Mean Delay in M/G/1 Queues with Head-of-Line Priority Service and Embedded Markov Chains
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Mean Delay in M/G/1 Queues with Mean Delay in M/G/1 Queues with Head-of-Line Priority ServiceHead-of-Line Priority Service
andandEmbedded Markov ChainsEmbedded Markov Chains
Wade Trappe
Lecture OverviewLecture Overview
Priority Service– Setup– Mean Waiting Time for Type-1– Mean Waiting Time for Type-2– Generalization– Summary
Embedded Markov Chain Technique– Nk inside of N(t)– The queue distribution for M/G/1
The PGF approach– Example: M/H2/1 Queue– Homework: Alternate Pollaczek-Khinchine Derivation
Consider a queue that handles K priority classes of customers– Type-k customers arrive as a Poisson Process with rate and service
times according to .
Customers are separated as they arrive and assigned to different “prioritized” queues
Each time the server is free:– Selects the next customer from highest priority, non-empty queue
– This is the “head of line” service
– We also assume that customers are not pre-empted!
Server Utilization for type-k customers is
Stability requirement:
M/G/1 Priority Services: Setup M/G/1 Priority Services: Setup
1K21
kkk XE
kkX
Priority Services: ExplanationPriority Services: Explanation
Server
Queue-1
Queue-2
1
1
22
1
Blue Type-1 Packets always have priority, but cannot preempt Type-2 packets
Our first goal will be to calculate the mean waiting time for Type-1 customers, i.e.
Assume a type-1 customer arrives and finds Nq1(t)=k1 type-1 already in the queue
Assume we use FCFS within each class
W1 consists of:– Residual time R of customer in server
– k1 service times of type-1 in queue
Thus:
We get the mean waiting time for type-1 as
Priority Services: Mean Waiting TimesPriority Services: Mean Waiting Times
1W
1q1 XENEREW1
1
l1
REW
Now, let us look at a type-2 customer. When he arrives he may find
– Nq1(t)=k1 type-1 already in the queue-1
– Nq2(t)=k2 type-2 already in the queue-2 Thus, W2 is the sum of:
– Residual service time R– k1 service times for type-1customers– k2 service times for type-2 customers– AND service times of any new type-1 customers!!!
We get the mean waiting time for type-2 as
Priority Services: Mean Waiting Times, pg 2Priority Services: Mean Waiting Times, pg 2
112q1q2 XEMEXENEXENEREW21
Priority Services: Mean Waiting Times, pg 3Priority Services: Mean Waiting Times, pg 3
Here, M1 is the number of customers of type-1 that arrive during type-2’s waiting period
By Little’s Theorem applied to the queues:
The mean number of type-1 arrivals during seconds is
Substitute to get
2122112 WWWREW
22q
11q
WNE
WNE
2
1
2W
211 WM
Priority Services: Mean Waiting Times, pg 4Priority Services: Mean Waiting Times, pg 4
Solve for :
Generalizing gives
Now, we’re left with finding E[R]!
121
1
1
21
112
11
RE
1
REWuse
1
WREW
2W
k211k21
k11
REW
Priority Services: Mean Waiting Times, pg 5Priority Services: Mean Waiting Times, pg 5
When a customer arrives, the one being serviced can be of any type! So R must account for this!
We will use the same formulation as
where is the aggregate Poisson Process rate.
To get E[X2] we use the fact that the fraction of type-k customers is :
Thus, we get
2k
22
21
2 XEXEXEXE k21
k1
k211k21
K
1j
2jj
k11
XE
W
2XE2
RE
k
Priority Services: Mean DelayPriority Services: Mean Delay
Now, calculating the mean delay for a type-k customer is easy!
From this, observe:– Class-k customers are affected by the behavior of lower
priority customers
kkk XWT
The Embedded Markov ChainThe Embedded Markov Chain
Let Nk be the number of customers found in the system just after the service completion of the k-th customer
Let Ak be the number of customers who arrive while the k-th customer is served
The Ak are i.i.d.
Under FCFS: Ak is independent of N1, N2, …, Nk-1
Nk is not independent of Ak
Nk-2 Nk-1 Nk
Ck-2 Ck-1 Ck
Time
N(t)
Ck-1
Ck-1
Ck
Ck
Ak-1=1 Ak=3
Ck+1
Xk-1 Xk
Service
Queue
Embedded Markov Chain, pg 2Embedded Markov Chain, pg 2
We may find the following recursion:
Where did first line come from?– If , then k-th customer finds the system empty, and on
departure he leaves behind those who have arrived during his service.
The second line:– The queue left by the k-th customer consists of the previous queue
decreased by one plus the new arrivals during his service.
Let U(x)=1 for x>0 and U(x)=0 for x<=0
Then we get
0Nif1AN
0NifAN
1kk1k
1kkk
0N 1k
k1k1kk ANUNN
Embedded Markov Chain, pg 3Embedded Markov Chain, pg 3
From
we can see that Nk is a Markov Chain since Nk depends on the past only through Nk-1
Nk is called the embedded Markov Chain
(Homework): The distribution for N(t) and Nk are the same:
– Step 1: Show, in M/G/1, distributions found by arriving customers is the same as that left by departing customers
– Step 2: Show, in M/G/1, the distribution of customers found by arriving customers is the same as the steady state distribution of N(t)
We will analyze Nk instead of N(t)
k1k1kk ANUNN
M/G/1 Queue DistributionM/G/1 Queue Distribution
To find the queue distribution, we use the probability generating function method
The PGF of the distribution {pn} is
Substitute and use independence to get
Thus
0n
N
k
nn
kzEzp)z(P lim
k1k1kk ANUNN
01
0k
1n
1nn0k
NUNA
k
p)z(PzpzA
zppzAzEzE)z(P 1k1kklim
zAz
)1z(zAp)z(P
k
k0
M/G/1 Queue Distribution, pg. 2M/G/1 Queue Distribution, pg. 2
If the service time of customer k is t, then Ak has a Poisson distribution with mean t
Now multiply by the fX(t), the pdf of the service time Xk, This gives the joint distribution.
Integrate out X to obtain the unconditional pdf of Ak
What does this remind you of?
Answer: Laplace Transforms!!!
!i
tetXiAP
it
kk
dttf!i
teiAPa X0
it
ki
M/G/1 Queue Distribution, pg. 3M/G/1 Queue Distribution, pg. 3
The Probability Generating Function of Ak is given by
Substitute into earlier result to get:
zf
dttfee
dttf!i
tze
zazEzA
X
X0
tzt
0iX0
it
0i
ii
Ak
k
zfz
)1z(zfp)z(P
X
X0
LaplaceTransformof fX(t)
M/G/1 Queue Distribution, pg. 4M/G/1 Queue Distribution, pg. 4
We need to now find p0!
To do this, look at limit as z goes to 1, and apply L’Hopital…
First, go back to
Observe that P(z=1)=1, Ak(z=1)=1
Thus
We used:
XEdtetftsf
1dttf0f
0
stXXds
d
0
XX
XE1
p
0f1
)0(fp1 0
'X
X0
zAz
)1z(zAp)z(P
k
k0
ShowThis!
M/G/1 Queue Distribution, pg. 5M/G/1 Queue Distribution, pg. 5
Thus,
This is exactly what we got for M/M/1 Queues!!!
Substitute into earlier result to get:
What do we do with this? Use it to find the probability distribution pn!
How? Equate terms… Best to see an example
1XE1p0
zfz
)1z(zf1)z(P
X
X
M/HM/H22/1 Queue/1 Queue
Consider the case where FX(t), the cdf, is a two-term hyper-exponential distribution:
Where do hyper-exponentials come from?– In practice, from situations where service times can be represented as a
mixture of exponential distributions…
– That is, suppose there are k types of customers, each occuring with a probability i (so sum = 1)
– Customer of type I has a service time exponentially distributed with mean 1/i
Then the density is
t2
t1X
21 ee1tF
k
1i
xiiX
iexf
M/HM/H22/1 Queue, pg. 2/1 Queue, pg. 2
Back to our 2-phase hyper-exponential problem…
The mean service time is given by
The Laplace Transform of fX(t) is:
We thus get:
2
2
1
11
ss
sf2
22
1
11X
ii2
2
1
1Xk /for
z11z11zfzA
M/HM/H22/1 Queue, pg. 3/1 Queue, pg. 3
The PGF of {pn} is
The probability distribution is obtained through partial-fraction expansion of P(z).
To do this, we must find the roots of the denominator… call these z1 and z2
Then P(z) has the form:
zz
zC
zz
zC)z(P
2
22
1
11
2211
2121212
21
21
where
1zz
z111zP
= Total Traffic Intensity
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