Exact Statistics of a Markov Chain trhough Reduction in Number of States: Marco Aurelio Alzate Monroy Universidad Distrital Francisco José de Caldas

Transcription

1 Exact Statistics of a Markov Chain trhough Reduction in Number of States: A Satellite On-board Switching Example Marco Aurelio Alzate Monroy Universidad Distrital Francisco José de Caldas

2 The general setting An irreducible discrete time Markov chain States, m = (m 1 1,,m 2 2,, m N N) ) T ` Z N Transition probabilities p(m,n) Equilibrium distribution π p (m) Some performance characteristic gp = cp( m) π p( m) m ` `

3 First simple example: mean shortest queue length λ m 1 0,3 λ 1,3 λ 2,3 λ 3,3 2 λ 0,2 λ 1,2 λ 2,2 λ 2 m 2 0,1 λ 1,1 m 1 m 2 0,1 1,1 λ 2 0,0

4 Discretization Normalize every transition rate with respect to λ+2 and let λ, λ, λ 2 1 λ + 2 λ = An idle server attends a ficticious client, which does not depart, and is interrupted by real clients (just typical normalization: the outgoing transition rates for each state add up to one, and the mean time spent in a state is one)

5 Equivalent discrete time Markov Chain 0,3 λ 1,3 λ 2,3 λ 3,3 2 λ 02 0,2 λ 12 1,2 λ 22 2,2 0,1 λ 1,1 λ 2 λ 2 These are not transition rates, but just transition probabilities 2 0,0

6 The proposed procedure Redirect one or more outgoing transitions from state m If a transition from m to n 1 is redirected to n 2, p ( mn, 2) p( mn, 2) + p( mn, 1 ) p( mn, ) 0 1 Denote the new transition probabilities by q(m,n), ) and the new performance characteristic by q q q m ` g = c ( m) π ( m)

7 For example: Truncate the buffers at T clients Redirect the transition from (T,T) to (T,T+1) as a transition from (T,T) to itself (reject the new arrival) 0,2 λ 1,2 λ 2,2 2 λ λ 0,1 λ 1,1 λ 2 g < g q p 2 0,0

8 Truncate the buffers at T clients 6 5 T = inf T = 12 T = 8 4 E[Qmin] ro = landa/mu

9 For example: Jump to keep closer queues 24 2,4 λ 34 3,4 λ 44 4,4 Redirect the transition from (m 1,m 1 +T) to (m 1-1, m 1 +T) as a transition from (m 1,m 1 +T) to (m 1,m 1 +T-1) 1,3 λ 2,3 λ 3,3 (If the difference between the two queues exceeds T, 2 a client jumps form longest to shortest queue) 2 2 λ λ 2 0,2 λ 1,2 λ 2,2 2 λ 0,1 λ 1,1 λ 2 g > g q p 2 0,0

10 Jump to keep closer queues 6 5 T = inf T = 2 4 E[Qmin] ro = landa/mu

11 Among the many possibilities, Which transitions should be redirected? Do they lead to an upper or lower bound? Can we prove it? How tight is this bound?

12 The proposed method Define v(t,m) as the total cost over a time interval of t frame periods, starting in state m v (0, m) = 0 + = + ( ) vt ( 1, m) c( m) p m, vt (, ) And, assuming ergodicity g as.. 1 = lim 1 v (, t ) t m t for every initial state m

13 Precedence among states We say that state m is more attractive than state (written m ) if v(t,m) v(t,) for all t 0. Then, for a lower bound, redirect to more attractive states; and, for an upper bound, redirect to less attractive states! Why?

14 Deleting less attractive states Let us redirect all transitions leading to state k as transitions leading to state j, where j k. Then q = p except for q ( mk, ) = 0 q( m, j) = p( m, j) + p( m, k) m ` So we just made state k a transient state

15 Then we can prove that ( m) ( m) v t, v t, t 0 (*) q p Indeed, (*) holds for t=0 because v q ( m) v ( m) 0, = 0, = 0 p But, if it holds fot t, it also holds for t+1 because + = + + v ( t 1, m) c( m) q( m,) v ( t, ) c( m) q( m,) v ( t, ) q q p by hypothesis

16 t+ = + t + t v ( t 1, m ) c ( m ) q ( m, ) v ( t, ) c ( m ) q ( m, ) v ( t, ) q q p But p ( m, ) if there were no redirections involving i q( m, ) = 0 if transitions leading to were redirected to some other state p ( m, ) + p ( m,k ) if transitions leading to k were redirected to So ( + 1, m) ( m) + ( m,) (, ) ( m, ) (, k) (, j) v t c p v t p k v t v t q p p p ( m) ( m,) (, ) ( 1, m) c + p vp t = vp t+ So (*) holds for every t Transition (m,k) were redirected as transition (m,j), with j k

17 So we have shown that 1 1 g = lim v ( t, m) lim v ( t, m) = g t t q q p p t t Redirecting transitions to more attractive states lead to a lower bound for the performance characteristic Similarly, redirecting transitions to less attractive states lead to an upper bound

18 The trick is how to determine the attractiveness relation between states! Use the same procedure above (without redirecting transitions) v p (0,j) = v p (0,k) = 0 Assume v p (t,j) v p (t,k) for some t Check from the recursion if it also holds for t+1 v + p t+ 1, j = c j + p( j, ) vp t, ( ) ( ) ( ) v p t+ 1, k = c k + p ( k, ) vp t, ( ) ( ) ( )

19 We made all this because We were interested in reducing the number of switching operations onboard a satellite Onboard Switch Onboard Switch The frame builder looks at these queues every T slots. If at least one of the queues has 1 or more packets, a frame is constructed with up to T packets taken from the longest queue.

20 The queueing model becomes Destination 0 p n T Frame Building Destination 0 Destination 1 Destination 0 p n T Destination 1 Earth Frame Building Onboard switch Destination 1 Space

21 In an earth station the Markov chain is Where f ( ( ) ( ) ) p j, j, k, k = pa k j + f j j k j + f j j (, ), (, ) ( ( ) ) T min Tm, 0 if m > m m, m = ( 0 min T, m ) T if m < m any of the above with equal probability if m = m ( ) ( ) Is the number of packets taken away by the frame builder when it finds (m 0, m 1 ) packets in the queues, and pa p j, k j+ k j 2 T j+ k = Pr A = = k j + k j j+ k 2 T j k 1 j = 0,1,...,2 T ( 1 p) 2 k = 0,1,...,2 T j is the probability of (j,k) arrivals in T slots 0 1

22 The average number of packets in the terrestrial queues is with g = c ( m ) π ( m ) m T 1 1 c( m) = [ 1 1 ] m f ( m) + E ( T t) At = T t= 0 ( ( )) ( ) = m + m min T, max m, m + T + 1 p so all we need is to find attractiveness relations among states

23 There are also equivalent states We say that state m is equivalent to state (written m ) if v(t,m) = v(t,) for all t 0.

24 So these transitions lead to exact solutions m1 = m0 m = min mod(, T),mod(, T) ( ) if m 1 <T if m 0=0 m 1 =0 else m 1 =T+mT 0 m 0 =0 end end Redirect transitions to ( 0, 1) as. transitions to (m 0,m 1 )

25 The resulting system can be easily solved through Matrix Geometric Methods B0 A0 0 0 B1 A1 A0 0 P = 0 A 2 A1 A0 0 0 A2 A1 Each submatrix becomes a [T(T+1)/2] x [T(T+1)/2] square one : Easily solved

26 An easier to solve approximation Eliminate a queue! If (m 0 0,,m 1 1) ) has an equivalent state of the form (0, 1 ), redirect all transitions. Otherwise, it has an equivalent states of the form, : = kt + b, b < T, k = 1,2,3,... ( ) { } redirect them to (0, 0 +kt+b) b (sometimes an upperbound, sometimes a lower bound just an approximation) The submatrix blocks are now just TxT

27 Average Number of Packets in an earth station Numb ber of packe ets Theory exact Theory lower bound Simulation 99.9% confidence interval T = Average number of packets in the terrestrial queues T=10 T=6 T= p

28 Variance of the number of packets in an earth station Numbe er of packets Theory exact Theory lower bound Simulation 99.9% 9% confidence interval T= Std.Deviation of the number of packets in the terrestrial queues T=10 T=6 T= p

29 Conclusions We find a systematic approach to find upper or lower bounds for the performance measures of impossible to solve highly dimensional infinite Markov chains. The method constructs a modified solvable Markov chain by redirecting some transitions between states in such a way that the bound can be guaranteed. We used the method to find exact analytical results for the average and the variance of the number of packets in the earth station queues of a satellite system that groups packets into frames for onboard switching.

30 GRACIAS!!(THANK YOU ) Questions? The only risk is wanting to stay