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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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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.
slide12
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
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
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
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.
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

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