Queuing Networks: Burke s Theorem, Kleinrock s Approximation, and Jackson s Theorem Wade Trappe
Lecture Overview Network of Queues Introduction Queues in Tandem roduct Form Solutions Burke s Theorem What is reversibility? Kleinrock s Approximation Quick Jackson Theorem
Network of Queues: Setup In many networking scenarios, a customer or packet must receive service from many servers before its final task is completed. Hence, departures from a queue might become arrivals at another queue. All that discussion we did in M/G/ queues becomes very important for networks of queues! Consider two queues in tandem: Departures from first queue become arrivals at second queue First queue s arrival process is oisson with rate λ and service time is exponential with rate µ > λ Service time at the second queue is exponential with rate µ 2 and independent of first server s service time How do we model these two queues together? What s the state and state diagram? > λ
Two Queues in Tandem Queue µ Queue2 µ 2 We need to keep track of N (t) and N 2 (t) to describe the state of the system!
Our state diagram must keep track of both N (t) and N 2 (t), and the many transitions that are possible Two Queues in Tandem: State Diagram 3 µ 2 µ µ µ 2 µ 2 µ µ 2 2 λ λ λ µ 2 µ µ 2 µ µ 2 µ µ 2 λ λ λ µ 2 µ µ 2 µ µ µ 2 µ 2 n 2 =0 λ λ λ n =0 2 3
Global Balance Equations for 2 Queues, pg Define (N (t), N 2 (t)) to be the state vector for the two queues in tandem Notice, now, that we have a Markov rocess in terms of the state vector Recall the global balance equations for M/M/ queues: λp 0 = µ p ( λ + µ ) p = λp + µ p, j =,2,... j j What this means is: At steady state, the amount entering a state is equal to the amount leaving the state. We similarly may find the global balance equations for this two queue system j+
Global Balance Equations for 2 Queues, pg 2 Global Balance Equations: Case λ N = 0, N Case 2: n>0, m=0 Case 3: m>0, n=0 Case 4: n>0, m>0 ( = 0) = µ ( N = 0, N ) 2 2 2 = ( λ + µ ) ( N = n, N = 0) = µ ( N = n, N = ) + λ( N = n, N 0) 2 2 2 2 = ( λ + µ ) ( N = 0, N = m) = µ ( N = 0, N = m + ) + µ ( N =, N = m ) 2 2 2 2 2 ( λ + µ + µ 2 ) ( N = n, N2 = m) = µ 2( N = n, N2 = m + ) + µ ( N = n +, N2 = m ) + λ( N = n, N = m) 2
Steady State MFs for Two M/M/ Queues We may show that the following joint probability mass function satisfies the global balance equations ( ) ( ) n N ( ) m = n, N2 = m = ρ ρ ρ2 ρ2 where ρ i =λ/µ i So, how do we get (N =n)? Easy, its just an M/M/! So (N =n) = (-ρ )ρ n for n=0,,2 How do we get (N 2 =m)? Answer: It s a marginal. Integrate out the joint pmf! Sum over all n to get: ( N ) ( ) m 2 = m = ρ2 ρ2 Check This!
Steady State MFs for Two Queues, pg 2 This looks interesting means ( ) ( ) n N ( ) m = n, N2 = m = ρ ρ ρ2 ρ2 ( N = n, N = m) = ( N = n) ( N m) 2 2 = Thus, the number of customers at queue and the number at queue 2 at a particular time are independent random variables! The steady state at queue 2 is the same as for an M/M/ queue with oisson arrival rate λ and exponential service time µ 2. Definition: A network of queues is said to have a product-form solution when the joint pmf of the number of customers at each queue is the product of the marginal pmfs of the number of customers at each queue.
Burke s Theorem Burke s Theorem is the fundamental result describing product form solutions Burke s Theorem: Consider an M/M/, M/M/m, M/M/infinity queuing system at steady state with arrival rate λ, then The departure process is oisson with rate λ; At each time t, the number of customers in the system N(t) is independent of the sequence of departure times prior to t. What Burke s Theorem implies: Two queue problem follows from Burke s Thm (arrivals to queue 2 are oisson with rate λ). Arrivals to queue 2 prior to time t are departures from queue prior to time t, thus Burke s theorem says queue- s departures (queue-2 s arrivals) are independent of N (t). N 2 (t) is determined by the sequence of arrivals from queue- prior to time t and independent of service times, then N (t) and N 2 (t) are independent as random variables. Note: N (t) and N 2 (t) are not independent as processes!
Example Application of Burke s Thm Consider the network of queues: /2 µ 2 λ µ /2 λ 2 µ 3 Here Queue is driven by a oisson process with rate λ,, and the departures are randomly routed to queues 2 and 3. Queue 3 has an additional, independent oisson arrival process with rate λ 2.
Example Application of Burke s Thm,, pg 2 Burke s Theorem says: N (t) and N 2 (t) are independent N (t) and N 3 (t) are independent Recall that the random split of a oisson process yields independent oisson processes Hence inputs to Queue 2 and Queue 3 are independent Input to Queue 2 is oisson with rate λ /2 Input to Queue 3 is oisson with rate λ /2 + λ 2 Thus k m ( N ) ( ) ( ) ( ) n (t) = k, N2(t) = m, N3(t) = n = ρ ρ ρ2 ρ2 ρ3 ρ3 where ρ =λ /µ, ρ 2 =λ /2µ 2, ρ 3 =(λ /2 +λ 2 )/µ 3. All queues are assumed to be stable.
Reversible Markov rocesses In order to prove Burke s theorem, we need the concept of the reversibility of a Markov process. A stationary Markov process X(t), with a countable state space (i.e. a Markov chain will do), is reversible if X(t) and Y(t)=X(-t) have the same joint distribution at arbitrarily chosen instants {t, t 2,, t N }. A necessary and sufficient condition for reversibility is p p = p i i, j where {p i } and {p ij } are the stationary probabilities and transition probabilities of X(t) This condition can be easily shown for M/M/ queues but we will show it in more general form In fact, it holds for any birth-death process, and N(-t) is statistically identical to N(t) j p j,i
Reversible Markov rocesses, pg 2 Time-Reversal Theorem: Let {X(t): t>=0} be a stationary Markov process with (infinitesimal) generator =[p ij ], and with initial distribution equal to stationary distribution. Then for all T>0, the time-reversed process is equivalent to a stationary ~ Markov process with (infinitesimal) generator ij given by: ~ p jp ji pij = p for all state pairs (i,j) { X(T t) :0 t T} i
Reversible Markov rocesses, pg 3 roof: Let Q(t)=[q ij (t)] denote the transition probabilities of X(t), [ N(t) = j N(0) i] q ij (t) = = We need to show X(T-t) is a Markov process with transition probabilities Q ~ (t) ~ q (t) = ij Then we obtain the necessary and sufficient condition by differentiating this and setting t=0. (Its an infinitesimal generator) p j q p ji i (t)
Reversible Markov rocesses, pg 4 Consider the interval (0,t+s] and divide it into (0,t] and (t,t+s], i.e. set T=t+s. The joint probability of the three random variables is Similarly we have X ~ (0) = X(t + s) X ~ (t) = X(s) X ~ (t + s) = = = ( X ~ (0) = i, X ~ (t) = i, X ~ (t + s) = i ) p = ( X(0) = i,x(s) = i,x(t + s) = i ) i q 0 2 (s)q 2 i2,i i, i0 (t) 2 0 X(0) ( X ~ (0) = i,x ~ (t) = i ) = ( X(s) = i,x(s + t) = i ) p i q i, i0 0 (t) 0
Reversible Markov rocesses, pg 5 The conditional probability = ( X ~ (t + s) = i X ~ (t) = i, X ~ (0) = i ) p i = = ~ q 2 ( X ~ (t + s) = i X ~ (t) = i ) i, j q i p 2 i,i (t) 2 2 Hence, we have shown that X ~ ( t) is a Markov process with ~ generator (after differentiation). 0
B-D D rocesses are Reversible It is now easy to show the following Time Reversibility of B-D rocesses: The stationary B-D process N(t) with generator and steady state probabilities p is a time-reversible Markov process. Thus, the time-reversal of the death process is a birth process. Burke s Theorem follows from this: Interdeparture times of the forward-time system are the interarrival times of the time-reversed system... Hence we have oisson with rate λ coming out of the system. Fix a time t, then the departures before time t from the forward system are arrivals after time t in the reverse system. Arrivals in reverse system are oisson, and thus arrivals in reverse system after time t do not depend on N(t) Consequently, departures after time t in forward system do not depend on N(t)
Step back to Network of Queues What we derived held true because the first queue was M/M/ and implicitly we assumed it had achieved steady state and independent service times between queues! The problem is more complicated when we have more general networks of queues. Again, consider two transmission lines in sequence. The arrivals to the first are oisson of rate λ, but all customers (packets) have deterministic and equal service times, i.e. we have an M/D/ queue. λ Queue µ Queue2 µ Average packet delay for first queue is given by ollaczek- Khinchine formula.
Network of Queues, M/D/ first queue The interarrival times of the second queue must be at least /µ Why? Now, each packet arriving at either queue takes /µ time to process. The first packet being finished by first queue is immediately sent to second server It takes at least another /µ amount of time for first queue to get and finish the next packet/customer. So, first packet will be finished by second server at or before the next packet arrives to second server. Result: No queue (waiting) at second system!
Two Queues, correlated service times Earlier we considered the service times independent of each other and independent of the arrival times. Reality: A big packet at the first system is probably still a big packet at the second system! Interarrival times at the second queue are strongly correlated with packet lengths! Long packets at first system will typically find the queue at the second server more empty Shorter packets from the first system will typically find the queue at the second server more busy because the second server is processing some prior big packet It is tough to find an analytical solution for joint pmf under dependence assumptions!
The Kleinrock Independence Approximation We have argued that in practice there is dependence upon the interarrival times and service times. Independence is lost after the first system! Reality hurt us, but reality provides us one more gift Reality: Real networks typically involve more than one stream of packets merging at a node The combination of multiple streams helps restore independence in many cases! This observation is due to Kleinrock. Kleinrock s Approximation: It is often appropriate to use M/M/ queues for each communication link when the arrivals at entry points are oisson, packet lengths are roughly exponentially distributed, network is dense and traffic is heavy.
Quick Look at Jackson s Theorem Many queuing networks, a packet/customer may visit a queue more than once. Burke s theorem does not apply! Typical example: Queue with feedback p α λ µ -p If the arrival rate is much less than departure rate, then net arrival process has a few, isolated external arrivals followed by a burst of feedback arrivals (dependent on packet length).
Jackson s Theorem, (Open) pg. Consider a queuing network consisting of M separate service stations, each with its own queue. Define the vector process: N t = N (t), N (t),, N (t) ( ) [ ] 2 L In an open queuing network, customers may arrive from an external source and eventually may leave the network (as depicted in previous slide). A closed network has no arrivals or departures from the system (total customers is fixed just they may move around) Assumptions: The rate of the source (birth process) is λ, and a customer goes to station i with probability q si. Service time at station k is exponential with rate µ k. Customers are routed according to a Markov chain: robability that a customer departing station i goes to station j is q ij. M
Jackson s Theorem, (Open) pg 2. Jackson s Decomposition Theorem: For an open queue as described, the joint distribution of the queue vector n(t) is given by: ( n) = ( N = n ) ( N = n ) L( N ) = 2 2 K n K where (N k =n k ) is the steady state pmf of an M/M/ system with arrival rate λ k and service rate µ k, i.e. n ( N = n ) = ρ ( ) k ρ k k k Where ρ k describes the utilization factor of system k. k