(复杂系统的性能评价与优化课件资料)Queueing_Theory_Fundamentals

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• Suppose in each day the operation starts after the part arrives. Then in version 1, in day 1, there is 0 part in the queue; in day 2, there is 1 part; in day 2, there are 2 parts, so and so forth. In day 10, there are 9 parts in the queue. In day 11, there is 0 part in the queue.
Queue length or Waiting time
0
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Operational knee
1Utilization,
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The impact of randomness
• Using the Pollaczek-Kinchin formula, we
can get the performance measure for
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Manufacturing example of Little’s Law
• Version 1: 10 people, 10 machines, 10 operations/part, one part arrival /day
• Everyone working on the same operation on all ten parts, (e.g. on day 1, all the 10 people work on operation No.1 for the 10 parts; on day 2, all work on operation No.2 for the 10 parts, etc.)
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Elements of Q-Network
• Notations and definitions • M/M/1 queue and others • Little’s Law • Burke’s theorem • Queueing Network fundamentals • Product form equation • QNA, MPX®
M/G/1 and M/D/1.
2
NQM/M/1
1
,W M/M/1
1
2
N Q M /D /121,W M /D /121
N Q M /D /1 1 2N Q M /M /1 ,W M /D /1 1 2 W M /M /1
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Some other extensions
• Discouraged arrivals vs. Poisson distribution • M/M/∞=>always exists new server for new
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•
The
size
at
of
e
shaded
are0taN: tdt
at
Ti e i1
lti m 0 tT N ttlti m d ta i 1 tT lti im ;0 t N a ttlti m d ta tltti m ;ata tn d lti m N etlti m 0 tltN imttet dt0
(1/).
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About N()
• Average number of customers in the
system N = /(1-) • What happens if 0?
• N 0
• What happens if 1?
• N +
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The hocky stick tradeoff
server numbers • M/M/m/K/M=>general case
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Little’s Law
1
n
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Total Average Arrival Average
rate
number of customers in
this arbitrarily
enclosed
subsystem
departures
N
Average Flow Time through the enclosed subsystem
T
L ittle 'sL a w :N T
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N(t)
t
• The last arrival within t will leave the system at (t+e). So a(t)=d(t+e).
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M/M/1 queue
P nP n 1 P n 1 , n 0
/= b u s y p r o b a b i l i t y o r u t i l i z a t i o n .
P nn1 ,n 0 ,1 ,2 ,...
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Performance measure
customer • M/M/m=>there are m servers in the service station • M/M/1/K=>finite storage • M/M/m/m=>loss systems • M/M/1//M=>finite customer population • M/M/∞//M=>finite customer population, infinite
• Average number of customers in the
system N = /(1-) • Average time in system = 1/(-) • Average waiting time W = N/
• Average time in system = average waiting time + average service time
• Thus N(t)=a(t)-d(t)=d(t+e)-d(t).
• In steady state, e=T (the average flow time through
the queueing system is T) and departure rate=arrival
rate=.
• So N(t)=d(t+e)-d(t)=(t+e)-(t)=e=T.