Service Engineering Class 13 QED (QD, ED) Queues Erlang-A ...
Advanced Markovian queues -...
Transcript of Advanced Markovian queues -...
Advanced Markovian queuesAdvanced Markovian queues
• Bulk input queue M[X]/M/1• Bulk service queue M/M[Y]/1Bulk service queue M/M /1• Erlangian queue M/Ek/1
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Bulk input queue M[X]/M/1Bulk input queue M[X]/M/1
• Batch arrival, Poisson process, arrival rate λ– number of customers in each arrival is an integer grandom number X
– Probability of X is c =Pr[X=n]Probability of X is cn=Pr[X=n]
• State transition rate diagram– X={1,2} as example, pp.118 of textbook
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Balance equation of M[X]/M/1Balance equation of M[X]/M/1
• Global balance equation0 ( ) 1
n
p p p c nλ μ μ λ= − + + + ≥∑11
0 1
0 ( ) , 1
0
n n n k kk
p p p c n
p p
λ μ μ λ
λ μ
+ −=
= + + + ≥
= − +
∑
• To solve these equations, use Z‐transforms of c and pp
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Balance equation of M[X]/M/1Balance equation of M[X]/M/1
• Multiply zn on both sides and sum all
0n
n n n nμλ μ λ∞ ∞ ∞ ∞
+ +∑ ∑ ∑ ∑∑
W h
0 1 1 1 10 n n n n
n n n n k kn n n n k
p z p z p z p c zzμλ μ λ −
= = = = =
= − − + +∑ ∑ ∑ ∑∑
• We have 0 00 ( ) [ ( ) ] [ ( ) ] ( ) ( )P z P z p P z p C z P z
zμλ μ λ= − − − + − +
• Thus, z
0 (1 )( ) | | 1p zP z zμ −= ≤( ) , | | 1
(1 ) [1 ( )]P z z
z z C zμ λ= ≤
− − −
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Average metrics of M[X]/M/1Average metrics of M[X]/M/1
• Let z=1, obtain the average metrics1 / [ ]p r rE Xρ λ μ ρ= − = =0
2 2
1 , / , [ ]
( [ ] [ ]) [ ]2(1 ) 2(1 )
p r rE X
r E X E X rE XL
ρ λ μ ρ
ρρ ρ
= = =
+ += =
− −2(1 ) 2(1 )ρ ρ
0(1 )qL L p L ρ= − − = −
/ , /q qW L W Lμ μ= =
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Example of M[X]/M/1Example of M[X]/M/1
• X is a constant K, apply the results we have1K K Kρ ρ ρ λ+ +1 ,
2(1 ) 2 1K K KL ρ ρ ρ λρρ ρ μ
+ += = =
− −
• What’s the difference from M/M/1 with Kλ?• Equivalent to M/M/1 with Kλ and a scaleEquivalent to M/M/1 with Kλ and a scale factor (K+1)/2
f– Worse performance metrics– Why? more bursty
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Example of M[X]/M/1Example of M[X]/M/1
• X is a geometric r.v. with parameter a1 (1 )( ) (1 ) n n z aC λ∞− −∑
• Thus
1
1
( )( ) (1 ) , 1 (1 )
n n
n
C z a a zaz a
ρμ=
= − = =− −∑
0 (1 )( ) p zP z μ −=Thus, ( )
(1 ) [1 ( )]
1(1 )
P zz z C z
az
μ λ
ρ
=− − −
⎡ ⎤= − −⎢ ⎥
• The steady‐state probability is
(1 )1 [ (1 ) ] 1 [ (1 )]z a a z a a
ρρ ρ⎢ ⎥− + − − + −⎣ ⎦
y p y1(1 )[ (1 ) ] (1 )n
np a a aρ ρ ρ−= − + − −
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Bulk service queue M/M[Y]/1Bulk service queue M/M[Y]/1
• Similar to M/M/1, except the server serves K customers at a time, called M/M[K]/1. K is a constant.
• Two model for different service mode• Two model for different service mode– Partial‐batch
• Start service no matter n<K
– Full‐batch• Start service until n>=k
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Partial batch model of M/M[K]/1Partial‐batch model of M/M[K]/1
• Global balance equation10 ( ) , 1n n K np p p nλ μ μ λ+= − + + + ≥
itt i ti f
1
0 1 2
( ) ,0 ...
n n K n
K
p p pp p p p
μ μλ μ μ μ
+ −
= − + + + +
• rewritten in operation form1[ ( ) ] 0K
nD D pμ λ μ λ+ − + + =
• root of characteristic equation above, we have1
1
Kn
n i ii
p C r+
=
=∑1
( 1) ( 1) 0 ( 1)[ ... ] 0
K
K K
r r rr r r rμ λ
μ μ μ λ−
− − − = ⇒
− + + + − =
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Partial batch model of M/M[K]/1Partial‐batch model of M/M[K]/1
• Only one root r0 within (0,1)0n
np Cr=
• With the law of total probability, C=1‐r0
• Similar to M/M/1 r instead of ρ0 0(1 ) n
np r r= −
Similar to M/M/1, r0 instead of ρ0 0,
1 (1 )r rLL W
r rλ λ= = =
− −1, q qL L W Wλ
μ μ= − = −
• For M/M/1, let K=1, how is the above result?0 01 (1 )r rλ λ− − μ μ
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M/M[K]/1 v s M/M/1M/M[K]/1 v.s. M/M/1
• formulas of metrics are similar– Formulas for distribution, L, W, Lq, Wq are similar, , , q, q
• similar to M/M/1 with service rate Kμi h f h i i i– r0 is the root of characteristic equation
– ρ=λ/Kμ, is the utilization factor of M/M/1– which one is bigger?
• Guess: r0 > ρ (because M/M[K]/1 is more bursty more 0 ρ ( / / ycrowded)
– Metrics of M/M[K]/1 is worse than those of M/M/1/ / / /
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Example of M/M[K]/1Example of M/M[K]/1
• Drive‐Thr Car wash– λ=20/h, μ=12/h, K=2/ , μ / ,
• Characteristic equation12 3 32 20 3 3 8 5 0 1 ( 3 69) / 612r3‐32r+20=3r3‐8r+5=0, roots are
• So, r0=0.884 and L=7.6, Lq=5.9, W=38min, Wq=33min
1, ( 3 69) / 6r r= = − ±
0 q q
• Compare the equivalent M/M/1 with λ=20 and μ=24ρ=5/6=0 833 and L=5 L =4 167W=15minW =12 5minρ=5/6=0.833 and L=5,Lq=4.167,W=15min,Wq=12.5min
• M/M[K]/1 is worse than M/M/1
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Erlangian queuesErlangian queues
• Exponential distribution – Memoryless property, simplify the analysis y p p y, p y y– Limitation, more general cases
E l di t ib ti f i ti d• Erlang distribution for service time and interarrival time– Sum of K‐stage exponential distributions– More general modelMore general model
• Other distributions– PH, Coxian, hyperexponential, etc.
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Erlang distributionErlang distribution
• k‐stage Erlang distribution, each stage is exponentially distributed with rate kμp y μ
• By convolution, pdf is 1( )( )
( 1)!
kk xk k xf x e
kμμ μ −
−=−
2
1 1 1[ ] , [ ] , XE X Var X ck kμ μ
= = =μ μ
kμ kμ kμkμ… …
1 2 i k
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Erlang distributionErlang distribution• Distribution with 2‐parameters,Distribution with 2 parameters, cx<1, to model smooth stochastic processprocess– Approximate empirical distribution, use the mean and variance to match μ and k
• If cx>1, we can use other If cx , we can use otherdistribution model, such as hyperexponential distributionyp p
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Phase Type distributionPhase‐Type distribution
• A generalization of concept of phases• PH distribution: the time to enter anPH distribution: the time to enter an absorbing state in Markov processF l E l 2 di ib i• For example, Erlang‐2 distribution:
μ μ 0μ μ−⎡ ⎤⎢ ⎥1 2 3 0
0 0 0Q μ μ⎢ ⎥= −⎢ ⎥
⎢ ⎥⎣ ⎦
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PH distributionPH distribution
• Coxian distributionμp1 μ 1 1(1 )p pμ μ μ− −⎡ ⎤
1 2 3μp1 μ 1 1(1 )
00 0 0
p pQ
μ μ μμ μ
⎡ ⎤⎢ ⎥= −⎢ ⎥⎢ ⎥⎣ ⎦
• Hyerexponential distributionμ(1-p1)
⎣ ⎦
Hyerexponential distribution1 μ1
1 10μ μ−⎡ ⎤⎢ ⎥
with an initial
23μ2
2 200 0 0
Q μ μ⎢ ⎥= −⎢ ⎥⎢ ⎥⎣ ⎦
distribution p(0)=(q,1-q)
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The calculation of pdf of PH distribution
• Use the C‐K equation and matrix calculation of Q
• Calculate the transient behavior of absorbing state p (t)state, pn(t). – pn(t)=P{T<t}, so pn(t) is cdf of PH and p’n(t) is pdf.
• Chalk writing, use hyperexponential distribution as exampledistribution as example– For any PH distribution, we can similarly analyze by its Q matrix of Markov processby its Q matrix of Markov process
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M/E /1 queueM/Ek/1 queue
• Service time is Erlang‐k distribution– k stage, each stage is exponential with kμg , g p μ
• State definition( i) d i i i i– (n,i): n customers and customer in service is in phase i, i=1,2,…,k. (k is the first phase and 1 is the l h )last phase)
• Global balance equation q
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Average metrics of M/E /1Average metrics of M/Ek/1
• Consider equivalent queue with state (n‐1)k+i, # of phase requestsp q
• Equivalent to M[k]/M/1, average # of customer is average # of phase requests 1kL ρ λ+is average # of phase requests
• So, ,
2 1L ρ ρ
ρ μ= =
−21 1k kW L Wρ ρλ+ +,
2 (1 ) 2 (1 )q q qW L Wk k
λμ ρ ρ
= = =− −
11 , q qW W L L ρμ
= + = +Can we use PASTA to do mean value analysis?
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Example of M/E /1Example of M/Ek/1
• Suppose a queuing system– λ=16/h, Poisson arrival/ ,– Mean service time 2.5min, std. deviation is 1 25min1.25min
• not exponentially distribution• Use Erlang k to approximate C =0 5=(1/k)0.5 k=4• Use Erlang‐k to approximate, CX=0.5=(1/k)0.5, k=4
– M/E4/1 queue, with λ=4/15/min, μ=0.4/min, ρ=2/3– So,
Lq=5/6, Wq=25/8minq q
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