Refer to the exhibit. The router has been configured with a super net to accommodate the requirements for 380 users on a Subnet. The requirement already considers 30% future growth. Which configuration verifies the IP subnet on router R4?
Refer to the exhibit. The router has been configured with a super net to accommodate the requirements for 380 users on a Subnet. The requirement already considers 30% future growth. Which configuration verifies the IP subnet on router R4?
To accommodate 380 users with a requirement for 30% future growth (which accounts for an additional 114 users), we need a subnet that supports at least 494 users. A /23 CIDR notation provides 512 IP addresses (including network and broadcast addresses), which is adequate. Therefore, a subnet mask of 255.255.254.0 is appropriate. The correct broadcast address for the subnet 10.7.54.0/23 is 10.7.55.255, and the usable IP address range is 10.7.54.1 to 10.7.55.254. Thus, the correct configuration is: Subnet: 10.7.54.0, Subnet mask: 255.255.254.0, Broadcast address: 10.7.55.255, Usable IP address range: 10.7.54.1 – 10.7.55.254.
Questions like this can be process of elimination. I highly recommend watching Subnetting Mastery playlist by Practical Networking on Youtube. You learn a very handy chart. Need 380 users. A /23 works. /23 is 254. So either C or D. Broadcast address is always odd. So D.
Answer C IP Address: 10.7.54.0 Network Address: 10.7.54.0 Usable Host IP Range: 10.7.54.1 - 10.7.55.254 Broadcast Address: 10.7.55.255 Total Number of Hosts: 512 Number of Usable Hosts: 510 Subnet Mask: 255.255.254.0 Wildcard Mask: 0.0.1.255 Binary Subnet Mask: 11111111.11111111.11111110.00000000 IP Class: B CIDR Notation: /23
You meant D right? Answer C has an incorrect broadcast address. You have listed the correct broadcast address 10.7.55.255, but chose option C with 10.7.54.255, which is not correct.
No, he is correct, the broadcast address is 10.7.55.255. Think about it, the last octet can only hold 256 bits, correct? Therefore, an address range of 10.7.54.0 - 10.7.54.255 wouldn't make any sense for a /23 network as there are 512 bits and that range only has room for 256. Therefore, the correct range is 10.7.54.0 - 10.7.55. 255. Think of it like this: 10.7.54.0 - 10.7.54.255 = 256 hosts 10.7.55.0 - 10.7.55.255 = 256 hosts 10.7.54.0 - 10.7.55.255 = 512 hosts
Mod: please delete, I read the comment too fast and thought he meant that C was correct.
We currently have 380 users in one network which means we need at-least /23 SM which gives 512 users in one network. The future growth is 30% of 380 = 114 additional users which makes our total to 393 users. using /23 SM means that the third quadrant will be divided into 2 hosts per network >> 256/2 = 128 total networks Working on the 3rd quadrant we will start from 10.7.0.0 to 10.7.1.255 we will work our way up till we reach 10.7.54.0 to 10.7.55.255
Discarding process is key. You need at min a /23 to satisfied 380 hosts. Subnet mask in 3rd octet is then .254 (so you can discard A and B). The IP range given for C and D is the same, so you can easily detect which one is the broadcast address by adding +1 to the last usable host. D is the final answer
D is right.
Answer D
D IS CORRECT
D is correct
Answer C is Correct