Exam 200-301 topic 1 question 34 discussion

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

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

calculate the required bits each floor 2^n - 2 > users -> n = 5 -> subnet each floor /27

According to ChatGPT the answer is B. It explains why like this: B. 192.168.0.0/23 as summary and 192.168.0.0/25 for each floor: A /23 subnet provides 512 IP addresses (510 usable IPs), which can cover multiple /25 subnets. A /25 subnet provides 128 IP addresses (126 usable IPs), which is more than enough for any of the floors. This option allows for efficient use of IP addresses, providing enough space for each floor and a summary network that covers them all.

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

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.

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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

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

here should be add vlan also, in this situation question was a little misunderstand, but its cisco tricky question