Networking Theory and Fundamentals - Lecture 1 potx

TCOM 501: Networking—Theory and Fundamentals University of Pennsylvania Class Notes Santosh S. Venkatesh c  1997 Delay Models in the Network Layer It is important in data communication settings to be able to characterise the delay that individual packets are subjected to in their sojourn through the system. Delays can be traced to four main causes. • Processing delay: This is the time it takes to process a frame at each subnet node and prepare it for retransmission. These delays are deter- mined by the complexity of the protocol and the computational power available at each subnet node. • Propagation delay: This is the time it actually takes a frame to prop- agate through the communication link. These delays are dictated by the distance or length of the communication pathway. They can be significant, for instance, in satellite links and in high-speed links when the propagation delay can be a significant fraction of the overall delay. • Transmission delay: This is the time it takes to transmit an entire frame, from first bit to last bit, into a communication link. These delays are dictated primarily by link speed, for example, a 9600 bps link occasions only half the delay of a 4800 bps link. • Queueing delay: This is the time a frame has to wait in a queue for ser- vice. These delays are occasioned by congestion at the subnet nodes. Propagation delays are determined by the physical channels and are inde- pendent of the actual traffic patterns in the link; likewise, processing delays are determined by the available hardware and again are not affected by traf- fic. We hence focus on the transmission and queueing delays endemic at any subnet node. The discussion ignores the possibility of retransmissions, which of course add to the overall delay. In practice, however, retransmis- sions are rare in many data networks so that this assumption may safely be 1 Class Notes Santosh S. Venkatesh c  1997 TCOM 501: Networking—Theory and Fundamentals University of Pennsylvania made. 1 Subnet node Packets arrive asynchronously Packets depart asynchronously Queues are intrinsic to packet-switched networks. In a generic packet-switching network, frames arrive asynchronously at subnet nodes in the communica- tion subnet and are processed and released according to some ser- vice protocol, for instance, a first- in, first-out (FIFO) or first-come, first-served (FCFS) protocol. Typically, a subnet node cannot handle all the traffic entering it simultaneously so that frames arriving at the node are buffered to await their turn for transmis- sion. In queueing parlance, the frames are “customers” awaiting “service” from the subnet node which is hence the “server.” The overall queueing delay for a frame (customer) is determined generally by the congestion at the subnet node (server) which is governed by packet arrival statistics and the service discipline in force. Arriving packets Departing packets Buffer Server Multi-server queue Single-server queue More specifically, node congestion is influenced by the following factors: • Arrival statistics embodied in the distribution of customer interarrival times τ. (The arrival rate λ = 1/ E(τ) will play a critical r ˆ ole in the sequel.) • Buffer size m—for instance, finite buffer systems (m<∞) in which customers are turned away if the queue is full, and infinite buffer sys- tems (m = ∞) in which customers are always accepted into the system. 1 Multi-access networks are the exception which proves the rule: retransmissions are the rule rather than the exception in multi-access settings. 2 TCOM 501: Networking—Theory and Fundamentals University of Pennsylvania Class Notes Santosh S. Venkatesh c  1997 • Number of servers n—for instance, single-server systems (n = 1), multi-server systems (n ≥ 2), and infinite-server systems (n = ∞). • Service statistics embodied in the distribution of customer service times X. (The mean service time E(X)=1/µ, in particular, will prove to be a particularly useful concept in the sequel.) Together, these determine the delay faced by each customer. A succinct notation has evolved to describe the various factors that affect the congestion and delays in a queueing environment. A generic queueing environment is described in the form A/B/c/d, where the first two descriptors A and B connote the arrival and service statistics, respectively, and the second two descriptors c and d (if present) connote the number of servers and the system capacity (buffer size or maximal number of cus- tomers in the system), respectively. The descriptors A and B are specified by the arrival and service statistics of the queueing discipline and may be specified, for instance, as M (exponential or memoryless distribution), E (Er- langian distribution), H (hyperexponential distribution), D (deterministic), or G (general distribution), to mention a few possibilities. The descriptors c and d take on positive integer values, allowing infinity as a possible value to admit limiting systems with an infinite number of servers and/or infinite storage capacity which are of theoretical interest. Little’s Theorem Consider a queueing environment which, after initial transients have died down, is operating in a stable steady state. N customers in system T time per customer (Closed) queueing environment in steady state Customer arrivals (rate λ) Customer departures The key parameters characterising the system are: • λ—the mean, steady state customer arrival rate. • N—the average number of customers in the system (both in the buffer and in-service). 3 Class Notes Santosh S. Venkatesh c  1997 TCOM 501: Networking—Theory and Fundamentals University of Pennsylvania • T—the mean time spent by each customer in the system (time spent in the queue together with the service time). It is tempting and intuitive to hazard the guess N = λT. This indeed is the content of Little’s theorem which holds very generally for a very wide range of service disciplines and arrival statistics. To motivate the result, consider a general, stable queueing envi- ronment which has customers arriving in accordance with some underlying arrival statistics and departing regulated by some given service discipline. Arrival, Departure Processes. At any instant of time t, let A(t) and D(t) denote the number of arrivals and departures, respectively, in the time in- terval [0, t]. The random processes A(t) and D(t), called the arrival and de- parture process, respectively, are governed by the probability distributions underlying customer arrivals and service provision, and provide a compre- hensive instantaneous description of the state of the system at any time t. t t N(t) = A(t) - D(t) = # customers in system at time t A(t) D(t) Sample Function -> Step Function Cadlag Continue ‘a droi limite ‘a gauch N(t) The adjoining figure illus- trates sample functions of the ar- rival and departure processes. Ob- serve that the sample functions of the processes A(t) and D(t) are step functions with jumps of unit height. Indeed, both processes are counting processes with the jump points indicating an arrival or departure. Observe further that the sample functions of the depar- ture process lag the correspond- ing sample functions of the ar- rival process. The random process N(t)=A(t)−D(t) then represents the instantaneous number of cus- tomers present in the queueing environment (both in-queue and in-service) at time t. Example 1 Transmission line. Consider a transmission line system with packets arriving every K seconds for transmission. T ra n s m i s s i o n l i n e Arriving packets Departing packets 4 TCOM 501: Networking—Theory and Fundamentals University of Pennsylvania Class Notes Santosh S. Venkatesh c  1997 Suppose that each packet requires a transmission time of aK seconds (a< 1) and that the propagation time for each packet is bK seconds. Viewing the transmission line as a server and packets as customers, customers arrive at a fixed rate λ = 1/K packets/second. Each customer spends a fixed amount of time in the system T =(a + b)K. The number of packets that enter the line before a given packet departs is hence N = T/K = a + b. This is the average number of packets in the line in the steady state. The arrival process A(t) is a fixed time function which has regular unit jumps every K seconds; the departure process D(t) is also a fixed time function with jump points regulated by the value a+b. Two cases, 0<a+b<1and 1<a+b<2 are shown. K 2K 3K 4K 0ψ<ψa + bψ<ψ1 K 2K 3K 4K 0 0 A(t) D(t) N(t) t t 0ψ<ψa + bψ<ψ2 K 2K 3K 4K K 2K 3K 4K 5K 5K 0 0 N(t) A(t) D(t) t t What kind of queueing discipline is this? Observe that the “cus- tomers” (i.e., packets) arrive at a fixed rate λ = 1/K, promptly go into service (i.e., have transmission initiated) as soon as they arrive regardless of how many packets are already in the line, Time Averages. For definiteness, consider a FIFO system where customers are served sequentially in order of arrival. Typical sample functions for the t T 1 T 2 T 3 T 4 T 5 T 6 D(t) A(t) arrival and departure process in such a system are shown alongside. The time average (up to t) of the in- stantaneous number of customers in the closed queueing environment under consideration is given by ^ N(t)= 1 t  t 0 N(τ) dτ. Now pick any instant t at which the system has just become empty so that N(t)=A(t)−D(t)=0. Let T i denote the time spent in the system by the ith customer as shown in the above figure for a FIFO system. Now observe 5 Class Notes Santosh S. Venkatesh c  1997 TCOM 501: Networking—Theory and Fundamentals University of Pennsylvania via the figure that the area under the curve (up to t) of the instantaneous number of customers in the system is identically equal to the area between the arrival and departure sample functions which, in turn, is comprised of A(t) rectangular blocks of unit height and widths T 1 , , T A(t) . More formally,  t 0 N(τ) dτ =  t 0  A(τ)−D(τ)  dτ = A(t)  i=1 T i . Consequently, we obtain ^ N(t)= 1 t A(t)  i=1 T i =  A(t) t  1 A(t) A(t)  i=1 T i  . The first quantity on the right-hand side may be identified with the average arrival rate of customers up to time t: ^ λ(t)= A(t) t . Likewise, the second term on the right-hand side may be identified with the average time spent by the first A(t) customers in the system: ^ T(t)= 1 A(t) A(t)  i=1 T i . Thus, we obtain ^ N(t)= ^ λ(t) ^ T(t). Arrivals Departures Queueing Environment Now suppose that, as t →∞, the various time averages tend to fixed steady state values ^ N(t) →N, ^ λ(t) → λ, and ^ T(t) → T. We then obtain Little’s formula N = λT where we interpret N as the steady state, average number of customers in the system, λ as the steady state arrival rate of customers into the system, and T as the steady state, average time spent in the system by each customer. 6 TCOM 501: Networking—Theory and Fundamentals University of Pennsylvania Class Notes Santosh S. Venkatesh c  1997 It may be remarked that the formulation, while derived for a FIFO system, is actually very general and in fact applies to a very wide spectrum of queueing environments and service disciplines. Ensemble Averages. Little’s theorem extends quite naturally to situations where the arrival and departure processes, A(t) and D(t), are specified by some underlying probability law. Indeed, suppose that at time t the number of customers N(t) in the system has distribution π n (t). Then, the expected number of customers in the system at time t is given by ¯ N(t)=E  N(t)  = ∞  n=0 nπ n (t). Now, for a large number of systems, the distribution of customers tends to a steady state or stationary distribution π n (t) →π n as t →∞. Then, ¯ N(t) → ¯ N = ∞  n=0 nπ n (t →∞). Likewise, the mean time spent in the system by a customer tends to a limit- ing value ¯ T i = E T i → ¯ T (i →∞), as does the instantaneous, mean customer arrival rate ¯ λ(t)= E  A(t)  t → ¯ λ (t →∞). A fundamental result known as the ergodic theorem allows us to relate these steady state ensemble averages to the time averages obtained from individ- ual sample functions: for a very wide range of situations, including most situations encountered in practice, N = ¯ N, T = ¯ T, and λ = ¯ λ with probability one. Thus, we can heretofore apply Little’s theorem with confidence to any closed queueing environment where we interpret the various quantities as steady state, long-term averages. Applications of Little’s Theorem The power of Little’s theorem is in its very wide applicability though care should be taken to make sure that it is applied in the context of a sta- ble, closed queueing environment where the mean arrival rate of customers matches the mean departure rate and the system is operating in the steady state. Some examples may serve to fix the result. 7 Class Notes Santosh S. Venkatesh c  1997 TCOM 501: Networking—Theory and Fundamentals University of Pennsylvania Example 2 Transmission line, reprise. Suppose, as before, that packets arrive at regular intervals of K sec- onds to a transmission line. As before, the transmission time for each packet is aK (where 0<a<1) and the propagation delay is bK (for some positive b). T ra n s m i s s i o n l i n e Arriving packets Departing packets The queueing environment consists of a single server (the trans- mission line). Packets arrive at a rate λ = 1/K packets/second, with each packet staying in the system a total of T = aK + bK seconds. Little’s the- orem hence shows that the steady-state average number of packets in the system is N = λT = a + b. Recall that the actual number of packets in the system is a periodically varying, deterministic function so that the number of customers in the system never converges, even in the limit of very large time, to the constant N (cf. Example 1). The long-term average number of customers in the system, however, tends to N. Example 3 Access-controlled highway entry. If, in the previous transmission line example, we view the opera- tions of transmission and propagation in reverse order, we obtain a related service discipline. Arrival rate of cars to highway access λγ Metering delay bK Merging delay aK In this alternative setting, customers arrive peri- odically every K seconds, i.e., the customer ar- rival rate is λ = 1/K, wait in queue for bK sec- onds, and are serviced and released in aK sec- onds. We may identify this queueing environment with an (idealised) access-controlled highway en- try system where traffic is permitted to enter a highway access road at a fixed rate of λ = 1/K cars per second. The cars wait in a queue of fixed size (fixed buffer size) and are released periodi- cally, one at a time, by a stop light metering sys- tem; the wait time in queue is bK seconds. Finally, on release, the car at the head of the queue moves at a fixed rate over the access road to merge with traffic in the highway; the time taken to traverse this segment is aK seconds. Observe that Little’s theorem applied to the queueing segment alone shows that the average number of customers waiting in queue is bK/K = b, while Little’s theorem applied to the service segment yields the (fractional) average number of customers being serviced at any instant aK/K = a. Ap- plying Little’s theorem to the entire system consisting of queue and server 8 TCOM 501: Networking—Theory and Fundamentals University of Pennsylvania Class Notes Santosh S. Venkatesh c  1997 yields the average number of customers in the system N =(aK + bK)/K = a + b which is the sum of the number of customers in the two segments of the service discipline, as it should be. Observe that, while the physical details are rather different, macro- scopically the two service disciplines are equivalent—at least in the average- sense, long term world view of Little’s theorem. Example 4 Airline counters. Consider a closed queueing environment in which customers enter- ing the system wait in a single queue for service from n service agents. The customer at the head of the queue proceeds for service to the first available server who immediately begins service for the new customer as soon as the previous customer has departed. The average service time for a customer is X. 1 n (2) (1) λγ Suppose that the the environment has a finite capacity and that, at any given moment, there are no more than N cus- tomers in the system. (For instance, we may consider an idealised airport counter type of environment in a room with fi- nite capacity or a fixed length serpentine queue with a fixed winding roped-off bor- der.) Suppose additionally that demand is such that any departing customer is instantly replaced by another customer. (All holiday airfares are advertised at half- price.) What is the average time a customer can expect to spend in the system once he enters it? Let λ denote the steady state mean arrival rate of customers into the system and T the average time spent in the system by each customer. Little’s theorem applied to the entire system (queueing environment A) yields N = λT , while an application of the theorem to the subsystem consisting only of the n servers (queueing environment B) yields n = λX. (Observe that, in steady state, the arrival rate to the system of servers must be exactly the same as the arrival rate into the system, else instability results.) It follows that the average time spent in the system by each customer is T = N/λ = N n X. (∗) Intuitive and satisfactory. Indeed, the form of the result is quite suggestive. Consider the following variation. Probabilistic airline counters. In a probabilistic variation, consider, as before, a finite capacity system which can accommodate at most N customers at a time with service being provided by n servers. The average service time per 9 Class Notes Santosh S. Venkatesh c  1997 TCOM 501: Networking—Theory and Fundamentals University of Pennsylvania customer is ¯ X. Suppose that there is constant demand so that the system is always full with each departing customer being promptly replaced by an arriving customer. In this version of the airport counter problem, each server has her own queue of customers. A customer entering the system joins a queue uniformly at random and awaits his turn for service. What is the average time spent in the system by a customer? (1) (2) 1 n λγ Consider the ith subsystem con- sisting of the ith server with her associ- ated queue and let N i , λ i , and T i denote the mean number of customers in the ith subsystem, the arrival rate into the sub- system, and the mean time spent in the subsystem by a customer, respectively. Then, Little’s theorem applied to the sub- system yields T i = N i /λ i . To determine the mean arrival rate into the ith subsys- tem, consider the queueing environment consisting of the ith server alone. In the steady state, suppose that the server is busy a fraction ρ i of the time or, equivalently, is idle for a fraction 1 − ρ i of the time. Little’s theorem then yields ρ i = λ i X. The quantity ρ i connotes the mean utilisation factor for server i in the steady state. How then does one go about estimating the mean utilisation factor? The ergodic theorem tells us that, with probability one in the steady state, the fraction of time that a server is idle is iden- tically the instantaneous probability (at an arbitrary point in time in the steady state) that the server is idle, i.e., has no customers in service. Write π 0 for the steady state instantaneous probability that the ith server is idle. It follows that ρ i = 1 − π 0 so that T i = N i X  ρ i . Now note that T i = T for each i as all servers have the same average customer delay. Furthermore, N i = N/n as, on average, each subsystem sees a fraction 1/n of the custom. Finally, observe that π 0 is identically the probability that each of the N cur- rent customers in the room have selected a server other than i. It follows that π 0 =  1 − 1 n  N . Consequently, T = N n X  1 −  1 − 1 n  N  . (∗∗) The ratio of the mean delay in (∗) to that of (∗∗) is precisely 1 ρ i = 1 1 −  1 − 1 n  N >1. Thus, the congestion independent assignment of customers to queues lead- ing to (∗∗) results in an added cost in total delay corresponding precisely 10 [...]... state n − 1 to state n, and, likewise, the only transition into the membrane is from state n to state n − 1 The ) hence yield balance equations ( (n ≥ 1) , πn 1 Pn 1, n = πn Pn,n 1 whence, by induction, we obtain n πn = πn 1 Pn 1, n Pi 1, i = π0 Pn,n 1 P i =1 i,i 1 We can determine π0 directly from the identity finally obtain n πn = Pi 1, i P i =1 i,i 1 ∞ ∞ n=0 πn = 1, whence we m Pi 1, i P m=0 i =1 i,i 1 (n ≥... is Poisson with parameter k 1 i =1 λi It is easy now to recursively specify the probability mass function of Sk : we have n P{Sk = n} = n P{Sk 1 = m, Xk = n − m} = m=0 n e−( 1 +···+λk 1 ) = m=0 = P{Sk 1 = m} P{Xk = n − m} m=0 e−( 1 +···+λk 1 +λk ) n! ( 1 + · · · + λk 1 )m m! n m=0 −( 1 +···+λk 1 +λk ) ( 1 =e e−λk λn−m k (n − m)! n ( 1 + · · · + λk 1 )m λn−m k m + · · · + λk 1 + λk )n , n! the last step... whereby the line first 11 TCOM 5 01: Networking Theory and Fundamentals University of Pennsylvania Class Notes Santosh S Venkatesh c 19 97 transmits some packets from user 1, then proceeds to transmit some packets from user 2, and continues likewise until it transmits some packets from user m The line then returns to user 1 in round-robin fashion and repeats the process (m) 1 (1) (2) λm Round robin switching... estimates Pnn = Q(0, 0 | 0) + o(δ) = 1 − λδ + o(δ) if n = 0, Q(0, 0 | n) + o(δ) = 1 − λδ − µδ + o(δ) if n ≥ 1, Pn 1, n = Q (1, 0 | n − 1) + o(δ) = λδ + o(δ), Pn,n 1 = Q(0, 1 | n) + o(δ) = µδ + o(δ), and, finally, when |m − n| ≥ 2, Q(a, d | m) = o(δ) Pmn = (a,d):a−d=n−m 34 TCOM 5 01: Networking Theory and Fundamentals University of Pennsylvania Class Notes Santosh S Venkatesh c 19 97 It is clear then that, up... customer in service departs 33 TCOM 5 01: Networking Theory and Fundamentals University of Pennsylvania Class Notes Santosh S Venkatesh c 19 97 and that the customer who takes his place does not within Ik is just µδe−µδ Consequently, P{Dk = 1 | Ak = 0, Nk 1 = m} = 1 − e−µδ µδe−µδ if m = 1, if m ≥ 2 It follows that Q(0, 1 | m) = (1 − e−µδ )e−λδ µδe−µδ e−λδ if m = 1, if m ≥ 2 Both cases reduce to the common... strictly increasing sequence of points in time and write Xk = X(sk ) − X(sk 1 ) for the number of arrivals 19 Class Notes Santosh S Venkatesh c 19 97 TCOM 5 01: Networking Theory and Fundamentals University of Pennsylvania in the time interval (sk 1 , sk ] Then { Xk , k ≥ 1 } is a sequence of independent, Poisson random variables, with Xk ∼ Po λ(sk − sk 1 ) (k ≥ 1) It is clear that the sample paths of a Poisson... the start of sub-interval Ik In this case Q (1, 0 | m) = P{Dk = 0 | Ak = 1, Nk 1 = m} P{Ak = 1 | Nk 1 = m} = P{Dk = 0 | Nk 1 = m} P{Ak = 1} = e−µδ λδe−λδ (m ≥ 1) , by a similar argument to the one invoked above: the number of arrivals is independent of the number of customers already in service, whence P{Ak = 1 | Nk 1 = m} = P{Ak = 1} = λδe−λδ by virtue of Poisson arrival statistics; and the event that... whence P{Dk = 0 | Ak = 1, Nk 1 = m} = P{Dk = 0 | Nk 1 = m} = P{Dk = 0 | Nk 1 = 1} = e−µδ for m ≥ 1 by virtue of the memoryless property of the exponential distribution 32 TCOM 5 01: Networking Theory and Fundamentals University of Pennsylvania Class Notes Santosh S Venkatesh c 19 97 Things get slightly more complex if there are no customers in the system at the start of the sub-interval Ik In this case,... Markov Chains Consider a sequence of random variables { Nk , k ≥ 0 } where each random variable takes on values only in the discrete set of points {0, 1, 2, } The random sequence {Nk } forms a Markov chain if, for every k, and every choice of nonnegative integers m0 , m1 , , mk 1 , mk = m and mk +1 = n, P{Nk +1 = n | Nk = m, Nk 1 = mk 1 , , N0 = m0 } = P{Nk +1 = n | Nk = m} = Pmn In words: in... intervals are independent In particular, if 0 ≤ t0 < t1 < · · · < tk and Xi = A(ti ) − A(ti 1 ) (1 ≤ i ≤ k), then the random variables { Xi , 1 ≤ i ≤ k } are independent Poisson random variables, Xi ∼ Po λ(ti − ti 1 ) For each n ≥ 1, let tn denote the arrival time of the nth customer and let τn = tn − tn 1 denote the time between the arrivals of the (n − 1) st and the nth customers The Poisson character of . m} = n  m=0  e −(λ 1 +···+λ k 1 ) (λ 1 + ···+ λ k 1 ) m m!  e −λ k λ n−m k (n − m)!  = e −(λ 1 +···+λ k 1 +λ k ) n! n  m=0  n m  (λ 1 + ···+ λ k 1 ) m λ n−m k = e −(λ 1 +···+λ k 1 +λ k ) (λ 1 + ···+ λ k 1 +. follows that π 0 =  1 − 1 n  N . Consequently, T = N n X  1 −  1 − 1 n  N  . (∗∗) The ratio of the mean delay in (∗) to that of (∗∗) is precisely 1 ρ i = 1 1 −  1 − 1 n  N > ;1. Thus, the congestion. time share a transmis- sion line. A switching mechanism is put into place whereby the line first 11 Class Notes Santosh S. Venkatesh c  19 97 TCOM 5 01: Networking Theory and Fundamentals University