Reliability Engineering

Slides:



Advertisements
Similar presentations
FAILURE MODE EFFECTIVE ANALYSIS
Advertisements

Chapter 12 Maintainability and Availability introduction Availability analysis of the system requires a knowledge: (1)How the components are functional.
Disk Arrays COEN 180. Large Storage Systems Collection of disks to store large amount of data. Performance advantage: Each drive can satisfy only so many.
Chapter 2 Solutions of Systems of Linear Equations / Matrix Inversion
Fault Tree Analysis Part 12 – Redundant Structure and Standby Units.
Chapter 8 Continuous Time Markov Chains. Markov Availability Model.
6. Reliability Modeling Reliable System Design 2010 by: Amir M. Rahmani.
Markov Analysis Jørn Vatn NTNU.
Oct. 2007State-Space ModelingSlide 1 Fault-Tolerant Computing Motivation, Background, and Tools.
Probability and Statistics with Reliability, Queuing and Computer Science Applications: Chapter 8 on Continuous-Time Markov Chains Kishor Trivedi.
Reliable System Design 2011 by: Amir M. Rahmani
Oct State-Space Modeling Slide 1 Fault-Tolerant Computing Motivation, Background, and Tools.
A. BobbioBertinoro, March 10-14, Dependability Theory and Methods 5. Markov Models Andrea Bobbio Dipartimento di Informatica Università del Piemonte.
CSE 221: Probabilistic Analysis of Computer Systems Topics covered: Pure death process Availability analysis (Sec , 8.4.1)
CSE 322: Software Reliability Engineering Topics covered: Software reliability models with explicit fault removal.
1 Spare part modelling – An introduction Jørn Vatn.
Part 2.  Review…  Solve the following system by elimination:  x + 2y = 1 5x – 4y = -23  (2)x + (2)2y = 2(1)  2x + 4y = 2 5x – 4y = -23  7x = -21.
System Reliability. Random State Variables System Reliability/Availability.
1 Logistics Systems Engineering Availability NTU SY-521-N SMU SYS 7340 Dr. Jerrell T. Stracener, SAE Fellow.
Write and graph a direct variation equation
7.3B – Solving with Gauss Elimination GOAL: Create Row echelon form and solve with back substitution. Row operations to create Row echelon form – 1.) Switch.
4. Dynamic reliability models Objectives Be able to find reliability of series, parallel, stand-by and shared load parallel systems, when the reliabilities.
Background on Reliability and Availability Slides prepared by Wayne D. Grover and Matthieu Clouqueur TRLabs & University of Alberta © Wayne D. Grover 2002,
Chapter 2 Machine Interference Model Long Run Analysis Deterministic Model Markov Model.
A. BobbioBertinoro, March 10-14, Dependability Theory and Methods 2. Reliability Block Diagrams Andrea Bobbio Dipartimento di Informatica Università.
3x – 5y = 11 x = 3y + 1 Do Now. Homework Solutions 2)2x – 2y = – 6 y = – 2x 2x – 2(– 2x) = – 6 2x + 4x = – 6 6x = – 6 x = – 1y = – 2x y = – 2(– 1) y =
Equations Reducible to Quadratic
Maintenance Policies Corrective maintenance: It is usually referred to as repair. Its purpose is to bring the component back to functioning state as soon.
System Reliability Modeling and Analysis-
 How do you know how long your design is going to last?  Is there any way we can predict how long it will work?  Why do Reliability Engineers get paid.
1 Component reliability Jørn Vatn. 2 The state of a component is either “up” or “down” T 1, T 2 and T 3 are ”Uptimes” D 1 and D 2 are “Downtimes”
Reliability Failure rates Reliability
Solving by Elimination Example 1: STEP 2: Look for opposite terms. STEP 1: Write both equations in Standard Form to line up like variables. STEP 5: Solve.
Stracener_EMIS 7305/5305_Spr08_ Systems Reliability Modeling & Analysis Series and Active Parallel Configurations Dr. Jerrell T. Stracener, SAE.
Section 5.3 Solving Systems of Equations Using the Elimination Method There are two methods to solve systems of equations: The Substitution Method The.
7.4. 5x + 2y = 16 5x + 2y = 16 3x – 4y = 20 3x – 4y = 20 In this linear system neither variable can be eliminated by adding the equations. In this linear.
Systems of Linear Equations in Two Variables. 1. Determine whether the given ordered pair is a solution of the system.
Solving Systems of Equations by Elimination. Standard and Objective A.REI.5 Prove that, given a system of two equations in two variables, replacing one.
REPAIRABLE SYSTEMS 1 AVAILABILITY ENGINEERING. A Single Repairable Component With Failure Rate λ and Repair Rate μ The component may exist in one of Two.
Fault Tree Analysis Part 11 – Markov Model. State Space Method Example: parallel structure of two components Possible System States: 0 (both components.
Reliability Engineering
Prof. Enrico Zio Availability of Systems Prof. Enrico Zio Politecnico di Milano Dipartimento di Energia.
Copyright 2007 Koren & Krishna, Morgan-Kaufman Part.4.1 FAULT TOLERANT SYSTEMS Part 4 – Analysis Methods Chapter 2 – HW Fault Tolerance.
Solving Linear Systems
Discrete-time Markov chain (DTMC) State space distribution
ECE 753: FAULT-TOLERANT COMPUTING
Equations Quadratic in form factorable equations
Chapter 12 Section 1.
All equations must be solved by the same values of x and y
Solving Linear Systems Algebraically
Solving Systems Equations in Three Variables
Modeling in the Time Domain
Solve Linear Equations by Elimination
State Space Method.
Dept. of Electrical & Computer engineering
Lesson 3.3 Function Notation
SIMULTANEOUS EQUATIONS 1
Solving systems using substitution
Reliability.
Expected Value and MTTF Calculation
Systems of Equations Solve by Graphing.
RELIABILITY Reliability is -
Equations Quadratic in form factorable equations
Example 2B: Solving Linear Systems by Elimination
6.3 Elimination with Multiplication!
Chapter 2 Machine Interference Model
The Substitution Method
Indicator 16 System of Equations.
LAPLACE TRANSFORMATION
Control System and Transfer Function
Presentation transcript:

Reliability Engineering Redundant Structure and Standby Units

Cold Standby, Perfect Switching, With Repairs

Possible States of a 2-Unit System with Cold Standby and Perfect Switching Unit A Unit B 4 O S 3 F 2 1

State Space Diagram 4 3 2 1

State Equations

Eliminating the Failed State

Laplace Transform Substitute s=0 Note that

Solution

Mean Time to Failure

Mean Time to Failure Take Laplace transform of R(t) Substitute s=0

Mean Time to Failure

Cold Standby, Perfect Switching, With Repairs, A Main Operating Unit

Possible States System Unit A (Main Unit) Unit B 4 O S 3 F 2 1

State Space Diagram 4 3

State Equations Where

Steady State Probabilities

Availability and Unavailability

Eliminate Failed State from State Equations Where

Treating State 0 as An Absorbing State Take Laplace transform and let s=0 Solution

Mean Times to Failure and to Repair Mean time to failure Mean time to repair

Cold Standby, Imperfect Switching, With Repairs, A Main Operating Unit

State Space Diagram 4 3

Steady State Probabilities

Availability and Unavailability

Mean Time to Failure

Partly-Loaded Standby, Perfect Switching, With Repairs, A Main Operating Unit

Possible States of a 2-Unit System with Partly-Loaded Standby and Perfect Switching Unit A Unit B 4 O S 3 F 2 1

State Space Diagram 4 3 1

Steady State Probabilities

L Spares, With Replacements and Repairs

State Space Diagram 2 2j 2L 1

Notation State 2j (j = 0, 1, …,L): A total of j spare units are in a repair queue, and (L-j) spares are normal. A failed unit in the system is being replaced by a normal spared unit, the system is working. State 2j+1 (j = 0, 1, …, L-1): A total of j spare units are in a repair queue, and (L-j) spares are normal. A failed unit in the system is being replaced by a normal spared unit, the system does not work. State 2L+1: All spares are in a repair queue. A failed unit in the system is under priority repair. This is a type of quasi-replacement.

Notation λ: Constant failure rate μ: Constant repair rate ε: Constant replacement rate

Steady-State State Equations

Steady-State Availability