TIME TO FAILURE AND ITS PROBABILITY DISTRIBUTIONS Lectures 3 and 4

Failure Data and Related Functions 200 bulbs have been subjected to a Life Test for 700 hours: I III IV Д F =  Reliability Failure Rate Time Interval  NF Failure Probability Failure Density Cumulative Failure Ratio Failures/hour Failures F(t)=NF/N ДNF /N f(t)=Д F/Д t R(t)=1-F(t) h(t)=f(t)/R(t-Д t)   1 100 70 0.35 0.0035 0.65 200 110 0.55 0.2 0.002 0.45 0.0031 300 140 0.7 0.15 0.0015 0.3 0.0033 400 165 0.825 0.125 0.00125 0.175 0.0042 500 185 0.925 0.1 0.001 0.075 0.0057 600 193 0.965 0.04 0.0004 0.035 0.0053 700 0.00035 0.0100 f(t) F(t) Failure Probability = NF / N f (t) Failure Probability Density Function = Д F/Д t = Д NF / (N Д t) R(t) Reliability = 1 – F(t) = 1 - NF / N =(N – NF) / N = NS / N 0.35 h (t) Failure Rate = Д NF / (NS Д t) 0..2 0.15 0.125 0.1 0.04 t

F(t) Failure Probability = NF / N f (t) Failure Probability Density Function = Д F/Д t = Д NF / (N Д t) R(t) Reliability = 1 – F(t) = 1 - NF / N =(N – NF) / N = NS / N h (t) Failure Rate = Д NF / (NS Д t) THEN AN IMPORTANT SPECIAL CASE THE FAILURE RATE IS CONSTANT As examples: All transistors, Diodes, Micro Switches, …

SUMMARY Three Basic Functions: Failure Probability Density Function Failure PDF Failure Rate Failure Probability Function Survival Probability Function or RELIABILITY

GENERAL FORMULA FOR MEAN TIME TO FAILURE The Time To Failure TTF of any component or system is a Random Variable with probability density function f (t) Accordingly, the Mean Time To Failure MTTF can be evaluated as follows: f (t) But t Substituting, MTTF Integrating by parts we get, Then finally,

GENERAL FORMULA FOR MEAN RESIDUAL LIFE f (t) t TO MRL

COMPONENTS WITH CONSTANT FAILURE RATE Case 1 We have already shown that In case of CONSTANT Failure Rate Therefore, the probability density function f (t) in case of Constant Failure Rate will be Is known as EXPONENTIAL Distribution This distribution f(t) In this case t

Case 2 COMPONENTS WITH INCREASING FAILURE RATE Example

COMPONENTS WITH DECREASING FAILURE RATE Case 3 COMPONENTS WITH DECREASING FAILURE RATE Example As a general rule

BATH TUB DISTRIBUTION and PRODUCT LIFECYCLE

Infant Mortality Useful Life Wear out Burn-in Hazard rate h (t) IFR DFR CFR time Start of Life Start of commissioning Start of Deterioration End of Life

Caused by Period Characterized by Remedy Burn-in Infancy Decreasing Failure rate DFR Design errors . Manufacturing defects: welding flaws, residual Stresses, contamination, Poor quality of materials Burn-in testing Quality assurance Useful Life Constant Failure Rate Chance Events, Acts of Nature, Human Errors Redundancy, Increasing Strength And Capacity Wear out Phase out Increasing Failure rate Fatigue, Corrosion, Aging, Wear of mech. parts, inadequate Maintenance and repair De rating, Proper maintenance, Proper Replacement policies

WEIBULL DISTRIBUTION It is a Universal distribution It fits all trends of Failure Rate by changing a single parameter β

Derivation of WEIBULL distribution Consider the following Failure Rate as function of time Failures / time As m=0, we get the case of Constant Failure Rate (CFR) As m>0, we get case of Increasing Failure Rate (IFR) As m<0, we get case of Decreasing failure rate (IFR) k is Constant Consider a Characteristic Time η as the time interval during which the mean number of failures is ONE Put m+1 = β, then we find h(t) as follows: As β =1, we get the case of Constant Failure Rate (CFR) As β > 1, we get case of Increasing Failure Rate (IFR) As β < 1, we get case of Decreasing failure rate (DFR)

η is the Characteristic Time Failure Rate h (t) β is the shape factor η is the Characteristic Time When β=1 the failure rate becomes constant and Weibull distribution turns to be Exponential Reliability Probability Density Function PDF f (t) PDF = t

Mean Time To Failure

Variance of Time To Failure

SUMMARY OF WEIBULL Distribution

SOLVED EXAMPLES

___________________________________________________ Example 1 The pdf of time to failure of a product in years is given Find the hazard rate as a function of time. Find MTTF If the product survived up to 3 years, find the mean residual life MRL Find the design life for a reliability of 0.9 ___________________________________________________ 1) Find the Reliability R(t) MRL f(t) 3 10 t 2) Find the Hazard rate h(t) 3) Find MTTF Since h(t) is Increasing function with time Then the product is in deteriorating h(t) 4) Find MRL t 10 5) Find Design Life at R=0.9

Example 2 Ten V belts are put on a life test and time to failure are recorded. After 1600 hours six of them failed at the following times: 476, 529, 550, 921, 1247 and 1522. Calculate MTTF, reliability of a belt to survive to 900 hours. Design life corresponding to Reliability of 0.94 based on: a) CFR b) Weibull distribution MTTF=(476+529+550+921+1247+1522+4*1600)/10 =1164.495 hrs Based on CFR CONTINUED

(1) Based on Weibull Variance=σ2=[(476-1164.945)2+(529-1164.945)2+(550-1164.945) 2+(921-1164.945)2 +(1247-1164.945)2+(1522-1164.945)2+4*(1600-1164.945)2)/(10-1) =158719.1 (2) Square (1) and divide (2)/(1), we find the following equation in ONE Unknown β. Then CONTINUED

η will be found from (1) as follows: The above equation is TRANSCENDENTAL and could be solved by Trial and Error Method or by Excel GOAL SEEK Using Excel GOAL SEEK, we find: β = 3.2 η will be found from (1) as follows: CONTINUED

TRIAL & ERROR Then β = 3.2 β Г(1+2/β) / [Г(1+1/β)]2 Required Value =1.117 β Г(1+2/β) / [Г(1+1/β)]2 1 2 2 1.273 3 1.1321 1.1176 3.2 Then β = 3.2

Example 2 Continued Then Notice the difference between Exponential and Weibull Distributions

f(t) F = 0.06 545.16 Time 72 F = 0 .06