Drag & Drop (IP Address)

A dental firm is redesigning the network that connects its three locations. The administrator gave the networking team 192.168.164.0 to use for addressing the entire network. After subnetting the address, the team is ready to assign the addresses. The administrator plans to configure ip subnet-zero and use RIP v2 as the routing protocol. As a member of the networking team, you must address the network and at the same time conserver unused addresses for future growth. With those goals in mind, drag the host addresses on the left to the correct router interface. Once of the routers is partially configured. Move your mouse over a router to view its configuration. Not all of the host addresses on the left are necessary.
DD1
How to Find?
Step 1: Find the Number of Hosts for each subnet mask.
Trust me! If you practice and practice, then the below formula will never go out of your head.
Network Bits Host Bits Formula Max. no. of Hosts
25 7 27-2 126 hosts
26 6 26-2 62 hosts
27 5 25-2 30 hosts
28 4 24-2 14 hosts
29 3 23-2 6 hosts
30 2 22-2 2 hosts


Step 2: Find the number of hosts and appropriate Network bits
Hosts Suitable place Network Bits Answer
11 hosts 14 hosts 28 192.168.164.166 /28
16 hosts 30 hosts 27 192.168.164.149 /27
6 hosts 6 hosts 29 192.168.164.178 /29


Step 3: Now the Serial Interface

We need only 2 IP address for Point-to-point Connection. Then suitable Network Bit were “30”.
But we have 2 IP addresses that have “/30”, 192.168.164.189 /30 and 192.168.164.188 /30.

Hint for Exam : The answer is the one that has ODD number in fourth octet. (Meaning .189)

Here goes the explanation:

Network Bits Subnet Mask
30 255.255.255.252


Binary for fourth part of the Subnet Mask (252) is 11111100 (6 1’s & 2 0’s)

If we convert 188 and 189 into binary format we will get

188 10111100

189
10111101


Check last 2 bits of 188 and 189

188 = 00 > Surely Network Address (Host bits were 0’s)
189 = 01 > Surely Assignable IP Address

DD2

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