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

2. Overview Motivations Timing diagrams Queueing Theory
Evaluation Techniques

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

4. Outline Motivations Timing diagrams Queueing Theory
Evaluation techniques

5. 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

6. 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

7. 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

8. 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

9. 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

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

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

12. 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 ?

13. 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

14. 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

15. 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

16. 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

17. 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

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

19. 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?

20. Answer Answer: N = λ x d = 5 packets time 1 2 3 4 5 6 7 8 9 10 11 121 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16

21. 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 ?

22. 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

23. 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))

24. 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

25. 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)

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

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

28. ? What do You Know? Customer arrival distribution5 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 ?

29. 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

30. Queueing System λ µ Queueing system: Parameters: Typical question:
Server
Queue λ
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?

31. 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

32. 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!

33. 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)

34. 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)

35. 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

36. 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)

37. 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

38. 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 ∞

39. 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

40. Overview Motivations Timing diagrams Queueing Theory
Evaluation Techniques

41. Evaluation Techniques
Measurements
gather data from a real network
e.g., ping
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

42. 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)