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?
A.
Subnet: 10.7.54.0 Subnet mask: 255.255.128.0 Broadcast address: 10.5.55.255 Usable IP address range: 10.7.54.1 ג€" 10.7.55.254
B.
Subnet: 10.7.54.0 Subnet mask: 255.255.255.0 Broadcast address: 10.7.54.255 Usable IP address range: 10.7.54.1 ג€" 10.7.55.254
C.
Subnet: 10.7.54.0 Subnet mask: 255.255.254.0 Broadcast address: 10.7.54.255 Usable IP address range: 10.7.54.1 ג€" 10.7.55.254
D.
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 ג€" 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
This is what I think; required users = 380
Considered 30% of 380 future growths = +114
Actual future number of users = 494 or more
Current IP: 10.7.54.0
Therefore, number of host bits needed for 494 users = 9 bits from 4th to 3rd octet ~/23 prefix
With the Prefix above; 2^9 = 512 host, Mask 255.255.254.0
Broadcast = 10.7.55.255, Usable IP 10.7.54.1 - 10.7.56.254
Answer = D
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
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