EE 122: Network Performance, Queueing Theory, Evaluation Computer Science Division Department of Electrical Engineering and Computer Science University of California, Berkeley, Berkeley, CA 94720-1776 September 7, 2004

Overview Motivations Timing diagrams Queueing Theory Evaluation Techniques

Motivations Understanding network behavior Improving protocols Verifying correctness of implementation Detecting faults Monitor service level agreements Choosing providers Billing

Outline Motivations Timing diagrams Queueing Theory Evaluation techniques

Definitions Link bandwidth (capacity): maximum rate (in bps) at which the sender can send data along the link Propagation delay: time it takes the signal to travel from source to destination Packet transmission time: time it takes the sender to transmit all bits of the packet Queuing delay: time the packet need to wait before being transmitted because the queue was not empty when it arrived Processing Time: time it takes a router/switch to process the packet header, manage memory, etc

Sending One Packet R bits per second (bps) Bandwidth: R bps Propagation delay: T sec T seconds P bits time Transmission time = P/R T Propagation delay =T = Length/speed_of_light 1/speed = 3.3 nsec in free space 4 nsec in copper 5 nsec in fiber

Sending one Packet: Examples P = 1 Kbyte R = 1 Gbps 100 Km, fiber => T = 500 usec P/R = 8 usec T >> P/R T P/R time T << P/R T P = 1 Kbyte R = 100 Mbps 1 Km, fiber => T = 5 usec P/R = 80 usec P/R time

Queueing The queue has Q bits when packet arrives  packet has to wait for the queue to drain before being transmitted Capacity = R bps Propagation delay = T sec P bits Q bits Queueing delay = Q/R T P/R time

Queueing Example P bits Q bits P = 1 Kbit; R = 1 Mbps  P/R = 1 ms Time (ms) Packet arrival 0.5 1 7 7.5 Delay for packet that arrives at time t, d(t) = Q(t)/R + P/R Time # bits in queue: Q(t) 1 Kb 0.5 Kb 1.5 Kb 2 Kb packet 1, d(0) = 1ms packet 2, d(0.5) = 1.5ms packet 3, d(1) = 2ms

Router: Store and Forward A packet is stored (enqueued) before being forwarded (sent) 10 Mbps 5 Mbps 100 Mbps 10 Mbps Receiver Sender time

Store and Forward: Multiple Packet Example 10 Mbps 5 Mbps 100 Mbps 10 Mbps Receiver Sender time

Cut-Through A packet starts being forwarded (sent) as soon as its header is received R1 = 10 Mbps R2 = 10 Mbps Receiver Sender Header time What happens if R2 > R1 ?

Throughput Throughput of a connection or link = total number of bits successfully transmitted during some period [t, t + T) divided by T Link utilization = throughput of the link / link rate Bit rate units: 1Kbps = 103bps, 1Mbps = 106bps, 1Gbps = 109bps [For memory: 1 Kbyte = 210 bytes = 1024 bytes] Some rates are expressed in packets per second (pps)  relevant for routers/switches where the bottleneck is the header processing

Delay Delay (Latency) of bit (packet, file) from A to B Jitter The time required for bit (packet, file) to go from A to B Jitter Variability in delay Round-Trip Time (RTT) Two-way delay from sender to receiver and back Bandwidth-Delay product Product of bandwidth and delay  “storage” capacity of network

Bandwidth-Delay Product Source Destination Window-based flow control: Send W bits (window size) Wait for ACKs (i.e., make sure packets reached destination) Repeat Throughput = W/RTT  W = Throughput x RTT Numerical example: W = 64 Kbytes RTT = 200 ms Throughput = W/T = 2.6 Mbps RTT W RTT RTT time

Delay: Illustration 1 1 2 Latest bit seen by time t at point 2 Sender Receiver Latest bit seen by time t at point 2 at point 1 Delay time

Delay: Illustration 2 1 2 Packet arrival times at 1 1 2 Sender Receiver Packet arrival times at 1 1 2 Packet arrival times at 2 Delay time

Overview Motivations Timing diagrams Queueing Theory Little Theorem M/M/1 Queue Evaluation Techniques

Little’s Theorem Assume a system at which packets arrive at rate λ(t) Let d(i) be the delay of packet i , i.e., time packet i spends in the system What is the average number of packets in the system? d(i) = delay of packet i λ(t) – arrival rate system Intuition: Assume arrival rate is λ = 1 packet per second and the delay of each packet is d = 5 seconds What is the average number of packets N in the system?

Answer Answer: N = λ x d = 5 packets time 1 2 3 4 5 6 7 8 9 10 11 12 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16

Little’s Theorem 1 2 Latest bit seen by time t d(i) B(t) time T Sender Receiver d(i) = delay of packet i B(t) = number of bits in transit (in the system) at time t d(i) B(t) time T What is the system occupancy, i.e., average number of packets in transit between 1 and 2 ?

Little’s Theorem 1 2 Latest bit seen by time t d(i) B(t) A= area time Sender Receiver d(i) = delay of packet i B(t) = number of bits in transit (in the system) at time t d(i) B(t) A= area time T Average occupancy = A/T

Little’s Theorem 1 2 Latest bit seen by time t P B(t) A= area time T Sender Receiver d(i) = delay of packet i x(t) = number of packets in transit (in the system) at time t A(m) P d(m-1) A(m-1) B(t) A= area time T A = A(1) + A(2) + … + A(m) = P*(d(1) + d(2) + … + d(m))

Little’s Theorem 1 2 Latest bit seen by time t P B(t) A= area time Sender Receiver d(i) = delay of packet i B(t) = number of bits in transit (in the system) at time t A(m) P d(m-1) A(m-1) B(t) A= area time Average occupancy T A/T = (P*(d(1) + d(2) + … + d(m)))/T = ((P*m)/T) * ((d(1) + d(2) + … + d(m))/m) Arrival rate Average delay

Little’s Theorem 1 2 Latest bit seen by time t λ(i) B(t) A= area time Sender Receiver d(i) = delay of packet i x(t) = number of packets in transit (in the system) at time t A(m) λ(i) A(m-1) d(m-1) B(t) A= area time T Average occupancy = (arrival rate) x (average delay)

Overview Motivations Timing diagrams Queueing Theory Little Theorem M/M/1 Queue Evaluation Techniques

Problem How many clerks do you need to handle all these people?

? What do You Know? Customer arrival distribution 5 10 15 20 25 30 35 40 45 50 service time (sec) # of customers that have been served in more than 20 sec but less than 25 sec Customer arrival distribution Customer service distribution 1 2 3 4 5 6 7 8 9 10 11 inter-arrival time (sec) # of customers that have arrived 5 sec after the previous customer ?

What I Assume? Basic probability knowledge: Let v be a random distributed variable that takes n distinct values, and takes value vi with probability pi. Then Expected value of v is E(v) = Σi=0 vi*pi Σi=0 pi = 1 Let a an b be two random distributed variables. Then E(a+b) = E(a) + E(b) n n

Queueing System λ µ Queueing system: Parameters: Typical question: Server Queue λ µ Queueing system: A queue at which packets (customers) arrive A server that processes the packets (customers) Parameters: λ : arrival rate x : service time, i.e., time to serve (transmit) a customer (packet) µ = 1/E(x) : service rate (E(x) : mean service time) Typical question: What is the time (delay) spent by a customer (packet) in the system?

M / M / 1 Queue Arrival rate (λ): Poisson process Exponential distributed service Poisson arrival M / M / 1 Queue One server Arrival rate (λ): Poisson process Service time (x): Exponentially distributed Compute average time (d) spent by packet in system Two key properties of M/M/1 queue that makes the analysis easy: Memoryless property of exponential distribution PASTA property of Poisson process Server Queue λ µ

Poisson Arrival Process and Exponential Distribution Arrival process where inter-arrival time is exponentially distributed Assume t1, t2, t3, …, tn are arrival instances of n packets Then inter-arrival time yi = ti+1 – ti is exponentially distributed Exponential distribution is memoryless, i.e., the future does not depend on the past Example: Assume a Poisson process with an arrival rate of λ = 5 pkts/sec Then, at any instance of time t, the expected time until the next packet arrives is 1/λ = 1/5 = 0.2sec, irrespective of when previous packet has arrived!

PASTA Principle 1/2 Problem: Assume a system characterized by a time dependent variable X(t). Compute E(X) using n measurements. When do you need to take these measurements? Does periodic measurement work? Example: assume you want to measure the following variable, but you don’t know period. What do you do? 1 2 3 4 5 6 7 8 X(t)

PASTA Principle 2/2 PASTA: Poisson Arrival See Time Averages Performing measurements at times drawn from a Poisson process  allow to compute the mean value of the measured parameter Example Let N(ti) be the number of packets in the queue at time ti (1 ≤ i ≤ n) If ti is given by a Poisson process then (Σni=1N(ti))/n = E(N)

M/M/1 Queue Notations: dn = x1 + x2 + … + xn + z pn = Prob(N(t) = n): probability that queue size is n at time t rn = Prob(Na(t) = n): number of packets found by a packet arriving at t xi = service (transmission) time for packet i z = remaining transmission time of a packet at the arrival of a new packet dn = delay of a packet finding n packets in the queue dn = x1 + x2 + … + xn + z

M/M/1 Queue Notations: dn = x1 + x2 + … + xn + z pn = Prob(N(t) = n): probability that queue size is n at time t rn = Prob(Na(t) = n): number of packets found by a packet arriving at t xi = service (transmission) time for packet i z = remaining transmission time of a packet at the arrival of a new packet dn = delay of a packet finding n packets in the queue because exp. distribution, E(x) = E(z) dn = x1 + x2 + … + xn + z E(dn) = n*E(x) + E(z) = (n+1)*E(x)

M/M/1 Queue Notations: dn = x1 + x2 + … + xn + z pn = Prob(N(t) = n): probability that queue size is n at time t rn = Prob(Na(t) = n): number of packets found by a packet arriving at t xi = service (transmission) time for packet i z = remaining transmission time of a packet at the arrival of a new packet dn = delay of a packet finding n packets in the queue because exp. distribution, E(x) = E(z) dn = x1 + x2 + … + xn + z E(dn) = n*E(x) + E(z) = (n+1)*E(x) d = Σn=0 E(dn)*rn = Σn=0 (n+1)*E(x)*pn ∞ ∞ because PASTA, pn = rn

M/M/1 Queue Notations: dn = x1 + x2 + … + xn + z pn = Prob(N(t) = n): probability that queue size is n at time t rn = Prob(Na(t) = n): number of packets found by a packet arriving at t xi = service (transmission) time for packet i z = remaining transmission time of a packet at the arrival of a new packet dn = delay of a packet finding n packets in the queue dn = x1 + x2 + … + xn + z E(dn) = n*E(x) + E(z) = (n+1)*E(x) d = Σn=0 E(dn)*rn = Σn=0 (n+1)*E(x)*pn = E(x)*(Σn=0(n*pn + pn)) = E(x)*(Σn=0n*pn)+ E(x)*(Σn=0 pn)) = E(x)*(E(n)+1) because exp. distribution, E(x) = E(z) because PASTA, pn = rn ∞

M/M/1 Queue d = E(x)*(E(n) + 1) By Little result: E(n) = λ*d By definition: E(x) = 1/µ λ/µ = u: system utilization d = (1/µ)/(1 – λ/µ) d = (1/µ)/(1 – u) d 1/µ 1 u

Overview Motivations Timing diagrams Queueing Theory Evaluation Techniques

Evaluation Techniques Measurements gather data from a real network e.g., ping www.berkeley.edu realistic, specific Simulations: run a program that pretends to be a real network e.g., NS network simulator, Nachos OS simulator Models, analysis write some equations from which we can derive conclusions general, may not be realistic Usually use combination of methods

Evaluation: Putting Everything Together Abstraction Reality Model Plausibility Derivation Tractability Prediction Conclusion Hypothesis Realism Tradeoff between plausibility, tractability and realism Better to have a few realistic conclusions than none (could not derive) or many conclusions that no one believes (not plausible)