Exam N10-008 topic 1 question 110 discussion

The standard Class C subnet mask 255.255.255.0 gives you 256 addresses. Borrowing one more bit from the network portion doubles the address capacity (512). Subtract 2 for gateway/broadcast and you're at 510. The answer is 255.255.254.0

Hopefully this explanation helps someone out: First, translate all of the subnet masks to CIDR notation A. 255.255.0.0 = /16 B. 255.255.252.0 = /22 C. 255.255.254.0 = /23 D. 255.255.255.0 = /24 Subtract 32 from the CIDR notation. A. 32 - 16 = 16 B. 32 - 22 = 10 C. 32 - 23 = 9 D. 32 - 24 = 8 Apply the Power of 2 from the differences A. 2^16 = 65,536 B. 2^10 = 1,024 C. 2^9 = 512 D. 2^8 = 256 Lastly subtract 2 to account for the reserved Network and Broadcast Address A. 65,536 - 2 = 65,534 B. 1,024 - 2 = 1,022 C. 512 - 2 = 510 D. 256 - 2 = 254 The answer is C. This is the smallest amount subnet that has 510 usable addresses and fits the criteria for what they are looking for.

2^9-2=510 so C is the answer WHy did I use 9? 0=8 bits 254=1 bit so, 9 is the host bit 2^n-2= Number of the host n: is the host bit

C. Looks good. The subnet mask C (255.255.254.0), which corresponds to a /23 subnet, can support up to 510 hosts. This is not sufficient for 412 devices considering network and broadcast addresses are not usable for devices. The subnet mask B (255.255.252.0), a /22 subnet, supports up to 1022 hosts, which is more than enough for 412 devices. Therefore, the original recommendation of subnet mask B (255.255.252.0) is the correct choice for accommodating 412 network-connected devices on the same subnet. Estimated correctness: 95%

borrowing one extra bit from 3rd octet (making a /23 network) will give you 510 hosts. this is equivalent to 255.255.254.0 netmask.

I like how GPT explains it: "To accommodate 412 network-connected devices on the same subnet while using the smallest possible subnet size, you need to calculate the appropriate subnet mask. The formula to calculate the number of usable host addresses in a subnet is 2^(n) - 2, where 'n' is the number of bits borrowed from the host portion of the IP address to create the subnet. You need at least 9 bits to accommodate 412 devices because 2^9 = 512, which is the smallest power of 2 that's greater than 412.However, we need to subtract 2 from the usable host addresses to account for the network address (all host bits set to 0) and the broadcast address (all host bits set to 1). So, the number of usable addresses = 2^9 - 2 = 510. The subnet mask would be 255.255.254.0 for a /23 subnet. This would give you a subnet that can accommodate 510 usable addresses, which should be sufficient for the 412 network-connected devices while keeping the subnet as small as possible. Correct answer: C. 255.255.254.0

Don't understand it one bit

It's kind of simple to come from /24 network which makes 2^8-2 = 254 , /23 will make 2^9-2 = 510 hosts, which is correct answer. /23 is 254 That needs to be remembered

To accommodate 412 network-connected devices, we need to select a subnet mask that provides at least 412 IP addresses. The smallest subnet mask that can achieve this is 255.255.254.0, which provides 510 IP addresses (512 - 2 reserved addresses). Therefore, the correct answer is C.

255.255.254.0 is correct, I'm not sure why everyone is saying B. They need 412 devices, /23 is 512 possible devices. How do you figure 1024 from /22 is more efficient? C is the only correct answer if you want to limit it to as few addresses as possible.

B. 255.255.252.0 The number of IP addresses in a subnet is determined by the number of bits used for the host portion of the IP address. The subnet mask used in this scenario should be as small as possible to accommodate the 412 devices. A subnet mask of 255.255.252.0 has 22 bits for the host portion of the IP address, which can support up to 1022 IP addresses (2^22 - 2). A. 255.255.0.0 has 16 bits for the host portion of IP address which support up to 65534 IP addresses. C. 255.255.254.0 has 23 bits for the host portion of the IP address, which can support up to 510 IP addresses D. 255.255.255.0 has 24 bits for the host portion of IP address which support up to 254 IP addresses Therefore, using a subnet mask of 255.255.252.0 is the most efficient option for accommodating the IP addresses for 412 devices.

An class C give you at most 256 addresses including network ID and broadcast address. In this case, we need more than 256 addresses while being smallest thus we looking at class B of at least 2 host size. With that said, option D is out because it is class C. Moving to the next smallest which is 255.255.254.0. Using wildcard which is reverse of CIDR and in this case 255 - 254 is 1 which means is 0.0.0.1. It also means it is 2 host size which give us 512 addresses. Why is it "2", because to get decimal 1 you will need an octet of 0000 0001 and applying the host formula of 2^1 = 2.