Post on 02-Jan-2016
description
Lecure 7 : Loss networks
1 Motivation: multiple services, single cell / link -- HSCSD
2 Generalised stochastic knapsack
3 Admission control
4 GSM network / PSTN network
5 Model, Equilibrium distribution
6 Blocking probabilities
7 Reduced load approximation
8 Monte-Carlo summation
9 Summary and exercises
Plan for today
1 multiple services, single cell / link -- HSCSD
2 Generalised stochastic knapsack
3 Admission control
4 GSM network / PSTN network
5 Model, Equilibrium distribution
6 Blocking probabilities
7 Reduced load approximation
8 Monte-Carlo summation
9 Summary and exercises
University of Twente - Stochastic Operations Research 33
GSM /HSCSD: High Speed Circuit Switched Data
Multiple types (speech, video, data)circuit switched: each call gets number of channels
GSM speech: 1 channel data: 1 channel (CS, data rate 9.6 kbps)
GSM/HSCSD speech: 1 channel
data: 1≤ b,...,B ≤ 8 channels (technical requirements, data rate 14.4 kbps)
Call accepted iff minimum channel requirement b is met: loss system
Up / downgrading:
data calls may use more channels (up to B) when other services are not using these channels
video: better picture quality, but same video lengthdata: faster transmission rate, thus smaller transmission time
HSCSD characteristics (LB)
University of Twente - Stochastic Operations Research 32
Plan for today
1 multiple services, single cell / link -- HSCSD
2 Generalised stochastic knapsack
3 Admission control
4 GSM network / PSTN network
5 Model, Equilibrium distribution
6 Blocking probabilities
7 Reduced load approximation
8 Monte-Carlo summation
9 Summary and exercises
number of resource units C
number of object classes K
class k arrival rate (n)
class k mean holding time (exp) (n)
class k size
state (number of objects)
state space
object of class k accepted only if
state of process at time t
stationary Markov process
transition rates
Generalised stochastic knapsack: model
University of Twente - Stochastic Operations Research 31
k
k/1
kb
),...,( 1 Knnn
}:{ 0 CnbnS K N
nbCbk
))(),...,(()( 1 tXtXtX K
}0),({ ttX
kk
kkk
ennn
ennnbCbnnnq
')(
')(1)()',(
1' enn n
1' enn
2' enn
2' enn
)(1 n
)(2 n
)(2 n)(1 n
Theorem 1:
For the generalised stochastic knapsack, a necessary and sufficient condition for
reversibility of is that
for all
for some function . Moreover, when such a function exists, the
equilibrium distribution for the generalised stochastic knapsack is given by
Generalised stochastic knapsack: equilibrium distribution
University of Twente - Stochastic Operations Research 30
)(
)(
)(
)(
n
en
en
n k
kk
k
))(),...,(()( 1 tXtXtX K
KkTSn k ,...,1,\
),0[: S
Snn
nn
Sn
,)(
)()(
Proof:
We have to verify detailed balance:
If exists that satisfies the last expression, then satisfies detailed balance. As the
right hand side of this expression is independent of the index k it must be that the
condition of the theorem involving is satisfied. Conversely, assume
that the condition involving is satisfied. Then
is the equilibrium distribution.
Generalised stochastic knapsack: equilibrium distribution
University of Twente - Stochastic Operations Research 29
)(
)(
)(
)(
)()()()(
),()(),()(
n
en
en
n
enennn
nenqenennqn
k
kk
k
kkkk
kkk
),0[: S
Snn
nn
Sn
,)(
)()(
),0[: S
Stochastic knapsack
Finite source input
State space constraints
Generalised stochastic knapsack: examples
University of Twente - Stochastic Operations Research 28
kkkk ennnbCbn ')(1)(
kkkk ennnn ')(
kkkkkkk ennMnnMn ')(1)()(
kkkk ennnn ')(
kkkkkk ennCnnbCbn ');(1)(
kkkk ennnn ')(
)(1!
)(1
Cnbn
nk
nk
K
k
k
knk
k
kK
k n
Mn
1
)(
);(1!
)(1
kkk
nk
K
k
CnCnbn
nk
Plan for today
1 multiple services, single cell / link -- HSCSD
2 Generalised stochastic knapsack
3 Admission control
4 GSM network / PSTN network
5 Model, Equilibrium distribution
6 Blocking probabilities
7 Reduced load approximation
8 Monte-Carlo summation
9 Summary and exercises
Admission control
University of Twente - Stochastic Operations Research 27
Admission class k whenever sufficient room Complete sharing
Simple, but
may be unfair (some classes monopolize the knapsack resources)
may lead to poor long-run average revenue (admitted objects may not contribute to revenue)
admission policies: restrict access even when sufficient room available
calculate performance under policy
determine optimal policy
Coordinate convex policies
In general: Markov decision theory
nbCbk
Admission policy
Transition rates
Recurrent states
Examples: complete sharing
trunk reservation
Stochastic knapsack under admission control
University of Twente - Stochastic Operations Research 26
nk
nknfk statein rejected class0
statein accepted class1)(
kkk
kkkk
ennn
ennnbCbnfnnq
'
')(1)()',(
}1,0{:),...,( 1 Sffff kK
otherwise0
1)( k
k
bCnbnf
otherwise0
1)( kk
k
tbCnbnf
}:{)( CnbnSfS
trunk reservation admits class k object iff after admittance at least resource units
remain available
Stochastic knapsack under trunk reservation
University of Twente - Stochastic Operations Research 25
2
0
1
2
4
2
1
21
t
t
bb
K
Ckt
otherwise0
1)( kk
k
tbCnbnf
Not reversible: so that equilibrium distribution usually not available in closed form
Coordinate convex set
Coordinate convex policy: admit object iff state process remains in
Stochastic knapsack coordinate convex policies
University of Twente - Stochastic Operations Research 24
kk ennf iff1)(
Theorem:Under the coordinate convex policy f the state process
is reversible, and
kk ennnS 0,:
nnG
nk
nk
K
kff
k
,!
1)(
1
))(),...,(()( 1 tXtXtX Kf
Note: Not all policies are coordinate convex, e.g. trunk reservation
Complete sharing: always admit if room available
Complete partitioning: accept class k iff
Threshold policies: accept class k iff
Coordinate convex policies: examples
University of Twente - Stochastic Operations Research 23
S
},...,1,,:{
)1(
Kkb
CnCnbn
Cbnb
Cnb
k
kk
k
kkk
CCC
b
C
b
C
Cnb
K
K
K
kkk
...
},...,0{...},...,0{
)1(
1
1
1
1C
2C
S
1C
2C
S
revenue in state n
long run average revenue
example: long run average utilization
long run average throughput
Intuition:optimal policy in special cases
blocking obsolete complete sharing
complete partitioning with
where is the optimal solution of the knapsack problem
If integer, where maximizes per unit revenue then
Coordinate convex policies: revenue optimization
University of Twente - Stochastic Operations Research 22
011
N subject to max
kkk
K
kkk
K
k
sCsbsr
kk
K
k
nrnr
1
)(
)()()( nnrfW fn
kk br
kkr
kk 0
kk
*kkk sbC
*/k
bC
),...,( **1 Kss
kk br /*k
otherwise 0
for / ** kkbr
s kkk
number of coordinate convex policies is finite
thus for each coordinate convex policy f compute W(f)
and select f with highest W(f)
infeasible as number of policies grows as
show that optimal policy is in certain class
often threshold policies
then problem reduces to finding optimal thresholds
in full generality: Markov decision theory
Coordinate convex policies: optimal policies
University of Twente - Stochastic Operations Research 21
)...( 1 KCCO
},...,1,,:{ Kkb
CnCnbn
Cbnb
k
kk
k
1C
2C
S
Plan for today
1 multiple services, single cell / link -- HSCSD
2 Generalised stochastic knapsack
3 Admission control
4 GSM network / PSTN network
5 Model, Equilibrium distribution
6 Blocking probabilities
7 Reduced load approximation
8 Monte-Carlo summation
9 Summary and exercises
Loss networks
University of Twente - Stochastic Operations Research 20
So far: stochastic knapsack equilibrium distribution blocking probabilities throughput admission control coordinate convex policy -- complex state space
model for multi service single link multi service multiple links / networks
PSTN / ISDN
University of Twente - Stochastic Operations Research 18
Link: connection between switchesRoute: number of linksCapacity Ci of link iCall class: route, bandwidth requirement per link
Stochastic knapsack: special case = model for single linkBut: is special case of generalised stochastic knapsack
C3C1
C7
C5
C4
C6C2
Plan for today
1 multiple services, single cell / link -- HSCSD
2 Generalised stochastic knapsack
3 Admission control
4 GSM network / PSTN network
5 Model, Equilibrium distribution
6 Blocking probabilities
7 Reduced load approximation
8 Monte-Carlo summation
9 Summary and exercises
number of links J
capacity (bandwidth units) link j
number of object classes K
class k arrival rate
class k mean holding time (exp)
bandwidth req. class k on link j
Route
Class k admitted iff bandwidth units free in each link
Otherwise call is blocked and cleared
Admitted call occupies bandwidth units in each link for duration of its holding time
Loss network : notation
University of Twente - Stochastic Operations Research 17
k
k/1
jkb
1' enn n
1' enn
2' enn
2' enn
12
22n11` n
jC
},...,1{ JRk
jkb kRj
kRjjkb
C3C1
C7C5
C4
C6C2
Set of classes
set of classes that uses link j
state
state space
class k blocked
Loss network : notation
University of Twente - Stochastic Operations Research 16
),...,( 1 Knnn
},...,1,:{ 0 JjCnbnS jkjkk
K jK
N
},...,1{ KK
}{ kRjk :KK j
)(}:{ 0 jkK bACAnnS N
} some,:{ jCbnbSnT jjkjk
jK
})(,:{ 0 CenACAnnT kK
k N /
state of process at time t
stationary Markov process
aperiodic,irreducible
equilibrium state
utilization of link j
long run fraction of blocked calls
long run throughput
unconstrained cousin (∞ capacity)
Poisson r.v. with mean
unconstrained cousin of utilization
Loss network : notation
University of Twente - Stochastic Operations Research 15
))(),...,(()( 1 tXtXtX K
}0),({ ttX
},Pr{1 kjkjjk RjbCUB
),...,( 1 KXXX
kjkk
j XbUj
K
:
kjkk
j XbUj
K
:
PASTA
][)1( kkkkk XEBTH Little
),...,( 1 KXXX
kkk /
Equilibrium distribution
University of Twente - Stochastic Operations Research 14
Theorem: Product form equilibrium distribution
Blocking probability of class k call
The Markov chain is reversible, PASTA holds, and the equilibrium distribution (and the blocking probabilities) are insensitive.
PROOF: special case of generalised stochastic knapsack!
SnCU
nX
nGnX
k
nk
K
k
k
,
}Pr{
}Pr{
!
1}Pr{
1
Sn
nG
k
nk
K
kSn
k
,!1
!
!1
}Pr{
}Pr{
1
1
1\\
n
n
nX
nX
B nK
Sn
nK
TSn
Sn
TSnk
kk
V, C vectors
Plan for today
1 multiple services, single cell / link -- HSCSD
2 Generalised stochastic knapsack
3 Admission control
4 GSM network / PSTN network
5 Model, Equilibrium distribution
6 Blocking probabilities
7 Reduced load approximation
8 Monte-Carlo summation
9 Summary and exercises
Computing blocking probabilities
University of Twente - Stochastic Operations Research 13
Direct summation is possible, but complexity
Recursion is possible (KR), but complexity
Bounds for single service loss networks ( )
For link j in loss network: probability that call on link j is blocked
load offered to link j facility bound
Call of class k blocked if not accepted at all links
product bound
kjkk
kk
jk
j
C
k
jC
jjjj b
k
CCErL
j
j
KK j
,
!/
!/),(
0
)),(1(1 jjRj
k CErBk
)...( 1 JCKCO
)...( 1 JCKCO
}1,0{jkb
Plan for today
1 multiple services, single cell / link -- HSCSD
2 Generalised stochastic knapsack
3 Admission control
4 GSM network / PSTN network
5 Model, Equilibrium distribution
6 Blocking probabilities
7 Reduced load approximation
8 Monte-Carlo summation
9 Summary and exercises
Computing blocking probabilities: single service networksReduced load approximation
University of Twente - Stochastic Operations Research 12
Facility bound
Part of offered load blocked on other links :
probability at least one unit of bandwidth available in each link in
reduced load
approximation: blocking independent from link to link
reduced load approximation
existence, uniqueness fixed pointrepeated substitutionaccuracy
€
L j = Er(k∈K j
∑ ρ k
i∈Rk \{ j}
∏ (1− Li),C j ), j =1,...,J
Bk =1−j∈Rk
∏ (1− L j ), k =1,...,K
),)(( jkkk
j CjtErL
jK
),( jkk
j CErL
jK
kk
jK
)( jtk }{\ jRk
)( jtkk
)1()(}\{
ijRi
k Ljtk
Reduced load approximation: existence and uniqueness
University of Twente - Stochastic Operations Research 11
Notation:
Theorem There exists a unique solution to the fixed point equation
Proof The mapping is continuous, so existence from Brouwer’s theorem
is not a contraction!
))(),...,((:)(
)),1((:)(
),...,(:
1
}\{
1
LTLTLT
CLErLT
LLL
J
jijRi
kk
j
J
k
jK
*L )(LTL
JJT ]1,0[]1,0[:
T
Reduced load approximation: uniqueness
University of Twente - Stochastic Operations Research 10
Notation:
inverse of Er for capacity : value of such that
is strictly increasing function of B
Therefore is strictly convex function of B for
Proof of uniqueness:consider fixed point and apply :
Define
which is strictly convex.
Thus, if is solution of
Then is unique minimum of over
writing out the partial derivatives yields (*)
)(LTL
€
Er−1j (L j ) :=
k∈K j
∑ ρ k
i∈Rk \{ j}
∏ (1− Li) (*)
jC)(1 BEr j ),( jCErB
dzzEr j
B
)(1
0
]1,0[B
jEr 1
JL ]1,0[
€
ψ(L) :=k∈K j
∑ ρ k
i∈Rk
∏ (1− Li) +j=1
J
∑0
L j
∫ Er−1j (z)dz
JL ]1,0[* JiL
L
i
,...,1,0)(
ψ
*LJ]1,0[ψ
Reduced load approximation: repeated substitution
University of Twente - Stochastic Operations Research 9
Start
Repeat
Theorem Let
Then
Thus
Proof is decreasing operator:
Thus
JL ]1,0[
,...2,1),(:
:1
0
mLTL
LLmm
LLLLLLL
LLLLLLL
L
nn
nnnn
*122
0222*32121
0
,,
)1,...,1()0,...,0(
)1,...,1(
€
T(L) < T(L') if L'< L componentwise
€
Tj (L) := Er(k∈K j
∑ ρ k
i∈Rk \{ j}
∏ (1− Li),C j )
T
€
T 2n (L) < T 2n (L') and
T 2n +1(L) > T 2n +1(L') if L'< L componentwise
Reduced load approximation: accuracy
University of Twente - Stochastic Operations Research 8
Corollary
so that
Proof
Indeed from reduced load approximation does not violate the upper bound
€
L j* ≤ Er(
k∈K j
∑ ρ k,C j )
€
Bk =1−j∈Rk
∏ (1− L j*) ≤1−
j∈Rk
∏ Er(k∈K j
∑ ρ k,C j ), k =1,...,K
€
Tj2(1...1) = T(0...0) = Er(
k∈K j
∑ ρ k,C j )
kB
)1,...,1()0,...,0( 02*121 LLLLL nn
)0,...,0(2 TL
€
Tj (L) := Er(k∈K j
∑ ρ k
i∈Rk \{ j}
∏ (1− Li),C j )
Plan for today
1 multiple services, single cell / link -- HSCSD
2 Generalised stochastic knapsack
3 Admission control
4 GSM network / PSTN network
5 Model, Equilibrium distribution
6 Blocking probabilities
7 Reduced load approximation
8 Monte-Carlo summation
9 Summary and exercises
Monte Carlo summation
University of Twente - Stochastic Operations Research 7
For performance measures: compute equilibrium distribution
for blocking probability of class k call
In general: evaluate
difficult due to size of the state space : use Monte Carlo summation
€
Pr{X = n} = k=1
K
∏ ρ knk
nk!
n∈S
∑k=1
K
∏ ρ knk
nk!
, n ∈ S
€
Bk = n∈Tk
∑l =1
K
∏ ρ ln l
nl !
n∈S
∑l =1
K
∏ ρ ln l
nl !
€
n∈U
∑l =1
K
∏ ρ ln l
nl !
Monte Carlo summation
University of Twente - Stochastic Operations Research 6
Example: evaluate integral
Monte Carlo summation:draw points at random in box abcd# points under curve / # points is measure for surface //// = value integral
MethodLet
draw n indep. Samples
Then unbiased estimator of
with 95% confidence interval
Powerful method: as accurate as desired
a b
b
a
dxxf )(
cd
))((1let and indep,),0(),( XfYZcUYbaUXdd
nZZ ,...,1
n
ZZZ n
...1
b
a
dxxfabc
)()(
1
€
Z ±1.96 S2(n) /n[ ]
Monte Carlo summation
University of Twente - Stochastic Operations Research 5
For blocking probabilities
ratio of multidimensional Poisson( ρ) distributed r.v.
Let then
Estimate enumerator and denominator via Monte Carlo summation:
for
confidence interval?
€
Bk = n∈Tk
∑l =1
K
∏ ρ ln l
nl !
n∈S
∑l =1
K
∏ ρ ln l
nl !
e
e
) Poisson( d
X })(Pr{
})(Pr{
SX
TXB k
k
€
draw Vi from Poisson( ρ ), i =1,...,n iid
Let g(Vi) =1(Vi ∈U)
Eg(V ) = Pr{X(ρ )∈U} =n∈U
∑l =1
K
∏ ρ ln l
nl !e−ρ l
SUTU k and
Monte Carlo summation
University of Twente - Stochastic Operations Research 4
For blocking probabilities
Estimate enumerator and denominator via Monte Carlo summation:
confidence interval? Harvey-Hills method (acceptance rejection method HH)
unbiased estimator!
€
Bk = n∈Tk
∑l =1
K
∏ ρ ln l
nl !
n∈S
∑l =1
K
∏ ρ ln l
nl !
e
e
succes ascount also sample if
count sample if
ignore sample if
iid ,...,1 ), Poisson( from draw
k
i
T
S
S
niV
})(|)(Pr{})(Pr{
})(Pr{SXTX
SX
TXB k
kk
k
ikii
BVEg
SVTVVg
)(
)|(1)(
Plan for today
1 multiple services, single cell / link -- HSCSD
2 Generalised stochastic knapsack
3 Admission control
4 GSM network / PSTN network
5 Model, Equilibrium distribution
6 Blocking probabilities
7 Reduced load approximation
8 Monte-Carlo summation
9 Summary and exercises
Summary loss network models and applications:
University of Twente - Stochastic Operations Research 3
Loss networks - circuit switched telephone wireless networks Markov chain equilibrium distribution blocking probabilities throughput admission control Evaluation of performance measures: recursive algorithms reduced load method Monte Carlo summation GSM/HSCSD
GSM network architecture
RADIO INTERFACE
BTSBSC
(G)MSCPSTN/ISDNMS
TELEPHONE
C3C1
C7C5
C4
C6C2
Consider a HSCSD system carrying speech and voice. Let C=24. For speech and video
load increasing from 0 to ∞ and minimum capacity requirement for speech of 1 channel
and for video ranging from 1 to 4 channels, investigate the optimality of complete
partitioning versus complete sharing for long run average utilization and long run average
througput. To this end, first show that for complete partitioning the long run average
revenue is given by
so that the optimal partitioning is obtained from
Exercises:
University of Twente - Stochastic Operations Research 2
011
N,0 subject to )(max
kkk
K
kkk
K
k
CCCCCCW
)(
)(
)(
)(1
1
00
1
00
1kkk
K
k
k
n
j
b
C
n
k
n
jk
b
C
nk
K
k
CWr
j
jn
rCW
kk
k
k
kk
k
k
7
Exercises: University of Twente - Stochastic Operations Research 1
8 For the stochastic knapsack show (directly from the equilibrium distribution) that
Prove the elasticity result:
KkBXE kkkk ,...,1/)1(][
KkjBB
j
k
k
j ,...,1,
References:
University of Twente - Stochastic Operations Research 0
Ross, sections 5.1, 5.2, 5.5 (reduced load method) Kelly: lecture notes (on website of Kelly)
Kelly F.P. Kelly Loss networks, Ann. Appl. Prob 1, 319-378, 1991
HH C. Harvey and C.R. Hills Determining grades of service in a network. Presented at the 9th International Teletraffic Conference, 1979.
KR J.S. Kaufman and K.M. Rege Blocking in a shared resource environment with batched
Poisson arrival processes. Performance Evaluation, 24, 249-263, 1996
Exercises: rules
In all exercises: give proof or derivation of results(you are not allowed to state something like: proof as given in class, or proof as in book…) motivate the steps in derivation
hand in exercises week before oral exam, that is based on these exercises
All oral exams (30 mins) for this part on the same day, you may suggest the date…
Hand in 5 exercises: 1 and 4 and select 2 or 3 and 5 or 6, and 7 or 8
Each group 2 persons: hand in a single set of answersG1: 1, 4, 2, 5, 7G2: 1, 4, 2, 6, 8G3: 1, 4, 3, 5, 8
Each group 3 persons: hand in a single set of answersG1: 1, 4, 2, 5, 7G2: 1, 4, 3, 6, 8