Exam 200-301 topic 1 question 476 discussion

We need to be able to put 600 hosts 3 different dedicated subnets. So not 300 existing servers in one subnet and 300 new servers in the next subnet, belonging to production only for example. The subnets must be dedicated. /22 is the only solution for each subnet.

300 additional server with 3 connections = 900 connects each connection will have it's own subnet = 3 subnets 900 connections divided by 3 subnets = 300 connections per subnet /24 = 254 connection/ip's /23 = 510 connections/ip's /23 = 255.255.254.0 So A is the correct answer

B is correct We have 600 Servers. Each Server have 3 Networks. Each Network has a different Subnet. Now we must configure only one Network of the 3 Networks. This Network has to accommodate 600 Servers (600 IPs).

600 servers with production, backup and management server : PRODUCTION : 600 address required BACKUP : 600 address required SERVER : 600 address required 255.255.254.0 give 512 address 255.255.252.0 give 1024 address So B is correct

My explanation why B is correct: from 300 servers to 600. Each server requires 3 network connections to support production, backup, and management traffic. - It states they will upgrade from 300 to 600, meaning 300 extra servers will be installed, each server will require 3 network connections, 300x3=900 : - 300 extra servers - 900 network connections/ip addresses 10.10.10.1/22 allows for 1024 hosts

Each server requires a dedicated connection for each subnet. Therefore we need a total of 1800 host connections (600 x 3) across the 3 subnets. Since, /22 --> 10 host bits --> 1024 - 2 = 1022 hosts /21 --> 11 host bits --> 2048 - 2 = 2046 hosts /21 fullfills the minimum IP address wasting by being the longest prefix length that supports the requirement. /21 --> 255.255.248.0 Matches none...

A is correct We have 600 Servers. Each Server have 3 Networks. Each Network has a different Subnet. Now we mus configure only one Network of the 3 Networks. This Network has to accommodate 600 Servers.

Non of the answers

B is correct 3 x 300 = 900 255.255.252.0 /2

It's a tricky question. Keep in mind that here we already have 300 existing servers (and we don’t need new IPs for them) but we’re adding 300 servers (to make the count to 600), so we need 300 new IPs for those new servers, divided in 3 subnets. Simply put, we need 3 new subnets where each subnet can accommodate 300 IP addresses. So, if we consider subnet 255.255.254 (or /23), it will give us 2^9 (=512) IP addresses i.e., 510 usable IP addresses, and 2^15 (23 – 8 = 15) subnets, which is sufficient for the given scenario. So correct answer is A.

Need 300 more servers, each 3 connections for: PROD, bck, mgm "The router configuration for the production network must be configured first using a subnet in the 10.0.0.0/8 network" "For the PRODuction" -> 300 IP needed, bck and mgm are DIFFERENT subnets. /23 -> A

Production Network had 300 servers. Now it has 600. /23 gives 510 /22 gives 1022 = 255.255.252.0

We aren't considering all 3 subnets. Just one subnet, for the production network: "Each server requires 3 network connections to support production, backup, and management traffic. Each connection resides on a different subnet. The router configuration for THE PRODUCTION NETWORK MUST BE CONFIGURED FIRST using a subnet in the 10.0.0.0/8 network ..." Therefore we only budget for the 300 hosts in the production network.

The question asks to add 300 hosts in three subnets, it doesn't make sense to mess with the settings of the servers that already exist (including SWs, Servers, and the whole network). In the case he asked to segment without waste (different from summarizing routes), that is, he would need to deliver to sub-interfaces (router-on-stick) or SVIs (SW L3), then the letter A 10.10.10.1 255.255.254.0 (/ 23) is correct, for the "Production Network" 10.10.10.0 - 10.10.11.255 (512-2 Hosts) would serve the additional 300 hosts. Example 10.10.12.0/23 for "Backup" and 10.10.14.0/23 for "Management"... on sub-interfaces or SVIs (SW L3).

There are already 300 servers divided in 3 subnets. The requirement here is to take up the number of servers from 300 to 600 i.e. increase it by "300" and each of those new 300 servers will need 3 different subnets. Therefore, the requirement is to have 3 new subnets and 900 (300 server x 3 subnets) IP addresses. Only /22 (255.255.252.0) fulfill this requirement by giving us 1024 IP addresses in each subnet.

The answers is B.

/23 can holds 500 hosts aprox, /22 reachs 1000 hosts. B is correct