Lecure 7 : Loss networks

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


3 Admission control







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)


Plan for today



3 Admission control







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


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


)(

)(

)(

)(

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


)(

)(

)(

)(

)()()()(

),()(),()(

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


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



3 Admission control







Admission control


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


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


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


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


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


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


)...( 1 KCCO

},...,1,,:{ Kkb

CnCnbn

Cbnb

k

kk

k

1C

2C

S

Plan for today



3 Admission control







Loss networks


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


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



3 Admission control







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


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



),...,( 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



))(),...,(()( 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


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



3 Admission control







Computing blocking probabilities


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



3 Admission control







Computing blocking probabilities: single service networksReduced load approximation


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


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


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


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


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



3 Admission control







Monte Carlo summation


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 !



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[ ]



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



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



3 Admission control







Summary loss network models and applications:


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:


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:


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

Lecure 7 : Loss networks

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