Post on 21-Dec-2015
Chapter 5
The Queue M/G/1
2
M/G/1
Arrival
Service
GeneralPoisson
teta )(
1 srver
arbitraryxb )(
3
Bus Paradox
If uniform
最少 0 分，最多 10 分；平均 5 分鐘
If Poisson(or Exponential)
人任意時刻來到 bus stop, 平均要 wait 多久 ?
(depends on bus 來的 distribution)
min 10xbusstop
10
1
10 minbus arrival
customerarrival
10 min
10 minbus arrival
customerarrival 10 min 10 min
4
X: “special” interarrival time we pick Y: residual life(waiting time as above) Let f(x) be the pdf of interval lengths (mn = nth monent)
fx(x) be the pdf of the interval we randomly pick
f(x) be the pdf of residual life Y
X(life)
Y(residual life)age
5
1
1010
00
0 x
)()()(
)(m )(
)(1)(
1)(f
n)(assumptio )()(
)( , )(
m
xxfxkxfxf
kmdxxxfkdxxxflet
dxxxfkdxxkxf
dxx
xkxfxf
xfxxf
x
x
x
成正比會跟
61
**
1
*
1
11
1
)(1)(11)(ˆ
)(1)(ˆ
)(ˆ])(1[)(
)(
0for ],[]|[
],[
0 ,]|[
Sm
SF
S
SF
SmSF
m
yFyf
dyyfyFm
dydx
m
xxf
x
dy
dxm
xxf
x
dy
xydxxXxPdxxXxdyyYyP
dxxXxdyyYyP
xyx
yxXyYP
y
yyx
7
1
2
1
1
n
n
2
using1
life residual theofmonent th -n is r
length internal theofmonent th -n is mLet
m
mtimeresidualmeanr
l rule) L'Hospita, ()m(n
mr
n
nn
102
,102
10m10
1
lExponentia
52
,10
10 ticDeterminis
:
1
21
22
1
1
212
2
1
m
mr
m
m
mr
m
m
example
8
Method of Embedded(Imbedded) Markov
Chain
dk = P[departure leaves behind k in system]
rk = P[arrival finds k in system upon his arrival]
pk = P[k in system at random point of time]
time
departure instants
9
Example: D/D/1
If arrivals are Poisson, then pk = rk
If system states change by ± 1, then dk = rk
For M/G/1 pk=rk=dk
Deterministic
time
N(t)
1
departure
p r d
,,, k
kx
k,
p
,,, k
k, r
,,, k
k, d
k
k
k
320
1,
01
210
01
210
01
x
10
Proof
1. Let A(t, t + △ t) = arrival occurs in interval (t, t + △ t)
)()(
])([
],([
])([])(|),([lim
],([
)],( ,)([lim
)],(|)([lim
0
0
0
tPtR
ktNP
tttAP
ktNPktNtttAP
tttAP
tttAktNP
tttAktNP(t)P
time t] system atfinds k inP[arrival (t)R
e t]tem at timP[k in sys(t)P
kk
t
t
tk
k
k
11
2.
rk=dkk k+1
balance
k customers
12
M/G/1GeneralPoisson
kkk
t
drpSBxb
eta
)()(
)(*
1 srver
1,,, 2 xxx b 2
kk
kk
d, then rte by change staif system
r then ps Poisson,if input i
1
13
x
x
m
mr
)m(n
mr
)life (r residual monent ofn
x
B(x)(x)b
(x)bife pdf residual l
nn
nth
22
1
1ˆ
ˆ
2
1
21
1
1
21
2
srvice]in him findshat t
arrivalan delay willservice
foundcustomer a that E[timeLet W0
xλρ)(Oρ
x
x
2 2
Delay 的時間有人被 served
沒被 serve 的機率
14
Imbeded Markov Chain
Let cn be the nth customer to enter system
qn be the number of customers left behind by the departure of cn
time
cn-2 cn-1 cn cn+1
qn left behind
15
]n x arrival iP[ P
]n x arrival iP[ P
me of cservice ti where x
]xal during P[no arriv, PGiven q
d,Pdd
i]j|qP[qNeed P
][PP],,,d[dd
imediscrete tte state, discreov Chain. is a Markq
)rp(dk]P[q
k]P[q
n
n
nn
nn
kk
nnij
ij
n
kkknn
n
102
101
100
0
1
10
2
1
0
1
lim
16
Let vn = number of arrivals during xn
P[vn+1 = k] = αk
10
210
210
210
00
0
P
)()(
)(!
)(
)(]~|~[
]~[
~
*
0
)1(
0
0
ZBdxxbe
dxxbk
ex
dxxbxxkvP
kvP
vtate s, nndent of s is indepev
xZ
xk
k
n
17
)(B
)(
)()ek!
x)((
))(!
)(()(
~
*
0
)1(
0
x-
0k
k
00
00
Z
dxxbe
dxxbZ
Zdxxbk
exZV
k]ZvP[ZαV(Z)Let α
xZ
k
k
kxk
k
k
k
kk
transformZ
k
xZe
18
xv
xx
dZ
Zd
Zd
ZdB
dZ
dV(Z)v E[v]
)(B) v(Z
λZ):(λBcheck V(Z)
ZZ
*
*
)()(
)(
)(
)(
1011
1
*
1
arrival rate
mean service time
進入的個數
222)2()0(
*
1
2
2
*2
1
2
22
)(B
)(
)(
)()(
x
dZ
Zd
Zd
ZBd
dZ
ZVdvv
ZZ
xxv 222
19
server
queue
cn cn+1
τn
cn cn+1
τn+1
cn cn+1
tn
xnxn+1
qn+1qn
vn+1
nnk
nnk
nn
nnn
nn
thn
y ct behind bs left lef# customerqd
during x arrivals # customervα
me f cservice tixb(x)
time)erarrival (ττta(t)
ime of c:arrival tτ
customer:nc
int1
20
11
02
nn
n
vq
: qCase
server
queue
cn+1
cn cn+1
cn+1
xn+1
qn+1qn=0
vn+1
0
0
1
11
1 if qn v
if qvqq
,n
n, nn
n
0
01
111
1
nnn-n
n
n
n
, if qvqq
rve c
, departure c
: qCase
se 會被便馬上被只要
當第 cn+1 離開時發現
原來的 進來的
21
ise, otherw
,,,, klet
0
3211Δk
k
△k
-2 -1 0 1 2 3
1
ρvs busy] P[server is busy]P[system i
]qP[]qP[]qP[]E[Δv
vv]vE[
dk]q: P[q vΔqq
)rd (Nkdkqq
vΔqq
vΔqq
kpkd
q
k
qn
kkk
nn
nqnn
nqnn
n
n
~
~
01
11
11
0~0~10~0
~~
~~~
lim
22
)1(2
)1(2)1(2
2
222
222
222
) (
2
2222
)(2d222
222
112
122
1
112
1222
1
vNq
vvq
vqvqvqq
vqvqvqq
vqvqvqq
nletq
xvvdZ
ZV
qqtake n
nqnnnnqnn
nqqnnnnqnn
nn
nnn
趨於取期望值兩邊平方求
nq nq
23
1
1
1
2/
)1(2
0
0
2
22
WW
WxT
xxT
TN
xNq
Pollaczek-Khinchin
MEAN-VALUE Formula
P-K Mean-Value Formula
24][)()(
][][][)(
][][][
)(
][]~[)()(lim
][)(][
)(][
1
1
11
~
0
0
11
111
11
nqn
nnqnnnqn
nnqnnnqnn
nnqnn
n
n
q
n
vqvq
n
vqvqq
vqq
nqnn
kkk
q
k
kn
n
qn
k
kn
nn
ZEZVZQ
ZEZEZZEZQ
ZZEZEZE
ZZ
vqq
ZQdpr
ZEZkqPZQZQ
ZEZQZkqP
ZQkqP
kd
)(ZV
25
Z
qPZQqPZVZQ
nlet
Z
qPZQqPZVZQ
qPZQZ
qP
kqPZqP
kqPZqPZZE
n
nnnn
nnn
kn
kn
kn
kkn
qnqn
]0~[)(]0~[)()(
1)(
]0[)(]0[)()(
]0[)(1
]0[
][]0[
][]0[][
1
1
1
1
0 0
26
kdZZB
ZZBZQ
-ρty]ver P[empSingle ser
)Q(her ] from eitqP[
ZBZV
ZZV
qPZZV
ZZV
ZqP
ZV
)(
)1)(1()()(
1
110~
)()(
)(
]0~[)1()(
)(1
)1
1](0~[)(Q(Z) Q(Z),
**
*
解
Pollaczek-Khinchin Transform equ.
27
example
kk
x
d
Z
ZZZ
Z
ZZZ
Z
ZZ
Z
ZZQ
eS
SB
)1(
1
1
)1()1(
)1)(1()1)(1(
)1)(1()(
)(
2
*
1// MM
28
P-K Mean-
Value
Formula
1
2/
1
2/
1
2/
1
22
2
20
xNq
xxWxT
xWW
2
2
2
2
0x
xx
xW
time waiting)(yw
timesystem )(ys
kd
29
example
1
1
)1
1(1
1212
1
2 ,
1
1
/1
2
22
T
xx
T
1// MM
)1(2
)1(
)1(
)()1(2
2
22
22
2
2222
2
b
b
cxW
xx
xx
xxxx
xW
30
)-2(1
xW
0
:1//
1
-1
xW
1
:1//
2b
2b
c
DM
x
c
MM
02
1
1//
1//1//
DD
MMDM
W
WW
W
ρ
M/M/1
M/D/1
D/D/1
31
)()(
)()(
FIFO) (assume
customer nfor timesystem:
)()(
*
*
th
*
ZSZQ
ZSZQ
s
ZBZV
n
n
n
server
queue
cn
cn
xn
vn::Poisson λcn
server
queue
cn
cn
sn
qn left behind
Poisson λcnfirst come first serve
32
SZ
ZSLet
ZZB
ZZBZQZS
1
)(
)1)(1()()()(
***
)(
)1()()(
1)(
)1)(11()()(
***
*
**
SBS
SSBSS
SSB
S
SBSS
33
)(
)1()(
)()()(
)()()(
~~~
**
***
SBS
SSW
SWSBSS
ywybys
wxs
wxs nnn
34
example
yeys
SS
S
S
SSS
SSB
)1(
*
*
)1()(
)1(
)1()1()(
)(
1// MM
y
s(y)
ye )1()1(
35
yeyuyw
SSW
S
S
SSS
SS
SS
SSW
)1(0
*
2*
)1()()1()(
)1(1)1()(
)1(
))(1()1)(()1()(
system idle
system busy
w(y)
ye )1()1( 1
y
36
0)(
0
*
*
*
****
)(ˆ)1()(
))(ˆ()1(
)(ˆ1
1)(
)(11
1
))(1
(
)1(
)(
)1()(
kk
k
k
k
ybyw
SB
SBSW
xSSB
xSB
xS
S
SBS
SSW
ice timeidual servpdf of res
(x)b(S)B* ˆˆ
convolved itself k times
37
Stages of method
1. Series/Parallel
2. d = dP
3. Supplementary Variables
4. Imbedded Markov Chain
38
U(t) = unfinished work in system at time t
= Time required to empty the system measured from time t, if no new customers are allowed to enter after tU(t)
t0 τ1τ1+x1
x1
x2
WAVG
Virtual Waiting Time (only for FCFS)
Busy Periody
Idle Period
I
WFCFS=WLCFS=WX