Reliability Engg Lect 3

Binomial Distribution (p+q)2 = p2 + 2pq + q2 (p+q)3 = p3 + 3p2q + 3pq2 + q3

= nCrr)!-(n r!

n!

Required conditions:- number of trials (n) must be known - all trials are independent- each trial results in either a success (p) or failure (q) i.e. p + q = 1- values for p and q are constant for each trial

Probability of exactly r failures (and n-r successes),

Pr = nCr p(n-r) qr

Coefficients of Binomial Distribution

No. of combinations of r successes (or failures) in n trials = = nCrr)!-(n r!

n!

If ‘n’ is small, Pascal’s Triangle:

1

1 1 n=1 (p+q)1 = 1p + 1q

1 2 1 2 (p+q)2 = 1p2 + 2pq + 1q2

1 3 3 1 3 (p+q)3 = 1p3 + 3p2q + 3p2q + 1q2

1 4 6 4 1 4

1 5 10 10 5 1 5

Example Consider the case in which the probability of success in a single trial is ¼ and four trials are to be made. Evaluate the individual and cumulative probabilities of success in this case.

n = 4, p = ¼ , q = 1 – ¼ = ¾

(p+q)4 = p4 + 4p3q + 6p2q2 + 4pq3 + q4

Number of Cumulative successes failures Individual probability probability

0 4 q4 = (3/4) 4 = 81/256 81/256 1 3 4pq3 = 4(1/4)(3/4) 3 = 108/256 189/256 2 2 6p2q2 = 6(1/4) 2 (3/4) 2 = 54/256 243/256 3 1 4p3q = 4(1/4) 3 (3/4) = 12/256 255/256 4 0 p4 = (1/4) 4 = 1/256 256/256

= 1

Probability Distributions Probability Distribution Function

0.0

0.2

0.4

0.6

0.8

1.0

0 1 2 3 4# of Heads, x

Cu

m. P

rob

ab

ility

, F(x

)

Probability Density Function

0.0

0.1

0.2

0.3

0.4

0 1 2 3 4# of Heads, x

Pro

bab

ilit

y, f(

x)

Moments of Binomial Distribution:

Expected # of successes, E(x) = n.pwhere,n = # of trials or samplesp = probability of successq = probability of failure

Std. Deviation, σ(x) = √ n.p.q

Example

A product is claimed to be 90% free of defects. What is the expected value and standard deviation of the number of defects in a sample of 4?

Here, n = 4p(defect) = 0.1

E(defects) = n.p= 4 x 0.1 = 0.4

σ(defects) = √ npq

= √ 4 x 0.1 x 0.9 = 0.6

Binomial Distribution Application

Binomial Distribution can be used to determine the probability of all possible outcomes in a system. Only a restricted number of outcomes (and their probabilities) are usually of interest in engineering applications, e.g. outcomes that lead to system failure.

Example:It is known that, in a certain manufacturing process, 1% of the products are defective. If the a customer purchases 50 of these products selected at random, what is the probability that he receives 2 or less defective products?

n = 50p(defect) = 0.01 p(no defect) = 1 – 0.01 = 0.99

P(2 or less defects) = 0.0756 + 0.3056 + 0.6050 = 0.9862

P(2 defective) = 50C2 (0.01) 2 (0.99) 48 = 0.0756P(1 defective) = 50C1 (0.01) 1 (0.99) 49 = 0.3056P(0 defective) = 50C0 (0.01) 0 (0.99) 50 = 0.6050

Mutually exclusive events

Example: The manufacturing company has a policy of replacing, free-of-charge, all defective products that are purchased. If the product manufacturing cost is $10 per unit and each product is sold for $15, how much profit is made?

p(defect) = 0.01For 1000 products, n = 1000Expected # of defects, E(defects) = n.p = 10

Therefore, 1010 products must be manufactured to sell 1000 products.

Manufacturing cost = $10 x 1010 = $10,100Income = $15 x 1000 = $15,000

Profit = $15,000 - $10,100 = $4900Profit per unit = $4900/1000 = $4.90

Example:

If the company decides to increase the manufacturing cost to $10.05 per unit in order to decrease the probability of defects to 0.1%,

p(defect) = 0.001For 1000 products, n = 1000Expected # of defects, E(defects) = n.p = 1

Therefore, 1001 products must be manufactured to sell 1000 products.

Manufacturing cost = $10.05 x 1001 = $10,060.05Income = $15 x 1000 = $15,000

Profit = $15,000 - $10,060.05 = $4939.95Profit per unit = $4939.95/1000 = $4.94

Effect of Redundancy Consider a system consisting of 4 identical components, each having a failure probability of 0.1.

q = 0.1 (p = 0.9) n = 4 (p+q)4 = p4 + 4p3q + 6p2q2 + 4pq3 + q4

System state Individual probability all components working p4 = (0.9) 4 = 0.6561

3 working, 1 failed 4p3q = 4(0.9) 3 (0.1) = 0.2916

2 working, 2 failed 6p2q2 = 6(0.9) 2 (0.1) 2 = 0.0486

1 working, 3 failed 4pq3 = 4(0.9)( 0.1) 3 = 0.0036

all components failed q4 = (0.1) 4 = 0.0001

= 1

Consider 4 criteria

System reliability, R

all components required for success (no redundancy)

0.6561

3 components required for success (partial redundancy)

0.6561+0.2916 = 0.9477

2 components required for success (partial redundancy)

0.6561+0.2916+0.0486 = 0.9963

1 component required for success (full redundancy)

0.6561+0.2916+0.0486+0.0036 = 0.9999

Effect of Redundancy A pump has a failure probability of 0.1. Compare the reliability of the cooling system with and without redundancy.

q = 0.1 (p = 0.9)

coolingSystem

pump 1

coolingSystem

pump 1

pump 2

pump 1

coolingSystem

pump 1

pump 2

Reliability, R = p = 0.9Unreliability, Q = q = 0.1

R = p1 + p2 + p1 x p2 = 0.9 + 0.9 - 0.9 x 0.9 = 0.99Q = q1 x q2 = q2 = 0.12 = 0.01

Q = q1 x q2 x q3= q3 = 0.13 = 0.001R = 1 - Q = 0.999

System with Derated States

Two-state systems: (1) success or up state (2) failure or down state

Many systems can also reside in derated (or partial output) states,e.g. process plant, power generation plant

One of the most useful application of Binomial Distribution is to evaluate engineering systems with de-rated states.

- build COPT using Binomial Distribution- use COPT to evaluate system reliability

COPT: capacity outage probability table

System with Derated States

Consider a generation plant with two 10 MW units, each having a forced outage rate (probability of failure) of 10%.

p = 0.9, q = 0.1, n = 2

(p + q)2 = p2 + 2pq + q2 Binomial Distribution:

Units Out Cap Out (MW) Cap In (MW) Probability Cum. Prob0 0 20 0.81 11 10 10 0.18 0.192 20 0 0.01 0.01

Capacity Outage Probability Table:

If the generation plant operates to supply a 15 MW load,what is the probability of load loss (system failure)?

Probability of load loss = Loss of Load Probability (LOLP) = 0.19

Expected # of days of load loss = 0.19 x 365 = 69.35 days/yrLoss of Load Expectation (LOLE)

System with Derated States

Consider a generation plant with two 10 MW units, each having a forced outage rate (probability of failure) of 10%.

If the generation plant operates to supply a 15 MW load,what is the Expected Load Loss (ELL)?

p = 0.9, q = 0.1, n = 2

Capacity Outage Probability Table:

Expected Load Loss (ELL) = 1.05 MW

Units Out Cap Out (MW) Cap In (MW) Load Loss (MW) Probability Col.4 x Col.50 0 20 0 0.81 01 10 10 5 0.18 0.902 20 0 15 0.01 0.15

1.05

Example

A generating plant is to be designed to satisfy a constant 10 MW load.

Four alternatives are being considered:

a) 1 x 10 MW unitb) 2 x 10 MW unitsc) 3 x 5 MW unitsd) 3 x 3.33 MW units

The probability of unit failure (unavailability or FOR) is assumed to be 0.02.

FOR or U = 0.02Availability of each unit, A = 0.98

Capacity Outage Probability Tables

units capacity

(MW) Binom. Individual cum. out out in Distr. prob prob

(a) 1 x 10 MW

0 0 10 A 0.98 1.00 1 10 0 U 0.02 0.01

(b) 2 x 10 MW

0 0 20 A2 0.9604 1.0000 1 10 10 2AU 0.0392 0.0396 2 20 0 U2 0.0004 0.0004

(c) 3 x 5 MW

0 0 15 A3 0.941192 1.000000 1 5 10 3A2U 0.057624 0.058808 2 10 5 3AU2 0.001176 0.001184 3 15 0 U3 0.000008 0.000008

(d) 4 x 3.33 MW

0 0 13.33 A4 0.92236816 1.00000000 1 3.33 10 4A3U 0.07529536 0.07763184 2 6.66 6.66 6A2U2 0.00230496 0.00233648 3 10 3.33 4AU3 0.00003136 0.00003152 4 13.33 0 U4 0.00000016 0.00000016

LOLP = 0.01LOLE = 365 x 0.01 = 3.65 d/yr

LOLP = 0.0004LOLE = 0.15 d/yr

LOLP = 0.001184LOLE = 0.43 d/yr

LOLP = 0.00233648LOLE = 0.85 d/yr

Expected Load Loss

ELL = 0.2 MW = 200 kW

Capacity (MW) Load Loss Prob Li x pi Out In Li (MW) pi (MW) (a) 1 x 10 MW 0 10 0 0.98 0 10 0 10 0.02 0.2 0.2 (b) 2 x 10 MW 0 20 0 0.9604 0 10 10 0 0.0392 0 20 0 10 0.0004 0.004 0.004 (c) 3 x 5 MW 0 15 0 0.941192 0 5 10 0 0.057624 0 10 5 5 0.001176 0.00588 15 0 10 0.000008 0.00008 0.00596 (d) 4 x 3.33 MW 0 13.33 0 0.92236816 0 3.33 10 0 0.07529536 0 6.67 6.67 3.33 0.00230496 0.00768320 10 3.33 6.67 0.00003136 0.00020907 13.33 0 10 0.00000016 0.00000160 0.00789387

ELL = 0.004 MW = 4.00 kW

ELL = 0.00596 MW = 5.96 kW

ELL = 0.00789 MW = 7.89 kW

Comparative Analysis

On the basis of the expected load loss only the alternative b is most reliable System LOLE D/YEAR INVESTMANT

1X10MW 3.65 1 P U

2X10MW 0.15 2

3X5MW 0.43 1.5

4X3.33MW 0.85 1.33

On the basis of investment alternative c can be adopted though it have higher LOLE

Effect of unit unavailabilitySystem ELL (kW) at Different Unit FOR’s

2% 4% 6%) (a) 1 x 10 MW 200 400 600 (b) 2 x 10 MW 4.00 16.00 36.00 (c) 3 x 5 MW 5.96 23.68 52.92 (d) 4 x 3.33 MW 7.89 31.12 69.09

These result indicate that the ELL and any reliability index is very Sensitive to an increase in value of unavailability (FOR)

System with Non-identical ComponentsAll components must be identical to apply the Binomial Distribution.If components of a system have non-identical capacities:

- Units with identical capacities are grouped together- COPT is developed for each group- COPT for different groups are combined, one at a time- Final COPT for the system is used for reliability evaluation

A pumping station has 2 x 20 t/hr units, each having an unavailability of 0.1, and 1 x 30 t/hr unit with an unavailability of 0.15. Calculate the capacity outage probability table for this plant.

COPT for 2 x 20 t/hr units: COPT for 1 x 30 t/hr unit:Units Out

Cap Out (t/hr)

Cap In (t/hr) Prob

Units Out

Cap Out (t/hr)

Cap In (t/hr) Prob

0 0 40 0.81 0 0 30 0.851 20 20 0.18 1 30 0 0.152 40 0 0.01

System with Non-identical Components

Combining COPTs:

Each cell contains: Capacity In / probability

40 / 0.81 20 / 0.18 0 / 0.0130 / 0.85 70 / 0.6885 50 / 0.153 30 / 0.0850 / 0.15 40 / 0.1215 20 / 0.027 0 / 0.0015

1 x 30 t/hr unit

2 x 20 t/hr units

Overall System COPT:Cap In (t/hr)

Cap Out (t/hr) Prob

70 0 0.688550 20 0.153040 30 0.121530 40 0.085020 50 0.02700 70 0.0015

Example

A 60 MW generation plant consists of one 20 MW and four 10 MW units. The plant must satisfy a load demand of 45 MW. What is the Loss of Load Probability (LOLP) and the Expected Load Loss (ELL)? Each 10 MW unit has an unavailability (or Forced Outage Rate, FOR) of 0.1, and the 20 MW unit has a FOR of 0.15.

Mixture of the exponential Distribution

• Let i= denotes the failure rate of the items coming from the plant I for i=1,2 …. . The items are mixed up before they are sold . The fraction of p is coming from plant 1 and rest (1-p) from plant 2

• I f we pick one item randomly the survivor function of the item is

• Mean Time to Failure MTTF •

Mixture of the exponential Distribution

• The Failure Rate function

The failure rate function is seen to be decreasing . If λ1 λ2 early failure should have failure rate close to λ1. After a while all the week components have failed we are left with components with lower failure rate λ2

Reliability Block Diagram • A reliability block diagram is success- oriented network

describing function of the system .

• It shows the logical connection of the components needed to fulfill the specified system function

• If the system has more than one function , each function must be considered individually and separate reliability block diagram has to be established for each system function

• Reliability block diagram are suitable for the system of non repairable components and where the order in which failures occurs doesn’t matter

• When the system are repairable and the order in which failures occurs are important Markov Method are more suitable

Reliability Block Diagram

• For the system functioning ith component must functioning . This doesn’t mean component i function in all accepts . It mean one ore specified set of the function is achieved

1 2 3 i n

Structure Function •The state of component i (i= 1,2,… n) can be described by binary variable xi where

X={x1, x2 , x3…..xn} is called state vector .By knowing the state of n components we knom whether the system is functioning or not

State of system (x) =( x1, x2 , x3…..xn)

This (x) is called the structure function

Series Structure

• The structure function of a series structure is

• A system that is functioning if and only if all of its n components are functioning is called series structure

• The survivor function of a series structure is

1 2 3 i n

Series Structure

• The failure rate function zs (t) of a series structure( of independent components) is equal to the sum of the failure rate functions of the individual components:

Series Structure

• Consider a series structure of n independent components with constant failure rates i for i = 1, 2, . . . , n. The survivor function is

• Failure rate of the system s=1+2 +……… n

Parallel Structure

• A system is functioning if at least one of its n components is functioning is called parallel structure

• The structure function of the Parallel structure is

n

i

1

2

Parallel Structure

• The survivor function of a non repairable parallel structure of independent components is

• When all the components have constant failure rates zi(t) = i, for i = 1, 2, . . . , n, then

•  

Parallel Structure

• Consider a parallel structure of two independent components with failure rates 1 and 2 respectively. The survivor function is

Parallel Structure

K out of N system

• A system that is functioning if at least k of n components are functioning is called k out of N structure(KOON)

• The structure function of a k-out of n structure may be written as

• When all the n components have identical reliabilities pi(t) = p(t) , the variable is binomially distributed (n,p(t)

• In this case

K out of N system

•A system has three components 1, 2 ,3 •2003 system can be represented as

•Structure Function

•Since xi is binary variables so xik =xi

1 2

2 3

1 3

MTTF of k-out-of-n Structures

Pivotal Decomposition

• Conditional Probability approach

• p(system success or failure )= p(system success or failure if component x is good).p(x is good)+ p(system success or failure if component x is bad).p(x is bad)

• By pivotal decomposition the structure function may be written

(X(t)) = Xi(t) · (1i,X(t)) + (1 − Xi(t)) · (0i,X(t))

• = Xi(t) · [ (1i,X(t)) − (0i,X(t))] + (0i,X(t))

Pivotal Decomposition

• Consider the system represented by the reliability block diagram.

• Assume that the components of the system are independent with the following function probabilities (reliabilities): p1 = 0.90, p2 = 0.95, p3 = 0.85, p4 = 0.90, p5 = 0.80.

• Find the system reliability ps using Pivotal

• Sol

• Considering ith component is good and bad both the structure function is expressed as

• Assume component 2 doesn’t work then the reliability can be reduced as

1 1 0i i i iX x x x x

Pivotal Decomposition

• Assume component 2 works then the component 1 becomes irrelevant component the reliability can be reduced as shown in below

•

• If the system components have independent probabilities (reliabilities p1 = 0.90, p2 = 0.95, p3 = 0.85, p4 = 0.90, p5 = 0.80.)

• Hence substituting the values of reliability =0.9949

0 3 1. 4 5i x x x x x

1 3 4 5i x x x x

223 4 5 1 3 1. 4 5X x x x x x x x x x

223 4 5 1 3 1. 4 5p X p p p p p p p p p

• Critical Component • The ith component is critical for the system if the

system is in such a state that system is functioning if and only that ith component is functioning

43

Let R=P [Success] Q=P [Failure]

R+Q=1

Series Systems

n

1ii

21s

R

RRR

2121

21

21

Ss

QQQQ

)Q)(1Q-(1-1

RR1

R1Q

21

product rule of reliability

Reliability Network Modeling

44

Example 4.1:

A system consists of 10 identical components, all of which must work for system success. What is the system reliability if each component has a reliability of 0.95?

Component reliability, R = 0.95

Number of components, n = 10

Using Product Rule of Reliability,

System Reliability, Rs = Rn

= (0.95)10 = 0.5987

45

Series System

Assume each component has a reliability of 0.9.

Number of Components Reliability1

2

3

4

5

10

20

50

0.9

0.81

0.729

0.6561

0.59049

0.348678

0.121577

0.005154

46

Parallel Redundant Systems

21s QQQ

2121

21

21

Ss

RRRR

)R)(1R-(1-1

QQ1

Q1R

2

1

product rule of unreliability

47

Example 4.6:

A system is to be designed with an overall reliability of 0.999 using components having individual reliabilities of 0.7. What is the minimum number of components that must be connected in parallel?

System Reliability, Rs = 0.999

Component reliability, R = 0.7

Number of components, n = ?

System Unreliability, Qs = 1 - Rs = 1 – 0.999 = 0.001

Component Unreliability, Q = 1 – R = 1 – 0.7 = 0.3

Using Product Rule of Unreliability, Qs = Qn i.e. 0.001 = (0.3)n

therefore, n = ln (0.001) / ln(0.3) = 5.74

since n is an integer, n = 6

48

Parallel System

Number of Components Reliability

1

2

3

4

5

0.9

0.99

0.999

0.9999

0.99999

49

Series/Parallel Systems

4

2 3

1

Redundant

Rs = ]RRRRR[RR 4324321

network reduction technique

50

Evaluate the unreliability of the system if all components have a reliability of 0.8.

Q8 = Q7 . Q5

= (1 - R7). Q5

= (1 - R1 . R2 . R6). Q5

= [1 - R1 . R2 . (1 - Q6)]. Q5

= [1 - R1 . R2 . (1 - Q3 . Q4)]. Q5

Q8 = [1 – 0.8 x 0.8 (1 – 0.2 x 0.2)]x 0.2 = 0.07712

51

Network Modelling – complex systems

A C

B D

E

System success requires continuity from input to output.

input output

Evaluation Techniques:- Conditional probability approach- Cut set method- Tie set method- Connection matrix technique- Tree diagrams

52

Cut Set Method

Cut Set:

set of system components which, when failed, causes system failure

set of components which if removed from the network separate the input from the output

Cut Sets:AB, CD - 2nd orderABE, ABC, ABD, ABE, AED, BEC, CDA, CDB, CDEABCD, ABCE, ABDE, ACDE, BCDEABCDE – 5th order

A C

B D

E

Minimal Cut Set:

any cut set which does not contain any other cut set as a subset

all components of a minimal cut set must fail in order to cause system failure

Minimal Cut Sets:AB, CD - 2nd orderAED, BEC – 3rd order

53

Cut Set Method contd..

P(C1) = QA.QB

P(C2) = QC.QD

P(C3) = QA.QD.QE

P(C4) = QB.QC.QE

QS = P(C1 C2 C3 C4)

= P(C1) + P(C2) + P(C3) + P(C4) - P(C1 C2) - P(C1 C3) - P(C1 C4) - P(C2 C3) - P(C2 C4) - P(C3 C4) + P(C1 C2 C3) + P(C1 C2 C4) + P(C1 C3 C4) + P(C2 C3 C4) - P(C1 C2 C3 C4)

Minimal Cut Sets:C1 -> AB A and B in parallel, since both must fail for system failureC2 -> CDC3 -> AED C1, C2, C3 and C4 in series, since all 4 must be successful for system successC4 -> BEC

C

D

A

B

A

D

E

B

C

E

C1 C2 C3 C4P(C1 C2) = P(C1).P(C2) = QA.QB QC.QD

P(C1 C3) = P(C1).P(C3) = QA.QB QD.QE

P(C1 C4) = P(C1).P(C4) = QA.QB QC.QE

.. P(C3 C4) = P(C1 C2 C3) = P(C1 C2 C4) = P(C1 C3 C4) = P(C2 C3 C4) = P(C1 C2 C3 C4) = QA.QB QC.QD .QE

54

Cut Set Method contd..

QS = QA.QB + QC.QD + QA.QD.QE + QB.QC.QE - QA.QB QC.QD - QA.QB QD.QE - QA.QB QC.QE

- QA.QC QD.QE - QB.QC QD.QE - QA.QB QC.QD .QE

+ 4 QA.QB QC.QD .QE

- QA.QB QC.QD .QE

QS = QA.QB + QC.QD + QA.QD.QE + QB.QC.QE - QA.QB QC.QD - QA.QB QD.QE - QA.QB QC.QE

- QA.QC QD.QE - QB.QC QD.QE + 2QA.QB QC.QD .QE

If QA = QB = QC = QD = QE = Q, then

QS = 2Q2 +2Q3 – 5Q4 + 2Q5

If Q = 0.01, QS = 0.00020195 RS = 0.99979805

55

Advantages of Cut Set Method:- cut sets identify ways in which a system may fail- approximation can be used to simplify evaluation- can be easily programmed on a computer

1st Approximation

QS = P(C1 C2 C3 C4)

= P(C1) + P(C2) + P(C3) + P(C4) - P(C1 C2) - P(C1 C3) - P(C1 C4) - P(C2 C3) - P(C2 C4) - P(C3 C4) + P(C1 C2 C3) + P(C1 C2 C4) + P(C1 C3 C4) + P(C2 C3 C4) - P(C1 C2 C3 C4)

P(C1) + P(C2) + P(C3) + P(C4)

QA.QB + QC.QD + QA.QD.QE + QB.QC.QE

If QA = QB = QC = QD = QE = Q, then QS = 2Q2 +2Q3

If Q = 0.01, QS = 0.000202, RS = 0.999798 n

For ‘n’ cut sets, QS = P(Ci) i=1

2nd Approximation

Neglect higher order cut sets (events with very low probabilities)

1st order: none2nd order: AB, CD3rd order: ADE, BCE

Neglecting cut sets higher than 2nd orderQS P(C1 C2) P(C1) + P(C2)

QA.QB + QC.QD

If QA = QB = QC = QD = QE = Q, then QS = 2Q2

If Q = 0.01, QS = 0.0002, RS = 0.9998

56

Consider:

2

3

1

Cuts Min Cuts Probability

1,3

2,3

1,2,3

1,3

2,3

---

Q1Q3

Q2Q3

-Qs ~ Q1Q3+Q2Q3

Complete Equation:

3213231

21213S

QQQQQQQ

]QQQ[QQ Q

57

Example:

Calculate the reliability of the system below if all the individual components have a reliability of 0.95.

58

Probability Density Function, f(x) (see p. 43) - plot of Probability vs. Random Variable x

(Cumulative) Probability Distribution Function, F(x) = ∫f(x).dx f(x) = dF(x) dx

For discrete distribution, F(x) = ∑f(x)

In reliability evaluation, the random variable is usually time (t)

Failure Density Function, f(t)

(Cumulative) Failure Distribution Function, Q(t) -- Probability of Failure

Survivor Function R(t) = 1 – Q(t) -- Probability of Success

Hazard Rate, (t) = f(t) / R(t) = # of failures per unit time # of components exposed to failure

Probability Distributions

59

Total number of samples = N0

Number of failures at time t = Nf (t)

Number of survivors at time t = Ns (t)

Number of failures in interval t = Nf (t)

Failure Density Function, f(t) = Nf (t) / N0

Failure Distribution Function, Q(t) = Nf (t) / N0

Probability of Failure

Survivor Function R(t) = Ns (t) / N0

Probability of Success

Hazard Rate, (t) = f(t) / R(t)

Example using Discrete Distribution

60

Bathtub Curve

Typical Electric Component Hazard Rate as a Function of Age

De-Bugging Normal operating Or useful life

Wear out

Region 1 Region 2 Region 3

Operating Life

Haz

ard

rat

e

61

Exponential Distribution

SystemUp

SystemDown

R(t) = Q(t) = 1 - f(t) =

is the constant failure rate

λte λte λte

If t << 1 then, Q(t) ~ t and R(t) ~ 1 - t

Mean or Expected Value of Exponential Distribution = 1/Or

Mean Time to Failure, MTTF = 1/

62

Series Systems

21

tλ-

)tλ(λ-

tλtλ

21s

i

21

21

e

e

ee

RRR

63

Parallel Systems

2

1

)tλ(λ-tλtλS

2121s

2121 eeeR

RRRRR

64

Complex Systems

*Develop the basic equations

*Substitute

λt

λt

e1Q(t)

eR(t)

Redundancy

Two way to ensure higher system reliability

• use components of higher reliability in critical places

•Introduce redundancy in those places Redundancy

•In an entity, the existence of more than one means for performing a required function [IEC50(191)]

Existence of means, in addition to the means which would be sufficient for a functional unit to perform a required function or for data to represent information [IEC61508, Part 4]

Redundant items may be classified as:

•Active (warm) redundancy

•Passive redundancy

•Cold redundancy

•Partly loaded redundancy

Passive Redundancy or perfect switching

• Ti time to failure item i for i=1,2…n

• Life time of the whole stand by system

• Mean time to failure

• For Two stand by system n=3

• For Two stand by system n=3

Cold Standby, Imperfect Switching

• Consider a system of n = 2 components with failure rates 1 and 2. Let the probability of successful switching be 1 − p. The system may survive (0, t] in two disjoint ways:– Item 1 does not fail in (0, t] (i.e., T1 > t).

– Item 1 fails in a time interval (, + d] where 0 < < t.

– The switch S is able to activate item 2. Item 2 is activated at time and does not fail in the time interval (, t].

– The probability of event 1 is Pr(T1 > t) = e−1t