Tag: Ethernet Frame Structure

Ethernet in Networking | Practice Problems

Ethernet in Networking-

 

Before you go through this article, make sure that you have gone through the previous article on Ethernet.

 

We have discussed-

  • Ethernet is one of the standard LAN technologies used to build wired LANs.
  • Ethernet uses bus topology in which all the stations are connected to a half duplex link.
  • Ethernet uses CSMA / CD as an access control method.

 

In this article, we will discuss practice problems based on Ethernet.

 

PRACTICE PROBLEMS BASED ON ETHERNET-

 

Problem-01:

 

Which of the following characteristic is most basic to LAN?

  1. Bit rate
  2. Delay x Bandwidth Product
  3. Geographical distance
  4. Cost

 

Solution-

 

  • Geographical distance is the basic criteria on which networks are classified.
  • On the basis of geographical distance, networks are classified as LAN, MAN, WAN.
  • Thus,Option (C) is correct.

 

Problem-02:

 

On an Ethernet LAN when a collision is detected, the sending station-

  1. continues to send the transmission
  2. temporarily quits the transmission
  3. notifies the destination of an error
  4. permanently quits the transmission

 

Solution-

 

  • Ethernet uses CSMA / CD as access control method.
  • On detecting a collision, the sending station temporarily quits the transmission.
  • Transmitting station waits for Back Off time and then tries again.
  • Thus, Option (B) is correct.

 

Problem-03:

 

Ethernet implements _________ service for its operation.

  1. connection oriented
  2. connection less
  3. Both A and B
  4. Either A or B

 

Solution-

 

REMEMBER

  • Connection oriented service involves allocation of the dedicated resources.
  • Connection less service does not involve allocation of dedicated resources.
  • TCP and Virtual Circuits are connection oriented services.
  • IP, Ethernet and Token Ring are connection less services.
  • Datagrams are connection less, that is why IP is connection less.

 

When an Ethernet frame is sent,

  • Destination is never expected to reserve the buffer or any other resource for the incoming frame.
  • The data is simply dumped at the destination side.
  • So, it is connectionless.
  • Thus, Option (B) is correct.

 

Problem-04:

 

The collision domain of Fast Ethernet is limited to ______ meters.

  1. 2.5
  2. 25
  3. 250
  4. 2500

 

Solution-

 

  • Collision domain defines the number of stations that can get involved in the collision when connected to a LAN.
  • In the given question, collision domain refers to maximum distance a LAN can run to detect the collisions.
  • Ethernet uses CSMA / CD as access control method.

 

In CSMA / CD, condition to detect collisions is-

Distance <= (Length x speed) / (2 x bandwidth)

 

On substituting the values, we get the value of distance.

 

REMEMBER

  • For normal Ethernet, collision domain = 2500 meters.
  • For Fast Ethernet, collision domain = 250 meters.
  • For Gigabit Ethernet, collision domain = 25 meters.

 

Thus, Option (C) is correct.

 

Problem-05:

 

The efficiency of Ethernet-

  1. increases when propagation delay and transmission delay are low
  2. increases when propagation delay and transmission delay are high
  3. increases when propagation delay is low and transmission delay is high
  4. increases when propagation delay is high and transmission delay is low

 

Solution-

 

  • Efficiency of Ethernet = 1 / ( 1 + 6.44a) where a = Tp / Tt.
  • Thus, Option (C) is correct.

 

Problem-06:

 

What is the baud rate of the standard 10 Mbps 802.3 LAN?

  1. 20 mega baud
  2. 10 mega baud
  3. 25 mega baud
  4. 40 mega baud

 

Solution-

 

LAN uses Manchester Encoding Technique where-

Baud rate = 2 x Bit rate

 

For 10 Mbps,

Baud rate

= 2 x 10 mega baud

= 20 mega baud

Thus, Option (A) is correct.

 

Problem-07:

 

Consider a 10 Mbps Ethernet LAN that has stations attached to a 2.5 km long coaxial cable. Given that the transmission speed is 2.3 x 108 m/sec, the packet size is 128 bytes out of which 30 bytes are overhead, find the effective transmission rate and maximum rate at which the network can send data.

 

Solution-

 

Given-

  • Bandwidth = 10 Mbps
  • Distance = 2.5 km
  • Transmission speed = 2.3 x 108 m/sec
  • Total packet size = 128 bytes
  • Overhead = 30 bytes

 

Calculating Transmission Delay-

 

Transmission delay (Tt)

= Packet size / Bandwidth

= 128 bytes / 10 Mbps

= (128 x 8 bits) / (10 x 106 bits per sec)

= 1024 / 107 sec

= 102.4 μsec

 

Calculating Propagation Delay-

 

Propagation delay (Tp)

= Distance / Speed

= 2.5 km / (2.3 x 108 m/sec)

= (2.5 x 103 m) / (2.3 x 108 m/sec)

= 1.08 x 10-5 sec

= 10.8 μsec

 

Calculating Value of ‘a’-

 

a

= Tp / Tt

= 10.8 μsec / 102.4 μsec

= 0.105

 

Calculating Efficiency-

 

Efficiency(η)

= 1 / (1 + 6.44 x a)

= 1 / (1 + 6.44 x 0.105)

= 1 / 1.67

= 0.59

= 59%

 

Calculating Maximum Rate-

 

Maximum rate or Throughput

= Efficiency x Bandwidth

= 0.59 x 10 Mbps

= 5.9 Mbps

 

Calculating Effective Transmission Rate-

 

Effective transmission rate

= Throughput x (128-30 / 128)

= 5.9 Mbps x (98 / 128)

= 0.77 x 5.9 Mbps

= 4.52 Mbps

 

Problem-08:

 

The following frame transition diagram shows an exchange of Ethernet frames between two computers, A and B connected via a 10BT Hub. Each frame sent by computer A contains 1500 B of Ethernet payload data, while each frame sent by computer B contains 40 B of Ethernet payload data. Calculate the average utilization of the media during this exchange.

 

 

  1. 10%
  2. 1.7%
  3. 20%
  4. 15.2%

 

Solution-

 

Calculating Data Sent By Computer A in One Frame-

 

Given-

  • Each frame sent by computer A contains 1500 B of Ethernet payload data.
  • This is divided as: 20 bytes of IP Header + 20 bytes of TCP Header + 1460 bytes of data.

 

So, Total bytes sent by computer A in one frame

= Preamble + SFD + Ethernet Header + Ethernet Payload + CRC

= 7 bytes + 1 byte + 14 bytes + 1500 bytes + 4 bytes

= 1526 bytes

 

Calculating Data Sent By Computer A in 0.6 Seconds:

 

Computer A sends 8 frames in 0.6 seconds.

So, Total bytes sent by computer A in 0.6 seconds

= 8 x 1526 bytes

= 12208 bytes

 

Calculating Data Sent By Computer B in One Frame-

 

Given-

  • Each frame sent by computer B contains 40 B of Ethernet payload data.
  • This is divided as: 20 bytes of IP Header + 20 bytes of TCP Header + 0 bytes of data.
  • Since minimum data in the payload field of Ethernet must 46 bytes. So, extra 6 bytes are padded.

 

So, Total bytes sent by computer B in one frame

= Preamble + SFD + Ethernet Header + Ethernet Payload + CRC

= 7 bytes + 1 byte + 14 bytes + (40 bytes + 6 bytes) + 4 bytes

= 72 bytes

 

Calculating Data Sent By Computer B in 0.6 Seconds-

 

Computer B sends 4 frames in 0.6 seconds.

So, Total bytes sent by computer B in 0.6 seconds

= 4 x 72 bytes

= 288 bytes

 

Calculating Total Data Sent in 0.6 Seconds:

 

Total data flow that takes place in 0.6 seconds

= Total data sent by computer A in 0.6 seconds + Total data sent by computer B in 0.6 seconds

= 12208 bytes + 288 bytes

= 12496 bytes

= 99968 bits

 

Calculating Throughput-

 

Throughput

= Amount of data that flows per second

= 99968 bits / 0.6 seconds

= 166613.33 bits/sec

 

Calculating Utilization-

 

Throughput = Efficiency x Bandwidth

So, Efficiency or Utilization

= Throughput / Bandwidth

= (166613.33 bits per sec) / 10 Mbps

= 0.017

= 1.7%

 

Thus, Option (B) is correct.

 

Problem-09:

 

Ethernet adaptor receives all frames and accepts-

  1. Frames addressed to its own address
  2. Frames addressed to the multicast or broadcast address
  3. Frames if it has been placed in promiscuous mode
  4. All of the above

 

Solution-

 

In a bus topology Ethernet,

  • Ethernet Adaptor enables a computer to access an Ethernet Network.
  • If one station sends a frame to other station, then other stations & Ethernet Adaptor also receives that frame.
  • But they accept only those frames which are destined for them.
  • Ethernet Adaptor accepts all those frames which are addressed to its own address or broadcast address or multicast address (if it is present in that multicast group)
  • Network administrator may set the network in promiscuous mode.
  • This is done to monitor the activities going on in the network.
  • So, if Ethernet Adaptor is set in promiscuous mode, it receives and accepts all the frames.
  • Thus, Option (D) is correct.

 

Next Article- Types of Switching | Circuit Switching

 

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Ethernet in Networking | Ethernet Frame Format

Local Area Network-

 

  • A Local Area Network (LAN) is a network of computers.
  • It is confined to a small area which may be a room, building or a group of buildings.
  • A LAN may be wired, wireless or a combination of the two.

 

LAN Technologies-

 

Standard technologies used to build a wired LAN are-

 

 

  1. Ethernet
  2. Token Ring

 

In this article, we will discuss about Ethernet and its Frame Format.

 

Ethernet-

 

  • Ethernet is one of the standard LAN technologies used for building wired LANs.
  • It is defined under IEEE 802.3.

 

Characteristics-

 

Point-01:

 

  • Ethernet uses bus topology.
  • In bus topology, all the stations are connected to a single half duplex link.

 

 

Point-02:

 

  • Ethernet uses CSMA / CD as access control method to deal with the collisions.

 

Point-03:

 

  • Ethernet uses Manchester Encoding Technique for converting data bits into signals.

 

Point-04:

 

  • For Normal Ethernet, operational bandwidth is 10 Mbps.
  • For Fast Ethernet, operational bandwidth is 100 Mbps.
  • For Gigabit Ethernet, operational bandwidth is 1 Gbps.

 

Ethernet Frame Format-

 

IEEE 802.3 defines the following Ethernet frame format-

 

 

1. Preamble-

 

  • It is a 7 byte field that contains a pattern of alternating 0’s and 1’s.
  • It alerts the stations that a frame is going to start.
  • It also enables the sender and receiver to establish bit synchronization.

 

2. Start Frame Delimiter (SFD)-

 

  • It is a 1 byte field which is always set to 10101011.
  • The last two bits “11” indicate the end of Start Frame Delimiter and marks the beginning of the frame.

 

NOTES

  • The above two fields are added by the physical layer and represents the physical layer header.
  • Sometimes, Start Frame Delimiter (SFD) is considered to be a part of Preamble.
  • That is why, at many places, Preamble field length is described as 8 bytes.

 

3. Destination Address-

 

  • It is a 6 byte field that contains the MAC address of the destination for which the data is destined.

 

4. Source Address-

 

  • It is a 6 byte field that contains the MAC address of the source which is sending the data.

 

5. Length-

 

  • It is a 2 byte field which specifies the length (number of bytes) of the data field.
  • This field is required because Ethernet uses variable sized frames.

 

NOTES

  • The maximum value that can be accommodated in this field = 216 – 1 = 65535.
  • But it does not mean maximum data that can be sent in one frame is 65535 bytes.
  • The maximum amount of data that can be sent in a Ethernet frame is 1500 bytes.
  • This is to avoid the monopoly of any single station.

 

The following three fields collectively represents the Ethernet Header

  • Destination Address (6 bytes)
  • Source Address (6 bytes)
  • Length (2 bytes)

Thus, Ethernet Header Size = 14 bytes.

 

6. Data-

 

  • It is a variable length field which contains the actual data.
  • It is also called as a payload field.
  • The length of this field lies in the range [ 46 bytes , 1500 bytes ].
  • Thus, in a Ethernet frame, minimum data has to be 46 bytes and maximum data can be 1500 bytes.

 

Minimum Length of Data Field

 

  • Ethernet uses CSMA / CD as access control method to deal with collisions.
  • For detecting the collisions, CSMA / CD requires-

Minimum length of data packet = 2 x Propagation delay x Bandwidth

  • Substituting the standard values of Ethernet, it is found that minimum length of the Ethernet frame has to be 64 bytes starting from the destination address field to the CRC field and 72 bytes including the Preamble and SFD fields.
  • Therefore, minimum length of the data field has to be = 64 bytes – (6+6+2+4) bytes = 46 bytes

 

Maximum Length of Data Field

 

  • The maximum amount of data that can be sent in a Ethernet frame is 1500 bytes.
  • This is to avoid the monopoly of any single station.
  • If Ethernet allows the frames of big sizes, then other stations may not get the fair chance to send their data.

 

7. Frame Check Sequence (CRC)-

 

  • It is a 4 byte field that contains the CRC code for error detection.

 

Advantages of Using Ethernet-

 

  • It is simple to understand and implement.
  • Its maintenance is easy.
  • It is cheap.

 

Limitations of Using Ethernet-

 

Point-01:

 

  • It can not be used for real time applications.
  • Real time applications require the delivery of data within some time limit.
  • Ethernet is not reliable because of high probability of collisions.
  • High number of collisions may cause a delay in delivering the data to its destination.

 

Point-02:

 

  • It can not be used for interactive applications.
  • Interactive applications like chatting requires the delivery of even very small amount of data.
  • Ethernet requires that minimum length of the data must be 46 bytes.

 

Point-03:

 

  • It can not be used for client server applications.
  • Client server applications require that server must be given higher priority than clients.
  • Ethernet has no facility to set priorities.

 

Token Ring overcomes these limitations of Ethernet.

 

Important Concept-

 

For data transmission-

  • TCP segment sits inside the IP datagram payload field.
  • IP datagram sits inside the Ethernet payload field.

 

 

Next Article- Practice Problems On Ethernet

 

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