Tag: Subnet Mask Examples

Subnet Mask | Subnet Mask Use

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.

Subnet Mask | Practice Problems

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

 

Get more notes and other study material of Computer Networks.

Watch video lectures by visiting our YouTube channel LearnVidFun.

Subnet Mask | How to Calculate Subnet Mask

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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