1. TIME TO FAILURE AND ITS PROBABILITY DISTRIBUTIONS
Lectures 3 and 4

2. 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
0.175
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
0.0100
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

3. 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, …

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

5. 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,

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

7. 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

8. Case 2
COMPONENTS WITH INCREASING FAILURE RATE
Example

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

10. BATH TUB DISTRIBUTION and PRODUCT LIFECYCLE

11. 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

12. 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

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

14. 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)

15. 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

16. Mean Time To Failure

17. Variance of Time To Failure

18. SUMMARY OF WEIBULL Distribution

19. SOLVED EXAMPLES

20. ___________________________________________________
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

22. 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=( *1600)/10 = hrs
Based on CFR
CONTINUED

23. (1)
Based on Weibull
Variance=σ2=[( )2+( )2+( ) 2+( )2
+( )2+( )2+4*( )2)/(10-1) =
(2)
Square (1) and divide (2)/(1), we find the following equation in ONE Unknown β.
Then
CONTINUED

24. 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

25. 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

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

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