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?
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?
The engineer must add a subnet to accommodate 20 new users. To determine the correct subnet, we need to ensure enough addresses are available while minimizing wasted addresses. A subnet mask of /28 provides 16 addresses (14 usable), which is insufficient for 20 users. Therefore, a subnet mask of /27, which provides 32 addresses (30 usable), is appropriate. The only option with a /27 mask is 10.10.225.32 255.255.255.224, making it the correct choice. This option allocates 32 addresses, sufficient for the 20 users, and fits within the existing subnet structure.
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
Thank you
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
Sorry .128
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.
It is quite confusing especially if you are not really careful 10.10.225.48 is a host in network 10.10.225.32-10.10.225.63 as broadcast
D CORRECT :) BLOCKS SIZE = 2^#OF ZEROS IN WORK OCTET 2^5=32 10.10.225.0 / 10.10.225.32 / 10.10.225.64/ 10.10.225.96 >>> ETC >>>>
The keyword is which “network” technically C is a host ip and D is the network id so D is correct
Why not C?
got it!
Because 10.10.225.48 255.255.255.224 is not a valid network ID. the valid network IDs for a /27 network are 0, 32, 64, 96, 128 etc in the 4th octet.
It is not valid because by vlsm they are subnetting from 28 bit to 28 bit leaving a 27 bit network and finally using a 26 bit network.
Definitely D
D is correct!
C is a host address we can't use it as a subnet but D is the network address So we can use it
D is correct
/27 = 32 host /27 magic number 256-224= 32 10.10.225.0 10.10.225.32 10.10.225.64 ....
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).
That is a good explanation for this question.
C - write. Because D - do not contain 20 propered host`s: IP range D = 15 (63-48)
I apologize)). Confuse myself! D - write. Because C - do not contain 20 propered host`s: IP range C = 15 (63-48)
Simple maths - D!