Let's solve the problem step by step: ## Step 1: Label the System Let’s label the blocks for clarity: ``` A (0.8) ──┬── B (0.9) ──┬ │ │ ├─ C (0.9) ───┤ │ │ D (0.9) ──┴── E (0.8) ──┘ ``` - A = 0.8 - B = 0.9 - C = 0.9 - D = 0.9 - E = 0.8 There are two parallel paths: - Upper path: A → B - Middle path: C - Lower path: D → E ## Step 2: Find All Tie-Sets (Minimal Path Sets) A tie-set (path set) is a set of components whose *functioning* ensures system success. **Path 1:** A → B **Path 2:** C **Path 3:** D → E ## Step 3: Write Reliability for Each Path - Path 1: Reliability = P(A) × P(B) = 0.8 × 0.9 = **0.72** - Path 2: Reliability = P(C) = **0.9** - Path 3: Reliability = P(D) × P(E) = 0.9 × 0.8 = **0.72** ## Step 4: System Reliability Using Tie-Set (Path Set) Method System works if **at least one path works**. Let \( R_1, R_2, R_3 \) be the reliabilities of each path. System Reliability, \( R_{sys} \): \[ R_{sys} = P(\text{at least one works}) = 1 - P(\text{all fail}) \] Probability all fail: - Path 1 fails: \( 1 - 0.72 = 0.28 \) - Path 2 fails: \( 1 - 0.9 = 0.1 \) - Path 3 fails: \( 1 - 0.72 = 0.28 \) All paths fail: \( 0.28 \times 0.1 \times 0.28 = 0.00784 \) So, \[ R_{sys} = 1 - 0.00784 = \boxed{0.9922} \] --- ### Step 5: Verify Using Cut-Set Method A minimal cut-set is a set whose *failure* causes system failure. Let's find all minimal cut-sets. #### Cut-Set 1: (A and B) fail #### Cut-Set 2: (C) fails #### Cut-Set 3: (D and E) fail System fails if **all three parallel paths fail**: - Path 1 fails: either A or B fails (\(1 - 0.8 \times 0.9 = 1 - 0.72 = 0.28\)) - Path 2 fails: C fails (\(1 - 0.9 = 0.1\)) - Path 3 fails: either D or E fails (\(1 - 0.9 \times 0.8 = 1 - 0.72 = 0.28\)) System fails if all three fail simultaneously: \[ P_{fail} = 0.28 \times 0.1 \times 0.28 = 0.00784 \] Thus, \[ R_{sys} = 1 - 0.00784 = \boxed{0.9922} \] --- ## **Final Answer:** \[ \boxed{R_{sys} = 0.9922} \] This matches for both tie-set and cut-set methods.

Let's analyze this reliability problem step-by-step to determine the system's overall reliability. --- ### Step 1: Understand the system configuration Based on the diagram, the system has **three parallel paths** from input to output: - **Path 1 (Top Path):** Components with reliabilities 0.8 and 0.9 in series - **Path 2 (Middle Path):** Single component with reliability 0.9 - **Path 3 (Bottom Path):** Components with reliabilities 0.9 and 0.8 in series --- ### Step 2: Calculate the reliability of each path - **Path 1:** \[ R_{1} = 0.8 \times 0.9 = 0.72 \] - **Path 2:** \[ R_{2} = 0.9 \] - **Path 3:** \[ R_{3} = 0.9 \times 0.8 = 0.72 \] --- ### Step 3: Find the probability that **at least one** path functions Since the paths are in parallel, the system functions if **any** path functions. The probability that **all paths fail**: \[ P_{\text{all fail}} = (1 - R_{1}) \times (1 - R_{2}) \times (1 - R_{3}) = (1 - 0.72) \times (1 - 0.9) \times (1 - 0.72) \] \[ = 0.28 \times 0.1 \times 0.28 = 0.00784 \] Thus, \[ \boxed{ R_{\text{system}} = 1 - P_{\text{all fail}} = 1 - 0.00784 = \boxed{0.99216} } \] --- ### **Final answer:** ### **The system reliability is approximately \(\boxed{0.9922}\).** --- ### **Summary:** - **Method:** Using the parallel path reliability formula. - **Result:** System reliability ≈ 0.9922, meaning the system is highly reliable. Let me know if you'd like further explanation on cut-set methods or diagrammatic reasoning!