1. Fault-Tolerant Computing Systems #5 Reliability and Availability2
Pattara Leelaprute
Computer Engineering Department
Kasetsart University

2. Reliability and Availability
The probability that a system survives till time t (it has not fail till t)
Availability
The probability that a system works properly at time t

3. Preliminaries of Probability
Discrete sample space:
Tossing a coin
{head, tail} sample space
Continuous sample space:
How long the pc stays up after reboot
{t | t>0} sample space
Random variable
A function mapping each element of sample space to a real number
Ex. heads=1, tails=0

4. Preliminaries Random variable
A function mapping each element of sample space to a real number
CDF (Cumulative distributed function) [สะสมเพิ่มขึ้น]
F(t) = Pr [X ≤ t]
Pr : probability that the system has gone down by time t
Pdf (Probability density function)
f(t) = dF(t) / dx
Expected Value, Mean
E[X] = 0 t f(t)dt (X≥0)
Average outcome of the random experiment expect value, mean of a random variable

5. Exponential Distribution
The most commonly used distribute function in reliability modeling.
CDF
F(t) = 1 – e-lt
pdf
f(t) = l e-lt
Mean
1/l
Memoryless property
Y = X – t
Gt(y) = Pr [Y ≤ y | X > t ] = 1 – e-ly
Distribute of remaining life of a component does not depend on how long it has been working.
The component does not AGE !
(remaining life of X does not depend on the time that has passed)
f(t) = 2e-2t
F(t) = 1 – e-2t

6. Reliability R(t) Reliability
The probability that a system survives till time t
R(t) = Pr [X > t]
= 1 – F(t)
X : Random probability variable X which represents a time to failure of the system (the life of the system)
R(t): represents probability that the system survives till time t
F(t) = exponential distribution
F(t) = 1 – e-2t
R(t) = e-2t t
time 0 X
time t

7. Reliability R(t) Reliability R(t) = Pr [X > t] = 1 – F(t) t time 0
The system is initially working
R() = 0
No system has infinite lifetime
F(t) = exponential Distribution
R(t) = reliability
F(t) = 1 – e-2t
R(t) = e-2t t
time 0 X
time t

8. Failure Rate
Probability that fault will occur in an interval time [t, t+Dt] =
f(t)Dt
Probability that fault will occur in time [t, t+Dt]
f(t)Dt / R(t)
Probability of occurrence of fault at time [t, t+Dt], when the system is working properly at t
Failure Rate
f(t) / R(t)
f(t) = probability of fault
F(t) = exponential distribution
R(t) = reliability
f(t) = 2e-2t
R(t) = e-2t
F(t) = 1 – e-2t
[t, t+Dt]

9. Bathtub Curve Failure Rate Bathtub Curve
f(t) / R(t)
Bathtub Curve
General Failure Rate observed from the empirical data collected from mechanical and electronic component
When lifetime of a system F(t) is exponential distribution，it has a constant Failure Rate (see previous slide)
2.constant failure rate
1.Initial stage:
Inherit defects
faulty design
3.last stage:
faults caused by age

10. Availability The probability that system works properly at time t
Availability is a measure that is frequently used for describing the behavior of the system
*If the system has no repair or replacement, availability is equal to reliability R(t)
Because, R(t) = the probability that no failures have occurred during the whole period (0,t)
fails
repairs
fails
repairs
Operational
Under repair
Operational t Xi
Xi+1
Xi+2 Ui
Ui+1

11. MTTF (Mean Time To Failure)
E[X] = 0 t f(t)dt = 0 R(t)dt
The expected value of the probability variable X which represents time till fault occurs in the system
When R(t) = e-lt (X is exponential distribution)
Failure Rate = l
MTTF = 1 / l
time 0
expected value

12. MTTR (Mean Time To Repair)
MTTR = E [ Ui ]
The expected value of the random variable Ui
which represents the downtime for i th repair or replacement
When R(t) = e-lt
Repairing Rate = m
MTTR = 1 / m t Xi
Xi+1
Xi+2 Ui
Ui+1

13. Availability Instantaneous availability (ทันทีทันใด)
A(t) = Pr [probability that the component is functioning correctly at t ]
Steady-State Availability (general meaning)
A = limt→∞ A(t)
fails
repairs
fails
repairs t Xi
Xi+1
Xi+2 Ui
Ui+1

14. Availability When Xi, Ui is exponential distribution
FXi(t) = 1 – e-lt, FUi(t) = 1 – e-mt
Instantaneous Availability
A(t) = (m + le-（l + m)t ) /(m + l)
Steady-State Availability
A = limt→∞ A(t) = m /(m + l) t Xi
Xi+1
Xi+2 Ui
Ui+1

15. Availability and MTTF / MTTR
MTTR (mean time to repair)
MTTR = E [ Ui ]
The expected value of Ui which is the random variable that represents the downtime for i th repair or replacement
MTTF (mean time to failure)
MTTF = E [ Xi ]
The expected value of Xi which is the random variable that represents the duration of the i th function period.
Steady-State Availability
A = MTTF / (MTTF+MTTR)
= m /(m + l) 　
(when Xi,Ui is the exponential distribution of parameter l,m) t Xi
Xi+1
Xi+2 Ui
Ui+1

16. Reliability Block Diagrams
Represents the logical structure of a system with regard to
how the reliability of its component affects the system reliability
Series Structure
Every component has to be functioning
Decreasing reliability
Parallel Structure
At least one component is operational
Increasing reliability R1 R2 R1 R2
R = R1*R2 Ra
R = 1– (1– R1)*(1– R2)
F(t)=1– (1– F1(t))*(1– F2(t)*F3(t))
R2=0.9
Ra = 1– (1- R2)*(1- R3)
= 1-(1-0.9)*(1-0.8)
= 1-(0.1*0.2) = 0.98
Rtotal = R1 * Ra
= 0.95*0.98
= 0.931
R1=0.95
R3=0.8

17. Reliability Block Diagrams
Represents the logical structure of a system with regard to
how the reliability of its component affects the system reliability
F(t) = distribution function for failure time (unreliability) F1 F1 F2
Series Structure F2
F(t)=1– (1– F1(t))*(1– F2(t))
Parallel Structure
F(t)=F1(t)*F2(t)
F(t)=1– (1– F1(t))*(1– F2(t)*F3(t))
F2=0.001
F2(t)*F3(t) = 0.001*0.002 =
F(t) = 1– (1- F1(t))( )
= 1-( )( )
= 1-(0.997)( )
= = 0.006
F1=0.003
F3=0.002

18. Example of Calculation

19. Reliability & Failure Rate
Calculate the reliability (R) within 2 years for the system that has the failure rate = 0.3 (case/year)
R(t) = e-lt
Failure rate (l) = 0.3
R(2) = e-0.3 * 2
= e-0.6
= =

20. MTTF What is MTTF of this system ? (100 + 120 + 140) / 3 = 120
fails
repaired
4hr
100hr
3hr
120hr
2hr
140hr t
( ) / 3 = 120
(hr / case)

21. MTTF & Failure Rate Failure rate (l) = 1 / MTTF = 1 / 40 = 0.025
What is the failure rate of the system which has MTTF = 40 hr/case
fails
repairs
4hr
30hr
3hr
40hr
2hr
50hr t
Failure rate (l) = 1 / MTTF
= 1 / 40
= 0.025
(case/hr)

22. MTTR What is MTTR of this system ? (3+ 2 + 4) / 3 = 3 (hr / case)
fails
repairs
4hr
100hr
3hr
120hr
2hr
140hr t
( ) / 3 = 3
(hr / case)

23. MTTR & Repairing Rate Repairing rate (m) = 1 / MTTR MTTR = 1 / m
What is the MTTR of the system which has repairing rate of 0.2 (case/hr)
Repairing rate (m) = 1 / MTTR
MTTR = 1 / m
= 1 / 0.2
= 5
(hr/case)

24. Availability & MTTF & MTTR
What is Availability of this system ?
fails
repairs
4hr
100hr
3hr
120hr
2hr
140hr t
A = MTTF / (MTTF+MTTR)
MTTF = ( ) / 3 = 120 (hr/case)
MTTR = ( ) / = 3 (hr/case)
A = 120 / 120+3
= 120/123 =

25. Fault Trees Si=kn ( )F(t)i(1-F(t))n-i
Pictorial representation of the combination of events that can cause the occurrence of an undesirable event (failure).
Staring point (of tree) is the definition of a single undesirable event (failure).
An event is reduced to a combination of low-level events by means of logic gates.
F(t) = probability of the occurrence of failure event
(function of F(t) is CDF)
0: Normal, 1: Fails
TMR
Failure
2-out-of-3 S1 S2 S3
OR gate
AND gate
k-out-of-n
Si=kn ( )F(t)i(1-F(t))n-i n i

26. Fault Tree Model & Reliability Block Model
The structure that shows when the system is functioning.
Fault Tree Model
The structure that shows when the system has failed
The output of the top event is a logic 1
Fault Tree
Failure P1 M2 M3 or
and M1 P2
Reliability block diagram
0: Normal, 1: Fails
2 processors (P)
3 memory module (M)

27. Example of Fault Tree 0: Normal, 1: Fails Failure
F1(t)*(1-(1-F2(t))*(1-F3(t)))
and
1-(1-F2(t))*(1-F3(t)) or S1 S2 S3

28. Fault Trees (Basic) Fault Tree when there is no repeated component
Failure distribution for the component is independent
Reliability block diagram
Fault Tree
Failure P1 M2 M3 or
and M1 P2
F(t) = ??
2 processors (P)
3 memory module (M)
The system is operational if at least one processor and one memory module are operational.

29. Fault Trees (Advance) We have to use factoring technique !
Fault Tree when there is repeated component
Failure distribution for the component is not independent
Suppose that instead of all three memory modules being shared between the two processors,
one of the memory modules (M3) is shared and the other two are private, one for each processor.
Failure
and or M1 M3 M2 P1 P2
What it reliability block diagram ?
We have to use factoring technique !

30. Factoring Fa(t) F(t) Fb(t) F(t) = FM3(t)*Fa(t) + (1-FM3(t))*Fb(t)
M3 has failed
Fa(t)
Failure
0: Normal, 1: Fails
and
F(t)
Failure or or
and P1 M1 P2 M2 or or
M3 has not failed
Fb(t)
Failure
and P1 P2
and
and P1 M1 M3 P2 M2 M3
F(t) = FM3(t)*Fa(t) + (1-FM3(t))*Fb(t)
Multiply the result for each case by the probability that case happens,
then add the products.