Communication Systems Engineering Episode 2 Part 2 pptx

Eytan Modiano Slide 1 16.36: Communication Systems Engineering Lecture 17/18: Delay Models for Data Networks Eytan Modiano Eytan Modiano Slide 2 Packet Switched Networks Packet Network PS PS PS PS PS PS PS Buffer Packet Switch Messages broken into Packets that are routed To their destination Eytan Modiano Slide 3 Queueing Systems • Used for analyzing network performance • In packet networks, events are random – Random packet arrivals – Random packet lengths • While at the physical layer we were concerned with bit-error-rate, at the network layer we care about delays – How long does a packet spend waiting in buffers ? – How large are the buffers ? Eytan Modiano Slide 4 Random events • Arrival process – Packets arrive according to a random process – Typically the arrival process is modeled as Poisson • The Poisson process – Arrival rate of λλ λλ packets per second – Over a small interval δδ δδ , P(exactly one arrival) = λλ λλ δδ δδ P(0 arrivals) = 1 - λλ λλ δδ δδ P(more than one arrival) = 0 – It can be shown that: P(narrivalsinintervalT)= − () ! λ λ Te n nT Eytan Modiano Slide 5 The Poisson Process P(narrivalsinintervalT)= − () ! λ λ Te n nT n = number of arrivals in T It can be shown that, E[n] = T E[n ] = T + ( T) = E[(n-E[n]) ] = E[n ]-E[n] = T 22 2222 λ λλ σλ Eytan Modiano Slide 6 Inter-arrival times • Time that elapses between arrivals (IA) P(IA <= t) = 1 - P(IA > t) = 1 - P(0 arrivals in time t) = 1 - e -λλ λλ t • This is known as the exponential distribution – Inter-arrival CDF = F IA (t) = 1 - e -λλ λλ t – Inter-arrival PDF = d/dt F IA (t) = λλ λλ e -λλ λλ t • The exponential distribution is often used to model the service times (I.e., the packet length distribution) Eytan Modiano Slide 7 Markov property (Memoryless) • Previous history does not help in predicting the future! • Distribution of the time until the next arrival is independent of when the last arrival occurred! PT t t T t PT t oof PT t t T t Pt T t t PT t edt edt e e ee t t tt t t t t tt t t tt (|)() Pr : (|) () () | | () ≤+ > = ≤ ≤+ > = <≤+ > == − − = −+ − + − ∞ − + −∞ −+ ∫ ∫ 00 00 00 0 0 0 0 0 0 0 0 λ λ λ λ λ λ λ −− − − =− = ≤ λ λ λ () () () t t t e ePTt 0 0 1 Eytan Modiano Slide 8 Example • Suppose a train arrives at a station according to a Poisson process with average inter-arrival time of 20 minutes • When a customer arrives at the station the average amount of time until the next arrival is 20 minutes – Regardless of when the previous train arrived • The average amount of time since the last departure is 20 minutes! • Paradox: If an average of 20 minutes passed since the last train arrived and an average of 20 minutes until the next train, then an average of 40 minutes will elapse between trains – But we assumed an average inter-arrival time of 20 minutes! – What happened? • Answer: You tend to arrive during long inter-arrival times – If you don’t believe me you have not taken the T Eytan Modiano Slide 9 Properties of the Poisson process • Merging Property Let A1, A2, … Ak be independent Poisson Processes of rate λλ λλ 1, λλ λλ 2, …λλ λλ k • Splitting property – Suppose that every arrival is randomly routed with probability P to stream 1 and (1-P) to stream 2 – Streams 1 and 2 are Poisson of rates Pλλ λλ and (1-P)λλ λλ respectively A = A is also Poisson of rate = ii ∑∑ λ λ P 1-P λP λ(1−P) λ 1 λ 2 λ k λ i ∑ Eytan Modiano Slide 10 Queueing Models • Model for – Customers waiting in line – Assembly line – Packets in a network (transmission line) • Want to know – Average number of customers in the system – Average delay experienced by a customer • Quantities obtained in terms of – Arrival rate of customers (average number of customers per unit time) – Service rate (average number of customers that the server can serve per unit time) server Queue/buffer Customers [...]... Eytan Modiano Slide 20 (λ / µ ) D = 1/ µ + (µ − λ ) M/M/1 formula Circuit (tdm/fdm) vs Packet switching Average Packet Service Time (slots) Average Service Time 100 TDM with 20 sources Ideal Statistical Multiplexing (M/D/1) 10 1 0 0 .2 0.4 0.6 0.8 Total traffic load, packets per slot Eytan Modiano Slide 21 1 Delay formulas • M/G/1 λX 2 D=X+ 2( 1 − λ / µ ) • • Eytan Modiano Slide 22 λ/µ µ−λ M/D/1 D=X+... Modiano Slide 19 µ W = T - 1/µ = 2. 5 minutes How many customers in the restaurant? – • Service rate = µ = 60/0.5= 120 customers per hour µ T = 1/µ−λ = 1/( 120 -100) = 1 /20 hrs = 3 minutes ρ = λ/µ = 5/6 Packet switching vs Circuit switching 1 λ/N 1 2 3 λ/N N 1 2 3 N TDM, Time Division Multiplexing Each user can send µ/N packets/sec and has packet arriving at rate λ/N packets/sec 2 λ/N N (λ / µ ) D= N/µ + (µ... Little’s theorem: N = λT – – Can be applied to entire system or any part of it Crowded system √ long delays On a rainy day people drive slowly and roads are more congested! Eytan Modiano Slide 12 Proof of Little’s Theorem A(t) A(t), B(t) T4 T3 T2 B(t) T1 t1 t2 t3 t4 • • • • • A(t) = number of arrivals by time t B(t) = number of departures by time t ti = arrival time of ith customer Ti = amount of time... =1 Ti = A(t )T A( t ) Application of little’s Theorem • Little’s Theorem can be applied to almost any system or part of it • Example: Customers server Queue/buffer 1) The transmitter: DTP = packet transmission time – Average number of packets at transmitter = λDTP = ρ = link utilization 2) The transmission line: Dp = propagation delay – Average number of packets in flight = λDp 3) The buffer: Dq =... Gives the probability that a caller finds all circuits busy PB = Eytan Modiano Slide 23 (λ / µ )m / m! (λ / µ )n / n! ∑n = 0 m Erlang B formula • Used for sizing transmission line – – How many circuits does the satellite need to support? The number of circuits is a function of the blocking probability that we can tolerate Systems are designed for a given load predictions and blocking probabilities (typically... small) • Example – Arrival rate = 4 calls per minute, average 3 minutes per call – How many circuits do we need to provision? Depends on the blocking probability that we can tolerate Circuits 20 15 7 Eytan Modiano Slide 24 PB 1% 8% 30% ... Modiano Slide 21 1 Delay formulas • M/G/1 λX 2 D=X+ 2( 1 − λ / µ ) • • Eytan Modiano Slide 22 λ/µ µ−λ M/D/1 D=X+ Service (transmission) time (LHS) Queueing delay (RHS) M/M/1 D=X+ Delay components: λ/µ 2( µ − λ ) Use Little’s Theorem to compute N, the average number of customers in the system Blocking Probability • A circuit switched network can be viewed as a Multi-server queueing system – – • M/G/m/m...Analyzing delay in networks (queueing theory) • Little’s theorem – – • Relates delay to number of users in the system Can be applied to any system Simple queueing systems (single server) – – • M/M/1, M/G/1, M/D/1 M/M/m/m Poisson Arrivals => – • λ = arrival rate in packets/second Exponential service time => – Eytan Modiano Slide 11 (λT)n e − λT P( n arrivals in interval... general service times M/D/1 – Eytan Modiano Slide 15 Poisson arrivals, exponential service times Poisson arrivals, deterministic service times (fixed) Markov Chain for M/M/1 system λδ 1−λδ 0 λδ 1 µδ λδ λδ 2 µδ k µδ µδ • State k => k customers in the system • P(I,j) = probability of transition from state I to state j – As δ => 0, we get: P(0,0) = 1 - λδ, P(j,j) = 1 - λδ −µδ P(j,j+1) = λδ P(j,j-1) = µδ P(I,j) . arrivals in T It can be shown that, E[n] = T E[n ] = T + ( T) = E[(n-E[n]) ] = E[n ]-E[n] = T 22 22 22 λ λλ σλ Eytan Modiano Slide 6 Inter-arrival times • Time that elapses between arrivals (IA) P(IA. Eytan Modiano Slide 1 16.36: Communication Systems Engineering Lecture 17/18: Delay Models for Data Networks Eytan Modiano Eytan Modiano Slide 2 Packet Switched Networks Packet Network PS PS PS PS PS PS PS Buffer Packet Switch Messages. average amount of time since the last departure is 20 minutes! • Paradox: If an average of 20 minutes passed since the last train arrived and an average of 20 minutes until the next train, then