m/g/1 variants and priority queue -...

Cheng-Fu Chou, CMLab, CSIE, NTU

M/G/1 variants and Priority Queue

Cheng-Fu Chou

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


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)


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


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)


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


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.


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)


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


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


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


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


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


P. 14

By using the same graphical argument

x1

x1

x2 xM(t)

rt

time t v1 v2


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


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


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


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


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.


P. 20

q

x

spB

sqB

zzB

zzBzQ

T

T

TTT

lr

ll

rll

and )(1

)(B where

)(

)1)(1()()(

*

**

T

*

*


P. 21

)1(2N So,

.)1(2

N M/G/1,For (c)

22

22

T

TT

x

x

r

lr

r

lr


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


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


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

QQ

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


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


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)


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.


P. 28

k11-k1

k1

k

1

11

11

1

1k1

k

k

k

1

2

k

k1

k

k

-......1-......1

-......11

,1

11

-...--1

R1T .3

2R where,

-...--1

R .2

1 1.

rrrr

rrm

r

rm

rrrm

l

rr

m

k

k

k

k

i

i

k

i

ii

R

T

R

T

T

x

m/g/1 variants and priority queue -...

Documents