Subnetting and Assigning Addresses Figure 2. Addressing Design Scenario You are
ID: 3633306 • Letter: S
Question
Subnetting and Assigning Addresses
Figure 2. Addressing Design Scenario
You are the network administration for the small network in Figure 2. It consists of your headquarters location with a LAN with 60 hosts, remote office R1 with 10 hosts, and remote office R2 with 30 hosts. You have been assigned the address space 192.163.3.0/25. From this space you will need to create the subnets for each of the sites on your network and the two WAN links. Once you find the subnets assign addresses to the Routers interfaces and PCs using the following guidelines: Assign the first address from LAN subnets to the router interface connected to that LAN. Assign the second IP address in the LAN subnet to the PC on the LAN. Assign the first Address from WAN subnets to the HQ router end of the WAN link and the second address to the remote router interface. Use the steps below to help you work through it.
Create the subnets:
Address space: 192.163.3.0/25
Write it out in binary form identifying the host and network then use the space below in table as a scratch pad to help you in subnetting as needed. Remember to be efficient you usually want to start by creating the largest subnet you need first and then continue subnetting to find the smaller subnets in succession.
Network Host
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
Question Answer
23 What is the HQ subnet (address/mask)?
24 What is the R1 subnet (address/mask)?
25 What is the R2 subnet (address/mask)?
26 What is the subnet for the HQ to R1 WAN (address/mask)?
27 What is the subnet for the HQ to R2 WAN (address/mask)?
Fill in the address and subnet mask for the device interfaces in the table below using the assignment rules described above.
Device Interface Address Mask
HQ Router Fa0/0 28 39
S0/0/0 29 40
S0/0/1 30 41
R1 Router Fa0/0 31 42
S0/0/0 33 43
R2 Router Fa0/1 34 44
S0/0/0 35 45
PC1 Ethernet 36
PC2 Ethernet 37
PC5 Ethernet 38
Explanation / Answer
Dear... Given numbering systems:192.163.3.0 is a decimal representation of four binary octets. And been given a /25 to work with, that's 128 addresses in decimal IP/mask format it's 192.163.3.0 mask 255.255.255.128 And it is needed 60 addresses on the Lan, and down to a /26 with 64 addresses. 1. For Remote R2, you'll halve the remaining space to a /27 with 32 addresses
2. For R1, you'll halve what's left, to a /28 with 16 addresses, and then split the remainder into the two WAN links Take each octet and convert it to binary:
192 is 1 hundreds + 9 tens + 2 ones. A binary octet is binary to eight places: 128s, 64s, 32s, 16s, 8s, 4s, 2s, 1s for each place respectively. 192 is 11000000 or 128+64. 163 is 10100011 or 128+32+2+1. 3 is 00000011 or 2+1. This means the network address in binary is:
11000000.1100011.0000011.00000000 and a subnet mask of 25 tells us which bits are relevant for traffic that would need to be sent to a router and looks like this: 11111111.11111111.10000000.00000000. It's probably easier to keep track of the addressing if you take the larger subnets from the high side, so you'd begin with 192.163.3.64/26 (netmask 255.255.255.192), then 192.163.3.32/27 (netmask 255.255.255.224)
This means any traffic that only uses IP address spaces defined by the 23 least significant bits is inside the router and can be sub-netted further as needed. I hope that helps. You really have to grasp numbering systems before you can understand IP networking.
In a netmask, when the mask bit is 1, the corresponding address bit is part of the net/subnet address, and while 0, is available as an address.
The / format is another representation, with classic class A, B and C being /8, /16 and /24 - encompassing whole octets, while other values such as /25 use bits from the top of the next octet.