please help A network administrator has the address range 192.168.10.0/24 to add

Question

please help

A network administrator has the address range 192.168.10.0/24 to address the networks shown. How many bits are in the network part of the address and how many are in the host part of the address? How many networks are there in the figure? Using basic subnetting methods (without VLSM), all the subnets need to be the same size How many bits would you need to borrow to have enough networks? How many bits would be left in the host part? How many available hosts could you have on each network?

Explanation / Answer

Variable Length Subnet Masking (VLSM) is a way of further subnetting a subnet. Using Variable Length Subnet Masking (VLSM) we can allocate IP addresses to the subnets by the exact need. Variable Length Subnet Masking (VLSM) allows us to use more than one subnet mask within the same network address space. If we recollect from the previous lessons, we can divide a network only into subnets with equal number of IP addresses. Variable Length Subnet Masking (VLSM) allows to create subnets from a single network with unequal number of IP addresses.

Example: We want to divide 192.168.10.0, which is a Class C network, into four networks, each with unequal number of IP address requirements as shown below.

Subnet A : 126 IP Addresses.
Subnet B : 62 IP Addresses.
Subnet C : 30 IP Addresses.
Subnet D : 30 IP Addresses.

This type of division is not possible as described in previous lessons, since it divide the network equally, but is possible with Variable Length Subnet Masking (VLSM).

Original Network (Network to be sub-netted) %u2013 192.168.10.0/24

Divide the two networks equally with 128 IP Addresses (126 usable IP addresses) in each network using 255.255.255.128 subnet mask (192.168.10.0/25).

We will get two subnets each with 128 IP Addresses (126 usable IP addresses).

1) 192.168.10.0/25, which can be represented in binaries as below.

11000000.10101000.00001010.0 | 0000000
11111111.11111111.11111111.1 | 0000000

2) 192.168.10.128/25, which can be represented in binaries as below.

11000000.10101000.00001010.1 | 0000000
11111111.11111111.11111111.1 | 0000000

Divide second subnet (192.168.10.128/25) we got from the first division again into two Networks, each with 64 IP Addresses(62 usable IP Addresses) using 255.255.255.192 subnet mask.

We will get two subnets each with 64 IP Addresses (62 usable IP Addresses).

1) 192.168.10.128/26, which can be represented in binaries as below.

11000000.10101000.00001010.1 | 0 | 000000
11111111.11111111.11111111.1 | 1 | 000000

2) 192.168.10.192/26

11000000.10101000.00001010.1 | 1 | 000000
11111111.11111111.11111111.1 | 1 | 000000

Divide 192.168.10.192/26 Network again into two Networks, each with 32 IP Addresses (30 usable IP addresses) using 255.255.255.224 subnet mask

We will get two subnets each with 32 IP Addresse (30 usable IP addresses).

1) 192.168.10.192/27, which can be represented in binaries as below.

11000000.10101000.00001010.11 | 0 | 00000
11111111.11111111.11111111.11 | 1 | 00000

2) 192.168.10.224/27, which can be represented in binaries as below.

11000000.10101000.00001010.11 | 1 | 00000
11111111.11111111.11111111.11 | 1 | 00000

Now we have split the 192.168.10.0/24 network into four subnets using Variable Length Subnet Masking (VLSM), with unequalnumber of IP addresses as shown below. Also note that when you divide a network using Variable Length Subnet Masking(VLSM), the subnet masks are also different.

1) 192.168.10.0 - 255.255.255.128 (126 (128-2) usable IP Addresses)
2) 192.168.10.128 - 255.255.255.192 (62 (64-2) usable IP Addresses)
3) 192.168.10.192 - 255.255.255.224 (30 (32-2) usable IP Addresses)
4) 192.168.10.224 - 255.255.255.224 (30 (32-2) usable IP Addresses)

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