A corporate office uses four floors in a building.
✑ Floor 1 has 24 users.
✑ Floor 2 has 29 users.
Floor 3 has 28 users.
✑ Floor 4 has 22 users.
Which subnet summarizes and gives the most efficient distribution of IP addresses for the router configuration?
Given the user distribution across the four floors: 24 users on Floor 1, 29 users on Floor 2, 28 users on Floor 3, and 22 users on Floor 4, we need to allocate IP addresses efficiently to avoid wasting addresses while accommodating the number of users per floor. Each floor requires a subnet that can handle at least the maximum number of users on any floor, which is 29. A /27 subnet provides 32 IP addresses (30 usable addresses), which is sufficient for each floor. Therefore, a /27 subnet for each floor is appropriate. Summarizing all subnets for the entire building, we have a total of 103 users. A /25 subnet provides 128 IP addresses (126 usable addresses), which accommodates the total number of users. Hence, the correct answer is the summary of 192.168.0.0/25 and 192.168.0.0/27 for each floor.
write this down first thing in the exam: /32 1 /31 2 /30 4 /29 8 /28 16 /27 32 /26 64 /25 128 /24 256 /23 512 /22 1024 /21 2048
love you
please, what's it's significance?
Number of addresses per subnet
network summary each floor, max user to each floor=30<=2^H-2 H=5, will give N=3 therefore /27 For Network Summary, Total Users,= 103 103<=2^H-2 H=7 will give N=1 Therefore/25,,, i hope u got ans in short, tht is C
Where did N came from? I’m lost….
N is the Network, U is the usable host, H is the host
8-7=1 BRO
4 floors = 4 subnets. And you have a total of 103 users. How many bits do you need to have 103 addresses? You need 7 bits: (2^7 – 2) = 126 addresses. Starting from the 192.168.0.0 subnet that you’re given, you must use a /25 subnet mask: 255.255.255.1xxxxxxx = 255.255.255.128 How many bits do you need to configure 4 subnets? You need 2 bits: (2^2) = 4 subnets. You have to borrow the two bits from the host ID. This way, the subnet mask, which is a /25 now, becomes a /27: 255.255.255.111xxxxx = 255.255.255.224 There are 5 bits remaining on the host ID. You have (2^5 – 2) = 30 addresses, and it fits the subnet on which you have the most users (floor 2). You started with a 192.168.0.0/25 subnet and you ended up with a 192.168.0.0/27 subnet. Answer C is correct.
16 addresses per floor is not enough so 32 per floor is needed simply count from /32=1 /31=2 /30=4....../27=32 /27 per floor is the answer
That's how I got the answer quickly :)
We are Keeping in mind not to waste IPs. /27 gives us 30 maximum hosts per subnet (per floor) /25 gives us 126 maximum hosts per subnet (Total no. hosts in the building) C. 192.168.0.0/25 as summary and 192.168.0.0/27 for each floor
Subnet Mask: 128 192 224 240 Hosts: 128 64 32 16 /Cider 25 26 27 28 /27 is the smallest number that will meet the number of hosts required for all the floors
total user 103, must be /25 Mask as for all floors, /27 Mask for each floor, refer to 2^5 = 32 - NW ID and Broad cast address, total usable IP 30.
C is correct /27 mask will give 30 host for each subnet (2^5-2)=30
/25 gives us 126 maximum hosts per subnet (Total no. hosts in the building) C. 192.168.0.0/25 as summary and 192.168.0.0/27 for each floor
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here should be add vlan also, in this situation question was a little misunderstand, but its cisco tricky question
C is the one believe me :)
2 to the power of 5 is 32 address . 32 -2 ( Network add and broadcast add ) = 30 on subnet -- that`s hosts on subnet. We need 3 bits for netw and 5bits for host - thats /27
Can we keep tables during the exam?
before the exam this should be in your head group size 128, 64, 32, 16, 8, 4, 2, 1 subnet 128, 192, 224, 240, 248, 252, 254, 255 CIDR /25 /26 /27 /28 /29 /30 /31 /32 e.g 10.1.1.55/28 from the table above the subnet is 255.255.255.240 the ip add has a group size of 16 usable ip add is the group size you subtract 2 {16-2} correct answer is C
max = 29 => 2^5-2=30 => borrow 5 bit => 32-5 = 27 therefore /27 is corect
The answer is c
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