# Subnet Mask | Practice Problems

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.

## Problems-01 to 09:

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

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:

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:

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:

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:

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:

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:

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:

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:

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:

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-

• 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-

• 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.

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