Stationary Analysis of Fluid Level Dependent Bounded Fluid Models


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1 Stationary Analysis of Flui Level Depenent Boune Flui Moels M. Gribauo [1], M. Telek [2] [1] Dipartimento i Informatica, Università i Torino, Torino, Italy [2] Department of Telecommunications, Technical University of Buapest, Buapest, Hungary Abstract In stochastic flui moels the rift at which the flui level changes in the flui buffer an the generator of the unerlying process might epen on the iscrete state of the system an on the flui level itself. In this paper we analyze the stationary behaviour of finite buffer Markov flui moels in which the rift an the generator of the unerlying continuous time Markov chain (CTMC) epens on both of these parameters. Especially, the case when the rift changes sign at a given flui level is consiere. This case requires a particular treatment, because at this flui level probability mass might evelop. When ealing with sign changes, new problems that were not aresse in previous works arises in the solution process. The set of stationary equations is provie an a transformation of the unknowns is applie to obtain a solvable system escription. Numerical examples introuce the behaviour of flui systems with various iscontinuities an sign changes of the rift. Key wors: Stochastic flui moel, stationary istribution. 1 Introuction There is a wie range of performance analysis problems in which the system behaviour is Markovian over a mixe iscrete an continuous state space This work is partially supporte by the ItalianHungarian bilateral R&D program an by OTKA grant n. T
2 an there are several stochastic moels evelope for the analysis of these systems. Examples of such moels are Markov rewar moels, Markov moulate Brownian motion an stochastic flui moels. The case when the variation of the continuous system parameter is boune (at least from one sie) is commonly referre to as flui moel. The complexity of flui moels epens on several moel features among which one of the most important is the epenency structure of the iscrete an continuous part of the moel. In case of several real life examples the iscrete state process of the system form a continuous time Markov chain (CTMC) on its own (inepenent of the continuous part) an the continuous system variable (commonly referre to as flui level) evolves accoring to the iscrete system state [10]. In these cases the marginal istribution of the iscrete system state can be analyze inepenent of the flui part of the moel. In this paper we consier the more complex case when the iscrete an the continuous part of the moel are mutually epenent, such that none of them can be analyze inepenently [7, 3]. Further classifying moel feature is the kin of epenence of the flui increment on the iscrete an the continuous part of the system. In first orer flui moels there is a eterministic relation between the iscrete state an the flui increment [1, 10], while in secon orer flui moels this relation is stochastic (the iscrete state etermines the mean an the variance of the flui increment) [3, 12]. The analytical escription of flui moels is provie by a set of partial ifferential equations (PDEs) an associate bounary conitions. For nontrivial cases the symbolic solution of the system equations is not available. Instea, the transient analysis is carrie out with numerical solution of the PDEs consiering the bounary conitions an the initial istribution of the system [7, 12]. The availability of an initial istribution, to start the numerical solution, is a significant avantage of the transient analysis of flui moels compare with their stationary analysis. In case of stationary analysis we can eliminate the erivative of the time variable, which simplifies the PDEs to orinary ifferential equations (ODEs), but there is no initial conition available for the numerical solution of the ODEs. It is a significant rawback of the stationary analysis of flui moels. To obtain the initial conitions of the ODEs a set of equations is compose base on the bounary equations, the solution of the ODE an a normalizing conition. For example, in case of flui level inepenent transition matrix an flui rift vector, the solution of the ODE is assume to be exponential an the coefficients of the exponential solution are calculate from the bounary equations an the normalizing conition [10]. Unfortunately, this effective solution approach is not applicable when the transition matrix or the flui rift vector epens 2
3 on the flui level. One of the most important result of this paper is that the nature of the consiere problem inhibits the use of previously applie moel escription with flui ensity an mass probability an it requires the introuction of new measures to analyze the system. Conventional techniques, like the one propose in [6], applie in this context, prouce systems of equations that cannot be solve. The rest of the paper is organize as follows. Section 2 compare the results propose in this paper with the one previously presente in the literature. Section 3 introuces the consiere flui moel an its transient escription over the continuous regions. Section 4 provies the bounary equations to escribe the behaviour at iscontinuities an bounaries. The number of obtaine equations an unknowns are consiere in section 5. A moifie set of equations is introuce in section 6 an the properties of the obtaine set of equations are provie in 7. Section 8 iscusses the case when there are states with zero flui rate an section 9 provies the normalizing conitions. A set of numerical examples emonstrate the applicability of the propose metho in section 10 an the paper is conclue in section Relate works There are several papers ealing with the transient analysis of flui level epenent transition matrix an flui rift vector [12, 7, 3], but rather few results available for the stationary analysis of these systems mainly ue to the mentione lack of an initial conition. The approach to compose the set of equation characterizing the initial conition without using the assumption on the exponential solution of the set of ODEs was provie in [6], but the first real nonexponential ODE system was introuce in [5], because [6] still assume flui level inepenent transition matrix an flui rift vector. There is a meaningful restriction applie in [5]. The elements of the flui rift vector o not change sign. This restriction results that the flui level istribution is continuous (i.e., there is no probability mass of the flui istribution) between the flui bouns. The transient behaviour of the case when the flui rift changes sign in a particular state was stuie in [3]. In this paper we use slightly ifferent assumptions (not the iscontinuity, but the sign change of the flui rift function separates continuous pieces an the behaviour of the isolating state is ifferent in this work) an provie the stationary analysis of first orer flui moels with mutually epenent continuous an iscrete parts. An analytical solution to flui moels with state epenent flow rate has 3
4 been propose in [8]. In that case, the system was analyze both in transient an steay state, using Laplace Transforms. With respect to that work, we restrict our analysis to steay state, but we allow the backgroun process to epen on the flui level, an consier sign changes. Flui moel with flui epenent rates, were also aresse in [9]. In that case, a system with two flui variables, with interepenent flui rates where consiere. Our work will focus only on systems with a single flui variable, but will allow them to be riven by a more complex stochastic process. The problem of ealing with states with zero flow rate was also consiere in [2] an in [11]. Here we will use a similar solution to remove zero flow rate states from a moel. 3 Stationary escription of flui moels with single finite flui buffer The Z(t) = {M(t), X(t); t 0} process represents the state of a flui moel with single flui buffer, where M(t) S is the (iscrete) state of the environment process an X(t) [0, B] is the flui level of the flui buffer at time t. S enotes the finite set of states of the environment an B the maximum flui level. The flui level istribution might have probability masses at particular flui levels an it is continuous between these levels. We efine ˆπ j (t, x) an ĉ j (t, x) to escribe the continuous part an the probability masses of the transient flui istribution as follows ˆπ j (t, x) = lim 0 P r(m(t) = j, x X(t) < x + ) ĉ j (t, x) = P r(m(t) = j, X(t) = x), an, assuming the system converges to a single stationary solution, the stationary flui ensity function an flui mass function are π j (x) = lim t ˆπ j (t, x) an c j (x) = lim t ĉ j (t, x). Since we are only consiering boune systems, stability is not an issue. However, the system may have some transient states, an thus the stationary solution may epen on the initial state of the moel. On the continuous intervals of the flui level istribution, π(x) = {π j (x)}, satisfies [10] π(x)r(x) = π(x)q(x), (1) x where matrix Q(x) is the transition rate matrix of the environment process when the flui level is x, an the iagonal matrix R(x) = iag < r j (x) > is 4,
5 compose by the flui rates r j (x), j S 1. The flui rate etermines the rate at which the flui level changes when the environment is in state j an the flui level is x, i.e., X(t) = r t j(x) when X(t) = x an M(t) = j. To avoi nonergoic moel behaviour we assume that Q(x) is a boune irreucible generator matrix, R(x) is a boune iagonal matrix for x [0, B], such that ɛ < r j (x) < ξ or r j (x) = 0, where ɛ is a small positive number an ξ is the upper boun of the rift 2 an Z(t) = {M(t), X(t)} is irreucible in the sense that there is a vali trajectory from any state {i, x}; i S, x [0, B] to any state {j, y}; j S, y [0, B] with a positive probability. In the following, we will use the notation x an x + to ientify the left an right limit of a function to point x respectively. In particular we will efine: x + = lim h 0 +(x + h), x = lim h 0 +(x h). The cases when r j (x) = 0 or when r j (x ) an r j (x + ) has ifferent sign result in bounaries of the flui istribution. The bounary conitions of eq. (1) are iscusse in section 4. Let 0 = x 0 < x 1 < x 2 <... < x n 1 < x n = B enote the set of bounaries, where the flui istribution might have probability mass. In the first part of this paper we assume that r j (x) 0, j S for x i < x < x i+1. The case when j S such that r j (x) = 0 for x i < x < x i+1 is iscusse in section 8. To obtain the solution of (1) between x i an x i+1 we rearrange it to where π(x) = π(x) x B(x) = ( Q(x) x R(x) ( Q(x) ) x R(x) R 1 (x). For x i < x < x i+1, the solution of (2) can be written as ) R 1 (x) = π(x)b(x), (2) π(x) = π(x + i )W(x i, x), (3) where W(x i, x i ) = I an W(x i, x) is the solution of x W(x i, x) = W(x i, x)b(x). (4) 1 The fact that the generator of the environment process epens on the flui level prevents us to apply the solution presente, e.g., in [4]. 2 The assumption that ɛ < r j (x) < ξ or r j (x) = 0 results that a iscontinuity is associate with all sign change. 5
6 Various solution methos of eq. (4) are iscusse in [5]. In the present paper we neglect the iscussion about this kin of inhomogeneous ODEs an refer intereste reaers to [5]. In the special case when Q(x) an R(x) are level inepenent in the (x i, x i+1 ) range, i.e., for x (x i, x i+1 ), Q(x) = Q i an R(x) = R i then B(x) = Q i R 1 i an W(x i, x) = e (x x i)q i R 1 i. 4 Bounaries an iscontinuities When at least one element of matrix R(x) have iscontinuity with sign change at level x i we ivie the analysis into pieces at these x i levels. We assume that the number of pieces are finite. Let n 1 the number of possible iscontinuities between 0 an B an x 1, x 2,..., x n 1 the points of iscontinuities. x 0 = 0 is the lower boun an x n = B is the upper boun of the flui level. Note that, the R(x i ) values have to be efine at x 0, x 1,..., x n as well, because they characterize the moel behaviour at the iscontinuities in the subsequent analytical escription. 4.1 Managing iscontinuities Discontinuity of matrix Q(x) o not generate particular problems in the application of the technique. Discontinuity of matrix R(x) instea can prouce probability masses or ambiguous situations if the rate changes sign. In particular, for each iscontinuity x i four ifferent situations are possible (See for example [3]). Figure 1 represents those cases. These cases can be further subivie in other subcases, epening on the value that R(x) assumes on the iscontinuity. In particular, following the terminology given in [3]: a) an b): (Emitting states) are states where the sign oes not change. Depening on the value of the rate at the iscontinuity point, the probability flux simply flows from one sie of the iscontinuity to the other an no probability mass is forme (if the rate at the iscontinuity point has the same sign), or the flow is stoppe an some probability mass buils up (if the rate is zero or has the opposite sign). Probability mass coming from other states simply as to the flux or to the mass epening on the sign of the rate at the iscontinuity point. Figure 2a) shows a physical representation of the case where the sign oes not change at the environment of x i, e.g., r(x i ) > 0, r(x+ i ) > 0 an r(x i) > 0. Figure 2b) show the case when the sign of r(x i ) equals with the sign of r(x+ i ), but the sign r(x i) is ifferent, e.g., r(x i ) > 0, r(x+ i ) > 0 an r(x i) 0. In the figures the horizontal move of the ball represents the change of the flui level at the flui place. 6
7 x i x i x i+1 x i1  x i  x i+1 a) b) + + x i1 x i  x i+1 x i1  x i x i+1 c) ) Figure 1: The four possible iscontinuity situations c): (Absorbing states) when the sign changes from positive to negative, the probability accumulates aroun the iscontinuity point, creating thus some probability mass. Probability mass coming from other states, as to the probability mass generate into this state. ): (Insulating states) when the sign changes from negative to positive, things becomes a little more complicate. The system is in a situation of unstable equilibrium. No probability mass is generate ue to the probability flow. Probability mass coming from other states may either flow to the left, to the right, or stan still forming a probability mass in this state. Placing probability mass in this iscontinuity is like placing a ball on the top of a hill (as shown in Figure 3). It may either fall on one sie, on the other, or stan still in equilibrium. Which of the three possible behavior will be chosen, epens on the value of the rate at the iscontinuity point. We will now present the equations that in a generic state j couple the probability ensities at the beginning π j (x + i ) an at the en π j(x i+1 ) of each continuous piece (x i, x i+1 ), with the probability masses c j (x i ) in all of the four previous cases. Case a): positive emitting states r j (x + i ) > 0 an r j(x i ) > 0. ( ) 0 = c j (x i )q jj (x i ) + α j (x i) π j (x i )r j(x i ) + k j c k (x i )q kj (x i ), 7
8 r(x) r(x) r(x i )>0 r(x i )<=0 x i a) b) x x i x Figure 2: The two possible behavior in the Emitting state ( π j (x + i )r j(x + i ) = α> j (x i ) π j (x i )r j(x i ) + ) c k (x i )q kj (x i ) k j, where αj rel (x i ) = 1{r j (x i ) rel 0} with 1{.} being the inicator function an rel stans for one of the following relations: <, =, >,,. E.g., α j (x i) = 1 if r j (x i ) 0 an it is 0 otherwise. The first equation efines the probability mass in state j at flui level x i. It is zero if the probability flows through level x i (i.e. r j (x i ) > 0). If it is not the case the probability mass is equal to the sum of the probability masses coming from neighboring states plus the one that flows from the left piece. The secon equation relates the initial probability ensity of piece i with the final probability ensity of piece i 1. If r j (x i ) > 0 the ifference of the two is coming from the probability masses that other neighboring states may have at flui level x i. If r j (x i ) 0 the flui level is stoppe at x i in state j, hence π j (x + i ) = 0. These equation can be erive by observing the rate conservation law. Case b): negative emitting states r j (x + i ) < 0 an r j(x i ) < 0. ( ) π j (x i )r j(x i ) = α< j (x i ) 0 = c j (x i )q jj (x i ) + α j (x i) ( π j (x + i )r j(x + i ) + k j π j (x + i )r j(x + i ) + k j ) c k (x i )q kj (x i ) c k (x i )q kj (x i )., The equations are ientical to the one presente for case a), except for the presence of the minus sign, which is require since the rates are negative. 8
9 r(x i )=0 r(x i )<0 r(x i )>0 Figure 3: The three possible behavior in the Insulating state Case c): absorbing states r j (x + i ) < 0 an r j(x i ) > 0. c j (x i )q jj (x i ) π j (x + i )r j(x + i ) + π j(x i )r j(x i ) + k j c k (x i )q kj (x i ) = 0. In this case we have only one equation that can be use to compute the probability mass c j (x i ) from the probability at the bounaries of the two pieces that are on the two sies of the iscontinuity x i. Note that we have an expression equal to 0 since q jj (x) is negative by efinition. Case ): isolating states r j (x + i ) > 0 an r j(x i ) < 0. 0 = c j (x i )q jj (x i ) + α j = (x i ) c k (x i )q kj (x i ), k j k j π j (x + i )r j(x + i ) = α> j (x i ) c k (x i )q kj (x i ), π j (x i )r j(x i ) = α< j (x i ) c k (x i )q kj (x i ). k j In this case we have three equations: one for the probability mass, an one for the probability ensity at each sie of the bounary. The probability mass coming from the other states is irecte along one of those three cases accoring to the flow rate at the iscontinuity point. 9
10 4.2 Bounary conition Bounary conition, at 0 an at the upper bounary B can be seen as special cases of iscontinuities. In this case only two ifferent cases per boun arises, epening on the irection of the flui flow. (In contrast with the cases of iscontinuities we use capital letters to enote the cases of bounaries.) Case A): absorbing states The first case is the one in which the flui flow is irecte towars the boun, that is r j (0 + ) < 0 (for the lower boun), or r j (B ) > 0 (for the upper boun). In this case, inepenently on the sign of the rate at the iscontinuity, probability mass buils up at the boun. In this case we can characterize the boun by a single equation: c j (0)q jj (0) π j (0 + )r j (0 + ) + k j c k (0)q kj (0) = 0, for the lower boun, an for the upper boun: c j (B)q jj (B) + π j (B )r j (B ) + k j c k (B)q kj (B) = 0. Note that the equation is almost ientical to the one for the absorbing state in the intermeiate iscontinuities. Case B): emitting states The secon case is the one in which the flui flow is irecte in the opposite irection with respect to the boun, that is r j (0 + ) > 0 (for the lower boun), or r j (B ) < 0 (for the upper boun). In this case we may have two ifferent behaviors epening on the sign of the rate at bounary. No probability mass will be forme if sign(r j (0)) = sign(r j (0 + )), that is α j > (0) = 1, an if sign(r j(b))sign(r j (B )), that is α j < (B) = 1. Probability mass will instea buil up if the sing of the rate at the bounary is zero or opposite to the flui flow next to the bounary. In any case we will have two equations per bounary, that are: 0 = c j (0)q jj (0) + α j (0) c k (0)q kj (0) k j π j (0 + )r j (0 + ) = α j > (0) c k (0)q kj (0), k j, for the lower boun, an for the upper boun: 0 = c j (B)q jj (B) + α j (B) c k (B)q kj (B) k j, 10
11 π j (B )r j (B ) = α j < (B) c k (B)q kj (B) k j. Note that also in this case the equations are basically ientical to the one for the emitting states in the intermeiate iscontinuities. 5 The number of equations an unknowns If the total number of pieces that composes functions Q(x) an R(x) is n, an the number of iscrete states of the moel is m (m = S ), then the total number of unknown is (3n + 1)m. In particular, each iscontinuity has associate a possible probability mass. Since the number of iscontinuities is n + 1, this means that there will be (n + 1)m unknowns: one for each iscontinuity an state. For each continuous segment, we will be intereste in the probability ensity at the beginning of the segment an the one at the en of the segment (2n unknowns). Multiplying this number by the number of states an aing the number of probability masses we obtain (3n + 1)m unknowns. Theorem 1. The number of equations presente in Section 3 an 4 is equivalent with the number of unknowns, i.e., there are (3n + 1)m equations to etermine the (3n + 1)m unknowns. To prove the theorem we nee the following lemma. Lemma 2. The number of equations (N e()) given by the arguments of Section 4 for the unknowns associate with state j an the initial an final ensity values of the first i (i < n) pieces (i.e., π j (0 + ), π j (x 1 ), π j (x + 1 ),..., π j (x i ), π j(x + i )) is Ne(i) = 2i + 2 if the sign of the rate of the last piece is positive (r j (x + i ) > 0), an Ne(i) = 2i + 1 if it is negative (r j (x + i ) < 0). Proof: By simply counting the equations for each case. For i = 0 it is true, since for the lower bounary we have one equation (Ne(0) = 1) if r j (0 + ) < 0 an two equations (Ne(1) = 2) if r j (0 + ) > 0. By counting the number of equations that are given in each case we have that: Number of equations per case. r j (x + i 1 ) < 0 r j(x + i 1 ) > 0 r j (x + i ) < 0 2 (Case b) 1 (Case c) r j (x + i ) > 0 3 (Case ) 2 (Case a) 11
12 Note that since r j (x) has a constant sign over a piece, this implies that sign(r j (x + i 1 )) = sign(r j(x i )). If we express the number of equations Ne(i+1) at piece i + 1 as a function of the number of equations at piece i, we have that: Ne(i) = Ne(i 1) + 1 = 2i + 1 = 2i + 1 if r j (x + i 1 ) > 0 an r j(x + i ) < 0, Ne(i) = Ne(i 1) + 2 = 2i = 2i + 1 if r j (x + i 1 ) < 0 an r j(x + i ) < 0, Ne(i) = Ne(i 1) + 2 = 2i + 2 = 2i + 2 if r j (x + i 1 ) > 0 an r j(x + i ) > 0, Ne(i) = Ne(i 1) + 3 = 2i = 2i + 2 if r j (x + i 1 ) < 0 an r j(x + i ) > 0. Proof of Theorem 1 : There are n m equations obtaine by the escription of the flui behavior over the continuous pieces (equation (3)), which relates the vectors {π j (x + i 1 )} with {π j(x i )}. We still nee to show that the bounary conitions presente in section 4 provies 2n + 1 further equations for every state. Using Lemma 2 we have that Ne(n 1) = 2n if r j (x + n 1) > 0 an Ne(n 1) = 2n 1 if r j (x + n 1) < 0. Consiering that the number of equations at the upper boun (B) is 1 if r j (x n ) > 0 (absorbing upper bounary) an it is 2 if r j (x n ) < 0 (emitting upper bounary) the total number of equations associate with state j is 2n Moifie system of equations Theorem 1 shows that the number of unknowns an the number of equations presente in Section 3 an 4 are equal. Unfortunately, the set of equations presente in Section 3 an 4 has only a trivial solution (c j (x i ) = π j (x i ) = π j (x + i ) = 0) in general cases3. To overcome this ifficulty we introuce a moifie system of equations to escribe the same flui system. We efine the probability flux ϕ j (x) an the probability mass flux j (x) as ϕ j (x) = π j (x)r j (x), an j (x) = c j (x)q jj (x), an to eal with the probability mass flux we efine matrix A(x) = {a jk (x)} as q jk (x) if j k, a jk (x) = q jj (x) (5) 0 if j = k. 3 Actually, this fact resulte in the major ifficulty in our way to fin the stationary solution of general flui systems. This problem is not present in flui moels with single piece, that is why it is not mentione in previous stuies of the subject. 12
13 6.1 Continuous pieces Base on equation (1) the probability flux satisfies ( ) x ϕ(x) = ϕ(x) R 1 (x)q(x). (6) We can use this equation to relate the probability flux at the beginning of piece i, ϕ(x + i 1 ), to the probability flux at the en of the same piece, ϕ(x i). In particular, we efine the flux transition matrix V(x + i 1, x i ) such that: ϕ(x i ) = ϕ(x+ i 1 ) V(x+ i 1, x i ). (7) By inserting equation (7) into equation (6), we can etermine the flux transition matrix by solving the following matrix ifferential equation up to x = x i : ( ) x V(x+ i 1, x) = V(x+ i 1, x) R 1 (x)q(x), (8) with initial conition V(x + i 1, x+ i 1 ) = I. In the special case when Q(x) an R(x) are level inepenent in the (x + i 1, x i ) range, i.e., for x (x+ i 1, x i ), Q(x) = Q i an R(x) = R i then 6.2 Discontinuities V(x + i 1, x) = e(x x i 1)R 1 i Q i. Case a): positive emitting states r j (x + i ) > 0 an r j(x i ) > 0. ( ) j (x i ) = α j (x i) ϕ j (x + i ) = α> j (x i ) ( ϕ j (x i ) + k ϕ j (x i ) + k k (x i )a kj (x i ) k (x i )a kj (x i ) ), (9). (10) Case b): negative emitting states r j (x + i ) < 0 an r j(x i ) < 0. ( j (x i ) = α j (x i) ϕ j (x + i ) + ) k (x i )a kj (x i ), (11) k ( ϕ j (x i ) = α< j (x i ) ϕ j (x + i ) + ) k (x i )a kj (x i ). (12) k 13
14 Case c): absorbing states r j (x + i ) < 0 an r j(x i ) > 0. j (x i ) = ϕ j (x + i ) + ϕ j(x i ) + k k (x i )a kj (x i ). (13) Case ): isolating states r j (x + i ) > 0 an r j(x i ) < 0. j (x i ) = α j = (x i ) k (x i )a kj (x i ), (14) k ϕ j (x + i ) = α> j (x i ) k (x i )a kj (x i ), (15) k ϕ j (x i ) = α< j (x i ) k (x i )a kj (x i ). (16) 6.3 Bounary conitions Case A): absorbing states r j (0 + ) < 0 (for the lower boun), or r j (B ) > 0 (for the upper boun). k j (0) = ϕ j (0 + ) + k j (B) = ϕ j (B ) + k k (0)a kj (0), (17) k (B)a kj (B). (18) Case B): emitting states r j (0 + ) > 0 (for the lower boun), or r j (B ) < 0 (for the upper boun). j (0) = α j (0) k (0)a kj (0) k ϕ j (0 + ) = α j > (0) k (0)a kj (0) k, (19), (20) j (B) = α j (B) k (B)a kj (B) k ϕ j (B ) = α j < (B) k (B)a kj (B) k, (21). (22) 14
15 6.4 Properties of W(x i, x) an V(x i, x) Accoring to the basic rules of the infinitesimal generator of CTMCs the rowsum of Q(x) ( k S Q jk(x)) must be 0 for x 0, j S. One of the main ifferences between (2) an (6) comes from the fact that Q(x) is multiplie with R 1 (x) from the right in (2), an from the left in (6). The matrix multiplication of Q(x) with the R 1 (x) iagonal matrix from the right multiplies the columns of Q(x) with the associate rift values. This transformation moifies the rowsum. In contrast, the multiplication of Q(x) with R 1 (x) from the left multiplies the rows of Q(x) with the associate rift values, hence the rowsum of the prouct remains 0 for all x. We emphasize an important consequence of this property in the following theorem. Theorem 3. The aggregate probability flux, j S ϕ j(x), remains constant in each continuous intervals, (x + i 1, x i ), an the rowsum of matrix V(x+ i 1, x i ) is 1. Proof: Let 1 be the column vector of ones. The rowsum of matrix R 1 (x)q(x) is 0, hence x ϕ(x) 1 = ϕ(x) R 1 (x) Q(x) 1 = 0, }{{} 0 i.e., the ifferential quation (6) preserves the aggregate probability flux in each infinitesimal intervals in (x + i 1, x i ) an so in the whole continuous interval (x + i 1, x i ). The theorem on the rowsum of matrix V(x + i 1, x i ) can be prove by contraiction. Since it has alreay been prove that the probability flux is constant over a piece, let us consier ϕ(x + i 1 ) 1 = ϕ(x i ) 1 = c. If V(x + i 1, x i ) 1 1, then we woul have: c = ϕ(x i ) 1 = ϕ(x+ i 1 )V(x+ i 1, x i ) 1 ϕ(x+ i 1 ) 1 = c. 7 Set of equations In this section we compose a matrix representation of the set of linear equation escribing the behaviour of the flui system. Basically, we collect the constant coefficients presente in the scalar equations above in a matrix T such that the row vector of the unknowns ψ is the solution of the equation ψt = 0. We procee in two steps. First we represent equations (9)(22) an then equation (7). Figure 4 shows the general structure of the coefficient matrices efine by equations (9)(22), an of vector ψ. 15
16 Figure 4: General structure of the coefficient matrix Consiering the restriction that the flui rate is nonzero between iscontinuities we have = 12 ifferent cases with respect to the sign of the flui rate on the left (r j (x i )), on the right (r j(x + i )) an at the ith iscontinuity (r j (x i )). The sign of the rate at the iscontinuities (r j (x i )) are represente by the α (x i ) function in the above equations. The ifferent possible signs of the flui rate on the left (r j (x i )) an on the right (r j(x + i )) of the iscontinuity are consiere in cases a), b), c) an ). At iscontinuity x i we ivie the set of states into the following isjoint subsets: S a (i), S (i) b, S c (i) an S (i) accoring to cases a) to ). I.e., j S a (i) if r j (x i ) > 0 an r j (x + i ) > 0; j S(i) b if r j (x i ) < 0 an r j(x + i ) < 0; j S(i) c if r j (x i ) > 0 an r j (x + i ) < 0; j S(i) if r j (x i ) < 0 an r j(x + i ) > 0. Accoring to this ivision of states at iscontinuity x i matrix A(x i ) an V(x + i 1, x i ) are subivie as follows A aa A ab A ac A a V aa V ab V ac V a A = A ba A bb A bc A b A ca A cb A cc A c, V = V ba V bb V bc V b V ca V cb V cc V c, A a A b A c A V a V b V c V 16
17 where the epenence on the particular iscontinuity is neglecte for notational simplicity. Using this ivision of states the structure of the coefficient matrix at iscontinuity x i epens on the sign of the rate at the iscontinuity (r j (x i )) as it is summarize in Figure 5. In the figure, matrix 0 uu inicates that equations (9)(22) o not efine the associate quantity an matrix I uu is the unity matrix of size S u for u {a, b, c, }. Matrix A uu enotes A uu = I uu + A uu (for u {a, b, c, }). if r j (x i ) > 0 if r j (x i ) < 0 if r j (x i ) = 0 Figure 5: Coefficient matrix at x i We place the coefficients provie by equation (7) into the ile columns of matrix T, i.e., into the columns of the 0 uu matrices The S a (i), S (i) b, S(i) c, S (i) partition of S is associate with iscontinuity x i. Due to the efini 17
18 tion of the iscontinuities the rift (r j (x)) can not change sign in the continuous (x + i 1, x i ) interval, hence (as pointe out before) sign(r j(x i )) = sign(r j (x + i 1 )). Base on this property the partitioning at x i 1 (S u (i 1), u {a, b, c, }) is such that j S a (i) j S c (i) j S (i 1) a j S (i 1) a (i 1) S, j S (i) b (i 1) S, j S (i) j S (i 1) b j S (i 1) b (i 1) S (i 1) S c, c, i.e., S a (i) (i) S c = S a (i 1) (i 1) S an S (i) (i) b S = S (i 1) (i 1) b S c. Figure 6 shows a part of matrix T with the items obtaine from equations (9)(22) an with the iscontinuity epenent partitioning of the states. u = S u (i 1) (u {a, b, c, }) enotes the state partitioning accoring to the sign of the rift aroun x i 1 an matrices I uv, V uv (u, v {a, b, c, }) are obtaine from matrices I uv, V uv with a reorering of states accoring to the state partitioning at x i 1. Figure 6: A block of matrix T with iscontinuity epenent state partitioning Finally, Figure 7 introuces the final structure of the T matrix at level x i. The figure contains all nonzero entries associate with level x i. The state partitioning at level x i 1 is enote with u (as before) an at level x i+1 is enote with u (u {a, b, c, }). Matrices I uv an V uv are efine accoring to the relevant state partitioning. The matrix structure in Figure 7 alreay suggests the following essential property of matrix T. Theorem 4. The set of equations (9)(22) an (7) is linearly epenent, an the ψt = 0 equation has nontrivial solution. 18
19 Figure 7: The final structure of the T matrix at x i (with r j (x i ) < 0) Proof: The rowsum of matrices A(x i ) an matrices V(x + i 1, x i ) (i {1,..., n}) is 1 base on eq. (5) an Theorem 3. By the escribe construction of matrix T, the rowsum of T is 0 in all above mentione possible cases. The theorem is a irect consequence of the rowsum of matrix T. Note that the introuction of the probability flux (ϕ(x)) instea of the flui ensity (π(x)) results in the use of matrix V(x + i 1, x i ) instea of matrix W(x + i 1, x i ), an the introuction of the probability mass flux ((x)) instea of the probability mass (c(x)) results in the use of matrix A(x i ) instea of matrix Q(x i ). We have to remark however, that Theorem 4, oes not ensure that the imension of the solution space of ψt = 0 is equal to 1, an thus that ψ can be uniquely etermine using some normalizing conition. However, if the system is irreucible an has a unique stationary istribution inepenent of its initial state, this can be etermine using the propose proceure. 8 Extension to states with zero rate In the previous sections we have always consiere r j (x) 0 for all the continuous piece. In many practical situation however, there are cases where r j (x) = 0 for a piece x i < x < x i+1. When there are this kin of zero states, both the continuous part an the bounary conitions changes accoringly. 19
20 8.1 Consiering zero states between the iscontinuities Let us enote with π 0 (x) the probability ensity of the states j where r j (x) = 0 (zero states), an with π (x) the probability of the other states (nonzero states). We can rewrite equation (1) as: ( ) π (x)r (x) x = π (x)q (x) + π 0 (x)q 0 (x), (23) 0 = π (x)q 0 (x) + π 0 (x)q 00 (x), where Q (x), Q 0 (x), Q 0 (x), Q 00 (x) an R (x) refers to the submatrices of Q(x) an R(x) that have elements corresponing to the zero or nonzero states accoring to their superscript. From equation (23) we obtain: π 0 (x) = π (x)q 0 (x)q 00 1 (x), (24) ( ) [ ] π (x)r (x) = π (x) Q (x) Q 0 (x)q 00 1 (x)q 0 (x). x Note that Q(x) = Q (x) Q 0 (x)q 00 1 (x)q 0 (x), is the generator of the iscrete state process restricte to the nonzero states at flui level x. Let R(x) = R (x) an by changing π(x) to ϕ(x) we obtain: ( Q(x)) x ϕ (x) = ϕ (x) R 1 (x) The solution of equation (25) can also be expresse in the form. (25) ϕ (x i ) = ϕ (x + i 1 ) V(x + i 1, x i ). (26) In the special case when Q(x) an R(x) are level inepenent in the (x i 1, x i ) range, i.e., for x (x i 1, x i ), Q(x) = Q i an R(x) = R i then V(x i 1, x) = e (x x 1 i 1) R i We can compute the actual solution of π(x) from ϕ (x) using the following equations: Q i. π (x) = ϕ (x) R 1 (x), (27) π 0 (x) = ϕ (x) R 1 (x)q 0 (x)q 00 1 (x). 20
21 8.2 Zero states at the iscontinuities When one of the pieces aroun iscontinuity x i might have a zero state the set of cases consiere in section 6 nees to be extene with the following ones: Discontinuities Case e): r j (x + i ) = 0 an r j(x i ) < 0. j (x i ) = α j (x i) k (x i )a kj (x i ), (28) k ϕ j (x + i ) = 0, (29) ϕ j (x i ) = α< j (x i ) k (x i )a kj (x i ). (30) Case f): r j (x + i ) = 0 an r j(x i ) > 0. k j (x i ) = ϕ j (x i ) + k k (x i )a kj (x i ), (31) ϕ j (x + i ) = 0. (32) Case g): r j (x + i ) < 0 an r j(x i ) = 0. j (x i ) = ϕ j (x i ) + k k (x i )a kj (x i ), (33) ϕ j (x i ) = 0. (34) Case h): r j (x + i ) > 0 an r j(x i ) = 0. j (x i ) = α j (x i) k (x i )a kj (x i ), (35) k ϕ j (x + i ) = α> j (x i ) k (x i )a kj (x i ), (36) ϕ j (x i ) = 0. (37) Case i): r j (x + i ) = 0 an r j(x i ) = 0. k j (x i ) = k k (x i )a kj (x i ), (38) ϕ j (x + i ) = 0, (39) ϕ j (x i ) = 0. (40) 21
22 Bounary conitions Case C): r j (0 + ) = 0. j (0) = k k (0)a kj (0), (41) Case D): r j (B ) = 0. ϕ j (0 + ) = 0. (42) j (B) = k k (B)a kj (B), (43) ϕ j (B ) = 0. (44) These equations can be obtaine as special cases of the equations in section 6 by omitting the flux towars the piece with zero state. 8.3 Number of equations Theorem 5. The number of equations presente in this section is equivalent with the number of unknowns, i.e., (3n + 1)m. To prove the theorem we nee the following notations an lemma. Let m 0 i be the number of zero states in piece i. In piece i the matrix equation (26) represents m m 0 i equations. We associate each of these equations with a ifferent nonzero state of piece i. Let Ne j (i) be the number of equations available for etermining the unknowns associate with state j at the iscontinuities x 0,..., x i. These unknowns are j (0), ϕ j (0 + ), ϕ j (x 1 ), j (x 1 ), ϕ j (x + 1 ),..., ϕ j (x i ), j(x i ), ϕ j (x + i ). This way Ne j (i) contains the bounary an iscontinuity equations associate with pieces 0,..., i an one equation of (26) for all pieces on the left of x i where j is a nonzero state. Lemma 6. For 0 i < the number of equations are Ne j (i) = 3i + 2, if r j (x + i ) > 0, Ne j(i) = 3i+2, if r j (x + i ) = 0 an Ne j(i) = 3i+1, if r j (x + i ) < 0. Proof of Lemma 6: For i = 0 the lemma is true. The number of equations at iscontinuity x i is: r j (x + i r j (x + i r j (x + i Number of equations per case. r j (x + i 1 ) < 0 r j(x + i 1 ) = 0 r j(x + i 1 ) > 0 ) < 0 2 (Case b) 2 (Case g) 1 (Case c) ) = 0 3 (Case e) 3 (Case i) 2 (Case f) ) > 0 3 (Case ) 3 (Case h) 2 (Case a) 22
23 Assuming the lemma vali for iscontinuity x i 1, we have: Ne j (i) = Ne j (i 1) = 3i + 1 if r j (x + i 1 ) < 0 an r j(x + i ) < 0, Ne j (i) = Ne j (i 1) = 3i + 2 if r j (x + i 1 ) < 0 an r j(x + i ) = 0, Ne j (i) = Ne j (i 1) = 3i + 2 if r j (x + i 1 ) < 0 an r j(x + i ) > 0, Ne j (i) = Ne j (i 1) + 2 = 3i + 1 if r j (x + i 1 ) = 0 an r j(x + i ) < 0, Ne j (i) = Ne j (i 1) + 3 = 3i + 2 if r j (x + i 1 ) = 0 an r j(x + i ) = 0, Ne j (i) = Ne j (i 1) + 3 = 3i + 2 if r j (x + i 1 ) = 0 an r j(x + i ) > 0, Ne j (i) = Ne j (i 1) = 3i + 1 if r j (x + i 1 ) > 0 an r j(x + i ) < 0, Ne j (i) = Ne j (i 1) = 3i + 2 if r j (x + i 1 ) > 0 an r j(x + i ) = 0, Ne j (i) = Ne j (i 1) = 3i + 2 if r j (x + i 1 ) > 0 an r j(x + i ) > 0. where the number of aitive equations escribing iscontinuity x i an the relation of the flux of nonzero states at the beginning an the en of piece i (equation (26)) are treate separately. Proof of Theorem 5 : Applying Lemma 6 for the last but one iscontinuity (x n 1 ) an aing the bounary equations at B, we have Ne j (n) = Ne j (n 1) = 3n + 1 if r j (x + n 1) < 0, Ne j (n) = Ne j (n 1) + 2 = 3n + 1 if r j (x + n 1) = 0, Ne j (n) = Ne j (n 1) = 3n + 1 if r j (x + n 1) > Set of linear equations Using equation (26) an the iscontinuity an bounary equations presente in this section we can compose a similar system of equations as before (ψt = 0), but the structure of the ψ an the T matrix is much more complex since we nee to introuce 3 subsets of states (with positive, negative an zero rift) which results in 9 cases (a to i) to consier at each iscontinuity, in contrast with the 4 cases (a, b, c, ) iscusse in Section 7. Without further explanation we conclue that the main equation set obtaine with the presence of zero states remains solvable in a similar way. We only emonstrate the applicability of the approach for the case with zero states via our implementation which automatically generates the T matrix an solves the linear system also in this case. 9 Normalizing conition The normalizing conition is obtaine from the flui istribution as follows 23
24 1 = j S = j S n c j (x i ) + n j S i=1 n j (x i ) q jj (x i ) + j S i=0 i=0 xi π j (x)x, x i 1 (45) n xi ϕ j (x) x. x i 1 r j (x) (46) Unfortunately, the normalizing conition can not be evaluate base on the V(x i 1, x i ) matrices, but it also requires the evaluation of the xi x i 1 π j (x)x integral. When some of the states have associate zero rift in a continuous piece (6) is no longer vali an the integrals of the probability ensities are calculate base on (24) instea of (1). The normalizing conition can also be expresse using a normalizing vector η. In this case the normalizing conition becomes: ψη T = 1. Vector η can be constructe from the efinition of ψ an equations (45) an (46). In particular, from (26) an (27) we may compute for each pieces (x i 1, x i ): xi xi π (x)x = ϕ (x + i 1 ) + V(x i 1, x) R 1 (x)x, x i 1 x i 1 xi x i 1 π 0 (x)x = ϕ (x + i 1 ) Introucing we have j S U i = U 0 i = xi xi xi i=1 x i 1 V(x + i 1, x) R 1 (x)q 0 (x)q 00 1 (x)x. + V(x i 1, x) R 1 (x)x, x i 1 xi x i 1 π j (x)x = Using the row vectors η (ϕ) i x i 1 V(x + i 1, x) R 1 (x)q 0 (x)q 00 1 (x)x, xi x i 1 π (x)x 1 + xi = ϕ (x + i 1 ) [ U i 1 + U 0 i 1 ]. = ( U i 1 + U 0 i 1 ) T, an η () i = we can buil the normalizing vector η as follows: x i 1 π 0 (x)x 1 = { } 1, j S, q jj (x i ) η = [ η () 0, η (ϕ) 1, 0 0 () 1, 0, η }{{}}{{}}{{} 1, η (ϕ) 2, 0 0 () 2, 0,..., η }{{}}{{}}{{} n ], (47) }{{} (0) ϕ(0 + ) ϕ(x 1 ) (x 1 ) ϕ(x + 1 ) ϕ(x 2 ) (B) 24
25 where 0 0 i an 0 are the row vectors with as many zero components as the number of zero states in piece i, an the number of states, respectively. The uner braces inicate the ϕ vector elements associate with the particular η vector elements. Note that the nonzero η elements are associate with the iscontinuities (η () i ) an the states of nonzero rate at the lower bounary of continuous pieces (η (ϕ) i ). 10 Numerical example To emonstrate the solution metho introuce in the previous sections we evaluate various versions of the highspee network moel presente in [4]. Source K Source 1 feeback ata Buffer with level epenent loss mechanism ata loss feeback channel Figure 8: The multiplexer moel The moel is compose by K ientical 2 state sources an a buffer (see Figure 8). The sources generate ata with J (1,..., J) ifferent priorities. The traffic behaviour of [ each source is governe ] by a 2state Markov chain α(x) α(x) with generator G(x) =, where x, the flui level, represents the queue length in the buffer. When the Markov chain of a source is in β(x) β(x) state i at queue length x the source generates priority j ata at rate λ (j) i (x). To reuce the probability of buffer overflow the buffer rops lower priority ata, when the buffer level is high. The B 1, B 2,..., B J 1 levels etermine the ropping mechanism. When the buffer content is higher than B j the buffer rops all incoming priority j ata. In state i an flui level x the actual rate of a source is λ i (x) = (x). λ (j) i j:x>b j Case I: When the source behaviour (G(x) an λ (j) i (x)) is inepenent of 25
26 [ α α the buffer level, i.e., G(x) = β β ] an λ (j) i (x) = λ (j) i, the obtaine moel is ientical with the one in [4], hence we can compare the results obtaine there with the results of our solution metho. The basic set of moel parameters is the following: Moel parameters K 20 number of sources α 0.4 transition rate (OFF ON) β 1.0 transition rate (ON OFF) C channel capacity J 2 number of priority levels λ (1) 1 = λ (2) 1 0 ata generation in OFF state λ (1) low priority ata generation λ (2) high priority ata generation B 1.5 buffer size B threshol of ropping low priority ata B 2 B threshol of ropping high priority ata In the following cases we inicate only the moifie values with respect to this basic set of parameters. Figure 9 epicts the joint istribution of the buffer content an the system state when the parameters are taken from the basic parameter set B1=0.4 B1=0.6 B1=0.8 B1=1.0 B1= B1=0.4 B1=0.6 B1=0.8 B1=1.0 B1= a) 8 active users b) 10 active users Figure 9: Buffer content istribution with 8 an 10 active users (Case I) To compare the results with the one presente in [4], we also evaluate the loss probability of the low priority ata. Since the exact loss probability 26
27 values are not given in [4] we only present figure 10, which is suppose to be ientical with figure 4a of [4] an leave the verification for the reaer Lp State Figure 10: Loss probability of priority 1 ata Case II: Increasing the number of sources to K = 30, changing B = 1, C = an varying B 1 from 0.3 to 0.7 results the istribution presente in Figure B1=0.3 B1=0.4 B1=0.5 B1=0.6 B1= B1=0.3 B1=0.4 B1=0.5 B1=0.6 B1= e a) 12 active users b) overall istribution Figure 11: Buffer content istribution with 30 sources (Case II) Case I an II can also be evaluate with the metho presente in [4]. The next case instea can no longer be solve using the approach propose in [4]. Case III: In the following we assume a more sophisticate system behaviour. The sources aapt their behaviour to the actual ropping of the buffer, which becomes known for the sources via feeback signals (e.g., missing acknowlegements). There are two ways to ajust the source behaviour. 27
28 The source might change the ata generation rate an also the governing Markov chain. The proceure presente in the paper allows to consier both of these ajustments. First we { make the governing Markov chain buffer level 1.0 if x < B1 epenent such that α(x) = a 2 if B 1 < x < B, where a 2 varies between an 1. The rest of the moifie moel parameters with respect to the basic set are B = 1, B 1 = 0.7. Figure 12 provies the buffer content istribution in this case a2=1.0 a2=0.5 a2=0.2 a2= a2=1.0 a2=0.5 a2=0.2 a2= a) 8 active users b) overall istribution Figure 12: Buffer content istribution with aaptive source behaviour (Case III.) Case IV: Finally, we consier a moel behaviour which contains isolating state as well. (Isolating state is not allowe in [4].) In this version of the moel the ata accumulation rate epens on the buffer level, such that the{ channel capacity rops own when the buffer is heavily loae, if x < B1 C(x) =, where s2 varies between 10% an s if B 1 < x < B 100%. This may happen, for example, when the system rops packets at high buffer level. Similar to Case III, B = 1 an B 1 = 0.7. The istributions are given in Figure 13. Figure 13a reflects the expectations base on the physical unerstaning of the moel. The lower is the system capacity at high buffer levels the higher is the buffer content. (Curves closer to the horizontal axes means higher average buffer level.) The propose numerical proceure is compose by 3 main sets: calculation of V(x i, x i+1 ) matrices (eq. (8)), solving the linear system of equations ϕt = 0, 28
29 2.5e006 2e006 s2=100% s2=95% s2=90% s2=85% s2=75% s2=60% s2=50% s2=25% s2=10% s2=100% s2=95% s2=90% s2=85% s2=75% s2=60% s2=50% s2=25% s2=10% 1.5e e e a) 4 active users b) overall istribution Figure 13: Buffer content istribution with varying channel capacity (Case IV.) normalizing the solution (eq. (46)). The computational complexity of these steps are characterize by the following moel parameters. The carinality of matrix V(x i, x i+1 ) is m m, while the carinality of matrix T is m(3n + 1) m(3n + 1). Matrices V(x i, x i+1 ), 0 i < n are complete matrices an shoul be store in complete matrix form, while matrix T has a regular block structure which can be exploite to reuce memory usage an spee up computation. Equation (8) is solve n times. The possible computational methos for solving (8) are etaile in [6] an [5] together with their computational complexity. We applie Gauss elimination for solving the linear system. For all presente examples, the computation of the (x) an ϕ(x) vectors took less than a minute with our C implementation on regular PC. Accoring to our experiences the main limitation of the propose metho is not the complexity of the computation, but its numerical stability. Depening on the moel parameters the elements of the V(x i, x i+1 ) matrices might become extremely large (as it is iscusse in [5]) an the linear system becomes ill conitione. In some cases, the ivision of continuous pieces by the introuction of artificial iscontinuity points helps to improve the numerical properties (because it ecreases the large values in the V(x i, x i+1 ) matrices associate with the ivie continuous pieces). We observe that the elements of the V(x i, x i+1 ) matrices shoul be less than for all continuous pieces to obtain a reasonable solution with our implementation. 29
30 11 Conclusion The stationary solution of boune flui moels with flui level epenent rift an transition matrix is consiere in this paper. The set of equation to characterize the unknown quantities of the stationary istribution has a nonlinear epenence on the applie system measures. The commonly applie system measures, the flui ensity an the flui mass functions, result in overetermine system of equations in case of more than one continuous pieces. Due to the nonlinear epenence of the equation system on the applie system measures we coul prove that a given linear transformation of the system measures results in a solvable set of equations. We also implemente the computational metho base on the transforme system measures an provie numerical examples using the propose analysis metho. References [1] D. Anick, D. Mitra, an M. M. Sonhi. Stochastic theory of a atahanling system. Bell System Thechnical Journal, 61(8): , Oct [2] S. Asmussen. Stationary istributions for flui flow moels with or without brownian noise. Stochastic Moels, 11(4):21 49, [3] D.Y. Chen, Y. Hong, an K. S. Trivei. Secon orer stochastic flui flow moels with flui epenent flow rates. Performance Evaluation, 49(14): , [4] A. I. Elwali an D. Mitra. Statistical multiplexing with loss priorities in ratebase congestion control of highspee networks. IEEE Transaction on communications, 42(11): , Nov [5] R. German, M. Gribauo, G. Horváth, an M. Telek. Stationary analysis of FSPNs with mutually epenent iscrete an continuous parts. In International Conference on Petri Net Performance Moels PNPM 2003, pages 30 39, Urbana, IL, USA, Sep IEEE CS Press. [6] M. Gribauo an R. German. Numerical solution of boune flui moels using matrix exponentiation. In Proc. 11th GI/ITG Conference on Measuring, Moelling an Evaluation of Computer an Communication Systems (MMB), Aachen, Germany, Sep VDE Verlag. 30
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