Category: Computer Networks

Subnet Mask | Subnet Mask Use

Computer Networks

Subnet Mask-

 

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

 

We have discussed-

  • Subnet mask is a 32 bit address consisting of a sequence of 1’s and 0’s.
  • It is used to determine to which subnet the given IP Address belongs to.

 

In this article, we will discuss how subnet mask is used by the host assigned to it.

 

Concept To Know-

 

When any host connects to the internet, ISP provides following 4 things to the host-

  1. IP Address
  2. Default Gateway
  3. Subnet Mask
  4. DNS

 

1. IP Address-

 

  • ISP assigns an IP Address to the host so that it can be uniquely identified on the Internet.

 

2. Default Gateway-

 

  • Default router connected to the network in which the host is present is the default gateway for the host.

 

3. Subnet Mask-

 

  • Subnet mask is a 32 bit number that is assigned to the host.
  • It is used to determine to which network the given IP Address belongs to.

 

4. Domain Name Service (DNS)-

 

 

Subnet Mask Use-

 

Subnet mask is used to determine to which network the given IP Address belongs to.

 

  • Host use its subnet mask to determine whether the other host it wants to communicate with is present within the same network or not.
  • If the destination host is present within the same network, then source host sends the packet directly to the destination host.
  • If the destination host is present in some other network, then source host routes the packet to the default gateway (router).
  • Router then sends the packet to the destination host.

 

Example-

 

Consider-

  • There is a host A present in some network X.
  • There is a host B.
  • Host A wants to send a packet to host B.

 

Before transmitting the packet, host A determines whether host B is present within the same network or not.

 

 

Here,

  • Host A = Source host
  • Host B = Destination host

 

To determine whether destination host is present within the same network or not, source host follows the following steps-

 

Step-01:

 

  • Source host computes its own network address using its own IP Address and subnet mask.
  • After computation, source host obtains its network address with respect to itself.

 

Step-02:

 

  • Source host computes the network address of destination host using destination IP Address and its own subnet mask.
  • After computation, source host obtains the network address of destination host with respect to itself.

 

Step-03:

 

Source host compares the two results obtained in the above steps.

Then, following two cases are possible-

 

Case-01:

 

If the results are same,

  • Source host assumes that the destination host is present within the same network.
  • Source host sends the packet directly to the destination host.

 

Case-02:

 

If the results are different,

  • Source host assumes that the destination host is present in some other network.
  • Source host sends the packet via router to the destination host.

 

Important Points-

 

Note-01:

 

  • Each host knows only its own subnet mask.
  • It does not know the subnet mask of any other host.

 

Note-02:

 

  • The conclusion drawn by a host about the presence of other host within the same or different network might be wrong.

 

Note-03:

 

  • Consider host A draws some conclusion about host B.
  • Then, same conclusion might not be drawn by host B about host A.
  • Both the hosts have to perform the above procedure separately at their ends to conclude anything.

 

PRACTICE PROBLEMS BASED ON USE OF SUBNET MASK-

 

Problem-01:

 

Two computers C1 and C2 are configured as follows-

  • C1 has IP Address 203.197.2.53 and net mask 255.255.128.0
  • C2 has IP Address 203.197.75.201 and net mask 255.255.192.0

 

Which one of the following statements is true?

  1. C1 and C2 both assume they are on the same network
  2. C2 assumes C1 is on same network but C1 assumes C2 is on a different network
  3. C1 assumes C2 is on same network but C2 assumes C1 is on a different network
  4. C1 and C2 both assume they are on different networks

 

Solution-

 

At Computer C1-

 

C1 computes its network address using its own IP Address and subnet mask as-

203.197.2.53 AND 255.255.128.0

= 203.197.0.0

 

C1 computes the network address of C2 using IP Address of C2 and its own subnet mask as-

203.197.75.201 AND 255.255.128.0

= 203.197.0.0

 

Since both the results are same, so C1 assumes that C2 is on the same network.

 

At Computer C2-

 

C2 computes its network address using its own IP Address and subnet mask as-

203.197.75.201 AND 255.255.192.0

= 203.197.64.0

 

C2 computes the network address of C1 using IP Address of C1 and its own subnet mask as-

203.197.2.53 AND 255.255.192.0

= 203.197.0.0

 

Since both the results are different, so C2 assumes that C1 is on a different network.

Thus, Option (C) is correct.

 

Problem-02:

 

The subnet mask for a particular network is 255.255.31.0. Which of the following pairs of IP Addresses could belong to this network?

  1. 172.57.88.62 and 172.56.87.233
  2. 10.35.28.2 and 10.35.29.4
  3. 191.203.31.87 and 191.234.31.88
  4. 128.8.129.43 and 128.8.161.55

 

Solution-

 

Let the given two IP Addresses belong to Host A and Host B.

 

Checking Option (A)-

 

  • Host A IP Address = 172.57.88.62
  • Host B IP Address = 172.56.87.233

 

At Host A-

 

Host A computes its network address using its own IP Address and subnet mask-

172.57.88.62 AND 255.255.31.0

= 172.57.24.0

 

Host A computes the network address of Host B using IP Address of Host B and its own subnet mask-

172.56.87.233 AND 255.255.31.0

= 172.56.23.0

 

Since both the results are different, so host A assumes that host B is on a different network.

Thus, both can’t belong to the same network.

Hence, this option gets eliminated.

 

Checking Option (B)-

 

  • Host A IP Address = 10.35.28.2
  • Host B IP Address = 10.35.29.4

 

At Host A-

 

Host A computes its network address using its own IP Address and subnet mask-

10.35.28.2 AND 255.255.31.0

= 10.35.28.0

 

Host A computes the network address of Host B using IP Address of Host B and its own subnet mask-

10.35.29.4 AND 255.255.31.0

= 10.35.29.0

 

Since both the results are different, so host A assumes that host B is on a different network.

Thus, both can’t belong to the same network.

Hence, this option gets eliminated.

 

Checking Option (C)-

 

  • Host A IP Address = 191.203.31.87
  • Host B IP Address = 191.234.31.88

 

At Host A-

 

Host A computes its network address using its own IP Address and subnet mask-

191.203.31.87 AND 255.255.31.0

= 191.203.31.0

 

Host A computes the network address of Host B using IP Address of Host B and its own subnet mask-

191.234.31.88 AND 255.255.31.0

= 191.234.31.0

 

Since both the results are same, so host A assumes that host B is on the same network.

 

At Host B-

 

Host B computes its network address using its own IP Address and subnet mask-

191.234.31.88 AND 255.255.31.0

= 191.234.31.0

 

Host B computes the network address of Host A using IP Address of Host A and its own subnet mask-

191.203.31.87 AND 255.255.31.0

= 191.203.31.0

 

Since both the results are different, so host B assumes that host A is on a different network.

Thus, both can’t belong to the same network.

Hence, this option gets eliminated.

 

Checking Option (D)-

 

  • Host A IP Address = 128.8.129.43
  • Host B IP Address = 128.8.161.55

 

At Host A-

 

Host A computes its network address using its own IP Address and subnet mask-

128.8.129.43 AND 255.255.31.0

= 128.8.1.0

 

Host A computes the network address of Host B using IP Address of Host B and its own subnet mask-

128.8.161.55 AND 255.255.31.0

= 128.8.1.0

 

Since both the results are same, so host A assumes that host B is on the same network.

 

At Host B-

 

Host B computes its network address using its own IP Address and subnet mask-

128.8.161.55 AND 255.255.31.0

= 128.8.1.0

 

Host B computes the network address of Host A using IP Address of Host A and its own subnet mask-

128.8.129.43 AND 255.255.31.0

= 128.8.1.0

 

Since both the results are same, so host B assumes that host A is on the same network.

Thus, both the hosts assume that they belong to the same network.

Hence, Option (D) is correct.

 

To gain better understanding about Subnet Mask Use,

Watch this Video Lecture

 

Next Article- Internet Protocol Version 4 | IPv4 Header

 

Get more notes and other study material of Computer Networks.

Watch video lectures by visiting our YouTube channel LearnVidFun.

Classless Addressing | CIDR in Networking

Computer Networks

IP Addressing-

 

There are two systems in which IP Addresses are classified-

 

 

  1. Classful Addressing System
  2. Classless Addressing System

 

In this article, we will discuss about Classless Addressing System.

Learn about Classful Addressing System.

 

Classless Addressing-

 

  • Classless Addressing is an improved IP Addressing system.
  • It makes the allocation of IP Addresses more efficient.
  • It replaces the older classful addressing system based on classes.
  • It is also known as Classless Inter Domain Routing (CIDR).

 

CIDR Block-

 

When a user asks for specific number of IP Addresses,

  • CIDR dynamically assigns a block of IP Addresses based on certain rules.
  • This block contains the required number of IP Addresses as demanded by the user.
  • This block of IP Addresses is called as a CIDR block.

 

Rules For Creating CIDR Block-

 

A CIDR block is created based on the following 3 rules-

 

Rule-01:

 

  • All the IP Addresses in the CIDR block must be contiguous.

 

Rule-02:

 

  • The size of the block must be presentable as power of 2.
  • Size of the block is the total number of IP Addresses contained in the block.
  • Size of any CIDR block will always be in the form 21, 22, 23, 24, 25 and so on.

 

Rule-03:

 

  • First IP Address of the block must be divisible by the size of the block.

 

REMEMBER

 

If any binary pattern consisting of (m + n) bits is divided by 2n, then-

  • Remainder is least significant n bits
  • Quotient is most significant m bits

 

So, any binary pattern is divisible by 2n, if and only if its least significant n bits are 0.

 

Examples-

 

Consider a binary pattern-

01100100.00000001.00000010.01000000

(represented as 100.1.2.64)

  • It is divisible by 25 since its least significant 5 bits are zero.
  • It is divisible by 26 since its least significant 6 bits are zero.
  • It is not divisible by 27 since its least significant 7 bits are not zero.

 

CIDR Notation-

 

CIDR IP Addresses look like-

a.b.c.d / n

 

  • They end with a slash followed by a number called as IP network prefix.
  • IP network prefix tells the number of bits used for the identification of network.
  • Remaining bits are used for the identification of hosts in the network.

 

Example-

 

An example of CIDR IP Address is-

182.0.1.2 / 28

 

It suggests-

  • 28 bits are used for the identification of network.
  • Remaining 4 bits are used for the identification of hosts in the network.

 

PRACTICE PROBLEMS BASED ON CLASSLESS INTER DOMAIN ROUTING-

 

Problem-01:

 

Given the CIDR representation 20.10.30.35 / 27. Find the range of IP Addresses in the CIDR block.

 

Solution-

 

Given CIDR representation is 20.10.30.35 / 27.

 

It suggests-

  • 27 bits are used for the identification of network.
  • Remaining 5 bits are used for the identification of hosts in the network.

 

Given CIDR IP Address may be represented as-

00010100.00001010.00011110.00100011 / 27

 

So,

  • First IP Address = 00010100.00001010.00011110.00100000 = 20.10.30.32
  • Last IP Address = 00010100.00001010.00011110.00111111 = 20.10.30.63

 

Thus, Range of IP Addresses = [ 20.10.30.32 , 20.10.30.63]

 

Problem-02:

 

Given the CIDR representation 100.1.2.35 / 20. Find the range of IP Addresses in the CIDR block.

 

Solution-

 

Given CIDR representation is 100.1.2.35 / 20.

 

It suggests-

  • 20 bits are used for the identification of network.
  • Remaining 12 bits are used for the identification of hosts in the network.

 

Given CIDR IP Address may be represented as-

01100100.00000001.00000010.00100011 / 20

 

So,

  • First IP Address = 01100100.00000001.00000000.00000000 = 100.1.0.0
  • Last IP Address = 01100100.00000001.00001111.11111111 = 100.1.15.255

 

Thus, Range of IP Addresses = [ 100.1.0.0 , 100.1.15.255]

 

Problem-03:

 

Consider a block of IP Addresses ranging from 100.1.2.32 to 100.1.2.47.

  1. Is it a CIDR block?
  2. If yes, give the CIDR representation.

 

Solution-

 

For any given block to be a CIDR block, 3 rules must be satisfied-

 

Rule-01:

 

  • According to Rule-01, all the IP Addresses must be contiguous.
  • Clearly, all the given IP Addresses are contiguous.
  • So, Rule-01 is satisfied.

 

Rule-02:

 

  • According to Rule-02, size of the block must be presentable as 2n.
  • Number of IP Addresses in the given block = 47 – 32 + 1 = 16.
  • Size of the block = 16 which can be represented as 24.
  • So, Rule-02 is satisfied.

 

Rule-03:

 

  • According to Rule-03, first IP Address must be divisible by size of the block.
  • So, 100.1.2.32 must be divisible by 24.
  • 100.1.2.32 = 100.1.2.00100000 is divisible by 24 since its 4 least significant bits are zero.
  • So, Rule-03 is satisfied.

 

Since all the rules are satisfied, therefore given block is a CIDR block.

 

CIDR Representation-

 

We have-

  • Size of the block = Total number of IP Addresses = 24.
  • To have 24 total number of IP Addresses, total 4 bits are required in the Host ID part.
  • So, Number of bits present in the Network ID part = 32 – 4 = 28.

 

Thus,

CIDR Representation = 100.1.2.32 / 28

 

NOTE-

 

For writing the CIDR representation,

  • We can choose to mention any IP Address from the CIDR block.
  • The chosen IP Address is followed by a slash and IP network prefix.
  • We generally choose to mention the first IP Address.

 

Problem-04:

 

Consider a block of IP Addresses ranging from 150.10.20.64 to 150.10.20.127.

  1. Is it a CIDR block?
  2. If yes, give the CIDR representation.

 

Solution-

 

For any given block to be a CIDR block, 3 rules must be satisfied-

 

Rule-01:

 

  • According to Rule-01, all the IP Addresses must be contiguous.
  • Clearly, all the given IP Addresses are contiguous.
  • So, Rule-01 is satisfied.

 

Rule-02:

 

  • According to Rule-02, size of the block must be presentable as 2n.
  • Number of IP Addresses in given block = 127 – 64 + 1 = 64.
  • Size of the block = 64 which can be represented as 26.
  • So, Rule-02 is satisfied.

 

Rule-03:

 

  • According to Rule-03, first IP Address must be divisible by size of the block.
  • So, 150.10.20.64 must be divisible by 26.
  • 150.10.20.64 = 150.10.20.01000000 is divisible by 26 since its 6 least significant bits are zero.
  • So, Rule-03 is satisfied.

 

Since all the rules are satisfied, therefore given block is a CIDR block.

 

CIDR Representation-

 

We have-

  • Size of the block = Total number of IP Addresses = 26.
  • To have 26 total number of IP Addresses, 6 bits are required in the Host ID part.
  • So, Number of bits in the Network ID part = 32 – 6 = 26.

 

Thus,

CIDR Representation = 150.10.20.64 / 26

 

Problem-05:

 

Perform CIDR aggregation on the following IP Addresses-

128.56.24.0/24

128.56.25.0/24

128.56.26.0/24

128.56.27.0/24

 

Solution-

 

All the 4 given entities represent CIDR block in itself.

We have to now perform the aggregation of these 4 blocks.

 

Rule-01:

 

  • According to Rule-01, all the IP Addresses must be contiguous.
  • Clearly, all the IP Addresses are contiguous.
  • So, Rule-01 is satisfied.

 

Rule-02:

 

  • According to Rule-02, size of the block must be presentable as 2n.
  • Total number of IP Addresses = 28 + 28 + 28 + 28 = 22 x 28 = 210.
  • So, Rule-02 is satisfied.

 

Rule-03:

 

  • According to Rule-03, first IP Address must be divisible by size of the block.
  • So, 128.56.24.0 must be divisible by 210.
  • 128.56.24.0 = 128.56.00011000.00000000 is divisible by 210 since its 10 least significant bits are zero.
  • So, Rule-03 is satisfied.

 

Since all the 3 rules are satisfied, so they can be aggregated.

 

CIDR Representation-

 

We have-

  • Size of the block = Total number of IP Addresses = 210.
  • To have 210 total number of IP Addresses, 10 bits are required in the Host ID part.
  • So, Number of bits in the Network ID part = 32 – 10 = 22.

 

Thus,

CIDR Representation = 128.56.24.0/22

 

Problem-06:

 

Perform CIDR aggregation on the following IP Addresses-

200.96.86.0/24

200.96.87.0/24

200.96.88.0/24

200.96.89.0/24

 

Solution-

 

All the 4 given entities represent CIDR block in itself.

We have to now perform the aggregation of these 4 blocks.

 

Rule-01:

 

  • According to Rule-01, all the IP Addresses must be contiguous.
  • Clearly, all the IP Addresses are contiguous.
  • So, Rule-01 is satisfied.

 

Rule-02:

 

  • According to Rule-02, size of the block must be presentable as 2n.
  • Total number of IP Addresses = 28 + 28 + 28 + 28 = 22 x 28 = 210.
  • So, Rule-02 is satisfied.

 

Rule-03:

 

  • According to Rule-03, first IP Address must be divisible by size of the block.
  • So, 200.96.86.0 must be divisible by 210.
  • 200.96.86.0 = 200.96.01010110.00000000 is not divisible by 210 since its 10 least significant bits are not zero.
  • So, Rule-03 is unsatisfied.

 

Since all the 3 rules are not satisfied, so they can not be aggregated.

 

To gain better understanding about Classless Addressing,

Watch this Video Lecture

 

Next Article- Subnetting | Examples

 

Get more notes and other study material of Computer Networks.

Watch video lectures by visiting our YouTube channel LearnVidFun.

TCP Retransmission | TCP Duplicate ACK

Computer Networks

Three Way Handshake-

 

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

 

We have discussed-

  • TCP uses Three Way Handshake to establish a connection between the sender and receiver.
  • Connection establishment using Three Way Handshake involves the steps as shown-

 

 

In this article, we will discuss how a lost TCP segment is retransmitted.

 

TCP Retransmission-

 

After establishing the connection,

  • Sender starts transmitting TCP segments to the receiver.
  • A TCP segment sent by the sender may get lost on the way before reaching the receiver.
  • This causes the receiver to send the acknowledgement with same ACK number to the sender.
  • As a result, sender retransmits the same segment to the receiver.
  • This is called as TCP retransmission.

 

When TCP Retransmission Occurs?

 

When sender discovers that the segment sent by it is lost,

it retransmits the same segment to the receiver.

 

Sender discovers that the TCP segment is lost when-

  1. Either Time Out Timer expires
  2. Or it receives three duplicate acknowledgements

 

1. Retransmission After Time Out Timer Expiry-

 

Each time sender transmits a TCP segment to the receiver, it starts a Time Out Timer.

Now, following two cases are possible-

 

Case-01:

 

  • Sender receives an acknowledgement for the sent segment before the timer goes off.
  • In this case, sender stops the timer.

 

Case-02:

 

  • Sender does not receives any acknowledgement for the sent segment and the timer goes off.
  • In this case, sender assumes that the sent segment is lost.
  • Sender retransmits the same segment to the receiver and resets the timer.

 

Also Read- Time Out Timer

 

Example-

 

 

2. Retransmission After Receiving 3 Duplicate Acknowledgements-

 

  • Consider sender receives three duplicate acknowledgements for a TCP segment sent by it.
  • Then, sender assumes that the corresponding segment is lost.
  • So, sender retransmits the same segment without waiting for its time out timer to expire.
  • This is known as Early retransmission or Fast retransmission.

 

Example-

 

Consider-

  • Sender sends 5 TCP segments to the receiver.
  • The second TCP segment gets lost before reaching the receiver.

 

The sequence of steps taking place are-

 

  • On receiving segment-1, receiver sends acknowledgement asking for segment-2 next.

(Original ACK)

  • On receiving segment-3, receiver sends acknowledgement asking for segment-2 next.

(1st duplicate ACK)

  • On receiving segment-4, receiver sends acknowledgement asking for segment-2 next.

(2nd duplicate ACK)

  • On receiving segment-5, receiver sends acknowledgement asking for segment-2 next.

(3rd duplicate ACK)

 

Now,

  • Sender receives 3 duplicate acknowledgements for segment-2 in total.
  • So, sender assumes that the segment-2 is lost.
  • So, it retransmits segment-2 without waiting for its timer to go off.

 

 

NOTE

After receiving the retransmitted segment-2,

  • Receiver does not send the acknowledgement asking for segment-3 or 4 or 5.
  • Receiver sends the acknowledgement asking for segment-6 directly from the sender.
  • This is because previous segments have been already received and acknowledgements for them have been already sent (although wasted in asking for segment-2).

 

Important Points-

 

Point-01:

 

  • Consider time out timer expires before receiving the acknowledgement for a TCP segment.
  • This case suggests the stronger possibility of congestion in the network.

 

Point-02:

 

  • Consider sender receives 3 duplicate acknowledgements for the same TCP segment.
  • This case suggests the weaker possibility of congestion in the network.

 

Point-03:

 

  • Consider receiver does not receives 3 duplicate acknowledgements for the lost TCP segment.
  • In such a case, retransmission occurs only after time out timer goes off.

 

Point-04:

 

  • Retransmission on receiving 3 duplicate acknowledgements is a way to improve the performance over retransmission on time out.

 

Also Read- TCP Congestion Control

 

PRACTICE PROBLEM BASED ON TCP RETRANSMISSION-

 

Problem-

 

Consider a TCP connection in a state where there are no outstanding ACKs. The sender sends two segments back to back. The sequence numbers of the first and second segments are 230 and 290 respectively. The first segment was lost but the second segment was received correctly by the receiver.

Let X be the amount of data carried in the first segment (in bytes) and Y be the ACK number sent by the receiver.

The values of X and Y are-

  1. 60 and 290
  2. 230 and 291
  3. 60 and 231
  4. 60 and 230

 

Solution-

 

It is given that sender sends two segments where-

  • 1st segment contains the sequence number 230 and gets lost.
  • 2nd segment contains the sequence number 290 and is received correctly.

 

Amount Of Data Sent In First Segment-

 

Given-

  • Sequence number of 1st segment = 230
  • Sequence number of 2nd segment = 290

 

From here, range of sequence numbers contained in the 1st segment = [230,289].

 

Now,

  • Total number of sequence numbers contained in the 1st segment = 289 – 230 + 1 = 60.
  • TCP assigns one sequence number to each byte of data.
  • Thus, Amount of data contained in the first segment = 60 bytes.

 

ACK Number Sent By Receiver-

 

On receiving the 2nd segment,

  • Receiver sends the acknowledgement asking for the first segment only.
  • This is because it expects the 1st segment first.
  • Receiver keeps sending this ACK number until it receives the first segment correctly.
  • Thus, Acknowledgement number = Sequence number of the 1st segment = 230.

 

Thus, Option (D) is correct.

 

Next Article- TCP Connection Termination

 

Get more notes and other study material of Computer Networks.

Watch video lectures by visiting our YouTube channel LearnVidFun.

Subnet Mask | Practice Problems

Computer Networks

Subnet Mask-

 

Before you go through this article, make sure that you gone through the previous article on Subnet Mask.

 

We have discussed-

  • Subnet mask is a 32 bit number consisting of a sequence of 1’s and 0’s.
  • It is used to identify the subnet to which the given IP Address belongs.

 

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

 

PRACTICE PROBLEMS BASED ON SUBNET MASK-

 

Problems-01 to 09:

 

Consider the following subnet masks-

  1. 255.0.0.0
  2. 255.128.0.0
  3. 255.192.0.0
  4. 255.240.0.0
  5. 255.255.0.0
  6. 255.255.254.0
  7. 255.255.255.0
  8. 255.255.255.224
  9. 225.255.255.240

 

For each subnet mask, find-

  1. Number of hosts per subnet
  2. Number of subnets if subnet mask belongs to class A
  3. Number of subnets if subnet mask belongs to class B
  4. Number of subnets if subnet mask belongs to class C
  5. Number of subnets if total 10 bits are used for the global network ID

 

Solutions-

 

All the problems are solved below one by one-

 

Solution-01:

 

Given subnet mask is 255.0.0.0

So,

  • Number of Net ID bits + Number of Subnet ID bits = 8
  • Number of Host ID bits = 24

 

Part-A:

 

Since number of Host ID bits = 24, so-

 

Number of hosts per subnet = 224 – 2

 

Part-B:

 

If the given subnet mask belongs to class A, then number of Net ID bits = 8.

Substituting in the above equation, we get-

Number of Subnet ID bits = 8 – 8 = 0

Thus,

 

Number of subnets = 20 = 1

 

Thus, there will be only one single network.

 

Part-C:

 

  • First two octets of the subnet mask are not completely filled with 1’s.
  • So, given subnet mask can not belong to class B.

 

Part-D:

 

  • First three octets of the subnet mask are not completely filled with 1’s.
  • So, given subnet mask can not belong to class C.

 

Part-E:

 

  • First 10 bits of the subnet mask are not completely filled with 1’s.
  • So, given subnet mask can not use 10 bits for the Network ID.

 

NOTE-

 

  • 255.0.0.0 is the default mask for class A.

 

Solution-02:

 

Given subnet mask is 255.128.0.0

So,

  • Number of Net ID bits + Number of Subnet ID bits = 9
  • Number of Host ID bits = 23

 

Part-A:

 

Since number of Host ID bits = 23, so-

 

Number of hosts per subnet = 223 – 2

 

Part-B:

 

If the given subnet mask belongs to class A, then number of Net ID bits = 8.

Substituting in the above equation, we get-

Number of Subnet ID bits = 9 – 8 = 1

Thus,

 

Number of subnets = 21 = 2

 

Part-C:

 

  • First two octets of the subnet mask are not completely filled with 1’s.
  • So, given subnet mask can not belong to class B.

 

Part-D:

 

  • First three octets of the subnet mask are not completely filled with 1’s.
  • So, given subnet mask can not belong to class C.

 

Part-E:

 

  • First 10 bits of the subnet mask are not completely filled with 1’s.
  • So, given subnet mask can not use 10 bits for the Network ID.

 

Solution-03:

 

Given subnet mask is 255.192.0.0

So,

  • Number of Net ID bits + Number of Subnet ID bits = 10
  • Number of Host ID bits = 22

 

Part-A:

 

Since number of Host ID bits = 22, so-

 

Number of hosts per subnet = 222 – 2

 

Part-B:

 

If the given subnet mask belongs to class A, then number of Net ID bits = 8.

Substituting in the above equation, we get-

Number of Subnet ID bits = 10 – 8 = 2

Thus,

 

Number of subnets = 22 = 4

 

Part-C:

 

  • First two octets of the subnet mask are not completely filled with 1’s.
  • So, given subnet mask can not belong to class B.

 

Part-D:

 

  • First three octets of the subnet mask are not completely filled with 1’s.
  • So, given subnet mask can not belong to class C.

 

Part-E:

 

Given 10 bits are used for the Net ID part.

Substituting in the above equation, we get-

Number of Subnet ID bits = 10 – 10 = 0

Thus,

 

Number of subnets = 20 = 1

 

Thus, there will be only one single network.

 

Solution-04:

 

Given subnet mask is 255.240.0.0

So,

  • Number of Net ID bits + Number of Subnet ID bits = 12
  • Number of Host ID bits = 20

 

Part-A:

 

Since number of Host ID bits = 20, so-

 

Number of hosts per subnet = 220 – 2

 

Part-B:

 

If the given subnet mask belongs to class A, then number of Net ID bits = 8.

Substituting in the above equation, we get-

Number of Subnet ID bits = 12 – 8 = 4

Thus,

 

Number of subnets = 24 = 16

 

Part-C:

 

  • First two octets of the subnet mask are not completely filled with 1’s.
  • So, given subnet mask can not belong to class B.

 

Part-D:

 

  • First three octets of the subnet mask are not completely filled with 1’s.
  • So, given subnet mask can not belong to class C.

 

Part-E:

 

Given 10 bits are used for the Net ID part.

Substituting in the above equation, we get-

Number of Subnet ID bits = 12 – 10 = 2

Thus,

 

Number of subnets = 22 = 4

 

Solution-05:

 

Given subnet mask is 255.255.0.0

So,

  • Number of Net ID bits + Number of Subnet ID bits = 16
  • Number of Host ID bits = 16

 

Part-A:

 

Since number of Host ID bits = 16, so-

 

Number of hosts per subnet = 216 – 2

 

Part-B:

 

If the given subnet mask belongs to class A, then number of Net ID bits = 8.

Substituting in the above equation, we get-

Number of Subnet ID bits = 16 – 8 = 8

Thus,

 

Number of subnets = 28

 

Part-C:

 

If the given subnet mask belongs to class B, then number of Net ID bits = 16.

Substituting in the above equation, we get-

Number of Subnet ID bits = 16 – 16 = 0

Thus,

 

Number of subnets = 20 = 1

 

Part-D:

 

  • First three octets of the subnet mask are not completely filled with 1’s.
  • So, given subnet mask can not belong to class C.

 

Part-E:

 

Given 10 bits are used for the Net ID part.

Substituting in the above equation, we get-

Number of Subnet ID bits = 16 – 10 = 6

Thus,

 

Number of subnets = 26 = 64

 

NOTE-

 

  • 255.255.0.0 is the default mask for class B.

 

Solution-06:

 

Given subnet mask is 255.255.254.0

So,

  • Number of Net ID bits + Number of Subnet ID bits = 23
  • Number of Host ID bits = 9

 

Part-A:

 

Since number of Host ID bits = 9, so-

 

Number of hosts per subnet = 29 – 2

 

Part-B:

 

If the given subnet mask belongs to class A, then number of Net ID bits = 8.

Substituting in the above equation, we get-

Number of Subnet ID bits = 23 – 8 = 15

Thus,

 

Number of subnets = 215

 

Part-C:

 

If the given subnet mask belongs to class B, then number of Net ID bits = 16.

Substituting in the above equation, we get-

Number of Subnet ID bits = 23 – 16 = 7

Thus,

 

Number of subnets = 27

 

Part-D:

 

  • First three octets of the subnet mask are not completely filled with 1’s.
  • So, given subnet mask can not belong to class C.

 

Part-E:

 

Given 10 bits are used for the Net ID part.

Substituting in the above equation, we get-

Number of Subnet ID bits = 23 – 10 = 13

Thus,

 

Number of subnets = 213

 

Solution-07:

 

Given subnet mask is 255.255.255.0

So,

  • Number of Net ID bits + Number of Subnet ID bits = 24
  • Number of Host ID bits = 8

 

Part-A:

 

Since number of Host ID bits = 8, so-

 

Number of hosts per subnet = 28 – 2

 

Part-B:

 

If the given subnet mask belongs to class A, then number of Net ID bits = 8.

Substituting in the above equation, we get-

Number of Subnet ID bits = 24 – 8 = 16

Thus,

 

Number of subnets = 216

 

Part-C:

 

If the given subnet mask belongs to class B, then number of Net ID bits = 16.

Substituting in the above equation, we get-

Number of Subnet ID bits = 24 – 16 = 8

Thus,

 

Number of subnets = 28

 

Part-D:

 

If the given subnet mask belongs to class C, then number of Net ID bits = 24.

Substituting in the above equation, we get-

Number of Subnet ID bits = 24 – 24 = 0

Thus,

 

Number of subnets = 20 = 1

 

Part-E:

 

Given 10 bits are used for the Net ID part.

Substituting in the above equation, we get-

Number of Subnet ID bits = 24 – 10 = 14

Thus,

 

Number of subnets = 214

 

NOTE-

 

  • 255.255.255.0 is the default mask for class C.

 

Solution-08:

 

Given subnet mask is 255.255.255.224

So,

  • Number of Net ID bits + Number of Subnet ID bits = 27
  • Number of Host ID bits = 5

 

Part-A:

 

Since number of Host ID bits = 5, so-

 

Number of hosts per subnet = 25 – 2

 

Part-B:

 

If the given subnet mask belongs to class A, then number of Net ID bits = 8.

Substituting in the above equation, we get-

Number of Subnet ID bits = 27 – 8 = 19

Thus,

 

Number of subnets = 219

 

Part-C:

 

If the given subnet mask belongs to class B, then number of Net ID bits = 16.

Substituting in the above equation, we get-

Number of Subnet ID bits = 27 – 16 = 11

Thus,

 

Number of subnets = 211

 

Part-D:

 

If the given subnet mask belongs to class C, then number of Net ID bits = 24.

Substituting in the above equation, we get-

Number of Subnet ID bits = 27 – 24 = 3

Thus,

 

Number of subnets = 23 = 8

 

Part-E:

 

Given 10 bits are used for the Net ID part.

Substituting in the above equation, we get-

Number of Subnet ID bits = 27 – 10 = 17

Thus,

 

Number of subnets = 217

 

Solution-09:

 

Given subnet mask is 255.255.255.240

So,

  • Number of Net ID bits + Number of Subnet ID bits = 28
  • Number of Host ID bits = 4

 

Part-A:

 

Since number of Host ID bits = 4, so-

 

Number of hosts per subnet = 24 – 2

 

Part-B:

 

If the given subnet mask belongs to class A, then number of Net ID bits = 8.

Substituting in the above equation, we get-

Number of Subnet ID bits = 28 – 8 = 20

Thus,

 

Number of subnets = 220

 

Part-C:

 

If the given subnet mask belongs to class B, then number of Net ID bits = 16.

Substituting in the above equation, we get-

Number of Subnet ID bits = 28 – 16 = 12

Thus,

 

Number of subnets = 212

 

Part-D:

 

If the given subnet mask belongs to class C, then number of Net ID bits = 24.

Substituting in the above equation, we get-

Number of Subnet ID bits = 28 – 24 = 4

Thus,

 

Number of subnets = 24

 

Part-E:

 

Given 10 bits are used for the Net ID part.

Substituting in the above equation, we get-

Number of Subnet ID bits = 28 – 10 = 18

Thus,

 

Number of subnets = 218

 

Problem-10:

 

Consider default subnet mask for a network is 255.255.255.0. How many number of subnets and hosts per subnet are possible if ‘m’ bits are borrowed from HID.

  1. 2m , 2(HID-m) – 2
  2. 2m , 2(HID-m)
  3. 2m – 1, 2(HID-m) – 2
  4. 2m , (HID-m) – 2

 

Solution-

 

  • Subnet mask = 255.255.255.0
  • Number of bits borrowed from Host ID part = m
  • So, number of subnets possible = 2m
  • Number of bits available for Hosts = HID – m
  • So, number of hosts that can be configured = 2(HID – m) – 2

 

Thus, Option (A) is correct.

 

Problem-11:

 

If default subnet mask for a network is 255.255.255.0 and if ‘m’ bits are borrowed from the NID, then what could be its supernet mask?

  1. 255.255.(28-m – 1) x 2m.0
  2. 255.255.(28-m) x 2m.0
  3. 255.255.(28-m-1) x 2m-1.0
  4. 255.255.(28-m) x 2m-1.0

 

Solution-

 

Given-

  • Subnet mask = 255.255.255.0
  • m bits are chosen from the NID part.

 

Clearly, given subnet mask belongs to class C.

If m = 4, then the subnet mask = 255.255.11110000.0

Now, let us check all the options one by one.

 

Option-A:

 

Given-

  • Supernet mask = 255.255.(28-m – 1) x 2m.0
  • Third octet = (28-m – 1) x 2m

 

On substituting m = 4, we get-

Third octet

= 15 x 24

= (1111)2 x 24

= 11110000 (Performing Left shift by 4 places)

 

Yes, this is what the third octet should be.

Thus, Option (A) is correct.

 

Option-B:

 

Given-

  • Supernet mask = 255.255.(28-m) x 2m.0
  • Third octet = (28-m) x 2m

 

On substituting m = 4, we get-

Third octet

= 16 x 24

= (10000)2 x 24

= 100000000 (Performing Left shift by 4 places)

 

This can not be true because these are 9 bits and octet can be only 8 bits.

Thus, Option (B) is incorrect.

Similarly, other options are also incorrect.

 

Finally, Option (A) is the only correct option.

 

To watch video solutions and practice more problems,

Watch this Video Lecture

 

Next Article- Routing Table | Arrangement of Subnets

 

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Subnet Mask | How to Calculate Subnet Mask

Computer Networks

Subnetting in Networking-

 

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

 

We have discussed-

  • Subnetting is a process of dividing a single network into multiple sub networks.
  • The number of sub networks created depends upon the requirements.

 

 

Subnet Mask-

 

Subnet mask is a 32 bit number which is a sequence of 1’s followed by a sequence of 0’s where-

  • 1’s represent the global network ID part and the subnet ID part.
  • 0’s represent the host ID part.

 

How to Calculate Subnet Mask?

 

For any given IP Address, the subnet mask is calculated-

  • By setting all the bits reserved for network ID part and subnet ID part to 1.
  • By setting all the bits reserved for host ID part to 0.

 

Subnet Mask Examples-

 

Now, let us discuss some examples on how to calculate subnet mask for any given network-

 

Example-01:

 

Consider we have a network having IP Address 200.1.2.0.

 

Clearly, this IP Address belongs to class C.

 

In class C-

  • 24 bits are reserved for the Network ID part.
  • 8 bits are reserved for the Host ID part.

 

Subnet mask is obtained-

  • By setting the first 24 bits to 1.
  • By setting the remaining 8 bits to 0.

 

So, Subnet mask

= 11111111.11111111.11111111.00000000

= 255.255.255.0

 

Example-02:

 

Consider a single network having IP Address 200.1.2.0 is divided into 4 subnets as shown-

 

 

Now, let us calculate the mask subnet for each subnet.

 

For each subnet-

  • 24 bits identify the global network.
  • 2 bits identify the subnet.
  • 6 bits identify the host.

 

For each subnet, subnet mask is obtained-

  • By setting the first 26 bits to 1.
  • By setting the remaining 6 bits to 0.

 

So, Subnet mask

= 11111111.11111111.11111111.11000000

= 255.255.255.192

 

NOTE

In fixed length subnetting,

All the subnets have same subnet mask since the size of each subnet is same.

 

Example-03:

 

Consider a single network having IP Address 200.1.2.0 is divided into 3 subnets as shown-

 

 

Now, let us calculate the subnet mask for each subnet.

 

For Subnet A-

 

For subnet A-

  • 24 bits identify the global network.
  • 1 bit identify the subnet.
  • 7 bits identify the host.

 

For subnet A, subnet mask is obtained-

  • By setting the first 25 bits to 1.
  • By setting the remaining 7 bits to 0.

 

So, Subnet mask

= 11111111.11111111.11111111.10000000

= 255.255.255.128

 

For Subnet B And Subnet C-

 

For subnet B and subnet C-

  • 24 bits identify the global network.
  • 2 bits identify the subnet.
  • 6 bits identify the host.

 

For subnet B and subnet C, subnet mask is obtained-

  • By setting the first 26 bits to 1.
  • By setting the remaining 6 bits to 0.

 

So, Subnet mask

= 11111111.11111111.11111111.11000000

= 255.255.255.192

 

NOTE

In variable length subnetting,

All the subnets do not have same subnet mask since the size of each subnet is not same.

 

Use of Subnet Mask-

 

  • Subnet mask is used to determine to which subnet the given IP Address belongs to.
  • To know more, Read here.

 

Important Notes-

 

Note-01:

 

Default mask for different classes of IP Address are-

  • Default subnet mask for Class A = 255.0.0.0
  • Default subnet mask for Class B = 255.255.0.0
  • Default subnet mask for Class C = 255.255.255.0

 

Also Read- Classes of IP Address

 

Note-02:

 

  • Network size is the total number of hosts present in it.
  • Networks of same size always have the same subnet mask.
  • Networks of different size always have the different subnet mask.

 

Note-03:

 

  • For a network having larger size, its subnet mask will be smaller (number of 1’s will be less).
  • For a network having smaller size, its subnet mask will be larger (number of 1’s will be more).

 

PRACTICE PROBLEMS BASED ON SUBNET MASK-

 

Problem-01:

 

If the subnet mask 255.255.255.128 belongs to class C, find-

  1. Number of subnets
  2. Number of hosts in each subnet

 

Solution-

 

Given subnet mask

= 255.255.255.128

= 11111111.11111111.11111111.10000000

 

Since 25 bits contain the value 1 and 7 bits contain the value 0, so-

  • Number of Net ID bits + Number of Subnet ID bits = 25
  • Number of Host ID bits = 7

 

Now,

  • It is given that subnet mask belongs to class C.
  • So, Number of Net ID bits = 24.

 

Substituting in the above equation, we get-

Number of Subnet ID bits

= 25 – 24

= 1

 

Thus,

Number of subnets = 21 = 2

 

Since number of Host ID bits = 7, so-

 

Number of hosts per subnet = 27 – 2 = 126

 

Problem-02:

 

If a class B network has a subnet mask of 255.255.248.0, what is the maximum number of hosts per subnet?

  1. 1022
  2. 1023
  3. 2046
  4. 2047

 

Solution-

 

Given subnet mask

= 255.255.248.0

= 11111111.11111111.11111000.00000000

 

Since 21 bits contain the value 1 and 11 bits contain the value 0, so-

  • Number of Net ID bits + Number of Subnet ID bits = 21
  • Number of Host ID bits = 11

 

Since number of Host ID bits = 11, so-

 

Number of hosts per subnet = 211 – 2 = 2046

 

Thus, Option (C) is correct.

 

To gain better understanding about Subnet Mask,

Watch this Video Lecture

 

Next Article- Practice Problems On Subnet Mask

 

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