Refer to the exhibit. An engineer must add a subnet for a new office that will add 20 users to the network. Which IPv4 network and subnet mask combination does the engineer assign to minimize wasting addresses?
D is correct!
Find the subnet mask
*To have 20 User in a subnet We have to use /27 prefix
* So Host count for /27 prefix is (2^5-2)=30
* Subnet Mask for /27 prefix is (sum of Network bits (128+64+32)=224 , so 255.255.255.224
Find the network ID
*As per the /27 prefix each subnet has 30 host and 32 including network ID & Broadcast ID
* so first network ID is 10.10.255.0 and the second will be 10.10.255.32
I think there is a gap between the second and the third subnets, so we use .32 for the required network id, if more than 30, we need to use .64 as there are more space
A. 10.10.225.48 255.255.255.240
This is a /28 subnet. 4 bits in the host ID. You have (2^4 – 2) = 14 addresses. But you need 20 more IP addresses.
Wrong answer.
B. 10.10.225.32 255.255.255.240
This is a /28 subnet. 4 bits in the host ID. You have (2^4 – 2) = 14 addresses. But you need 20 more IP addresses.
Wrong answer.
C. 10.10.225.48 255.255.255.224
This looks like a /27 subnet. 5 bits in the host ID. You have (2^5 – 2) = 30 addresses. Could be the right answer, but there’s a mismatch between the subnet ID and the subnet mask.
If you perform the logical AND between the subnet ID and the subnet mask, you should obtain the subnet ID:
Subnet ID 00001010.00001010.11111111.00110000
Subnet mask 11111111.11111111.11111111.11100000
Result 00001010.00001010.11111111.00100000
Decimal 10.10.255.32
This is not the subnet ID. Wrong answer.
D. 10.10.225.32 255.255.255.224
This is a /27 subnet. 5 bits in the host ID. You have (2^5 – 2) = 30 addresses. No mismatches between subnet ID and subnet mask.
Correct answer.
Convert IP address and subnet mask to binary:
IP address: 10.10.225.48 becomes 00001010.00001010.11111111.00110000
Subnet mask: 255.255.255.224 becomes 11111111.11111111.11111111.11100000
Compare the first 27 bits:
Align the two binary strings and check if the first 27 bits (from left to right) are identical:
00001010.00001010.11111111.00110000 (IP address)
11111111.11111111.11111111.11100000 (Subnet mask)
The first 27 bits (highlighted) match perfectly.
10.10.225.48 255.255.255.224 Is a Valid Network ID: It might seem unusual at first glance, but it's indeed a valid network ID for a /27 subnet. Here's why:
The first 27 bits of 10.10.225.48 (in binary) match exactly with the first 27 bits of 255.255.255.224, fulfilling the requirement for a valid network ID within a /27 subnet.
This means the first three octets (10.10.225) and the first three bits of the fourth octet (001) are fixed for all IP addresses within this subnet, while the remaining 5 bits in the fourth octet can vary for host addresses.
Why D is currect answer because /27 networks = 2^3 = 8 networks
So because the /27 more than the maximum /24
We must divided the 256 over the number of networks 256/8=32
32-2 =30 ip for each network
i found this info about Spine and Leaf
In this two-tier topology, everything is one hop from everything else.
And leaf switches have no link between them.. but with spine swithc..
Talking about last octet here. Need 5 host bits to fit the desired usable IPs. Therefore, CIDR=/27 (32-5) which is dotted-decimal 224. Since 5 host bits are needed, that leaves the first 3 bits for network, so can fit only the 32 value (as the 48 value takes up 4 bits).
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