Find a new subnet ID must be part of Class A network 10.0.0.0, that the new subn
ID: 3725832 • Letter: F
Question
Find a 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.
The already used ranges are:
10.0.0.0 /24 (Broadcast: 10.0.0.255, Host.Min: 10.0.0.1, Host.Max: 10.0.0.254)
10.0.1.0 /25 (Broadcast: 10.0.1.127, Host.Min: 10.0.1.1, Host.Max: 10.0.0.126)
10.0.2.0 /26 (Broadcast: 10.0.2.63, Host.Min: 10.0.2.1, Host.Max: 10.0.2.62)
10.0.3.0 /27 (Broadcast: 10.0.3.31, Host.Min: 10.0.3.1, Host.Max: 10.0.3.30)
10.0.6.0 /28 (Broadcast: 10.0.6.15, Host.Min: 10.0.6.1, Host.Max: 10.0.6.14)
Explanation / Answer
Solution:
Based on the given ranges
10.0.0.0 /24 (Broadcast: 10.0.0.255, Host.Min: 10.0.0.1, Host.Max: 10.0.0.254)
10.0.1.0 /25 (Broadcast: 10.0.1.127, Host.Min: 10.0.1.1, Host.Max: 10.0.0.126)
10.0.2.0 /26 (Broadcast: 10.0.2.63, Host.Min: 10.0.2.1, Host.Max: 10.0.2.62)
10.0.3.0 /27 (Broadcast: 10.0.3.31, Host.Min: 10.0.3.1, Host.Max: 10.0.3.30)
10.0.6.0 /28 (Broadcast: 10.0.6.15, Host.Min: 10.0.6.1, Host.Max: 10.0.6.14)
The mask which will be still in class A will be /23 and /22
I would be using /23.
If the information about the number of hosts present in it would have been easy, but from these two masks, the lowest possible subnet id will be from /23.
Number of hosts possible in this mask are = (2^(32-23))-2 = 2^9 - 2 = 510 hosts
Range:
10.0.0.5 /23 (Broadcast: 10.0.1.255, Host.Min: 10.0.0.1, Host.Max: 10.0.1.254)
I hope this helps if you find any problem. Please comment below. Don't forget to give a thumbs up if you liked it. :)
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.