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Theory Queueing Original Material Prepared by: Professor James S. Meditch Lecturer: Biswanath Mukherjee Typesetter: Dr. Anpeng Huang A. Notation and terminology 1 cn = customer n n = arrival time of cn tn = n n 1 = interarrival time ~ t wn = waiting time for cn xn = service time for cn sn = system time for cn sn = wn + xn ~ w ~ x ~ s Arrival process Service process ~ A(t ) P[ t t ] dA(t ) a (t ) = dt 1 E[ ~ ] = t = t E[( ~ ) k ] = t k t B( x) P[ ~ x] x dB( x) b( x ) = dx 1 E[ ~ ] = x = x E[( ~ ) k ] = x k x 2 Laplace transform/moment generating fn. A ( s ) = a (t )e dt = E[e * st 0 ~ st ] E[ f (t )] B * ( s ) = E[e sx ] A *( k ) ~ d k A* ( s ) kk ( 0) s = 0 = ( 1) t k ds queueing system performance System defined by t and ~ ~ x Performance defined by N(t) = no. of customers in system at time t wn and sn t , N (t ) N (statistical equilibrium, stationarity) 3 N FN (k ) = P[ N k ] Pk = P[ N = k ] E[ N ] = N E[ z N ] = Q( z ) = Pk z k =0 k w ~ W ( y ) = P[ w y ] w( y ) = dW ( y ) dy ~ E[ w] = W E[e sw ] = W * ( s ) ~ s S ( y ) = P[~ y ] s s( y) = dS ( y ) dy E[~ ] = T s E[e ss ] = S * ( s ) ~ Other performance variables: I D G Nq = = = = idle period interdeparture time busy period no. of customers in queue 4 Notation for queueing systems A/B/m no. of servers M Er Hr exponential(Markovian) r-stage Erlangian R-stage hyperexponential D G deterministic general A/B/m/K/M B. General results 1. Utilization factor avg. rate at which work arrives R capacity of the system to do work C = fraction of system capacity in use (on the avg.) 0 <1 5 G/G/1 R = x C =1 = x = avg. no. arrivals / sec = avg. rate of service / sec G/G/m R = x C = m servers = = x m = avg. fraction of busy servers 1 , = avg. service time for each server m Stability requires 0 < 1 (except for D/D/m where 0 < 1) 6 2. System time ~=~+w ~ s x ~ E [~ ] = E [~ ] + E [w ] s x T = x +W 3. Little s result N = T N ,T N q = W (t ) = no. arrivals in (0, t) (t ) = no. departures in (0, t) N(t) = (t) - (t) 0 7 (t ) total time all customers have spent in the system up to time t Tt system time per customer averaged over all customers who arrived during (0, t) N t avg. no. of customers in the system during the interval (0, t) (t ) (t ) customers t t 0 sec (avg. arrival rate) (t ) = N ( s )ds customer - sec (t ) system time (sec/customers) (t ) customers (t ) (t ) avg. no. customers in system over (0, t) Nt = t (t ) N t = tTt Tt = 8 If = lim t and T = lim Tt exist, then N q = W N = Ns + Nq N s = x (= ) and we also get T = x + W N = T N q = N m , t t G /G /m : = m C. Poisson process independent Interarrival times : ti exponentially distributed, 1 ~ P[ t t ] = 1 e t , t 0, t = Pk (t ) = ( t ) t e , k = 0,1,K k! = Pr {k arrivals in ( 0,t]} k 9 1. Derivation Pr[1 arrival in t ] = t + O( t) Pr[0 arrival in t ] = 1 Pr[1 arrival in t ] = 1 t + O( t) process intensity (arrival rate) Pk (t + t ) = Pk (t )[1 t ] + Pk 1 (t ) t + O( t ) Pk (t + t ) Pk (t ) = Pk (t ) t + Pk 1 (t ) t + O( t ) dPk (t ) = Pk (t ) + Pk 1 (t ) dt k = 1,2, K P0 (t + t ) = P0 (t )[1 t ] + O( t ) dP0 (t ) = P0 (t ) dt P0 (0) = 1 10 2. Properties (t>=0) i. N (t ) = kPk (t ) = t k =0 = avg. arrival rate ii. iii. 2 N (t ) = t k =0 Q(z,t) = Pk (t ) z k = E[ z N (t ) ] = e t ( z 1) dQ ( z , t ) dz = t e t ( z 1) z =1 z =1 = t iv. A(t) = P[~ t ] = 1 e t t a (t ) = e t Interarrival times are exponentia lly distribute d 1 1 t= 2 = 2 D. Markov chains Chain: State space is discrete Ex. S = {0, 1, . . ., K} S = {0, 1, . . .} finite state infinite state 11 1. Discrete-time Markov chains Def . P[ X n = j X n 1 = in 1 ,L , X 1 = i1 , X 0 = i0 ] = P[ X n = j X n 1 = in 1 ] State probabilities i( n ) = P[ X n = i ] (n) i (n) i =1 = [ (n) 0 (n) 1 L] (One-step) transition probabilities Pij( n 1) = P[ X n = j X n 1 = i ] P ( n 1) = [ Pij( n 1) ] P j (n) ij = 1 for each i and all n 12 State equations ( n ) = ( n 1) P (n 1) n = 1,2,L ( 0 ) given Homogeneous chain Pij( n ) = Pij , independent of time n Pij = P[ X n = j X n 1 = i ] P = [ Pij ] P j ij = 1 for each i ( n ) = ( n 1) P ( n) = (0) P n ( 0) given n = 1,2, L i (n) i =1 If lim ( n ) = exists, then n = P equilibrium probabilities (" steady state" ) i i =1 13 Proof of (n) = (n-1) P Proof- Contd. (n) i = ( n 1) 0 P0 i + ( n 1) s ( n 1) 1 P1 i + ( n 1) 2 P2 i + + ...+ P si + . . . = P0 i P 1i P 2( n 1 ) . . . s( n 1 ) . . . ] 2 i ... P si i-th column of P ... = [ ( n 1) 0 ( n 1) 1 14 Example Discrete-time birth-death (arrival-departure) process with no waiting room. P(1 arrival at any time n) = a P(1 departure at any time n) = d s = {0,1} = [ 0 1] Example Contd. 1 a a P= d 1 d ( n ) = ( n 1) P n = P 0 = (1 a) 0 + d 1 1 = a 0 + (1 d ) 1 15 Example Contd. d 1 = a 0 1 = a 0 d 0 + 1 = 1 (1 + ) 0 = 1 0 = d a+d a d a 1 = a+d 2. Continuous-time Markov chains X (t) S, S = {0, 1, ..., N}, N 16 Contd. Def . For all n and any 0 t1 < t2 < ... < tn < tn+1 P[ X (tn+1 ) = j X (tn ) = in , ..., X (t1 ) = i1 ] = P[ X (tn+1 ) = j X (tn ) = in ] i , j S Th. = time in state Ei S is a r.v. with P[ i t ] = 1 e it a. General Case Transition probabilities pij ( s, t ) = P[ X (t ) = j X ( s ) = i ] P (t ) = [ pij (t , t + t )] Q (t ) = lim P(t ) I = transition rate matrix t t 0 p (t , t + t ) 1 qii (t ) = lim ii t t 0 pij (t , t + t ) i j qij (t ) = lim t t 0 qij (t ) = 0 for each i j t s If these limits do not exist, we do not have a continuous-time Markov chain 17 State Probabilities j (t ) = P[ X (t ) = j ] j S N (t ) = [ 0 (t ) 1 (t ) ... N (t )], d (t ) = (t ) Q (t ) dt (0) given j (t ) = 1 j (t ) = (0) exp[ Q( s )ds ] 0 t b. Homogenous case pij (t ) = pij ( s, s + t ) qij (t ) = qij = const. Q = [qij ] d (t ) = (t ) Q (t ) = (0) eQ t dt indep. of s 18 Homogenous case Contd. Steady state (if it exists) lim j (t ) = j independent of (0) : t Q=0 j =1 E. Continuous-time birth-death (arrival departure) processes (1) Continuous-time Markov chain dealing with a population of size N at time t Pk (t ) = P[ N (t ) = k ] S = {0, 1, ..., L} k S L (2) System state changes by at most one (up or down, or no change) in t 19 Contd. 3). Births and deaths independent 4). Transitions P[exactly 1 b in (t , t + t ) N (t ) = k ] = k t + O ( t ) P[exactly 1 d in (t , t + t ) N (t ) = k ] = k t + O( t ) P[exactly 0 b in (t , t + t ) N (t ) = k ] = 1 k t + O( t ) P[exactly 0 d in (t , t + t ) N (t ) = k ] = 1 k t + O ( t ) k = birth rate k = death rate Transitions - State Eqns. Pk (t + t ) = Pk (t ) pk ,k ( t ) + Pk 1 (t ) pk 1,k ( t ) + Pk +1 (t ) pk +1,k ( t ) k 1 P0 (t + t ) = P0 (t ) p00 ( t ) + P (t ) p10 ( t ) k = 0 1 P (t ) = 1 k =0 k L L 20 State Eqns. Contd. Pk (t + t ) = Pk (t )[1 k t + O( t )][1 k t + O( t )] + Pk 1 (t )[ k 1 t + O( t )] + Pk +1 (t )[ k +1 t + O( t )] P0 (t + t ) = P0 (t )[1 0 t + O( t )] + P (t )[ 1 t + O( t )] 1 State Eqns. Contd. dP (t) k = ( k + k )P (t) + k 1P 1(t) + k+1P +1(t) k k k dt k 1 dP (t) 0 = 0P (t) + 1P (t) 0 1 dt t 0 k =0 21 M/M/1 Queue k = arrival rate k = service rate Equilibrum behavior (t dPk ) =0 dt jobs / sec jobs / sec t , k = 0, 1, ... 1 . State dependent k and k ; Pk ( t ) p k 0 = ( k + k ) p k + k 1 p k 1 + k + 1 p k + 1 , 0 = 0 p 0 + 1 p1 k=0 k 1 k =0 pk = 1 p1 = 0 p 0 , p 2 = 0 1 p 0 , ... 1 1 2 k 1 i=0 pk = p0 p0 = i i +1 1 k 1 1+ k =1 i = 0 i i +1 22 2. Classical M/M/1 k = , k = , all k p k = p0 1 p0 = k 1+ k =1 k k 1 Classical M/M/1 Contd. 1 < 1, 1 + = = If k =1 k =0 1 = 0 <1 k k p0 = 1 pk = (1 ) k pk = (1 ) k k 0 23 N = k pk = (1 ) k k k =0 k =0 k k =0 k = = d d 1 k = d k =0 d 1 (1 ) 2 N= 1 k =0 2 N = ( k N ) 2 pk 2 N = (1 ) 2 Contd. 1 N = T T= 1 T = x +W = W= 1 +W = 1/ 1 / 1 =N 1 24 Contd. N q = W P[ N k ] = i 2 Nq = 1 p = (1 ) i=k i=k i P[ N k ] = k Above results are valid iff 0 < 1 State-Transition Rate Diagrams 1. Birth-death process Flow into Ek = k 1 Pk 1 (t ) + k +1 Pk +1 (t ) Notion of probability flow Flow out of Ek = ( k + k )Pk (t ) dP(t) k = Flow Ek - Flow of Ek into out dt = k-1P 1(t) + k+1P +1(t) ( k + k )P (t) k = 0,1,... k k k 25 Equilibrium Flow across boundary balanced dPk (t ) =0 Hence, Pk (t ) = Pk dt k Pk = k 1 Pk 1 Pk = k 1 Pk 1 k k = 1,2,... 2. M/M/1 ( k = , k = ) This leads to P1 = P0 2 Equilibrium P 2 = P = P0 1 M Pk = P0 subject to k P k = Pk 1 k = 1,2,... Pk = Pk 1 P k =0 k =1 26 1. M/M/m Queue Model = Equivalently, <1 m k = k 0 k m k m m Use state-dependent birth-death model to determine pk , k 0 p1 = 1 p0 p2 = p3 = pm = pm +1 = pm + 2 = p0 = 1 p1 = 2 2 p2 = 3 3 2 1 13 3! 2 p0 p0 p0 p0 1m m! 1 m 1 +1 k! m! 1 /m k =0 1 1 = m 1 (m ) k (m ) m 1 k! + m! 1 k =0 p0 = m 1 k 1 = 1 /m 1m +1 1m + 2 m!m 2 m!m p0 (mp) k p0 k! pk = m k m p0 m! 0 k m k m 27 2. M/M/m Queue system characteristics (mp) k p0 k pk = m ! k m p0 m! Where 0 k m k m p[m] = (m ) m p0 m!(1 ) N = m + 1 pm pm N s = m = p0 = 1 (m ) (m ) m 1 + k! m! 1 k =0 m 1 k Nq = 1 p[queueing ] = pk k =m pm T= N W= Nq x= Ns = 1 M/G/1 Queue Poisson arrival process: General service process: customers/sec B ( x), x, x k queueing discipline is FCFS 1. Pollaczek-Khinchin (P-K) relations x = E[ x] x 2 = E[ x 2 ] = x < 1 W= x2 T = x +W 2(1 ) N = + N = x + W 2 x 2 2(1 ) 28 Ex. = 0.4cust / sec x = 2 sec 1 13 x = x 2 dx = sec 2 21 3 2 3 = x = (0.4)(2) = 0.8 13 (0.4)( ) x 3 = 13 W= = 2(1 ) 2(0.2) 3 W = 4.33 sec 2 N = + W = 0.8 + 0.4( N = 2.53 13 ) 3 T = x + W = 2 + 4.33 T = 6.33 sec 2. P-K transform relations Q ( z ) = pk z , k k =0 a. b. pk = Z [Q( z )] 0 1 w( y ) = prob. density fn. of waiting time B * ( s ) = L[b( x)] = b( x)e sx dx Q ( z ) = B * ( z ) where B * ( z ) = B * ( s ) | s = z (1 )(1 z ) B * ( z ) z W * ( s ) = L[ w( y )] s (1 ) W * ( s) = s + B * ( s ) w( y ) = L 1[W * ( s )], s( y) = y 0 prob. density fn. of system time S ( s ) = L[ s ( y )] s (1 ) B* ( s) S * (s) = s + B * ( s ) s ( y ) = L 1[ S * ( s )] 29 S * ( s) = s (1 ) s + /( s + ) s + s (1 ) =2 s + s ( ) + (1 ) = s+ = s+ s ( y ) = ( )e ( ) y 3. Modeling a. Imbedded Markov Chain qn = vn = no. customers left behind by Cn xn no. customers which enter during q 1 + vn +1 qn > 0 qn +1 = n qn = 0 vn +1 {qn , n = 0,1,...} = Cont. time Markov chain b. Tagged job W = R + ( W ) x R = Residual work (residual life time) W = R + ( x)W = R + W (1 )W = R W = R /(1 ) 30 Residual Life - R Service process: x b( x ), x , x 2 z Random Arrival FZ ( x ) = P[ Z x| system busy at time of arrival] r = E[Z | system busy] Intuition: r = x 2 ? Wrong! Residual Life - Cont d Chances are higher that x arrives during longer periods f Z ( x) = density fn. for length of service period during which arrival occurs, given system busy f Z ( x) ( x) = Kx b(x) (x) 0 f Z (x) d(x) = K x b(x) dx = kx = 1, K = 0 1 x f Z (x) = r= x b(x) x 0 x2 x x b(x) d(x) = 2x 2 x 31 Residual Life - Cont d R = E[Z | system busy] E[system busy] x2 1 . x = x 2 2x 2 12 x x2 2 W= = (1 ) 2(1 ) = Markovian queueing Networks N-node interconnection of queueing systems queueing and service at each node Exponential service times at each node Model multi-service processes 1. Open Networks r22 r12 2 1 r11 1 r23 r21 r32 r31 3 r44 4 1 (r43 + r44 ) r43 r34 3 N nodes Each node single queue, mi servers Exponential service time 1 (r31 + r32 + r34 ) xi = 1 i sec 32 rij = P[that a job which completes service at node i will proceed next to node j ] N External arrivals Poisson process (indep.) i jobs/second 1 j =1 rij = P[that a job which completes service at node i will leave i =Avg. arrival rate at node i from both external sources ands other nodes (including itself) the network] r1i 1 i rii i i r2i 2 i = i + j = 1 j r ji N rNi N Flow Equations: = [ 1 2 ....... N ] = [ 1 2 ....... N ] R = [r ji ] = + R Jackson s theorem (1957) Let p(k 1 , k 2 ....., k N ) = p[k 1 jobs at node 1, k 2 jobs at node 2, ....., k N jobs at node N] and p i ( k i ) = p[k i jobs at node i] Then p(k 1 , k 2 ....., k N ) = p 1 ( k 1 ) p 2 ( k 2 )........... p N ( k N ) Each node in the network can be treated as an M / M / m i queue, m i 1 33 1 N-1 N 2 N = i=1 N i N = i=1 i N T= N Example: r11 r22 r12 2 1 r23 3 I/O CPU I/O r11 = 0.2 r22 = 0.4 r12 = 0.8 r23 = 0.6 1 1 = 01sec . 1 2 = 0.05 sec 1 3 = 0.9 sec = 1 job / sec 34 Node 1 1 = + r11 1 = 1 + 0.2 1 1 = 125 . 1 1 1 = = 0125 . N1 = = 01429 . 1 1 1 Node 2 2 = r12 1 + r22 2 = 0.8(125) + 0.4 2 2 = 16667 . . 2 2 = 2 = 0.0833 = 0.0909 N2 = 2 1 2 Node 2 3 = r22 2 = 0.6(16667) = 1 = . 3 3 = 3 = 0.9 N3 = = 0.9 3 1 3 N = N 1 + N 2 + N 3 = 9.2338 T= N = 9.23 sec 2. Closed Networks r22 2 r12 r21 1 r23 r32 r33 3 r24 r14 r42 r43 4 N nodes; K jobs circulating through the network No external arrivals or departures Each node single queue, m servers Exponential service time i xi = 1 i sec i = 0 N j=1 ij r = 1 for each i R = [rij ] N i=1 Ki = K = [ 1 2 .......... N ] (flow vector) = R 35 Gorjon and Newell (1967) p(k 1 , k 2 ....., k N ) = where (1) {x i } satisfy 1 N x ik i C G(K) i =1 i ( k i ) i x i = j =1 j x j r ji i = 1,2,....., N N (2) G(K) = k A C i =1 N x ik i i ( k i ) with k = (k 1 , k 2 ,......., k N ) and A = set of all k vectors for which k 1 + k 2 +.....+k N = k , k i ! (3) i ( k i ) = k i mi , m i ! m i Utilization at node i: i = k i mi k i mi xi i = 1,2, ..... , N mi Example: (Application-interactive computing) Node 1 Nodes 2,3, .N . . . . . K terminals . . . . k jobs Multi-Processor = T= N1 = avg. " think" time at each terminal (exp. distribution think times) K seconds T = avg. responce time 36

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