Matrix Geometric and Spectral Expansion Meh-thods
Harf Zatschler
Imperial College London
10/06/2004
Two solution mechanisms for a class of modulated queues
Blocking Region
L=8
t = 7
6
5
Repeating Region
4
3
b = 2
c=1
Filling Region
0
1
2
• Modulation process specifies arrival and service rates
Sample transitions within the repeating region
→
µ1 −
v j+1,1
→
λ1 −
v j,1
−
µ1 →
v j,1
→
q1,2 −
v j,1
→
λ1 −
v j−1,1
→
µ2 −
v j+1,2
→
q2,1 −
v j,2
→
λ2 −
v j,2
→
µ2 −
v j,2
→
λ2 −
v j−1,2
• Kolmogorov balance equations within the repeating region are of identical form
→
→
→
v
Λ−−
v
M=0
• −
v (Q − Λ − M) − −
j
j−1
j+1
PM →
v j+k Qk = 0
• General form: k=0 −
PM →
−
k=0 v j+k Qk = 0 - Spectral expansion solution
• Find the eigenvalues and eigenvectors, Q(ξ) =
PM
k=0
ξ k Qk
→
−
• Setting det Q(ξ) = 0 gives the eigenvalues, ψ i Q(ξi ) = 0 the eigenvectors.
P n∗
→
−
−
→
• Spectral Expansion representation now is v j = i=1 αi ξij ψ i where b ≤ j ≤ t
PM →
−
k=0 v j+k Qk = 0 - Matrix geometric solution
• Find a matrix F, that describes the changes in probability between levels
→
→
→
→
v jF ⇔ −
v j+i = −
v j Fi
• −
v j+1 = −
b ≤ j ≤ j + i ≤ t
→ →
−
→ j−b
−
−
• Start this expression from a vector f : v j = f F
b ≤ j ≤ t
• Finding F is done by using e.g. the Latouch-Ramaswami (LR) algorithm
→
j−
i=1 αi ξi ψ i
P n∗
− j−b
→
(SE) or f F
(MGM) - who wins?
?
good
bad
MGM
fast
many complex systems not soluble
popular
inaccurate for large finite systems
always works, can use generating
finding eigenvalues is expensive
functions for e.g. exact sojourn times
not very popular
SE
• Only difference is in the representation of the solution space.
• We fix the bad sides of MGM and then use it to speed up SE, so actually we’ll use both!
When does MGM not work?
P n∗
→ j−b
−
→
j−
−
→
→
−
Compare v j = i=1 αi ξi ψ i and v j = f F
N × N -matrices (usually) have N eigenvalues.
→
−
→ PN
−
Write f = i=1 βi ψ i , so that
N
− j−b X
→
→
j−b −
−
→
=
βi ξ i
ψi
vj = f F
i=1
If n∗ > N , MGM cannot work!
Fixing MGM by folding
Method broke because F contained too few eigenvalues/eigenvectors.
j
5
j’
4
Folding 2
adjacent
queue
lengths
3
2
2
1
0
1
1
0
1
2
2
3
4 N’
N
Now, the queue is shorter, but the number of modulation states,N , has doubled
We have n∗ ≤ N again!
Numerical Example
Infinite queue with batched arrivals, services and negative customers.
PM −
→
Q =0
v
Reminder: the form of the recurrence is
k=0

Q0 = 

Q1 = 

Q2 = 

Q3 = 
j+k
96614646928067997729
238418579101562500
− 4751540012855803167
59604644775390625
− 2580086226980701119681
238418579101562500
191645447185184061069
59604644775390625
9678887006187271051179
11920928955078125000
− 291427787455155927576
1490116119384765625
− 207483913894703404959
11920928955078125000
9503080025711606334
298023223876953125
k
2893401
− 2000000000000
124416243
4000000000000


20253807
2500000000000
743604057
− 8000000000000
176497461
− 40000000000000
2931015213
40000000000000
2893401
80000000000000
66548223
− 40000000000000






Spectral Expansion version
We find
λ
SE
=

0.054104017
− SE  0.68071170
→
=
ζ
0.32659553
0.11388376
−0.51971510
0.72848827
0.49875131
−0.99318108
0.86849535
−0.1300806
−0.028069268 −0.9900304
There are 4 eigenvalues within the unit circle.
Classic MGM can only represent 2 (= N ) of these.


Matrix Geometric version
We fold the queue by the factor 2.
The new queue length is
∞
2
= ∞ and N 0 = 2N .
LR algorithm quickly finds

−0.0894858330

 −0.00294677564

F̂ = 
 0.895824252

0.161185554
−0.0175654602

−0.0829948779
−0.0243799599
−0.0304039902
−0.0075404771
0.146933042
0.736698742
0.639410188
0.244510838

−0.0198735998 


0.208011207 

0.402124882
Matrix geometric version (part 2)
p
λF̂ = 0.054104017

0.68071170

→F̂ 
−
 0.32659553
ζ =
 0.03682923

0.01767013
0.11388376
−0.51971510
0.49875131
0.86849535

−0.99318108
−0.1300806
0.72848827
−0.028069268
−0.05918711
−0.49535095
0.082962983
−0.013999601

−0.9900304 


−0.1129744 

−0.8598368
Eigenvectors are too big, but the first two vector components match.
→SE p F̂ −
−
→
→F̂
−
, λ ζ SE )
In general ζ = ( ζ
Comparing the solutions
0.1
0.1
P<J=j,N=n>
0.05
P<J=j
0.05
1
1
2
0
n
2
30
25
20
15
j
10
5
0
n
0
3
4
14
12
10
8
6
j
4
2 0