Matrix analytic methods. Lecture 1: Structured Markov chains and their stationary distribution

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1 1/29 Matrix analytic methods Lecture 1: Structured Markov chains and their stationary distribution Sophie Hautphenne and David Stanford (with thanks to Guy Latouche, U. Brussels and Peter Taylor, U. Melbourne Université Libre de Bruxelles University of Western Ontario Cirm, Luminy. April 28, 2010

2 2/29 Matrix-Analytic Methods In the 1970s and 1980s Marcel Neuts proposed a class of techniques for analysing Markov chains with block-structured transition matrices that have become known as matrix-analytic methods. More recently, other classes of models have been analysed using similar techniques, such as fluid queues (see Lecture 2) branching processes (see Lecture 3) The interaction of mathematical analysis and physical insight has played an important role in the development of results in this area.

3 2/29 Matrix-Analytic Methods In the 1970s and 1980s Marcel Neuts proposed a class of techniques for analysing Markov chains with block-structured transition matrices that have become known as matrix-analytic methods. More recently, other classes of models have been analysed using similar techniques, such as fluid queues (see Lecture 2) branching processes (see Lecture 3) The interaction of mathematical analysis and physical insight has played an important role in the development of results in this area.

4 2/29 Matrix-Analytic Methods In the 1970s and 1980s Marcel Neuts proposed a class of techniques for analysing Markov chains with block-structured transition matrices that have become known as matrix-analytic methods. More recently, other classes of models have been analysed using similar techniques, such as fluid queues (see Lecture 2) branching processes (see Lecture 3) The interaction of mathematical analysis and physical insight has played an important role in the development of results in this area.

5 /29 Quasi-Birth-and-Death processes A QDB process is a two-dimensional Markov chain {(X k, ϕ k ), k N} where X k N is called the level, ϕ k {1, 2,, m} is called the phase, the only possible transitions from state (n, i) are to state (n + 1, j) (1 level up), (n, j) (the same level), (n 1, j) (1 level down)

6 Quasi-Birth-and-Death processes 4/29

7 5/29 Example: tandem queues λ µ 1 µ 2 C C < = maximum capacity of the second buffer n is number ahead of first server = level of the QBD j is number beyond first server = phase of the QBD.

8 6/29 Quasi-Birth-and-Death processes m m n n + 1 Transition probabilities: (A 1 ) ij to go up: (n, i) to (n+1, j) (A 0 ) ij to remain in same level: (n, i) to (n, j) (A 1 ) ij to go down: (n, i) to (n 1, j)

9 7/29 Stationary distribution Block-structured transition matrix: B A 1 0 A 1 A 0 A 1 P = 0 A 1 A 0... Transitions are homogeneous with respect to the levels π = [π 0, π 1, π 2, ] is the stationary probability vector of the QBD, where (π n ) i = lim k P[(X k, ϕ k ) = (n, i)]

10 /29 Stationary distribution Block-structured transition matrix: B A 1 0 A 1 A 0 A 1 P = 0 A 1 A 0... Transitions are homogeneous with respect to the levels π = [π 0, π 1, π 2, ] is the stationary probability vector of the QBD, where (π n ) i = lim k P[(X k, ϕ k ) = (n, i)]

11 /29 Stationary distribution Block-structured transition matrix: B A 1 0 A 1 A 0 A 1 P = 0 A 1 A 0... Transitions are homogeneous with respect to the levels π = [π 0, π 1, π 2, ] is the stationary probability vector of the QBD, where (π n ) i = lim k P[(X k, ϕ k ) = (n, i)]

12 8/29 Existence of the stationary distribution Let x be the solution to x(a 1 + A 0 + A 1 ) = x. Then the QBD is positive recurrent, null recurrent or transient according as x A 1 1 x A 1 1 is < 0, = 0 or > 0. The stationary probability vector π exists if and only if the QBD is positive recurrent.

13 8/29 Existence of the stationary distribution Let x be the solution to x(a 1 + A 0 + A 1 ) = x. Then the QBD is positive recurrent, null recurrent or transient according as x A 1 1 x A 1 1 is < 0, = 0 or > 0. The stationary probability vector π exists if and only if the QBD is positive recurrent.

14 8/29 Existence of the stationary distribution Let x be the solution to x(a 1 + A 0 + A 1 ) = x. Then the QBD is positive recurrent, null recurrent or transient according as x A 1 1 x A 1 1 is < 0, = 0 or > 0. The stationary probability vector π exists if and only if the QBD is positive recurrent.

15 9/29 Computation of the stationary distribution The stationary probability vector π satisfies π P = π, π 1 = 1 Partition the state space into E, E c [ ] [ P E P EE c π E π E c P E c E P E c ] [ = π E π E c ] One has that π E (P E + P EE c (I P E c ) 1 P E c E ) = π E π E c = π E P EE c (I P E c ) 1

16 9/29 Computation of the stationary distribution The stationary probability vector π satisfies π P = π, π 1 = 1 Partition the state space into E, E c [ ] [ P E P EE c π E π E c P E c E P E c ] [ = π E π E c ] One has that π E (P E + P EE c (I P E c ) 1 P E c E ) = π E π E c = π E P EE c (I P E c ) 1

17 9/29 Computation of the stationary distribution The stationary probability vector π satisfies π P = π, π 1 = 1 Partition the state space into E, E c [ ] [ P E P EE c π E π E c P E c E P E c ] [ = π E π E c ] One has that π E (P E + P EE c (I P E c ) 1 P E c E ) = π E π E c = π E P EE c (I P E c ) 1

18 Computation of the stationary distribution Let l(n) denote the level n of the QBD. Choose E = l(0), E c = {l(1), l(2), l(3),, } [ P = P E P E c E P EE c P E c ] = B A A 1 A 0 A 1 0 A 1 A 0 A 1 0 A 1 A 0. The equation π E c = π E P EE c (I P E c ) 1 becomes [ π 1 π 2 ] = π 0 [ A ] I A 0 A 1 A 1 I A 0 A 1 A 1 I A {z } N 10/29 We obtain for π 1 : π 1 = π 0 A 1 N 11.

19 Computation of the stationary distribution Let l(n) denote the level n of the QBD. Choose E = l(0), E c = {l(1), l(2), l(3),, } [ P = P E P E c E P EE c P E c ] = B A A 1 A 0 A 1 0 A 1 A 0 A 1 0 A 1 A 0. The equation π E c = π E P EE c (I P E c ) 1 becomes [ π 1 π 2 ] = π 0 [ A ] I A 0 A 1 A 1 I A 0 A 1 A 1 I A {z } N 10/29 We obtain for π 1 : π 1 = π 0 A 1 N 11.

20 Computation of the stationary distribution Let l(n) denote the level n of the QBD. Choose E = l(0), E c = {l(1), l(2), l(3),, } [ P = P E P E c E P EE c P E c ] = B A A 1 A 0 A 1 0 A 1 A 0 A 1 0 A 1 A 0. The equation π E c = π E P EE c (I P E c ) 1 becomes [ π 1 π 2 ] = π 0 [ A ] I A 0 A 1 A 1 I A 0 A 1 A 1 I A {z } N 10/29 We obtain for π 1 : π 1 = π 0 A 1 N 11.

21 Computation of the stationary distribution Let l(n) denote the level n of the QBD. Choose E = l(0), E c = {l(1), l(2), l(3),, } [ P = P E P E c E P EE c P E c ] = B A A 1 A 0 A 1 0 A 1 A 0 A 1 0 A 1 A 0. The equation π E c = π E P EE c (I P E c ) 1 becomes [ π 1 π 2 ] = π 0 [ A ] I A 0 A 1 A 1 I A 0 A 1 A 1 I A {z } N 10/29 We obtain for π 1 : π 1 = π 0 A 1 N 11.

22 Computation of the stationary distribution Let l(n) denote the level n of the QBD. Choose E = l(0), E c = {l(1), l(2), l(3),, } [ P = P E P E c E P EE c P E c ] = B A A 1 A 0 A 1 0 A 1 A 0 A 1 0 A 1 A 0. The equation π E c = π E P EE c (I P E c ) 1 becomes [ π 1 π 2 ] = π 0 [ A ] I A 0 A 1 A 1 I A 0 A 1 A 1 I A {z } N 10/29 We obtain for π 1 : π 1 = π 0 A 1 N 11.

23 11/29 Computation of the stationary distribution Then, for any n 1, choose E = {l(0), l(1),, l(n)}, E c = {l(n + 1), l(n + 2),, } By the same argument, we get n 0 π n+1 = π n A 1 N 11 = π 0 (A 1 N 11 ) n+1 (N 11 ) ij is the matrix of expected number of visits of state (n + 1, j), starting from (n + 1, i), before returning to {l(0), l(1),, l(n)}, for any n 0.

24 1/29 Computation of the stationary distribution Then, for any n 1, choose E = {l(0), l(1),, l(n)}, E c = {l(n + 1), l(n + 2),, } By the same argument, we get n 0 π n+1 = π n A 1 N 11 = π 0 (A 1 N 11 ) n+1 (N 11 ) ij is the matrix of expected number of visits of state (n + 1, j), starting from (n + 1, i), before returning to {l(0), l(1),, l(n)}, for any n 0.

25 1/29 Computation of the stationary distribution Then, for any n 1, choose E = {l(0), l(1),, l(n)}, E c = {l(n + 1), l(n + 2),, } By the same argument, we get n 0 π n+1 = π n A 1 N 11 = π 0 (A 1 N 11 ) n+1 (N 11 ) ij is the matrix of expected number of visits of state (n + 1, j), starting from (n + 1, i), before returning to {l(0), l(1),, l(n)}, for any n 0.

26 12/29 Stationary distribution In summarize, for all n 0, π n = π 0 R n, where R = A 1 N 11 and R ij is the expected number of visits to (1, j) starting from (0, i) before the first return to level 0. This is a matrix-geometric stationary distribution. To complete its characterization, it remains to characterize π 0 and R.

27 2/29 Stationary distribution In summarize, for all n 0, π n = π 0 R n, where R = A 1 N 11 and R ij is the expected number of visits to (1, j) starting from (0, i) before the first return to level 0. This is a matrix-geometric stationary distribution. To complete its characterization, it remains to characterize π 0 and R.

28 2/29 Stationary distribution In summarize, for all n 0, π n = π 0 R n, where R = A 1 N 11 and R ij is the expected number of visits to (1, j) starting from (0, i) before the first return to level 0. This is a matrix-geometric stationary distribution. To complete its characterization, it remains to characterize π 0 and R.

29 13/29 Characterization of π 0 Take again E = l(0). From π E (P E + P EE c (I P E c ) 1 P E c E ) = π E, we get π 0 (B + A 1 N 11 A 1 ) = π 0. From the normalization constraint π 1 = i=0 π i 1 = 1 we have π i 1 = π 0 i=0 i=0 R i 1 = π 0 (I R) 1 1 = 1. Thus π 0 satisfies { π0 (B + R A 1 ) = π 0 π 0 (I R) 1 1 = 1

30 13/29 Characterization of π 0 Take again E = l(0). From π E (P E + P EE c (I P E c ) 1 P E c E ) = π E, we get π 0 (B + A 1 N 11 A 1 ) = π 0. From the normalization constraint π 1 = i=0 π i 1 = 1 we have π i 1 = π 0 i=0 i=0 R i 1 = π 0 (I R) 1 1 = 1. Thus π 0 satisfies { π0 (B + R A 1 ) = π 0 π 0 (I R) 1 1 = 1

31 13/29 Characterization of π 0 Take again E = l(0). From π E (P E + P EE c (I P E c ) 1 P E c E ) = π E, we get π 0 (B + A 1 N 11 A 1 ) = π 0. From the normalization constraint π 1 = i=0 π i 1 = 1 we have π i 1 = π 0 i=0 i=0 R i 1 = π 0 (I R) 1 1 = 1. Thus π 0 satisfies { π0 (B + R A 1 ) = π 0 π 0 (I R) 1 1 = 1

32 13/29 Characterization of π 0 Take again E = l(0). From π E (P E + P EE c (I P E c ) 1 P E c E ) = π E, we get π 0 (B + A 1 N 11 A 1 ) = π 0. From the normalization constraint π 1 = i=0 π i 1 = 1 we have π i 1 = π 0 i=0 i=0 R i 1 = π 0 (I R) 1 1 = 1. Thus π 0 satisfies { π0 (B + R A 1 ) = π 0 π 0 (I R) 1 1 = 1

33 14/29 Matrix R and first passage probability matrix G Many expressions for R; for instance where R = A 1 [ I (A0 + A 1 G) ] 1 G ij is the probability of eventually moving to (0, j), starting from (1, i). (A 0 + A 1 G) ij is the probability of visiting (1, j) from (1, i) avoiding level 0. G is stochastic and it is the minimal nonnegative solution of X = A 1 + A 0 X + A 1 X 2

34 14/29 Matrix R and first passage probability matrix G Many expressions for R; for instance where R = A 1 [ I (A0 + A 1 G) ] 1 G ij is the probability of eventually moving to (0, j), starting from (1, i). (A 0 + A 1 G) ij is the probability of visiting (1, j) from (1, i) avoiding level 0. G is stochastic and it is the minimal nonnegative solution of X = A 1 + A 0 X + A 1 X 2

35 15/29 Formal manipulation X = A 1 + A 0 X + A 1 X 2 leads to X A 0 X A 1 X 2 = A 1 X = [ I (A 0 + A 1 X ) ] 1 A 1. (1) Other equations have been proposed, from decomposition A 1 + A 0 X + A 1 X 2 = F (X ) + G(X )X All may be solved by functional iteration, (1) is the most interesting for its efficiency and its ease of interpretation

36 15/29 Formal manipulation X = A 1 + A 0 X + A 1 X 2 leads to X A 0 X A 1 X 2 = A 1 X = [ I (A 0 + A 1 X ) ] 1 A 1. (1) Other equations have been proposed, from decomposition A 1 + A 0 X + A 1 X 2 = F (X ) + G(X )X All may be solved by functional iteration, (1) is the most interesting for its efficiency and its ease of interpretation

37 16/29 Interpretation of functional iterations Start with X 0 = 0 and iterate X n = [ I (A 0 + A 1 X n 1 ) ] 1 A 1 or X n = A 1 + A 0 X n + A 1 X n 1 X n n n + 1 n n 1 X n : process is allowed to move freely among n levels as n, restrictions disappear and X n G

38 17/29 Convergence rates Linear convergence with rates X n = [ I (A 0 + A 1 X n 1 ) ] 1 A 1 η = lim n G X n 1 n Actually, η is the Perron-Frobenius eigenvalue of R. When the QBD is positive recurrent, η < 1.

39 17/29 Convergence rates Linear convergence with rates X n = [ I (A 0 + A 1 X n 1 ) ] 1 A 1 η = lim n G X n 1 n Actually, η is the Perron-Frobenius eigenvalue of R. When the QBD is positive recurrent, η < 1.

40 8/29 Acceleration: keep it stochastic Instead of X 0 = 0, take X 0 = I with the same functional iteration X n = [ I (A 0 + A 1 X n 1) ] 1 A 1 The new sequence is well defined and X n corresponds to a new irreducible process which is forced to remain between levels 1 and n prove by induction that X n X n convergence rate strictly less than η, the P.F. eigenvalue of R. Total complexity O(m 3 ).

41 19/29 Newton s scheme X = A 1 + A 0 X + A 1 X 2 Newton s scheme: convergence to G, monotone and quadratic for any initial nonnegative matrix between 0 and G. Requires at each step the solution of T n H(T n 1 )A 1 T n H(T n 1 )A 1 = H(T n 1 )A 1, where H(X ) = [ I (A 0 + A 1 X ) ] 1 Sylvester equation, O(m 3 ) algorithm since 1992 (Gardiner, Laub, Amato and Moler) Total complexity O(m 6 ).

42 20/29 Logarithmic & cyclic reduction algorithms Two nearly identical, quadratically convergent, algorithms Logarithmic reduction: Latouche and Ramaswami, 1993 Cyclic reduction: Bini and Meini, 1995; slightly faster at each step, generalized to more complex Markov chains

43 21/29 Cyclic reduction The equation G = A 1 + A 0 G + A 1 G 2 is transformed into the linear system I A 0 A 1 0 A 1 I A 0 A 1. A 1 I A G G 2 G 3. = Apply an even-odd permutation to rows and columns. Perform one step of Gaussian elimination to remove the even-numbered blocks and obtain A

44 21/29 Cyclic reduction The equation G = A 1 + A 0 G + A 1 G 2 is transformed into the linear system I A 0 A 1 0 A 1 I A 0 A 1. A 1 I A G G 2 G 3. = Apply an even-odd permutation to rows and columns. Perform one step of Gaussian elimination to remove the even-numbered blocks and obtain A

45 21/29 Cyclic reduction The equation G = A 1 + A 0 G + A 1 G 2 is transformed into the linear system I A 0 A 1 0 A 1 I A 0 A 1. A 1 I A G G 2 G 3. = Apply an even-odd permutation to rows and columns. Perform one step of Gaussian elimination to remove the even-numbered blocks and obtain A

46 22/29 Cyclic reduction the same structure. I Â(1) 0 A (1) A (1) 1 I A (1) 0 A (1) 1 A (1) 1 I A (1) G G 3 G 5. = A Repeat as needed [ At nth step: G G 2n +1 G 2 2n +1 G 3 2n +1 ] T upper diagonal: A (n) 1 tends to 0 as n tends to. When A (n) 1 0, G (I Â(n) 0 ) 1 A 1.

47 22/29 Cyclic reduction the same structure. I Â(1) 0 A (1) A (1) 1 I A (1) 0 A (1) 1 A (1) 1 I A (1) G G 3 G 5. = A Repeat as needed [ At nth step: G G 2n +1 G 2 2n +1 G 3 2n +1 ] T upper diagonal: A (n) 1 tends to 0 as n tends to. When A (n) 1 0, G (I Â(n) 0 ) 1 A 1.

48 23/29 Other structured Markov chains GI /M/1-type Markov chains A GI /M/1-type Markov chain is a two-dimensional Markovian chain {(X k, ϕ k ), k N} where the only possible transitions from state (n, i) are to state (n + 1, j) (1 level up), (n, j) (the same level), (n l, j) (l levels down) for all 1 l n. Ã 0 A Ã 1 A 0 A 1 0 P = Ã 2 A 1 A 0 A 1 Ã 3 A 2 A 1 A

49 24/29 Solution for GI /M/1-type Markov Chains For a GI /M/1-type Markov chain, let x be the solution to [ ] x A 1 k = x. k=0 Then the chain is positive recurrent, null recurrent or transient according as [ ] x A 0 1 x (k 1)A 1 k 1 is < 0, = 0 or > 0. k=2 The stationary distribution π exists if and only if the chain is positive recurrent.

50 24/29 Solution for GI /M/1-type Markov Chains For a GI /M/1-type Markov chain, let x be the solution to [ ] x A 1 k = x. k=0 Then the chain is positive recurrent, null recurrent or transient according as [ ] x A 0 1 x (k 1)A 1 k 1 is < 0, = 0 or > 0. k=2 The stationary distribution π exists if and only if the chain is positive recurrent.

51 24/29 Solution for GI /M/1-type Markov Chains For a GI /M/1-type Markov chain, let x be the solution to [ ] x A 1 k = x. k=0 Then the chain is positive recurrent, null recurrent or transient according as [ ] x A 0 1 x (k 1)A 1 k 1 is < 0, = 0 or > 0. k=2 The stationary distribution π exists if and only if the chain is positive recurrent.

52 25/29 Stationary distribution of GI /M/1-type Markov Chains The stationary distribution π of a positive recurrent GI /M/1-type Markov chain of is such that π n = π 0 R n n 0, where the matrix R has exactly the same probabilistic interpretation as for a QBD, and the vector π 0 satisfies [ ] π 0 R k à 2 k = π 0, k=0 and a normalization constraint. This is the well-known Matrix-Geometric form of the stationary distribution.

53 25/29 Stationary distribution of GI /M/1-type Markov Chains The stationary distribution π of a positive recurrent GI /M/1-type Markov chain of is such that π n = π 0 R n n 0, where the matrix R has exactly the same probabilistic interpretation as for a QBD, and the vector π 0 satisfies [ ] π 0 R k à 2 k = π 0, k=0 and a normalization constraint. This is the well-known Matrix-Geometric form of the stationary distribution.

54 26/29 Matrix R for GI /M/1-type Markov chains The matrix R is the minimal nonnegative solution to the matrix equation X = X k+1 A k. k= 1 It is obtained using a duality property with M/G/1-type Markov chains: Determining R for a positive recurrent GI /M/1-type MC determining G for a transient M/G/1-type MC.

55 6/29 Matrix R for GI /M/1-type Markov chains The matrix R is the minimal nonnegative solution to the matrix equation X = X k+1 A k. k= 1 It is obtained using a duality property with M/G/1-type Markov chains: Determining R for a positive recurrent GI /M/1-type MC determining G for a transient M/G/1-type MC.

56 Other structured Markov chains M/G/1-type Markov chains An M/G/1-type Markov chain is a two-dimensional Markovian chain {(X k, ϕ k ), k N} where the only possible transitions from state (n, i) are to state (n + l, j) (l levels up) for all l 1, (n, j) (the same level), 7/29 (n 1, j) (1 level down). Ã 0 Ã 1 Ã 2 Ã 3 A 1 A 0 A 1 A 2 P = 0 A 1 A 0 A A 1 A M/G/1-type Markov chains do not have a matrix-geometric stationary distribution.

57 28/29 Matrix G for M/G/1-type Markov chains The matrix G is the minimal nonnegative solution to the matrix equation X = A k X k+1. k= 1 It may be obtained using for instance the generalization of the cyclic-reduction algorithm to M/G/1-type Markov chains.

58 8/29 Matrix G for M/G/1-type Markov chains The matrix G is the minimal nonnegative solution to the matrix equation X = A k X k+1. k= 1 It may be obtained using for instance the generalization of the cyclic-reduction algorithm to M/G/1-type Markov chains.

59 29/29 References Latouche and Ramaswami, Introduction to Matrix Analytic Methods in Stochastic Modeling, SIAM, 1999 Bini, Latouche and Meini, Numerical Methods for Structured Markov Chains, Oxford University Press, 2005 Latouche, Newton s iteration for non-linear equations in Markov chains, IMA Journal of Numerical Analysis, 1994, 14(4):