m/g/1 variants and priority queue -...
TRANSCRIPT
Cheng-Fu Chou, CMLAB, CSIE, NTU
P. 2
HW. 1 M/G/1 with bulk service
Consider an M/G/1 system with bulk service. Whenever the server becomes free, he accepts 2 customers from the queue into service simultaneously, or , if only one is on queue, he accepts that one; in either case, the service time for the group (of size 1 or 2) is taken from B(x). Let qn be the number of customers remaining after the nth service instant. Let vn be the number arrivals during the nth service. Define B*(s), Q(z), and V(z) as transforms associated with the random variables x, q, v as usual. Let r= lx/2
Cheng-Fu Chou, CMLAB, CSIE, NTU
P. 3
– Using the method of imbedded Markov chain, find E(q) in term of r, P(q=0) =p0.
– Find Q(z) in term of B*(.), p0, p1= P(q=1)
– Express p1 in terms of p0
Ans:
(a) E[q]= r + (2(1-p0)+l2E[x2]-4r2)/(4(1-r))
(b) Q(z)= B*(l-lz) (p0(1-z2)+p1z(1-z))/(B*(l-lz)-z2)
(c) p1 = 2(1-p0-r)
Cheng-Fu Chou, CMLAB, CSIE, NTU
P. 4
10q2,
101
2k
0
,2q2,
q2,
1nq2,n1n
k2,
1
1
1
1
p-2p - 2 ]E[ v So,
)p-p-2(1 p
k]2P[q 1]P[q
][]E[
But
v ]E[ - q q
expection taking
and n letting ,v- q q have We
2k0k
2k 2
function thegIntroducin
0
1 1
2 2
can write Clearly we (a)
n
k
k
nn
nnn
nnn
n
kqP
qv
qvq
qvq
q
Cheng-Fu Chou, CMLAB, CSIE, NTU
P. 5
HW2 M/G/1 (service time)
Consider an M/G/1 queueing system in which service is given as follows. Upon entry into service, a coin is tossed, which has probability p of giving Heads. If the result is Heads, the service time for that customer is 0 seconds. If Tails, his service time is drawn from the following uniform distribution: f(x)=1/(b-a), if a<x<b; otherwise f(x)=0
– Find the average service time x
– Find the variance of service time
– Find the expected waiting time
– Find W*(s)
Cheng-Fu Chou, CMLAB, CSIE, NTU
P. 6
M/G/1 with vacations
Consider a first-come-first-served M/G/1 queue with the following changes. The server serves the queue as long as someone is in the system. Whenever the system empties the server goes away on vacation for a certain length of time, which may be a random variable. At the end of his vacation the server returns and begins to serve customers again: if he returns to an empty system then he goes away on vacation. Let
– be the z-transform for the number
of customers awaiting service when the server returns from vacation to find at least one customer waiting
1
)(j
j
j zfzF
Cheng-Fu Chou, CMLAB, CSIE, NTU
P. 7
(a) Derive an expression which gives qn+1 in term of qn, vn+1 and j
(b) Derive Q(z) in term of p0
(c) Show that p0 = (1-r)/F1(1), r = lx
(d) Assume now that the service vacation will end whenever a new customer enters the empty system. For this case find F(z) and show that when we substitute it back into our answer for (b) then we arrive at the classical M/G/1 solution.
Cheng-Fu Chou, CMLAB, CSIE, NTU
P. 8 1nkf,1n
kf,
n1n1n
n
1nn 1nn
v1 q
So,
0k f
0k k Let
0. qfor v 1 - f q Thus,
system. in the 1 f are thereuntil servingbegin not
does andn on vacatio goesserver the0q If
system M/G/1 usual for the as
v1q q 0,q if Clearly, (a)
Cheng-Fu Chou, CMLAB, CSIE, NTU
P. 9 z - z)-(*B
F(z))-(1pz)-(*B Q(z)
z)-(*B V(z) and z - V(z)
F(z))-(1pV(z) Q(z)
]p-[Q(z) F(z)p
][ 0]]p[qE[z ]z[ and
]z[z
V(z) Q(z)
find wen Letting .qf,, vof nceindepenedeby
]]E[zE[z
]E[z ]E[z (z)Q
have We(b)
0
0
00
1k
f
n1n
1v
v1-q
1n
qf,
qf,
qnf,1n
1nqnf,1n
llll
ll
kqpzE
E
k
Cheng-Fu Chou, CMLAB, CSIE, NTU
P. 10 (1)F
-1 p
x-1
(1)Fp
1 - )(0)(-*B
(1))(-Fp
z - z)-(*B
F(z))-(1plim
V(z)
Q(z) 1
Then, rule). HospitalL' (using
1 zat eqn. above theevaluate we,p determine To
z - V(z)
F(z))-(1pV(z) Q(z)
have we(b), From (c)
(1)0
(1)
0
(1)
(1)
0
0
1
0
0
r
ll
ll
z
Cheng-Fu Chou, CMLAB, CSIE, NTU
P. 11
M/G/1for equation mK transfor-P theish whic
z-z)-(*B
z)-)(1-(1z)-(*B Q(z)
.-1 p1, (1)F and z F(z) So,
1.k for 0f and 1f case, In this (d)
0
(1)
k1
ll
rll
r
Cheng-Fu Chou, CMLAB, CSIE, NTU
P. 12
M/G/1 with vacations
At the end of each busy period, the server goes on vacation for some random interval of time
A new arrival to an idle system rather than going into service immediately, waits for the end of the vacation
x1 x2 x3 x4 v1 v2 v3 x5
Cheng-Fu Chou, CMLAB, CSIE, NTU
P. 13
Let v1, v2, …, be the duration of successive vacations taken by the server and they are i.i.d. r.v.
Observation
– A new arrival to the system, waits for the completion of the current service or vacation
o So, the waiting time formula W = R/(1-r) is still valid
– R is the mean residual time for completion of the service or vacation
Cheng-Fu Chou, CMLAB, CSIE, NTU
P. 14
By using the same graphical argument
x1
x1
x2 xM(t)
rt
time t v1 v2
Cheng-Fu Chou, CMLAB, CSIE, NTU
P. 15
Residual service time for an M/G/1 system with vacations
– M(t): # of services completed by time t
– L(t) : # of vacations completed by time t
*)(
2
1
)(2
1
2
11
2
11)(
1
2)(
1
)(
1
2
2)(
1
)(
1
2
0
tL
v
t
tL
M(t)
x
t
M(t)
vt
xt
drt
R
i
tL
i
tM
i
i
i
tL
i
tM
i
i
t
tt
Cheng-Fu Chou, CMLAB, CSIE, NTU
P. 16
v
vxλR W
v
vρ)(xλ * R
vv
2)1(21
2
1
2
L(t)
)-t(1 ,
1
t
L(t)
t
M(t)
taslimit Taking
22
22
rr
rr
l
Cheng-Fu Chou, CMLAB, CSIE, NTU
P. 17
M/G/1 with feedback queue
Consider an M/G/1 system in which a departing customer immediately joins the queue again with probability p, or departs forever with probability q = 1- p. Service is FCFS, and the service time for a returning customer is independent of his previous service time. Let B*(s) be the transform for the service time pdf and let B*T(s) be the transform for a customer’s total service time pdf.
(a) Find B*T(s) in term of B*(s), p and q
(b) Find QT(z)
(c) Find N, the average number of customer in the systsem
Cheng-Fu Chou, CMLAB, CSIE, NTU
P. 18
0
**1n*
0n
0
**
T
1n**
T
)]([)((s))B(
ips)resturn trn ly ps)p(exactreturn trin exactly |()(B
yields ditioning Uncon
(s))(B ps)return trin exactly |(B
have weps,return tri ofnumber on the ngConditioni (a)
n
nn
n
T
spBsqBqp
sBs
s
Cheng-Fu Chou, CMLAB, CSIE, NTU
P. 19
(b) In determining the number in the system, we may assume that a customer cycles backs directly into service instead of to the tail of the queue. This is allowed due to the “memoryless” selection of a new service time each time a customer returns in addition to the independence of the feedback decision. Thus, we may consider our queue as an M/G/1 system with B*T(s) as the transform for the service time.
Cheng-Fu Chou, CMLAB, CSIE, NTU
P. 20
q
x
spB
sqB
zzB
zzBzQ
T
T
TTT
lr
ll
rll
and )(1
)(B where
)(
)1)(1()()(
*
**
T
*
*
Cheng-Fu Chou, CMLAB, CSIE, NTU
P. 22
Priority Queue
M/G/1 system with n different priority classes
– class 1 > class 2 > class 3 >…
– Arrival rate: lk
– Mean service time: xk = 1/mk
– Second moment of service time: 2
ix
Cheng-Fu Chou, CMLAB, CSIE, NTU
P. 23
Nonpreemptive priority
A customer undergoing service is allowed to complete service without interruption even if a customer of higher priority arrives in the meantime. A separate queue is maintained for each priority class
– Goal: find an equation for average delay for each priority class
– Total n classes
– NQk: average number on queue for class k
– Wk: average queueing time for class k
– rk = lk/mk: system utilization for class k
– R: mean residual service time
Cheng-Fu Chou, CMLAB, CSIE, NTU
P. 24
Assume that overall system utilization is less than 1, i.e., r1
r2 ... rn < 1
)-ρ-ρ)(-ρ(
R W
-ρ-ρ
W ρR W
Wλμ
Nμ
Nμ
R W
ρ
RW,WλN
Nμ
R W
Q
Q
211
2
21
112
21
1
2
2
1
1
2
1
111
1
1
1
1
11
1
111
similarly,
1
result sLittle'By
1
Cheng-Fu Chou, CMLAB, CSIE, NTU
P. 25 )-...--)(1-...---(12
W
2
1 R
R timeservice residualmean theMoreover,
1 T
isk class ofcustomer per delay avg. The
)-...--)(1-...---(1
R W
1 k classespriority allfor similar is derivation The
k11-k21
1
2
k
1
2
k
k
k11-k21
k
rrrrr
l
l
m
rrrrr
n
i
ii
n
i
ii
k
x
x
w
Cheng-Fu Chou, CMLAB, CSIE, NTU
P. 26
HW
Consider a nonpreemptive system and 2 customer classes A and B with respective arrival and service rate lA, mA, and lB, mB .
If mA > mB show that the average delay per customer (avg. over both classes)
T = lATA + lBTB/(lA+lB) is smaller when class A with higher priority (class A > class B) than the case when class B with higher priority ( class B > class A)
Cheng-Fu Chou, CMLAB, CSIE, NTU
P. 27
Preemptive resume priority
Service of a customer is interrupted when a higher priority customer arrives and is resumed from the point of interruption once all customers of high priority have been served.