Questions

 

With a clear demonstration, explain the following questions with binary, octet, and dotted decimal addresses. 1. You have been assigned a network with the IP address 192.168.1.0/24. How many usable host addresses are available in this subnet? 2.Given the IP address 192.168.1.0/24, how many subnets can you create if you use a /25 subnet mask? 3.A network has the IP address 10.0.0.0/24. You need to create subnets with a /25 subnet mask. What are the subnet addresses and ranges for the first four subnets? 4.You have a network with the IP address 172.16.10.0/24. How many /25 subnets can you create from this network? 5.Given the IP address 192.168.1.0/24, what is the maximum number of /25 subnets you can create? 6.You have a network with the IP address 10.0.0.0/24. How many hosts are there in each /25 subnet? 7.A network administrator wants to divide the network 192.168.1.0/24 into eight /25 subnets. What are the subnet addresses and ranges for each subnet? 8.Given the IP address 172.16.0.0/24, what is the maximum number of /25 subnets you can create? 9.You have a network with the IP address 192.168.1.0/24. How many bits must be borrowed from the host portion to create eight /25 subnets? 10.A company has been assigned the network 10.0.0.0/24. They want to divide this network into /25 subnets. What are the subnet addresses and ranges for the first four subnets?

 

ANSWERS1

1. To determine the number of usable host addresses available in the subnet 192.168.1.0/24, we need to consider the subnet mask. In a /24 subnet, the subnet mask has 24 network bits and 8 host bits. Since the host bits are 8, we subtract 2 (network address and broadcast address) to find the number of usable host addresses.

Binary representation of the subnet mask: 11111111.11111111.11111111.00000000 Octet representation of the subnet mask: 255.255.255.0 Dotted decimal representation of the subnet mask: 255.255.255.0

Number of usable host addresses = 2^(number of host bits) - 2 = 2^8 - 2 = 256 - 2 = 254

Therefore, there are 254 usable host addresses available in the subnet 192.168.1.0/24.

1. Given the IP address 192.168.1.0/24, if we use a /25 subnet mask, we are borrowing 1 additional bit from the host portion to create subnets. This means we have 23 network bits and 9 host bits.

Binary representation of the /25 subnet mask: 11111111.11111111.11111111.10000000 Octet representation of the /25 subnet mask: 255.255.255.128 Dotted decimal representation of the /25 subnet mask: 255.255.255.128

To calculate the number of subnets, we need to determine the number of subnet bits (additional bits borrowed from the host portion). In this case, we have borrowed 1 bit (/25 - /24).

Number of subnets = 2^(number of subnet bits) = 2^1 = 2

Therefore, with a /25 subnet mask, you can create 2 subnets.

1. Given the network IP address 10.0.0.0/24 and the requirement to create subnets with a /25 subnet mask, we need to determine the subnet addresses and ranges for the first four subnets.

Binary representation of the /25 subnet mask: 11111111.11111111.11111111.10000000 Octet representation of the /25 subnet mask: 255.255.255.128 Dotted decimal representation of the /25 subnet mask: 255.255.255.128

Subnet 1: Subnet address: 10.0.0.0 Usable host range: 10.0.0.1 - 10.0.0.126 Broadcast address: 10.0.0.127

Subnet 2: Subnet address: 10.0.0.128 Usable host range: 10.0.0.129 - 10.0.0.254 Broadcast address: 10.0.0.255

Subnet 3: Subnet address: 10.0.1.0 Usable host range: 10.0.1.1 - 10.0.1.126 Broadcast address: 10.0.1.127

Subnet 4: Subnet address: 10.0.1.128 Usable host range: 10.0.1.129 - 10.0.1.254 Broadcast address: 10.0.1.255

Each subnet has a subnet address, usable host range, and broadcast address. The usable host range excludes the network and broadcast addresses.

1. To determine the number of /25 subnets that can be created from the network 172.16.10.0/24, we need to calculate the number of subnet bits. In a /24 subnet, we have 24 network bits and 8 host bits. When we use a /25 subnet mask, we borrow 1 additional bit for the subnet.

Number of subnet bits = Number of bits borrowed for subnetting = 1

Number of subnets = 2^(number of subnet bits) = 2^1 = 2

Therefore, you can create 2 /25 subnets from the network 172.16.10.0/24.

1. Given the IP address 192.168.1.0/24, if we use a /25 subnet mask, we are borrowing 1 additional bit from the host portion to create subnets. This means we have 23 network bits and 9 host bits.

To calculate the maximum number of /25 subnets, we need to determine the number of subnet bits (additional bits borrowed from the host portion). In this case, we have borrowed 1 bit (/25 - /24).

Number of subnets = 2^(number of subnet bits) = 2^1 = 2

Therefore, with the IP address 192.168.1.0/24, you can create a maximum of 2 /25 subnets.

1. Given the network IP address 10.0.0.0/24 and the requirement to create /25 subnets, we need to determine the number of hosts in each /25 subnet.

Binary representation of the /25 subnet mask: 11111111.11111111.11111111.10000000 Octet representation of the /25 subnet mask: 255.255.255.128 Dotted decimal representation of the /25 subnet mask: 255.255.255.128

To calculate the number of hosts in each /25 subnet, we subtract 2 (network address and broadcast address) from the total number of addresses in the subnet.

Number of hosts = 2^(number of host bits) - 2 = 2^7 - 2 = 128 - 2 = 126

Therefore, each /25 subnet in the network 10.0.0.0/24 can have 126 hosts.

1. To divide the network 192.168.1.0/24 into eight /25 subnets, we need to determine the subnet addresses and ranges for each subnet.

Binary representation of the /25 subnet mask: 11111111.11111111.11111111.10000000 Octet representation of the /25 subnet mask: 255.255.255.128 Dotted decimal representation of the /25 subnet mask: 255.255.255.128

Subnet 1: Subnet address: 192.168.1.0 Usable host range: 192.168.1.1 - 192.168.1.126 Broadcast address: 192.168.1.127

Subnet 2: Subnet address: 192.168.1.128 Usable host range: 192.168.1.129 - 192.168.1.254 Broadcast address: 192.168.1.255

Subnet 3: Subnet address: 192.168.2.0 Usable host range: 192.168.2.1 - 192.168.2.126 Broadcast address: 192.168.2.127

Subnet 4: Subnet address: 192.168.2.128 Usable host range: 192.168.2.129 - 192.168.2.254 Broadcast address: 192.168.2.255

Subnet 5: Subnet address: 192.168.3.0 Usable host range: 192.168.3.1 - 192.168.3.126 Broadcast address: 192.168.3.127

Subnet 6: Subnet address: 192.168.3.128 Usable host range: 192.168.3.129 - 192.168.3.254 Broadcast address: 192.168.3.255

Subnet 7: Subnet address: 192.168.4.0 Usable host range: 192.168.4.1 - 192.168.4.126 Broadcast address: 192.168.4.127

Subnet 8: Subnet address: 192.168.4.128 Usable host range: 192.168.4.129 - 192.168.4.254 Broadcast address: 192.168.4.255

Each subnet has a subnet address, usable host range, and broadcast address. The usable host range excludes the network and broadcast addresses.

1. Given the IP address 172.16.0.0/24, if we use a /25 subnet mask, we are borrowing 1 additional bit from the host portion to create subnets. This means we have 23 network bits and 9 host bits.

To calculate the maximum number of /25 subnets, we need to determine the number of subnet bits (additional bits borrowed from the host portion). In this case, we have borrowed 1 bit (/25 - /24).

Number of subnets = 2^(number of subnet bits) = 2^1 = 2

Therefore, with the IP address 172.16.0.0/24, you can create a maximum of 2 /25 subnets.

1. To create eight /25 subnets from the network 192.168.1.0/24, we need to determine the number of bits that must be borrowed from the host portion.

Number of bits borrowed for subnetting = log2(number of subnets) = log2(8) = 3 (since 2^3 = 8)

To create eight /25 subnets, we need to borrow 3 bits from the host portion.

Therefore, 3 bits must be borrowed from the host portion to create eight /25 subnets.

1. Given the network 10.0.0.0/24 and the requirement to divide it into /25 subnets, we need to determine the subnet addresses and ranges for the first four subnets.

Binary representation of the /25 subnet mask: 11111111.11111111.11111111.10000000 Octet representation of the /25 subnet mask: 255.255.255.128 Dotted decimal representation of the /25 subnet mask: 255.255.255.128

Subnet 1: Subnet address: 10.0.0.0 Usable host range: 10.0.0.1 - 10.0.0.126 Broadcast address: 10.0.0.127

Subnet 2: Subnet address: 10.0.0.128 Usable host range: 10.0.0.129 - 10.0.0.254 Broadcast address: 10.0.0.255

Subnet 3: Subnet address: 10.0.1.0 Usable host range: 10.0.1.1 - 10.0.1.126 Broadcast address: 10.0.1.127

Subnet 4: Subnet address: 10.0.1.128 Usable host range: 10.0.1.129 - 10.0.1.254 Broadcast address: 10.0.1.255

Each subnet has a subnet address, usable host range, and broadcast address. The usable host range excludes the network and broadcast addresses

 

ANSWERS2

 

1. The network 192.168.1.0/24 has 254 usable host addresses because the subnet mask /24 allows for 256 total addresses, but 2 addresses are reserved for network and broadcast addresses, leaving 254 usable host addresses.

2. Using a /25 subnet mask with the IP address 192.168.1.0/24, you can create 2 subnets because a /25 mask divides the original network into two equal halves.

3. The subnet addresses and ranges for the first four subnets of the network 10.0.0.0/24 with a /25 subnet mask are determined by incrementing the third octet of the network address:

   • Subnet 1: Subnet address: 10.0.0.0, Usable host range: 10.0.0.1 - 10.0.0.126
   • Subnet 2: Subnet address: 10.0.0.128, Usable host range: 10.0.0.129 - 10.0.0.254
   • Subnet 3: Subnet address: 10.0.1.0, Usable host range: 10.0.1.1 - 10.0.1.126
   • Subnet 4: Subnet address: 10.0.1.128, Usable host range: 10.0.1.129 - 10.0.1.254

4. From the network 172.16.10.0/24, you can create 2 /25 subnets because each /25 subnet requires a block of 128 addresses, and the original network has 256 addresses.

5. The maximum number of /25 subnets that can be created from the IP address 192.168.1.0/24 is 2 because a /25 mask divides the original network into two equal halves.

6. Each /25 subnet in the network 10.0.0.0/24 can accommodate 126 hosts because a /25 mask allows for a total of 128 addresses, but 2 addresses are reserved for network and broadcast addresses.

7. The subnet addresses and ranges for each of the eight /25 subnets in the network 192.168.1.0/24 are determined by incrementing the third octet of the network address:

   • Subnet 1: Subnet address: 192.168.1.0, Usable host range: 192.168.1.1 - 192.168.1.126
   • Subnet 2: Subnet address: 192.168.1.128, Usable host range: 192.168.1.129 - 192.168.1.254
   • Subnet 3: Subnet address: 192.168.2.0, Usable host range: 192.168.2.1 - 192.168.2.126
   • Subnet 4: Subnet address: 192.168.2.128, Usable host range: 192.168.2.129 - 192.168.2.254
   • Subnet 5: Subnet address: 192.168.3.0, Usable host range: 192.168.3.1 - 192.168.3.126
   • Subnet 6: Subnet address: 192.168.3.128, Usable host range: 192.168.3.129 - 192.168.3.254
   • Subnet 7: Subnet address: 192.168.4.0, Usable host range: 192.168.4.1 - 192.168.4.126
   • Subnet 8: Subnet address: 192.168.4.128, Usable host range: 192.168.4.129 - 192.168.4.254

8. The maximum number of /25 subnets that can be created from the IP address 172.16.0.0/24 is 8 because each /25 subnet requires a block of 128 addresses, and the original network has 256 addresses.

9. To create eight /25 subnets, you need to borrow 3 bits from the host portion of the IP address because each /25 subnet requires a block of 128 addresses, and 2^3 equals 8.

10. The subnet addresses and ranges for the first four /25 subnets in the network 10.0.0.0/24 are determined by incrementing the third octet of the network address:

    • Subnet 1: Subnet address: 10.0.0.0, Usable host range: 10.0.0.1 - 10.0.0.126
    • Subnet 2: Subnet address: 10.0.0.128, Usable host range: 10.0.0.129 - 10.0.0.254
    • Subnet 3: Subnet address: 10.0.1.0, Usable host range: 10.0.1.1 - 10.0.1.126
    • Subnet 4: Subnet address: 10.0.1.128, Usable host range: 10.0.1.129 - 10.0.1.254

 

 

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