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Leistungsanalyse Übung zu 5. Example 1 (contd.). The set of all possible values of X is { 1 , 2 , . . , n + 1 } and X = n + 1 for unsuccessful searches.

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### LeistungsanalyseÜbung zu 5

• The set of all possible values of X is {1, 2, . . , n + 1} and X = n + 1 for unsuccessful searches.

• Consider a random variable Y denoting the number of comparisons on a successful search. The set of all possible values of Y is {1, 2, . . , n}.

• Assume pmf of Y to be uniform over the range

• Thus, on the average, approximately half the table needs to be searched.

• Zipf’s law has been used to model the distribution of Web page requests.

• pY (i), the probability of a request for the i th most popular page is inversely proportional to i

• Assumption

• Web page requests are independent

• The cache can hold only m Web pages regardless of the size of each Web page.

• Adopting “least frequently used” removal policy, hit ratio h (m) -the probability that a request can find its page in cache-is given by (using Eq. (4.2) on p. 195 of text)

• Hit ratio increases logarithmically as a function of cache size.

• Except for the sign of s, the Laplace transform is the moment generating function used in mathematical statistics:

• Therth moment of X about the origin, if it exists, is given by the coefficient of (-sr)/r! in the Taylor series expansion of f*(s).

• If X denotes the time to failure of a system, then from a knowledge of the transform f*(s) we can obtain the system MTTF E[X], while it is more difficult to obtain the pdf f(t) and the reliability R(t).

• Example:

• Failure-time distribution is exponential with parameter λ:

• R(t) = P(X > t), X: Lifetime of a component

• Expected life time or MTTF is

• In general, kthmoment is,

• Simplified formula above can be derived using integration by parts and the fact that X is a non-negative random variable

• Series of components, component i lifetime is EXP(λi)

• Thus lifetime of the system is EXP with parameter

• and series system MTTF =

Series SystemMTTF (contd.)

• rv Xi : ith comp’s life time (arbitrary distribution)

• Case of weakest link. To prove above

• Parallel system: lifetime of ith component is rv Xi

• X = max{X1, X2, ..,Xn}

• If all Xi’s are EXP(λ), then,

• As n increases, MTTF increases

• and so does the Variance.

Variation of expected life with degree of parallel redundancy with each component having failure rate λ=10-6

Standby Redundancy redundancy with each component having failure rate

• A system with 1 component and (n-1) cold spares.

• If all Xi’s same EXP() X has Erlang distribution.

• TMR and ‘k of n’.

Triple Mode Redundancy (TMR) redundancy with each component having failure rate

• Assuming that the reliability of a single component is given by,

• we get:

• Comparing with expected life of a single component.

TMR redundancy with each component having failure rate (Continued)

• Thus TMR actually reduces (by 16%) the MTTF over the simplex system.

• Although TMR has lower MTTF than does Simplex, it has higher reliability than Simplex for “short” missions, defined by mission time t<(ln2)/λ.

EXP(3 redundancy with each component having failure rate )

EXP()

EXP(3)

EXP(2)

TMR and TMR/simplexas hypoexponentials

TMR/Simplex

TMR

Homework 1: redundancy with each component having failure rate

• Derive & compare reliability expressions for two component Cold, Warm and Hot standby cases.

• Also find MTTF in each case.

EXP( redundancy with each component having failure rate )

EXP()

Cold standby

X

Y

Spare state

EXP()

Active state

EXP()

• Assumptions:

• Detection & Switching perfect

• Spare does not fail

EXP( redundancy with each component having failure rate +)

EXP()

Warm standby:

• With Warm spare, we have:

• Time-to-failure in active state: EXP()

• Time-to-failure in spare state: EXP()

• 2-stage hypoexponential distribution

Warm standby redundancy with each component having failure rate

EXP(2 redundancy with each component having failure rate )

EXP()

Hot standby:

• With hot spare, we have:

• Time-to-failure in active state: EXP()

• Time-to-failure in spare state: EXP()

• 2-stage hypoexponential

Hot standby redundancy with each component having failure rate

Comparison graph: redundancy with each component having failure rate

The WFS Example redundancy with each component having failure rate

File Server

Computer Network

Workstation 1

Workstation 2

RBD for the WFS Example redundancy with each component having failure rate

Workstation 1

File Server

Workstation 2

R redundancy with each component having failure rate w(t): workstation reliability

Rf (t): file-server reliability

System reliability Rsys(t) is given by:

Note: applies to any time-to-failure distributions

RBD for the WFS Example (contd.)

RBD for the WFS Example (contd.) redundancy with each component having failure rate

• Assuming exponentially distributed times to failure:

• failure rate of workstation

• failure rate of file-server

• The system mean time to failure (MTTF) is

given by:

Homework 2: redundancy with each component having failure rate

• For a 2-component parallel redundant system

with EXP( ) and EXP( ) behavior, write down expressions for:

• Rp(t)

• MTTFp

Solution 2: redundancy with each component having failure rate

Homework 3 :Series-Parallel system (Example) redundancy with each component having failure rate

Example: 2 Control Channels and 3 Voice Channels

voice

control

voice

control

voice

Homework 3 redundancy with each component having failure rate (Contd.):

• Specialize formula to the case where reliability of control and voice are given as :

• Derive expressions for system reliability and system meantime to failure.

Control channels-Voice channels redundancy with each component having failure rate

Homework 4: redundancy with each component having failure rate

• Specialize the bridge reliability formula to the case where

Ri(t) =

• Find Rbridge(t) and MTTF for the bridge.

Bridge: conditioning redundancy with each component having failure rate

C1

C2

C3 fails

S

T

C1

C2

C4

C5

C3

S

T

C3 is working

C4

C5

C1

C2

S

T

Factor (condition)

on C3

C4

C5

Non-series-parallel block diagram

C1 redundancy with each component having failure rate

C2

S

T

C4

C5

Bridge: Rbridge(t)

When C3 is working

C1 redundancy with each component having failure rate

C2

S

T

C4

C5

Bridge: Rbridge(t)

When C3 fails

Bridge: R redundancy with each component having failure rate bridge(t)

Bridge: MTTF redundancy with each component having failure rate