This document discusses various performance metrics for computer networks including bandwidth, throughput, latency, jitter, and bandwidth-delay product. It provides examples to explain these terms and how they are calculated. Bandwidth refers to the capacity of a link to transmit data and is measured in bits per second. Throughput measures the actual rate of successful message delivery over a communication channel. Latency is the delay between when a packet is sent and received, including propagation, transmission, queueing, and processing delays. Jitter and bandwidth-delay product are also introduced.

1. CS – 8591
COMPUTER NETWORKS
Unit - I
INTRODUCTION AND
PHYSICAL LAYER
Ms. Angayarkanni S A
Assistant Professor, IT, RMKEC

2. Performance
 How GOOD the network is?

3. Performance metrics
Bandwidth
Throughput
Latency (Delay)
Bandwidth-Delay Product
Jitter

4. Bandwidth
 two contexts
 Bandwidth in hertz,
 the range of frequencies in a composite
signal or the range of frequencies that a
channel can pass.
 Bandwidth in bits per second,
 the speed of bit transmission in a
channel or link.

5. Example 1
 ABC Telephone
 The bandwidth of a subscriber line is 4
kHz for voice or data.
 The bandwidth of this line for data
transmission can be up to 56,000 bps
using a sophisticated modem to
change the digital signal to analog.
 BHz=4 KHz
 Bbps=56 Kbps

6. Bandwidth
 BHz Increases
 Bbps increases

7. Example 2
 ABC Telephone
 The bandwidth of a subscriber line is 4
kHz for voice or data.
◦ (8 KHz)
 The bandwidth of this line for data
transmission can be up to 112,000 bps
using a sophisticated modem to change
the digital signal to analog.
 BHz=8 KHz
 Bbps=112 Kbps

8. Throughput
 a measure of how fast we send data
through a network

10.  Bandwidth Bbps
◦ Potential measurement of a link
 Throughput Tbps
◦ Actual measurement of a data transfer
through a link

11. Example 3
 A network with bandwidth of 10 Mbps
can pass only an average of 12,000
frames per minute with each frame
carrying an average of 10,000 bits.
What is the throughput of this
network?
 Bbps= 10 Mbps
 No of frames = 12000 frames/minute
 1 frame = 10000bits

12. Example 3…
 Bbps=10 Mbps
 No of frames = 12000 frames/minute
 1 frame = 10000bits
 Throughput =
No of frames/ sec * No of bits in a
frame
 = (12000*10000)/60
 =2 * 106 bps = 2 Mbps
The throughput is almost one-fifth of the
bandwidth in this case.

13. Latency (Delay)
 How long it takes for an entire message
to completely arrive at the destination
from the time the first bit is sent out from
the source
 Latency = Propagation time +
transmission time + Queuing time +
Processing delay

14. Propagation time
 Measures the time required for a bit to
travel from source to destination
 Propagation speed  speed at which
a bit travels though the medium from
source to destination.
 Propagation time = distance/
propagation speed

15. Propagation time
 Depends on nature of signal and
medium
 Light
 Vacuum – 3 * 108 m/s
 Air – get reduced

16. Example 4
 What is the propagation time if the
distance between the two points is
12,000 km? Assume the propagation
speed to be 2.4 × 108 m/s in cable?
 Distance = 12000 km
 Propagation speed = 2.4 × 108 m/s

17. Example 4
 Distance = 12000 km
 Propagation speed = 2.4 × 108 m/s
 Propagation Delay =
Distance/Propagation speed
a bit can go over the Atlantic Ocean in only
50 ms if there is a direct cable between the
source and the destination.

18. Transmission time
 Time taken to put the data packet on
the transmission link is called
as transmission delay.
 Transmission speed the speed at
which all the bits in a message arrive
at the destination. (difference in arrival
time of first and last bit)
 Transmission = Message-size/
Bandwidth

19. Example
 What are the propagation time and the
transmission time for a 2.5-kbyte
message (an e-mail) if the bandwidth
of the network is 1 Gbps? Assume that
the distance between the sender and
the receiver is 12,000 km and that
light travels at 2.4 × 108 m/s?

20. Example …
 Message size =2.5 KB
 Bbps = 1 Gbps
 Distance = 12000km
 Speed = 2.4 * 10 8 m/s
 Propagation Delay = Distance/Propagation
speed
 = (12000*10000)/(2.4*108)
 = 0.05 s = 50 ms

21. Example …
 Message size =2.5 KB
 Bbps = 1 Gbps
 Distance = 12000km
 Speed = 2.4 * 10 8 m/s
 Transmission = Message-size/ Bandwidth
 = (2.5*1000 * 8)/(1*109)
 = 20* 10 -6
 = 0.020 ms

22. Example
 What are the propagation time and the
transmission time for a 5MB message
(an image) if the bandwidth of the
network is 1 Mbps? Assume that the
distance between the sender and the
receiver is 12,000 km and that light
travels at 2.4 × 108 m/s?

23. Example …
 Message size =2.5 KB
 Bbps = 1 Gbps
 Distance = 12000km
 Speed = 2.4 * 10 8 m/s
 Propagation Delay = Distance/Propagation
speed
 = (12000*10000)/(2.4*108)
 = 0.05 s = 50 ms

24. Example …
 Message size =2.5 KB
 Bbps = 1 Gbps
 Distance = 12000km
 Speed = 2.4 * 10 8 m/s
 Transmission = Message-size/ Bandwidth
 = (5*106 * 8)/(1*106)
 = 40 s

25. Inference
 Message small
 Tx time << Propagation time
 Message large
 Tx time >> Propagation time
 Msg size increases & BW is not
enough  Tx time is more dominant
then propagation time

26. Queuing time
 Time needed for each intermediate or
end device to hold the message
before it is processes
 Time spent by the data packet waiting
in the queue before it is taken for
execution
 Not a fixed factor
 Changes with network load
(congestion)

27. Processing delay
 Time taken by the processor to
process the data packet is called
as processing delay.
 It depends on the speed of the
processor.
 Processing of the data packet helps in
detecting bit level errors that occurs
during transmission.

28. Jitter
 refers to small intermittent delays
during data transfers.
 It can be caused by a number of
factors including network congestion,
collisions, and signal interference.
 Technically, jitter is the variation in
latency — the delay between when a
signal is transmitted and when it is
received

29. Jitter

30. Bandwidth delay product
No of bits in the line = 5 bits

31. Bandwidth delay product
No of bits in the line = 25 bits

32. Pipe

34. Performance metrics
 Bandwidth - capacity of the system
 Throughput - no. of bits that can be
pushed through
 Latency (Delay) - delay incurred by a
bit from start to finish
 Bandwidth-Delay Product- no of bits
in line

35.  Which of the following delay is faced
by the packet in travelling from one
end system to another?
a) Propagation delay
b) Queuing delay
c) Transmission delay
d) All of the mentioned
1.35
Q-A

36.  For a 10Mbps Ethernet link, if the
length of the packet is 32bits, the
transmission delay is ____________
(in microseconds)
a) 3.2
b) 32
c) 0.32
d) 320
1.36
Q-A

37.  For a 10Mbps Ethernet link, if the
length of the packet is 32bits, the
transmission delay is ____________
(in microseconds)
a) 3.2
b) 32
c) 0.32
d) 320
1.37
Q-A

38.  In the transfer of file between server
and client, if the transmission rates
along the path is 10Mbps, 20Mbps,
30Mbps, 40Mbps. The throughput is
usually ___________
a) 20Mbps
b) 10Mbps
c) 40Mbps
d) 50Mbps
1.38
Q-A

39.  In the transfer of file between server
and client, if the transmission rates
along the path is 10Mbps, 20Mbps,
30Mbps, 40Mbps. The throughput is
usually ___________
a) 20Mbps
b) 10Mbps
c) 40Mbps
d) 50Mbps
1.39
Q-A
The throughput
is generally the
transmission
rate of
bottleneck link.

40.  Propagation delay depends on
___________
a) Packet length
b) Transmission rate
c) Distance between the routers
d) Speed of the CPU
1.40
Q-A

41.  Propagation delay depends on
___________
a) Packet length
b) Transmission rate
c) Distance between the routers
d) Speed of the CPU
1.41
Q-A

42.  Propagation delay depends on
___________
a) Packet length
b) Transmission rate
c) Distance between the routers
d) Speed of the CPU
1.42
Q-A