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
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
GSM /HSCSD: High Speed Circuit Switched Data
33
University of Twente - Stochastic Operations Research
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
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 length
data: faster transmission rate, thus smaller transmission time
32
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
31
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Generalised stochastic knapsack: model
number of resource units
C
number of object classes
K
n'  n  e2
2 (n)
1 (n)
n'  n  e1
n
class k arrival rate
k (n)
class k mean holding time (exp)
1 /  k (n)
class k size
bk
state (number of objects)
n  (n1 ,..., nK )
state space
S  {n  N 0K : b  n  C}
object of class k accepted only if
bk  C  b  n
state of process at time t
X (t )  ( X1 (t ),..., X K (t ))
stationary Markov process
{ X (t ), t  0}
transition rates
1 (n)
n'  n  e1
2 (n)
n'  n  e2
k (n)1(bk  C  b  n) n'  n  ek
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 )  ( X1 (t ),..., X K (t )) is that
k (n)
 (n  ek )

 k (n  ek )
 ( n)
for some function
for all n  S \ Tk , k  1,..., K
 : S  [0, ). Moreover, when such a function 
exists, the
equilibrium distribution for the generalised stochastic knapsack is given by
 ( n)
 ( n) 
,
  ( n)
nS
nS
University of Twente - Stochastic Operations Research
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 )
 k (n  ek )
If

 (n  ek )
 ( n)
 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
 : S  [0, )is satisfied. Conversely, assume
 : S  [0, ) is satisfied. Then
condition of the theorem involving
that the condition involving
 ( n)
 ( n) 
,
  ( n)
nS
nS
is the equilibrium distribution.
29
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Generalised stochastic knapsack: examples
Stochastic knapsack
k (n)  k 1(bk  C  b  n) n'  n  ek
 k (n)  nk  k n'  n  ek
K
 ( n)  
k 1
 kn
1(b  n  C )
nk !
k
Finite source input
k (n)  (M k  nk )k 1(nk  M k ) n'  n  ek
 k (n)  nk  k n'  n  ek
K
 ( n)  
k 1
 M k  nk
   k
 nk 
State space constraints
K
k (n)  k 1(bk  C  b  n; nk  Ck ) n'  n  ek  (n) 

k 1
 k (n)  nk  k n'  n  ek
 kn
1(b  n  C; nk  Ck )
nk !
k
28
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
Admission control
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
27
University of Twente - Stochastic Operations Research
Stochastic knapsack under admission control
Admission policy
f  ( f1 ,..., f K )
1
f k ( n)  
0
f k : S  {0,1}
class k accepted in state n
class k rejected in state n
Transition rates
k f k (n)1(bk  C  b  n) n'  n  ek
q(n, n' )  
nk  k
n'  n  ek

Recurrent states
S ( f )  S  {n : b  n  C}
Examples: complete sharing
1 b  n  C  bk
f k ( n)  
otherwise
0
trunk reservation
1 b  n  C  bk  t k
f k ( n)  
otherwise
0
26
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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  t k
f k ( n)  
otherwise
0
Not reversible: so that equilibrium distribution usually not available in closed form
25
24
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Stochastic knapsack coordinate convex policies
  S : n  , nk  0  n  ek  
Coordinate convex policy: admit object iff state process remains in 
Coordinate convex set
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
 f ( n) 
Gf
K

k 1
 kn
k
nk !
,
n
University of Twente - Stochastic Operations Research
Coordinate convex policies: examples
Note: Not all policies are coordinate convex, e.g. trunk reservation
Complete sharing: always admit if room available
Complete partitioning: accept class k iff
C2

C1
S
S
bk (nk  1)  Ck
 C1 
 CK 
  {0,...,  }  ...  {0,..., 
}
 b1 
 bK 
C1  ...  C K  C
Threshold policies: accept class k iff
bk (nk  1)  Ck
b  n  bk  C
C2

C1
S
C 
  {n : b  n  C , nk   k , k  1,..., K }
 bk 
23
22
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Coordinate convex policies: revenue optimization
K
revenue in state n
r (n)   rk nk
long run average revenue
W ( f )   r (n) f (n)
k 1
n
rk  bk
rk   k
example: long run average utilization
long run average throughput
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
max  rk sk
k 1
*
K
subject to

k 1
bk sk  C
If C / bk * integer, where k maximizes per unit revenue rk
sk  N 0
rk / bk for k  k *
/ bk then s  
otherwise
 0
*
k
University of Twente - Stochastic Operations Research
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

C1
S
 Ck 
  {n : b  n  C , nk   , k  1,..., K }
 bk 
then problem reduces to finding optimal thresholds
in full generality: Markov decision theory
21
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
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
20
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PSTN / ISDN
C4
C1
C3
C5
C2
C6
C7
Link: connection between switches
Route: number of links
Capacity Ci of link i
Call class: route, bandwidth requirement per link
Stochastic knapsack: special case = model for single link
But: is special case of generalised stochastic knapsack
18
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
17
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C1
Loss network : notation
C4
C3
C2
number of links
J
capacity (bandwidth units) link j
Cj
number of object classes
K
class k arrival rate
k
class k mean holding time (exp)
1/ k
bandwidth req. class k on link j
b jk
Route
Rk  {1,..., J }
C5
C6
C7
n'  n  e2
2
1
n'  n  e1
n
`n11
Class k admitted iff b jk bandwidth units free in each link j  Rk
n'  n  e1
n2 2
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
Loss network : notation
Set of classes
K  {1,..., K }
set of classes that uses link j
K j  {k  K : j  Rk }
state
state space
n  (n1 ,..., nK )
S  {n  N 0K :

kK j
b jk nk  C j , j  1,..., J }
S  {n  N0K : An  C} A  (b jk )
class k blocked
Tk  {n  S :  b j n  b jk  C j , some j}
K j
Tk  {n  N 0K : An  C , A(n  ek ) / C}
16
15
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Loss network : notation
state of process at time t
X (t )  ( X1 (t ),..., X K (t ))
stationary Markov process
{ X (t ), t  0}
aperiodic,irreducible
equilibrium state
utilization of link j
X  ( X1 ,..., X K )
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 }
long run throughput
TH k  k (1  Bk )   k E[ X k ]
unconstrained cousin (∞ capacity)
Poisson r.v. with mean
unconstrained cousin of utilization
X   ( X 1 ,..., X  K )
 k  k /  k
U  j :

kK j
b jk X  k
PASTA
Little
14
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Equilibrium distribution
Theorem: Product form equilibrium distribution
1 K  k k Pr{ X   n}
Pr{ X  n}  

,

G k 1 nk ! Pr{U  C}
n
nS
G
nS
K

k 1
k n
,
nk !
k
nS
Blocking probability of class k call
Bk  1 

Pr{ X  n}

Pr{ X  n}
nS \Tk
nS
K
 1

nS \Tk  1
K

nS
 1
 n
n !
 n
n !

V, C vectors

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
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
Computing blocking probabilities
Direct summation is possible, but complexity
O( KC1...CJ )
Recursion is possible (KR), but complexity
O( KC1...CJ )
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

k 0
 j k / k!
,
j 

kK j
 k   b jk  k
kK
load offered to link j
facility bound
Call of class k blocked if not accepted at all links
Bk  1   (1  Er (  j , C j ))
jRk
product bound
13
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
12
University of Twente - Stochastic Operations Research
Computing blocking probabilities: single service networks
Reduced load approximation
Facility bound
L j  Er (   k , C j )
kK j
Part of offered load

kK j
L j  Er (  tk ( j )  k , C j )
 k blocked on other links :
kK j
tk ( j ) probability at least one unit of bandwidth available in each link in Rk \ { j}
 k tk ( j ) reduced load
tk ( j ) 
approximation: blocking independent from link to link

(1  Li )
iRk \{ j}
reduced load approximation
L j  Er(  k
k K j
existence, uniqueness fixed point
repeated substitution
accuracy

(1 Li ),C j ),
j 1,...,J
iR k \{ j}
Bk 1  (1 L j ),
j R k
k 1,...,K
University of Twente - Stochastic Operations Research
11
Reduced load approximation: existence and uniqueness
Notation:
L : ( L1 ,..., LJ )
T j ( L) : Er (   k
kK j

(1  Li ), C 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!
10
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Reduced load approximation: uniqueness
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

Er 1 j ( z )dz is strictly convex function of B for B  [0,1]
0
Proof of uniqueness:
consider fixed point L  T (L) and apply Er 1 j : Er1 j (L j ) :
Define L [0,1]
 
k
k K j
J
 (L) :
J
Lj
j1
0
   (1 L )   
k
k K j
i
iR k

which is strictly convex.
Thus, if L*  [0,1]J is solution of
Er1 j (z)dz
 ( L)
 0, i  1,..., J
Li

Then L* is unique minimum of  over [0,1]J
writing out the partial derivatives yields (*)
iR k \{ j}
(1 Li )
(*)
University of Twente - Stochastic Operations Research
9
Reduced load approximation: repeated substitution
Start
L  [0,1]J
T j (L) : Er(  k
k K j
Repeat L0 : L
Lm : T ( Lm1 ),
Theorem Let
m  1,2,...

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


(1 Li ),C j )
iR k \{ j}
University of Twente - Stochastic Operations Research
8
Reduced load approximation: accuracy
T j (L) : Er(  k
L j  Er(  k ,C j )
*
Corollary
k K j

(1 Li ),C j )
iR k \{ j}
k K j
Bk 1  (1 L j ) 1  Er(  k ,C j ),
j R k
j R k
k
K j
*
so that

Proof

k 1,...,K
T j (1...1)  T(0...0)  Er(  k ,C j )
2
k K j
(0,...,0)  L1  L2n1  L*  L2n  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
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
Monte Carlo summation
For performance measures: compute equilibrium distribution
k n
 n!
Pr{X  n}  k1 K k n ,
k
  n!
k
n S k1
K
k
k
nS
K

for blocking probability of class k call
Bk  n Tk
l 1
K
n S
l 1


In general: evaluate
K

n U
l 1
l n
nl !
l
l n
nl !
l
n

l
nl !
l
difficult due to size of the state space : use Monte Carlo summation

7
University of Twente - Stochastic Operations Research
Monte Carlo summation
Example: evaluate integral
c
d
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 Z ,..., Z
1
n
Then Z 
Z1  ...  Z n
n
b
1
f ( x)dx
unbiased estimator of c(b  a) 
a
with 95% confidence interval

Z 1.96 S 2 (n) /n
Powerful method: as accurate as desired

6
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Monte Carlo summation
K

For blocking probabilities
Bk  n Tk
l 1
K
n S
l 1

5
l n
n l ! e  
l n e  
nl !
l
l
ratio of multidimensional Poisson( ρ) distributed r.v.
d
Let X  Poisson(  ) then
Bk 

Pr{ X (  )  Tk }
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)
K
Eg(V )  Pr{X()  U}  

n U
l 1
l  
e
nl !
nl
l
for U  Tk
and
U S
confidence interval?
University of Twente - Stochastic Operations Research
Monte Carlo summation
K
For blocking probabilities

Bk  n Tk
l 1
K
n S
l 1

l n
n l ! e  
l n e  
nl !
l
l
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
if sample  S ignore
if sample  S count
if sample also  Tk count as succes
unbiased estimator!
g (Vi )  1(Vi  Tk | Vi  S )
Eg (V )  Bk
4
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
3
Summary loss network models and applications:
GSM network architecture
Loss networks - circuit switched telephone
wireless networks
TELEPHONE
Markov chain
equilibrium distribution
blocking probabilities
throughput
admission control
Evaluation of performance measures:
recursive algorithms
reduced load method
Monte Carlo summation
MS
(G)MSCPSTN/ISDN
BTS
BSC
RADIO
INTERFACE
GSM/HSCSD
C1
C2
C4
C3
C5
C6
C7
2
University of Twente - Stochastic Operations Research
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 
K
W (C )   rk
k 1

nk  0
nk 1
nk   k ( j )
j 0
 Ck 


 bk  nk 1
 
nk  0
K
  rkWk (Ck )
k 1
k ( j)
j 0
so that the optimal partitioning is obtained from
K
max  Wk (Ck )
k 1
K
subject to

k 1
Ck  C
0  Ck  C ,
Ck  N 0
University of Twente - Stochastic Operations Research
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

 k  j
j , k  1,..., K
1
University of Twente - Stochastic Operations Research
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
0
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