Local Area Networks Ethernet-Data Communication Systems-Assignment Solution, Exercises of Data Communication Systems and Computer Networks

This file contains solution to problems related Data Communication Systems. Mr. Prajin Ahuja assigned task at Birla Institute of Technology and Science. Its main points are: Ethernet, Preamble, Broadcast, Unicast, Standard, Fast, Gigabit, Possible, Frames, Address, Equipment, Propagation

Typology: Exercises

2011/2012

Uploaded on 07/26/2012

waania
waania 🇮🇳

4.4

(31)

119 documents

1 / 2

Toggle sidebar

This page cannot be seen from the preview

Don't miss anything!

bg1
1
CHAPTER 13
Local Area Networks: Ethernet
Solutions to Odd-Numbered Review Questions and Exercises
Review Questions
1. The preamble is a 56-bit field that provides an alert and timing pulse. It is added to
the frame at the physical layer and is not formally part of the frame. SFD is a one-
byte field that serves as a flag.
3. A multicast address identifies a group of stations; a broadcast address identifies
all stations on the network. A unicast address identifies one of the addresses in a
group.
5. A layer-2 switch is an N-port bridge with additional sophistication that allows
faster handling of packets.
7. The rates are as follows:
9. The common Fast Ethernet implementations are 100Base-TX, 100Base-FX, and
100Base-T4.
11. The common Ten-Gigabit Ethernet implementations are 10GBase-S, 10GBase-L,
and 10GBase-E.
Exercises
13. The bytes are sent from left to right. However, the bits in each byte are sent from
the least significant (rightmost) to the most significant (leftmost). We have shown
the bits with spaces between bytes for readability, but we should remember that
that bits are sent without gaps. The arrow shows the direction of movement.
01011000 11010100 00111100 11010010 01111010 11110110
Standard Ethernet: 10 Mbps
Fast Ethernet: 100 Mbps
Gigabit Ethernet:
Ten-Gigabit Ethernet: 1 Gbps
10 Gbps
docsity.com
pf2

Partial preview of the text

Download Local Area Networks Ethernet-Data Communication Systems-Assignment Solution and more Exercises Data Communication Systems and Computer Networks in PDF only on Docsity!

1

CHAPTER 13

Local Area Networks: Ethernet

Solutions to Odd-Numbered Review Questions and Exercises

Review Questions

  1. The preamble is a 56-bit field that provides an alert and timing pulse. It is added to the frame at the physical layer and is not formally part of the frame. SFD is a one- byte field that serves as a flag.
  2. A multicast address identifies a group of stations; a broadcast address identifies all stations on the network. A unicast address identifies one of the addresses in a group.
  3. A layer-2 switch is an N-port bridge with additional sophistication that allows faster handling of packets.
  4. The rates are as follows:
  5. The common Fast Ethernet implementations are 100Base-TX , 100Base-FX , and 100Base-T.
  6. The common Ten-Gigabit Ethernet implementations are 10GBase-S , 10GBase-L , and 10GBase-E.

Exercises

  1. The bytes are sent from left to right. However, the bits in each byte are sent from the least significant (rightmost) to the most significant (leftmost). We have shown the bits with spaces between bytes for readability, but we should remember that that bits are sent without gaps. The arrow shows the direction of movement. ← 01011000 11010100 00111100 11010010 01111010 11110110

Standard Ethernet: 10 Mbps Fast Ethernet: 100 Mbps Gigabit Ethernet: Ten-Gigabit Ethernet:

1 Gbps 10 Gbps

docsity.com

2

  1. The first byte in binary is 0100001 1. The least significant bit is 1. This means that the pattern defines a multicast address. A multicast address can be a destination address, but not a source address. Therefore, the receiver knows that there is an error, and discards the packet.
  2. The maximum data size in the Standard Ethernet is 1500 bytes. The data of 1510 bytes, therefore, must be split between two frames. The standard dictates that the first frame must carry the maximum possible number of bytes (1500); the second frame then needs to carry only 10 bytes of data (it requires padding). The follow- ing shows the breakdown: Data size for the first frame: 1500 bytes Data size for the second frame: 46 bytes (with padding)
  3. We can calculate the propagation time as t = (2500 m) / (200,000.000) = 12.5 μ s. To get the total delay, we need to add propagation delay in the equipment (10 μs). This results in T = 22.5 μ s.

docsity.com