Please answer with the given IPs. Your boss wants you to add one subnet to an ex
ID: 3725839 • Letter: P
Question
Please answer with the given IPs.
Your boss wants you to add one subnet to an existing design. The existing design already has five subnets:
10.0.0.0 /24
10.0.1.0 /25
10.0.2.0 /26
10.0.3.0 /27
10.0.6.0 /28
The boss cannot decide among five competing subnet masks to use for this next new subnet to add to the internetwork. However, the boss wants you to practice VLSM and plan the subnet ID that he would use for each of those four possible masks. He tells you that the new subnet ID must be part of Class A network 10.0.0.0, that the new subnet must not overlap with the original five subnets, and that the new subnet ID must be the numerically lowest possible subnet ID (without breaking the other rules). Pick the one subnet ID that you would plan to use based on each of the following mask choices by the boss: /24, /23, /22, /25
Explanation / Answer
We are given existing subnets as:
10.0.0.0 /24
10.0.1.0 /25
10.0.2.0 /26
10.0.3.0 /27
10.0.6.0 /28
Our aim is to find in each /25, /24, /23, /22 subnets what will be the least value of subnet id that doesn’t overlap with these values
Case 1: Assume we take /24.
In classless subnetting, we have 4 octects, representing the host and network id.
/24 means:
255.255.255.0 -> as all the 1st 24 bits will be 1:
The new subnets would be 10.0.0.0, 10.0.1.0, 10.0.2.0, 10.0.3.0, 10.0.4.0, 10.0.5.0, and so on. The reason is that the value will be incrementing by 1 in the 3rd octet.
It will overlap with 4 existing subnets: 10.0.0.0, 10.0.1.0, 10.0.2.0, 10.0.3.0, 10.0.4.0
Same value present in the question and what we have found.
So, the least value that doesnot overlap in this case is 10.0.5.0/24
Case 2: Assume we take /23.
/23 means: 11111111. 11111111. 11111110.00000000 i.e., 255.255.254.0
The subnets are 10.0.0.0, 10.0.2.0, 10.0.4.0, 10.0.6.0, 10.0.8.0, and so on, counting by 2 in the third octet-> as 2nd last bit that is 1 corresponds to value 2
It will overlap with 2 existing subnets 10.0.0.0, 10.0.6.0
Same value present in the question and what we have found.
So, the least value that doesnot overlap is 10.0.4.0/23 -> it is present in the subnet we found and not in the list of existing networks given.
Case 3: Assume we take /22.
/23 means: 11111111. 11111111. 11111100.00000000 i.e., 255.255.254.0
The subnets are 10.0.0.0, 10.0.4.0, 10.0.8.0, 10.0.12.0, 10.0.16.0, and so on, incrementing by 4 in the third octet
It will overlap with only 1 existing subnets 10.0.0.0
Same value present in the question and what we have found.
So, the least value that doesnot overlap is 10.0.4.0 /22 -> it is present in the subnet we found and not in the list of existing networks given.
10.0.4.0 is the correct answer here
Case 4: Mask /25
/25 means-> 255.255.255.128=>
11111111.11111111.11111111.10000000
The 1st 14 masks in /25 are:
10.0.0.0, 10.0.0.128 , 10.0.1.0, 10.0.1.128, 10.0.2.0, 10.0.2.128, 10.0.3.0, 10.0.3.128 , 10.0.4.0, 10.0.4.128, 10.0.5.0, 10.0.5.128 , 10.0.6.0, 10.0.6.128
The ones highlighted in bold are the ones present in the exsiting system. Hence the least possible value that doesnot overlap is 10.0.1.128/25
I hope the explanation is clear, as I have discussed from the very ba