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VLSM LAB MANUAL
Created by Sandup Kolwalkar, CCNA [email protected]
Edited by Virginia Phillips, CCNA,CCAI
VLSM vs. CIDR
VLSM is similar to CIDR Both recursively divide networks into small sub networks
Difference VLSM:
The recursion is performed on the address space previously assigned to an organization and is invisible to the global Internet.
CIDR:
CIDR permits the recursive allocation of an address blocked by an Internet Registry to a high-level ISP, to a mid level to a lower level ISP and finally to a private organization’s network.
The issues to be considered while designing a network
1) How many total subnets does the organization need today?
2) How many total subnets will the organization need in the future?
3) How many hosts are there on the organization’s largest subnet today?
4) How many hosts will there be on the organization’s largest subnet in
the future?
Solution to Problem 1: To support 26 hosts on a subnet, a minimum of 5 bits is needed in the host portion of the address. 5 bits result in 30 possible host addresses (2^5-2). The other 3 bits in the last octet can be added to the default 24-bit Class C mask. Thus, a 27-bit mask can be used to create the following subnets:-
Subnet # Subnet Address 0 192.168.15.0 /27 1 192.168.15.32 /27 2 192.168.15.64 /27 3 192.168.15.96 /27 4 192.168.15.128 /27 5 192.168.15.160 /27 6 192.168.15.192 /27
Sub-subnet 0 192.168.15.0 /30 Sub-subnet 1 192.168.15.4 /30 Sub-subnet 2 192.168.15.8 /30 Sub-subnet 3 192.168.15.12 /30 Sub-subnet 4 192.168.15.16 /30 Sub-subnet 5 192.168.15.20 /30 Sub-subnet 6 192.168.15.24 /30
To maximize the address space, the 192.168.150.0 /27 subnet is further subnetted using a 30-bit mask. This creates subnets that can be used on point-to-point links with minimal waste, because each subnet contains only 2 possible host addresses.
Solution to Problem 2: The network address given is 192.168.15.0. Since this is the only network address available it cannot be supernetted. Also the problem we are now facing is that one subnet requires 60 hosts and we require at least 4 subnets. Therefore at least 3 bits are needed for 4 subnets (2^3-2=6). We are therefore left with 5 bits which would give us a maximum of 30 hosts (2^5-2). We solve this problem by supernetting 2 subnets. A 27-bit mask can be used to create the following subnets:-
The 2 subnets viz. subnet #2 and #3 can be supernetted using a 26-bit mask thus giving a subnet 192.168.15.64 /26 which provides 62 (2^6-2) hosts on this subnet. To maximize the address space, the 192.168.15.0 /27 subnet is further subnetted using a 30-bit mask. This creates subnets that can be used on point-to-point links with minimal waste, because each subnet contains only 2 possible host addresses.
Subnet # Subnet Address 0 192.168.15.0 /27 1 192.168.15.32 /27 2 192.168.15.64 /27 3 192.168.15.96 /27 4 192.168.15.128 /27 5 192.168.15.160 /27 6 192.168.15.192 /27
Sub-subnet 0 192.168.15.0 /30 Sub-subnet 1 192.168.15.4 /30 Sub-subnet 2 192.168.15.8 /30 Sub-subnet 3 192.168.15.12 /30 Sub-subnet 4 192.168.15.16 /30 Sub-subnet 5 192.168.15.20 /30 Sub-subnet 6 192.168.15.24 /30
Solution to Problem 3: The CIDR address given is 192.168.24.0 /22. This corresponds to the following 4 subnets: 192.168.24.0 /24 192.168.25.0 /24 192.168.26.0 /24 192.168.27.0 /24 One of the network requires 400 hosts. We therefore combine the 2 subnets to give a bigger subnet which provides 510 hosts.(2^9-2) viz.192.168.24.0/24 and 192.168.25.0 /24 are combined to give 192.168.24.0 /23 subnet. The 192.168.26.0/24 subnet will give 254 (2^8-2) hosts and is therefore used for one of the subnets consisting of 200 hosts. The 192.168.27.0 /24 subnet can be further subnetted as follows:-
The subnet #0 and #1 would provide 62 hosts (2^6-2) per subnet and are therefore used for the networks consisting of 50 hosts. To maximize the address space, the 192.168.27.128 /26 subnet is further subnetted using a 30-bit mask. This creates subnets that can be used on point-to-point links with minimal waste, because each subnet contains only 2 possible host addresses.
Subnet # Subnet Address 0 192.168.27.0 /26 1 192.168.27.64 /26 2 192.168.27.128 /26 3 192.168.27.192 /26
Sub-subnet 0 192.168.27.128 /30 Sub-subnet 1 192.168.27.132 /30 Sub-subnet 2 192.168.27.136 /30 Sub-subnet 3 192.168.27.140 /30 -- -- -- -- Sub-subnet 14 192.168.27.184 /30
The CIDR address given is 192.68.30.0 /23.
}
} → This would give a 62 host subnet
Solution to Problem 4
192.168.30.0 /24 192.168.31.0 /24
192.168.30.0 /25 192.168.31.128 /25
192.168.31.0 /26
192.168.31.64 /26
192.168.128 /26
192.168.31.192 /26
192.168.30.0 /23
These 2 would provide 2 subnets of 126 hosts each
192.168.31.64 /27 192.168.31.96 /27
} 2 30 host subnets 192.168.31.128 /27 192.168.31.160 /27
} 2 30 host subnets
192.168.31.192 /27 192.168.31.224 /27
192.168.31.192 /30 192.168.31.196/30 192.168.31.200/30 . . . . . . . 192.168.31.220/30
Solution to Problem 5
192.168.48.0 /21
192.168.48.0 /23
192.168.50.0 /23
192.168.52.0 /23
192.168.54.0 /23
We require 2 subnets with 512 hosts, so we use a mask of 23 bits
192.168.52.0 /24 192.168.53.0 /24 } is used for 210 host subnet. Give 2 254 host subnets
} gives 2 126 host subnets 192.168.53.0 /25 192.168.53.128 /25
192.168.54.0 /24 192.168.55.0 /24
192.168.54.0 /28 192.168.54.16 /28 192.168.54.32 /28
… 192.168.54.240 /28
192.168.54.128 /30 192.168.54.132 /30 192.168.54.136 /30
… 192.168.54.140 /30
For 3 serial connections{
} give 2 510 hosts each
14 host subnets
2 host subnet
