Leistungsanalyse bung zu 5
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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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Leistungsanalyse Übung zu 5

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Leistungsanalyse bung zu 5

LeistungsanalyseÜbung zu 5


Example 1 contd

Example 1 (contd.)

  • 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


Example 1 contd1

Example 1 (contd.)

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


Example 2 zipf s law

Example 2-Zipf’s law

  • 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


Example 2 zipf s law contd

Example 2-Zipf’s law (contd.)

  • 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.


Moment generating property

Moment Generating Property

  • 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).


Moment generating property1

Moment Generating Property

  • Example:

    • Failure-time distribution is exponential with parameter λ:


Mttf computation

MTTF Computation

  • 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


Mttf computation series system

MTTF Computation-Series System

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

  • Thus lifetime of the system is EXP with parameter

  • and series system MTTF =


Series system mttf contd

Series SystemMTTF (contd.)

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

  • Case of weakest link. To prove above


Parallel system mttf computation

Parallel System-MTTF Computation

  • 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.


Leistungsanalyse bung zu 5

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


Standby redundancy

Standby Redundancy

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

  • System lifetime,

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

  • TMR and ‘k of n’.


Triple mode redundancy tmr

Triple Mode Redundancy (TMR)

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

  • we get:

  • Comparing with expected life of a single component.


Tmr continued

TMR (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.


Tmr and tmr simplex as hypoexponentials

EXP(3)

EXP()

EXP(3)

EXP(2)

TMR and TMR/simplexas hypoexponentials

TMR/Simplex

TMR


Homework 1

Homework 1:

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

  • Also find MTTF in each case.


Cold standby

EXP()

EXP()

Cold standby

X

Y

Lifetime in

Spare state

EXP()

Lifetime in

Active state

EXP()

  • Total lifetime 2-Stage Erlang

  • Assumptions:

    • Detection & Switching perfect

    • Spare does not fail


Warm standby

EXP(+)

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 standby1

Warm standby


Hot standby

EXP(2)

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 standby1

Hot standby


Comparison graph

Comparison graph:


The wfs example

The WFS Example

File Server

Computer Network

Workstation 1

Workstation 2


Rbd for the wfs example

RBD for the WFS Example

Workstation 1

File Server

Workstation 2


Rbd for the wfs example contd

Rw(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 contd1

RBD for the WFS Example (contd.)

  • 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

Homework 2:

  • For a 2-component parallel redundant system

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

    • Rp(t)

    • MTTFp


Solution 2

Solution 2:


Homework 3 series parallel system example

Homework 3 :Series-Parallel system (Example)

Example: 2 Control Channels and 3 Voice Channels

voice

control

voice

control

voice


Homework 3 contd

Homework 3 (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

Control channels-Voice channels


Homework 4

Homework 4:

  • Specialize the bridge reliability formula to the case where

    Ri(t) =

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


Bridge conditioning

Bridge: conditioning

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


Bridge r bridge t

C1

C2

S

T

C4

C5

Bridge: Rbridge(t)

When C3 is working


Bridge r bridge t1

C1

C2

S

T

C4

C5

Bridge: Rbridge(t)

When C3 fails


Bridge r bridge t2

Bridge: Rbridge(t)


Bridge mttf

Bridge: MTTF


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