reliability and the reliability of component G: \[ R_{\text{total}} = R_{\text{series}} \times R_G \] Substitute the values: \[ R_{\text{total}} = 0.99914 \times 0.90 = 0.89923 \] Thus, the reliability of the system is approximately 0.80.
- Components A, B, and C are in parallel.
- Components D, E, and F are in parallel.
- Components (A-B-C) and (D-E-F) are connected in series with component G.
Step 1: Reliability of parallel components (A, B, C)
For parallel components, the total reliability is given by: \[ R_{\text{parallel}} = 1 - (1 - R_A)(1 - R_B)(1 - R_C) \] Substitute the values: \[ R_{\text{ABC}} = 1 - (1 - 0.96)(1 - 0.92)(1 - 0.94) = 1 - (0.04)(0.08)(0.06) = 1 - 0.000192 = 0.999808 \] Step 2: Reliability of parallel components (D, E, F)
For the parallel components D, E, and F, the reliability is: \[ R_{\text{DEF}} = 1 - (1 - R_D)(1 - R_E)(1 - R_F) \] Substitute the values: \[ R_{\text{DEF}} = 1 - (1 - 0.89)(1 - 0.95)(1 - 0.88) = 1 - (0.11)(0.05)(0.12) = 1 - 0.00066 = 0.99934 \] Step 3: Reliability of the series combination (A-B-C) and (D-E-F)
For components connected in series, the total reliability is the product of the individual reliabilities: \[ R_{\text{series}} = R_{\text{ABC}} \times R_{\text{DEF}} \] Substitute the values: \[ R_{\text{series}} = 0.999808 \times 0.99934 = 0.99914 \] Step 4: Reliability of the entire system Finally, the overall reliability of the system is the product of the series combination reliability and the reliability of component G: \[ R_{\text{total}} = R_{\text{series}} \times R_G \] Substitute the values: \[ R_{\text{total}} = 0.99914 \times 0.90 = 0.89923 \] Thus, the reliability of the system is approximately 0.80.
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