# Stochastic Models for Telecommunication Systems

Document Sample

```					1 Motivation: multiple services, single cell / link -- HSCSD
2 Generalised stochastic knapsack
4 GSM network / PSTN network
5 Model, Equilibrium distribution
6 Blocking probabilities
8 Monte-Carlo summation
9 Summary and exercises
1 multiple services, single cell / link -- HSCSD
2 Generalised stochastic knapsack
4 GSM network / PSTN network
5 Model, Equilibrium distribution
6 Blocking probabilities
8 Monte-Carlo summation
9 Summary and exercises
University of Twente - Stochastic Operations Research   33

GSM /HSCSD: High Speed Circuit Switched Data
University of Twente - Stochastic Operations Research                    32

HSCSD characteristics (LB)

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

data calls may use more channels (up to B) when other services are not using these channels

video: better picture quality, but same video length
data: faster transmission rate, thus smaller transmission time
1 multiple services, single cell / link -- HSCSD
2 Generalised stochastic knapsack
4 GSM network / PSTN network
5 Model, Equilibrium distribution
6 Blocking probabilities
8 Monte-Carlo summation
9 Summary and exercises
University of Twente - Stochastic Operations Research                               31

n'  n  e2
Generalised stochastic knapsack: model
2 ( n )
number of resource units                    C
n'  n  e1
1 (n)
number of object classes                    K                                                    n'  n  e1
n
class k arrival rate                         k (n)
1 (n)
class k mean holding time (exp)            1 /  k (n)                               2 ( n)
class k size                                bk
state (number of objects)                   n  (n1 ,..., nK )                     n'  n  e2
state space                                 S  {n  N 0 : b  n  C}
K

object of class k accepted only if           bk  C  b  n
state of process at time t                   X (t )  ( X 1 (t ),..., X K (t ))
stationary Markov process                    { X (t ), t  0}
k (n)1(bk  C  b  n) n'  n  ek
transition rates                                q(n, n' )  
         k ( n)        n'  n  ek
University of Twente - Stochastic Operations Research                  30

Generalised stochastic knapsack: equilibrium distribution
Theorem 1:
For the generalised stochastic knapsack, a necessary and sufficient condition for
reversibility of X (t )  ( X 1 (t ),..., X K (t )) is that

k ( n )      (n  ek )
                        for all n  S \ Tk , k  1,..., K
 k (n  ek )      ( n)

for some function        : S  [0, ). Moreover, when such a function       exists, the
equilibrium distribution for the generalised stochastic knapsack is given by

 ( n)
 ( n)           ,           nS
  ( n)
nS
University of Twente - Stochastic Operations Research                      29

Generalised stochastic knapsack: equilibrium distribution
Proof:
We have to verify detailed balance:           (n)q (n, n  ek )   (n  ek )q(n  ek , n)

 (n)k (n)   (n  ek )  k (n  ek )

k ( n )           (n  ek )

 k (n  ek )          ( n)

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      : S  [0, )is satisfied. Conversely, assume
that the condition involving      : S  [0, ) is satisfied. Then
 ( n)
 ( n)             ,       nS      is the equilibrium distribution.
  ( n)
nS
University of Twente - Stochastic Operations Research                             28

Generalised stochastic knapsack: examples
Stochastic knapsack
k (n)  k 1(bk  C  b  n) n'  n  ek                      K
kn
(n)  
k

1(b  n  C )
 k (n)  nk  k n'  n  ek                                   k 1   nk !

Finite source input
k (n)  ( M k  nk )k 1(nk  M k ) n'  n  ek
K
M 
 (n)    k   knk
 
k 1  nk 
 k (n)  nk  k n'  n  ek

State space constraints
k (n)  k 1(bk  C  b  n; nk  Ck ) n'  n  ek (n) 
K
 kn

k

1(b  n  C; nk  Ck )
 k (n)  nk  k n'  n  ek                               k 1         nk !
1 multiple services, single cell / link -- HSCSD
2 Generalised stochastic knapsack
4 GSM network / PSTN network
5 Model, Equilibrium distribution
6 Blocking probabilities
8 Monte-Carlo summation
9 Summary and exercises
University of Twente - Stochastic Operations Research                      27

Admission class k whenever sufficient room      bk  C  b  n    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
University of Twente - Stochastic Operations Research                     26

Admission policy                       f  ( f1 ,..., f K )   f k : S  {0,1}
1 class k acceptedin state n
f k (n)  
0 class k rejected in state n

k f k (n)1(bk  C  b  n) n'  n  ek
Transition rates                      q(n, n' )  
           nk  k           n'  n  ek

Recurrent states                      S ( f )  S  {n : b  n  C}

1 b  n  C  bk
Examples: complete sharing            f k (n)  
0  otherwise
trunk reservation                    1 b  n  C  bk  tk
f k (n)  
0      otherwise
University of Twente - Stochastic Operations Research                     25

Stochastic knapsack under trunk reservation

trunk reservation admits class k object iff after admittance at least t k resource units
remain available                                                   C4
K 2
b1  b2  1
t1  0
t2  2

1 b  n  C  bk  tk
f k (n)  
0      otherwise

Not reversible: so that equilibrium distribution usually not available in closed form
University of Twente - Stochastic Operations Research                        24

Stochastic knapsack coordinate convex policies

Coordinate convex set       S : n  , nk  0  n  ek  
Coordinate convex policy: admit object iff state process remains in 

f k (n)  1 iff        n  ek  

Theorem:
Under the coordinate convex policy f
the state process
X f (t )  ( X 1 (t ),..., X K (t ))
is reversible, and
1      K
 kn

k

 f (n)                      ,     n
Gf    k 1   nk !
University of Twente - Stochastic Operations Research                      23

Coordinate convex policies: examples

Note: Not all policies are coordinate convex, e.g. trunk reservation

Complete sharing: always admit if room available           S
Complete partitioning: accept class k iff
bk (nk  1)  Ck
 C1              CK 
  {0,..., } ... {0,...,    }
C2                                                         b1              bK 
                                       C1  ... CK  C
C1        S
Threshold policies: accept class k iff
bk (nk  1)  Ck
C2                                             b  n  bk  C
C 
                                         {n : b  n  C , nk   k , k  1,...,K }
C1   S                                               bk 
University of Twente - Stochastic Operations Research                         22

Coordinate convex policies: revenue optimization
K

revenue in state n                                         r (n)   rk nk
k 1

long run average revenue                                 W ( f )   r (n) f ( n)
n

example: long run average utilization                      rk  bk
long run average throughput                      rk   k

Intuition:optimal policy in special cases
 k  0 k            blocking obsolete  complete sharing

k   k             complete partitioning with     C k  bk sk
*

*        *
where ( s1 ,..., sK ) is the optimal solution of the knapsack problem
K                             K
max  rk sk          subject to             bk sk  C       sk  N 0
k 1                         k 1
rk / bk for k  k *
*
If C / bk * integer, where k maximizes per unit revenue rk        / bk then s  
*
k
 0      otherwise
University of Twente - Stochastic Operations Research                      21

Coordinate convex policies: optimal policies

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 O(C1...CK )

show that optimal policy is in certain class
often threshold policies
b  n  bk  C
C2
 Ck 
  {n : b  n  C , nk   , k  1,...,K }
                                                                   bk 
C1   S

then problem reduces to finding optimal thresholds

in full generality: Markov decision theory
1 multiple services, single cell / link -- HSCSD
2 Generalised stochastic knapsack
4 GSM network / PSTN network
5 Model, Equilibrium distribution
6 Blocking probabilities
8 Monte-Carlo summation
9 Summary and exercises
University of Twente - Stochastic Operations Research      20

Loss networks
So far: stochastic knapsack
equilibrium distribution
blocking probabilities
throughput
coordinate convex policy -- complex state space

model for multi service single link

multi service multiple links / networks
University of Twente - Stochastic Operations Research   18

PSTN / ISDN
C4
C1           C3
C5

C2                      C6         C7

Call class: route, bandwidth requirement per link

Stochastic knapsack: special case = model for single link
But: is special case of generalised stochastic knapsack
1 multiple services, single cell / link -- HSCSD
2 Generalised stochastic knapsack
4 GSM network / PSTN network
5 Model, Equilibrium distribution
6 Blocking probabilities
8 Monte-Carlo summation
9 Summary and exercises
University of Twente - Stochastic Operations Research                               17
C4
C1          C3
Loss network : notation                                                                        C5
C2              C6         C7
capacity (bandwidth units) link j          Cj                                    n'  n  e2
number of object classes                    K                         2
class k arrival rate                        k
n'  n  e1
1
class k mean holding time (exp)             1/ k                                             n'  n  e1
n
bandwidth req. class k on link j            b jk
`n11
Route                                       Rk  {1,..., J }                      n2  2

Class k admitted iff b jk bandwidth units free in each link j  Rk              n'  n  e2

Otherwise call is blocked and cleared

Admitted call occupies b jk bandwidth units in each link j  Rk for duration of its
holding time
University of Twente - Stochastic Operations Research                        16

Loss network : notation
Set of classes                         K  {1,...,K}
set of classes that uses link j        K j  {k  K : j  Rk }
state                                   n  (n1 ,..., nK )

state space                            S  {n  N 0 :
K

kK j
b jk nk  C j , j  1,...,J }

S  {n  N0 : An  C} A  (b jk )
K

class k blocked                        Tk  {n  S :  b j n  b jk  C j , some j}
K j

Tk  {n  N 0 : An  C , A(n  ek )  C}
K
/
University of Twente - Stochastic Operations Research                  15

Loss network : notation
state of process at time t                  X (t )  ( X 1 (t ),..., X K (t ))
stationary Markov process                   { X (t ), t  0}
aperiodic,irreducible
equilibrium state                         X  ( X 1 ,..., X K )
utilization of link j                     U j :  b jk X k
kK j
long run fraction of blocked calls          Bk  1  Pr{U j  C j  b jk , j  Rk }   PASTA
long run throughput                         TH k  k (1  Bk )   k E[ X k ]        Little

unconstrained cousin (∞ capacity)            X   ( X 1 ,..., X  K )
Poisson r.v. with mean           k  k /  k
unconstrained cousin of utilization         U  j :      
kK j
b jk X  k
University of Twente - Stochastic Operations Research                             14

Equilibrium distribution
Theorem: Product form equilibrium distribution

1 K  k k Pr{X   n}                                                         k n
n                                                   K
Pr{X  n}  
k

    
,                           nS       G                   ,    nS
G k 1 nk ! Pr{U  C}                                          nS     k 1   nk !

Blocking probability of class k call
K
 n   


nS \Tk
Pr{X  n}           
nS \Tk  1    n !
Bk  1                           1                                                   V, C vectors

nS
Pr{X  n}

K
 n


nS     1    n !

The Markov chain is reversible, PASTA holds, and the equilibrium distribution (and the
blocking probabilities) are insensitive.

PROOF: special case of generalised stochastic knapsack!
1 multiple services, single cell / link -- HSCSD
2 Generalised stochastic knapsack
4 GSM network / PSTN network
5 Model, Equilibrium distribution
6 Blocking probabilities
8 Monte-Carlo summation
9 Summary and exercises
University of Twente - Stochastic Operations Research                                          13

Computing blocking probabilities

Direct summation is possible, but complexity                        O( KC1...C J )

Recursion is possible (KR), but complexity                          O( KC1...C J )

Bounds for single service loss networks ( b jk  {0,1} )

For link j in loss network: probability that call on link j is blocked

 jC / C j!
j

L j  Er(  j , C j )    Cj
,   j            k   b jk  k   load offered to link j

kK j        kK
 j k / k!                                         facility bound
k 0

Call of class k blocked if not accepted at all links

Bk  1   (1  Er(  j , C j ))
jRk                                                                    product bound
1 multiple services, single cell / link -- HSCSD
2 Generalised stochastic knapsack
4 GSM network / PSTN network
5 Model, Equilibrium distribution
6 Blocking probabilities
8 Monte-Carlo summation
9 Summary and exercises
University of Twente - Stochastic Operations Research                                              12

Computing blocking probabilities: single service networks

Facility bound      L j  Er(   k , C j )
kK j

Part of offered load            k blocked on other links :             L j  Er(  tk ( j )  k , C j )
kK j                                                           kK j

tk ( j ) probability at least one unit of bandwidth available in each link in Rk \ { j}

 k t k ( j ) reduced load

approximation: blocking independent from link to link                          tk ( j )      
iRk \{ j }
(1  Li )

reduced load approximation                    L j  Er(  k              (1 Li ),C j ),                  j 1,...,J
k K j     iR k \{ j}
existence, uniqueness fixed point
repeated substitution                                          Bk 1  (1 L j ),                          k 1,...,K
j R k
accuracy
University of Twente - Stochastic Operations Research            11

Reduced load approximation: existence and uniqueness
Notation:   L : ( L1 ,..., LJ )
T j ( L) : Er (   k                (1  Li ), C j )
kK j    iRk \{ j }

T ( L) : (T1 ( L),..., TJ ( L))

Theorem There exists a unique solution L* to the fixed point equation L  T (L)

Proof The mapping T : [0,1]J  [0,1]J is continuous, so existence from Brouwer’s theorem

T is not a contraction!
University of Twente - Stochastic Operations Research                                                      10

Notation:
Er 1 j ( B) inverse of Er for capacity C j : value of  such that B  Er (  , C j )
is strictly increasing function of B
B
Therefore    
0
Er 1 j ( z )dz is strictly convex function of B for B [0,1]

Proof of uniqueness:
consider fixed point L  T (L) and apply Er 1 j : Er1 j (L j ) :                           k                 (1 Li )   (*)
k K j       iR k \{ j}
Define L  [0,1]    J
J     Lj

 (L) :      (1 L )   
k                i              Er1 j (z)dz
k K j       iR k            j1   0

which is strictly convex.
 ( L)
Thus, if L*  [0,1]J is solution of    0, i  1,..., J
Li

Then L* is unique minimum of  over [0,1] J
writing out the partial derivatives yields (*)
University of Twente - Stochastic Operations Research                                         9

L  [0,1]J                                                   T j (L) : Er(  k
Start
       (1 Li ),C j )
k K j   iR k \{ j}
Repeat L0 : L

Lm : T ( Lm1 ),     m  1,2,...

Theorem Let       L  (1,...,1)
0

Then (0,..., 0)  L1  L2 n 1  L2 n 3  L*  L2 n  2  L2 n  L0  (1,...,1)

Thus L  L ,
2n
L2 n 1  L , L  L*  L
Proof
T is decreasing operator:          T(L)  T(L') if L' L componentwise
Thus    T 2n (L)  T 2n (L') and
T 2n 1(L)  T 2n 1(L') if L' L componentwise

University of Twente - Stochastic Operations Research                                      8

T j (L) : Er(  k            (1 Li ),C j )
L j  Er(  k ,C j )
*
Corollary                                                                      k K j    iR k \{ j}
k K j

Bk 1  (1 L j ) 1  Er(  k ,C j ),
*
so that                                                                 k 1,...,K
j R k          j R k

kK j


Tj (1...1)  T(0...0)  Er(  k ,C j )
2
Proof
                                   kK j

(0,..., 0)  L1  L2 n 1  L*  L2 n  L0  (1,...,1)

                                           L2  T (0,..., 0)

Indeed Bk from reduced load approximation does not violate the upper bound
1 multiple services, single cell / link -- HSCSD
2 Generalised stochastic knapsack
4 GSM network / PSTN network
5 Model, Equilibrium distribution
6 Blocking probabilities
8 Monte-Carlo summation
9 Summary and exercises
University of Twente - Stochastic Operations Research        7

Monte Carlo summation
For performance measures: compute equilibrium distribution
K
k n
 n!
k

Pr{X  n}  k1 K k n ,                   nS
k
  n!
k

n S k1      k                                    K
l n

l

nl !
for blocking probability of class k call               Bk  n Tk   l 1
K
l n

l

                                                                              nl !
n S   l 1

K
l n

l
In general: evaluate

n U       l 1
nl !

difficult due to size of the state space : use Monte Carlo summation

University of Twente - Stochastic Operations Research    6

Monte Carlo summation
Example: evaluate integral
d                           c
b

 f ( x)dx
a

a                 b
Monte Carlo summation:
draw points at random in box abcd
# points under curve / # points is measure for surface //// = value integral

Method d            d
Let   X U (a, b) Y U (0, c) indep, and let Z  1(Y  f ( X ))

draw n indep. Samples Z1 ,..., Z n
b
1
unbiased estimator of c(b  a) 
Then Z 
Z1  ... Z n                                    f ( x)dx
n                                        a

with 95% confidence interval        Z  1.96   S 2 (n) /n   
Powerful method: as accurate as desired
University of Twente - Stochastic Operations Research                      5

Monte Carlo summation
K
l n

l

For blocking probabilities
n Tk l 1      nl ! e    

Bk           K
l n e  

l         

n S    l 1   nl !
ratio of multidimensional Poisson( ρ) distributed r.v.
d
Pr{ X (  )  Tk }
Let X  Poisson(  ) then              Bk 
                   Pr{ X (  )  S}

Estimate enumerator and denominator via Monte Carlo summation:


draw Vi from Poisson( ), i 1,...,n                  iid
Let g(Vi ) 1(Vi  U)
for U  Tk   and   U S
K
l  
nl
Eg(V )  Pr{X()  U}                   e   l
confidence interval?
n U   l 1
nl !
University of Twente - Stochastic Operations Research                              4

Monte Carlo summation
K
l n

l

For blocking probabilities
n Tk l 1     nl ! e    

Bk           K
l n e  

l         

n S   l 1   nl !
Estimate enumerator and denominator via Monte Carlo summation:

confidence interval? Harvey-Hills method (acceptance rejection method HH)

Pr{ X (  )  Tk }
Bk                        Pr{ X (  )  Tk | X (  )  S}
Pr{ X (  )  S}

draw Vi      from Poisson(  ), i  1,...,n iid
unbiased estimator!
if sample  S ignore
if sample  S count                                                        g (Vi )  1(Vi  Tk | Vi  S )
if sample also  Tk count as succes                                 Eg(V )  Bk
1 multiple services, single cell / link -- HSCSD
2 Generalised stochastic knapsack
4 GSM network / PSTN network
5 Model, Equilibrium distribution
6 Blocking probabilities
8 Monte-Carlo summation
9 Summary and exercises
University of Twente - Stochastic Operations Research                            3

Summary loss network models and applications:
GSM network architecture
Loss networks - circuit switched telephone
wireless networks
TELEPHONE
MS
Markov chain                                                  (G)MSCPSTN/ISDN

equilibrium distribution                               BSC
BTS
blocking probabilities
throughput
Evaluation of performance measures:                 INTERFACE
recursive algorithms
Monte Carlo summation                                   GSM/HSCSD

C4
C1           C3
C5
C2                 C6        C7
University of Twente - Stochastic Operations Research                           2

Exercises:

7   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                Ck 
    
 bk             nk 1

K               nk   k ( j )              K
W (C )   rk                                          rkWk (Ck )
nk  0            j 0
 Ck 
k 1                                       k 1
 bk  nk 1

 
nk  0        j 0
k ( j)

so that the optimal partitioning is obtained from

K                                    K
max  Wk (Ck )        subject to                  Ck  C            0  Ck  C ,     Ck  N 0
k 1                                  k 1
University of Twente - Stochastic Operations Research   1

Exercises:
8 For the stochastic knapsack show (directly from
the equilibrium distribution) that
E[ X k ]  k (1  Bk ) /  k k  1,..., K
Prove the elasticity result:

B j  Bk
                 j , k  1,...,K
 k  j
University of Twente - Stochastic Operations Research            0

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 answers
G1: 1, 4, 2, 5, 7
G2: 1, 4, 2, 6, 8
G3: 1, 4, 3, 5, 8

Each group 3 persons: hand in a single set of answers
G1: 1, 4, 2, 5, 7
G2: 1, 4, 3, 6, 8

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