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Lecture 2 IEGR 459: Introduction to Logistics Management and Supply Chain James Ngeru Industrial and System Engineering Reliability • Definition: – The probability that a system or a product will preform in a satisfactory manner for a given period of time when used under specified operating conditions. • Note that the definition emphasizes on the following elements: – – – – Probability Satisfactory performance Time Operating conditions – Reliability is important as the following are dependent on it: • Availability – Maintainability Reliability • Reliability Function – R(t) = 1 – F(t) F(t) is the probability that the system will fail Reliability curve for the exponential distribution Reliability – Example • Find the failure rate of the component – Component 1 failed after 75 hours – Component 2 failed after 125 hours – Component 3 failed after 130 hours – Component 4 failed after 325 hours – Component 5 failed after 525 hours Solution: – Total Operating Time = 75+125+130+325+525 = 3805 – Failure rate, λ = 5/3805 = 0.001314 Reliability – Example 2 A system operational cycle. Failure Rate λ = Number of failures/Total Mission Time = 6/142 = 0.0422535 Mean Time Between Failure MTBF = 1/ λ = 23.6667 hours Reliability R(t) = e –t/m = e –λt = 0.002479 Reliability Nomograph. Example: If for the example in previous slide, what would be reliability of the system if it is expected to run for 10hrs? Find using the reliability nomogragh Join the points with straight line - 1st point- On MTBF scale = 23.667 - 2nd point - On operating time scale = 10hrs The systems reliability is where the line intersects the reliability scale approx.. 0.65 Check: RS = e -λt = e -t/m = e -10/23.667 = 0.65538 Reliability nomograph for the exponential failure distribution. Source:NAVAIR 00-65-502/NAVORD OD 41146, Reliability Engineering Handbook, Naval Air Systems Command and Naval Ordnance Systems Command. Typical failure-rate curve relationships. Failure-rate curve with maintenance (software application). Reliability - Series Network Reliability (RS) = (RA)(RB)(RC) • R(t) = e-λt • R(s) = e-(λ1 + λ2 + λ3 + …… + λn )t Example 1: Consider a system that includes transmitter (A), receiver (B) and power supply (C). If the transmitter reliability is 0.9712, the receiver reliability is 0.8521 and that of power supply is 0.9357, what is total system reliability if are all connected in series. Solution: RS = (0.9712)(0.8521)(0.9357) = 0.7743 Example 2: A system with four (A, B, C and D) subsystems configured in series consists is expected to run for 1000 hours. The MTBF for A = 6000hrs, B = 4500 hrs, C = 10,500 hrs and D = 3200 hrs. Determine the overall reliability of the series network. Solution: RS = e – (0.000797)(1000) Reliability - Parallel Network System 1 Reliability (Rs) = RA + RB – (RA)(RB) System 2 R(s) = 1- (1- RA) (1-RB )(1-Rc) If n identical components in parallel R(s) = 1 – (1-R)n Example 1: Consider a system of two identical components configured in parallel series, each with reliability of 0.95. What is the total system reliability Solution: RS = 0.95 +0.95 – (0.95)(0.95) = 0.9975 Example 2: Consider a system of three identical components configured in parallel, each with reliability of 0.95. What is the total system reliability Solution: RS = 1 – (1 - 0.95)3 = 0.999875 Some combined series–parallel networks. (Rs) = RA (RB + RC – RBRC) (Rs) = (RA + RB – RARB) (RC + RD – RCRD) Note In the book: Rs = [(1-(1-RA)(1-RB)][1-(1-RC)(1-RD)] = [1-(1-RA-RD+RARB)][1-(1-RC-RD +RCRD)] = (1-1+RA+RB-RARB)(1-1+RC+RD-RCRD) =(RA+RB-RARB)(RC+RD-RCRD) (Rs) = [1-(1-RA)(1-RB)(1-RC)]*RD*[(RE + RF-RERF)