Operational Laws

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Chapter 33

• Operational Laws

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Operational Laws

• Relationships that do not require any assumptions
  about the distribution of service times or
  interarrival time.
• Identified originally by Buzen (1976) and later
  extended by Denning and Buzen (1978)
• Operational gt directly measured.

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Operational Laws (2)

• Operationally testable assumptions
• gtassumptions that can be verified by
  measurements.
• For example, whether number of arrivals is equal
  to the number of completions?
• This assumption, called job flow balance, is
  operational testable.
• As set of observed service times is or is not a
  sequence of independent random variables which is
  not operationally testable.

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Operational Laws (3)

• Operational quantities Quantities that can be
  directly measured during a finite observation
  periods.

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Operational Law (4)
Black Box

• Arrival Departures
• T observation interval
• Ai number of arrivals.
• Ci number of completions
• Bi busy time

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Operational Laws (4)

• Arrival Rate ?i No. of Arrivals /Time
• Ai/T
• Throughput Xi No. of completions / Time
• Ci/T
• Utilization Ui Busy Time /Total Time
• Bi/T
• Mean service time Si Total time served/no.
  served
• Bi/Ci
• Relationships that hold in every observation
  period are called operational laws.

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Utilization Law

• Ui Bi/T Ci/T x Bi/Ci
• or
• Ui XiSi

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Utilization Law

• Operational laws are similar to the elementary
  laws of motion.
• For example,
• d1/2 a t2
• Notice that distance d, acceleration a and time
  t are operation quantities. There is no need to
  consider them as expected value of random
  variables or to assume a probability distribution
  for the.

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Example 33.1

• Consider a network gateway at which the packet
  arrive at a rate of 125 packets/s, and the
  gateway takes an average of 2 ms to forward them.
• Xi ?i 125 pack/s
• Si 0.002 s
• Ui 125 x 0.002 0.25 25
• This result is valid for any arrival or service
  process.
• Even if interarrival time and service time are
  not IID random variable with exponential
  distribution.

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Force Flow Law

• Relates the system throughput to individual
  device throughputs.
• In an open model, system throughput no. of jobs
  leaving the system per unit time.
• In a closed model, system throughput no. of jobs
  traversing the OUT to IN link per unit time

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Force Flow Law

• If observation period T is such that AiCi.
• gt Device satisfies the assumption of job flow
  balance.
• Each job makes Vi requests for ith device in the
  system.

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Force Flow Law (2)
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Force Flow Law (3)

• Ci C0 Vi
• or
• Vi Ci /C0
• Vi is called visit ratio.
• System throughput X
• Jobs completed/Time C0/T
• Throughput for ith device Xi
• Ci/T Ci/C0 x C0/T

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Force Flow Law (4)

• Xi Ci/T Ci/C0 x C0/T
• Xi X Vi
• This is the forced flow law.

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Force Flow Law (5)

• Combining the force flow law and the utilization
  law, we get
• Utilization of ith device Ui Xi Si
• X Vi Si
• or
• Ui X Di
• Here, Di Vi Si is the total service demand on
  the device for all visits of a jobs.
• The device with the highest Di has the highest
  utilization and is the bottleneck device.

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Example 33.2

• In a timesharing system, accounting log data
  produced the following profile for user program.
  Each program required five seconds of CPU time,
  make 80 I/O requests to disk A, and 100 I/O
  request to disk B. Average think-time of the
  users was 18 s. From the device specifications,
  it was determined that disk A takes 50 ms to
  satisfy an I/O request, and the disk B take 30 ms
  I/O request. With 17 active terminals, disk A
  throughput was observed to be 15.70 I/O
  request/s. We want to fine the system throughput
  and device utilization.

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Example 33.2 (i)

• five seconds of CPU time,
• Dcpu 5s
• make 80 I/O requests to disk A,
• VA80
• and 100 I/O request to disk B.
• VB100
• Average think-time of the users was 18 s.
• Z18 s
• disk A takes 50 ms to satisfy an I/O request,
• SA0.050s
• and the disk B take 30 ms I/O request.
• SB0.030s
• With 17 active terminals,
• N17
• disk A throughput to be 15.70 I/O request/s.
• XA15.70jobs/s

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Example 33.2 (ii)

• Dcpu 5s
• VA80
• VB100
• Z18 s
• SA0.050s
• SB0.030s
• N17
• XA15.70jobs/s

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Example 33.2 (iii)

• Since the job visit CPU before going to the disk
  or terminals, the CPU visit
• Vcpu VA VB 1 80 100 1 181
• Dcpu 5 s
• DA SA VA 0.050 x 80 4s
• DB SB VB 0.030 x 100 3s
• Using the force flow law
• X XA /VA 15.70/80 0.1963 jobs/s
• Xcpu X Vcpu 0.1963x 181 35.48 requests/s
• XB X VB 0.1963x 100 19.63 requests/s

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Example 33.2 (iv)

• Using the utilization law, the device
  utilizations are
• Ucpu X Dcpu 0.1963 x 5 98
• UA X DA 0.1963 x 4 78.4
• UB X DB 0.1963 x 3 58.8

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Transition Probabilities

• pijProbability a job moving to jth queue after
  service completion at ith queue.
• Visit ratios and transition probabilities are
  equivalent in the sense that given one we can
  always find the other.
• In a system with job flow balance.

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Transition Probabilities (2)

• Visit to the outside link
• Pi0Probability of a job exiting from the system
  after completion of service at the ith device.
• Dividing by C0 we get.

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Transition Probabilities (3)

• Since each visit to the outside link is defined
  as completion of the jobs, we have V01
• These are called visit ratio equations.
• In central server models, after completion of
  service at every queue, the jobs always move back
  to the CPU queue

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Transition Probabilities

• The above probabilities apply to exit and
  entrances from the system (i0), also.
  Therefore, the visit ratio equations become
• 1V1p1,0
• V11V2V3VM
• VjV1p1,jp1,j/p1,0 j2,3,,M
• The we can find the visit ratio by dividing the
  probability p1,j of moving to jth queue from CPU
  by the exist probability p1,0

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Example 33.3

• Consider the queuing network
• VA80
• VB100
• VCPU181
• After completion of service at the CPU, the
  probabilities of the job moving to disk A, disk B
  or terminal are 80/181, 100/181 and 1/181,
  respectively. Thus, the transition probability
  are 0.4420, 0.5525, and 0.005525

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Example 33.3 (ii)

• Given the transition probabilities, we can find
  the visit ratio by dividing these probabilities
  by the exit probability (0.005525).
• VA0.4420/0.00552580
• VB0.5525/0.005525100
• VCPU1VAVB180100181

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Littles Law

• Mean number in the device Arrival rate x mean
  time in the device
• Qi ?i Ri
• If the job flow is balanced, the arrival rate is
  equal to the throughput and we can write
• Qi Xi Ri

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Example 33.4

• The average queue length in the computer system
  (In Example 33.2 and Example 33.3) was observed
  to be 8.88, 3.19 and 1.40 job at the CPU, disk A
  and disk B, respectively. What were the response
  times of these devices?

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Example 33.4 (ii)

• In previous example, we know
• Xcpu 35.48 XA 15.70 XB 19.6
• Now, we have
• Qcpu 8.88 QA 3.19 QB 1.40
• Using Littles law, the device response times
  are
• RCPU QCPU /XCPU 8.88/35.480.250 s
• RA QA /XA 3.19/15.70 0.203 s
• RB QB /XB 1.40/19.60.071 s

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General Response Time Law

• There is one terminal per user and the rest of
  the system is shared by all users

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General Response Time Law (2)

• Applying Littles Law to the central subsystem
• QXR
• Here Qtotal number of jobs in the system
• R system response time
• X system throughput
• QQ1Q2QM QX1R1X2R2XMRM

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General Response Time Law (2)

• QX1R1X2R2XMRM
• Dividing both sides by X and using force flow
  law
• RV1R1V2R2VMRM
• This is called the general response time law,
  this law holds even if the job flow is not
  balanced.

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Example 33.5

• Let us compute the response time for the time
  for the timesharing system of previous example.
• For this system.
• Vcpu 181 VA 80 VB 100
• RCPU 0.250 s RA 0.203 s RB 0.071 s
• The system response time is
• R VCPURCPUVARAVBRB
• 181x0.250 80x0.203 100x0.071
• 68.6
• The system response time is 68.6 seconds

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Interactive Response Time Law

• If Zthink-time, RResponse time
• The total cycle time of requests is RZ,
• Each user generates about T/(RZ) requests in T
• If there are N users
• The system throughput X
• Total number of requests/Total Time
• N(T/(RZ))/T
• N/(RZ) or R(N/X) Z
• This is the interactive response time law.

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Example 33.6

• For the timesharing system of Example 33.2, we
  can compute the response time using the
  interactive response time law as follows
• X0.1963, N17, Z18
• Therefore
• R(N/X) Z 17/0.1963 -1868.6 s
• The is same as that obtained earlier in Example
  33.5

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Bottleneck Analysis

• From force flow law
• The device with highest total service demand Di
  has the highest utilization and is called the
  bottleneck device.
• Note Delay center can have utilizations more
  than one without any stability problem.
  Therefore, delay centers cannot be a bottleneck
  device.
• Only queuing centers used in computing Dmax.

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Bottleneck Analysis (2)

• The bottleneck device is the key limiting factor
  in achieving higher throughput.
• Improving the bottleneck device will provide the
  highest payoff in terms of system throughput.
• Improving other devices will have little effect
  on the system performance.
• Identifying the bottleneck device should be the
  first step in any performance improvement project.

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Bottleneck Analysis (3)

• Throughput and response times of the system are
  bound as follows
• X(N)ltmin1/Dmax, N/(DZ)
• and
• R(N)gtmaxD, N Dmax - Z)
• Here, D?Di is the sum of total service demands
  on all devices except terminals.
• These are known as asymptotic bounds.

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Proof

• The asymptotic bounds are based on the following
  observation
• The utilization of any device cannot exceed one.
  This put a limit on the maximum obtainable
  throughput.
• The response time of the system with N users
  cannot be less than a system with just one user.
  This put a limit on the minimum response time.
• The interactive repose time formula can be used
  to convert the bound on throughput to that on
  response time and vice versa.

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Proof (2)

• For the bottleneck device b we have.
• UbX Dmax.
• Since Ub cannot be more than one, we have
• Dmaxlt1
• Or
• X lt1/Dmax.

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Proof (3)

• With just one job in the system, there is no
  queuing and the system response time is simply
  the sum of the service demand
• R(1)D1D2DMD
• Here D is defined as the sum of all service
  demands. With more than one user there may be
  some queuing and so the response time will be
  higher. That is
• R(N)gtD

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Proof (4)

• R(N)gtD
• Applying the interactive response time law to the
  bounds
• R(N) N/X(N)-Z gtN Dmax Z
• and
• X(N) N/(R(N)Z) ltN (D Z)
• Combining these bounds we get the asymptotic
  bounds.

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Typical Asymptotic Bounds
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Typical Asymptotic Bounds (2)

• Both throughput and response time bounds consist
  of two straight lines.
• The point of intersection of the two lines is
  called the knee.
• For both response time and throughput, the knee
  occurs at the same value of number of users.

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Typical Asymptotic Bounds (3)

• The number of Jobs N at the knee is given by
• DN Dmax Z
• or
• N (DZ) / Dmax
• If the number of jobs is more than N, then we
  can say with certainty that there is queuing
  somewhere in the system.

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Typical Asymptotic Bounds (4)

• The asymptotic bounds can be easily explained to
  people who do not have any background in queuing
  theory or performance analysis.

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Example 33.7

• For the timesharing system considered in previous
  example
• DCPU 5 DA 4 DB 3 Z18
• D DCPUDADB 54312
• DmaxDCPU5
• The asymptotic bounds are
• X(N) ltminN/(DZ), 1/DmaxminN/30,1/5
• R(N) gtmaxD, NDmax-Zmax12, 5N-18

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Example 33.7 (ii)

• The knee occurs at
• 125 N 18
• N (1218)/56
• Thus, if there are more than 6 users on the
  system, there will certainly be queuing in the
  system.

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Example 33.8

• How many terminals can be supported on the
  timesharing system of previous example if the
  response time has be kept below 100 seconds?

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Example 33.8 (ii)

• Using the asymptotic bounds on the response time,
  we get
• R(N) gt max12, 5N-18
• the response time will be more than 100, if
• 5N 18 gt 100
• That is, if Ngt23.6
• The response time is bound to 100. Thus, the
  system cannot support more than 23 users if a
  response time of less than 100 is required.

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