CCNA1 8.2.1.4 Packet Tracer – Designing and Implementing a VLSM Addressing Scheme

Jun 4, 2016 | Comments

I’m currently taking the CCNA1 course offered by Cisco. I struggled a lot with this activity so I thought it would be good to share how I finally figured it out. If you’re a little lazy and just want the answers, click here to go straight to the addressing table or here to download the PDF. Be aware that the addresses may vary but the process is the same regardless.

I am only human and will make mistakes so do not hesitate to point out any errors!

Part 1: Examine the Network Requirements

Step 1: Determine the number of subnets needed.

You will subnet the network address 192.168.72.0/24. The network has the following requirstrongents:

  • ASW-1 LAN will require 7 host IP addresses
  • ASW-2 LAN will require 15 host IP addresses
  • ASW-3 LAN will require 29 host IP addresses
  • ASW-4 LAN will require 58 host IP addresses
How many subnets are needed in the network topology?

8

5 subnets are needed. If you look at the topology, there are 4 LANs (coloured in orange) and 1 serial connection between Building1 and Building2. Therefore, you need 5 subnets.

Step 2: Determine the subnet mask information for each subnet.

The original subnet mask of the network address is 255.255.255.0. This comes from the prefix length /24, which indicates that there are 24 bits set in the subnet mask. We will use this as the basis for subnetting.

11111111 11111111 11111111 00000000
255 255 255 00000000
a. Which subnet mask will accommodate the number of IP addresses required for ASW-1?

255.255.255.240 with a prefix length of /28.

First, calculate the number of host bits that will be able to contain at least 7 hosts. (2^n-2\

= 2^4 – 2\

= 14 usable >= 7 required)

14 is greater than 7, so this gives 4 bits are not set in the subnet mask.

255 255 255 240
128+64+32+16+8+2+1 128+64+32+16+8+2+1 128+64+32+16+8+2+1 128+64+32+16
11111111 11111111 11111111 11110000
How many usable host addresses will this subnet support?

14. This comes from the formula in the previous question.

b. Which subnet mask will accommodate the number of IP addresses required for ASW-2?

255.255.255.224 with a prefix length of /27. (2^n-2\

= 2^5 – 2\

= 30 usable >= 15 required)

255 255 255 224
11111111 11111111 11111111 11100000
How many usable host addresses will this subnet support?

30.

c. Which subnet mask will accommodate the number of IP addresses required for ASW-3?

255.255.255.224 with a prefix length of /27. (2^n-2\

= 2^5 – 2\

= 30 usable >= 29 required)

255 255 255 224
11111111 11111111 11111111 11100000
How many usable host addresses will this subnet support?

30.

d. Which subnet mask will accommodate the number of IP addresses required for ASW-4?

255.255.255.192 with a prefix length of /26. (2^n-2\

= 2^6 – 2\

= 62 usable >= 58 required)

255 255 255 224
11111111 11111111 11111111 11000000
How many usable host addresses will this subnet support?

62.

e. Which subnet mask will accommodate the number of IP addresses required for the connection between Building1 and Building2?

255.255.255.2552 with a prefix length of /30.

We can use one subnet for the WAN. Since there are only two routers involved, we just need two addresses for this subnet. (2^n-2\

= 2^2 – 2\

= 2 usable >= 2 required)

255 255 255 252
11111111 11111111 11111111 11111100

Part 2: Design the VLSM Addressing Schstronge

Step 1: Divide the 192.168.72.0/24 network based on the number of hosts per subnet.

a. Use the first subnet to accommodate the largest LAN.

192.168.72.0/26. The largest LAN is ASW-4 with 58 hosts. Subnet 192.168.72.0/24 into 192.168.72.0/26. This will give us 4 subnets ((2^2 = 4)) with 64 hosts per subnet.

The subnets are:

  • 192.168.72.0
  • 192.168.72.64
  • 192.168.72.128
  • 192.168.72.192

Since the subnets each contain 64 hosts, simple add 64 to the last octet. This method will not be as feasible for subnets with a large number of hosts. Another way is to convert everything to binary. Only the first 2 bits will change while the rstrongaining 6 bits stay the same.

192.168.72.0 110000.10101000.01001000.00000000
192.168.72.64 110000.10101000.01001000.01000000
192.168.72.128 110000.10101000.01001000.10000000
192.168.72.192 110000.10101000.01001000.11000000
b. Use the second subnet to accommodate the second largest LAN.

192.168.72.6427.. We are using the second subnet because we are reserving the first subnet for the ASW-4 network. The second largest LAN is ASW-3 with 29 hosts. Subnet 192.168.72.6226 into 192.168.72.6227. This will give 2 subnets ((2^1 = 2)) with 32 hosts per subnet. We use (2^1) because the base is /26 and /27 is only one bit longer.

The subnets are:

  • 192.168.72.64
  • 192.168.72.96
192.168.72.64 110000.10101000.01001000.0100000
192.168.72.96 110000.10101000.01001000.01100000
c. Use the third subnet to accommodate the third largest LAN.

192.168.72.9627. The third largest LAN is ASW-2 with 15 hosts. In the previous question, we already have 2 subnets that have 32 addresses each. The second subnet will be able to accomodate ASW-2. So we do not need to subnet further.

d. Use the fourth subnet to accommodate the fourth largest LAN.

192.168.72.12828. Subnet 192.168.72.12826 into 192.168.72.12828. This will give 4 subnets ((2^2 = 4)) with 16 hosts per subnet. We use (2^2) because the base is /26 and /28 is two bits longer.

The subnets are:

  • 192.168.72.128
  • 192.168.72.144
  • 192.168.72.160
  • 192.168.72.176
192.168.72.128 110000.10101000.01001000.10000000
192.168.72.144 110000.10101000.01001000.10010000
192.168.72.160 110000.10101000.01001000.10100000
192.168.72.176 110000.10101000.01001000.10110000
e. Use the fifth subnet to accommodate the connection between Building1 and Building2.

192.168.72.14530 and 192.168.72.14630. Subnet 192.168.72.14428 into 192.168.72.14430. This will give 4 subnets ((2^2 = 4)) with 2 hosts per subnet.

The subnets are:

  • 192.168.72.144
  • 192.168.72.148
  • 192.168.72.152
  • 192.168.72.156
192.168.72.144 110000.10101000.01001000.10010000
192.168.72.148 110000.10101000.01001000.10010100
192.168.72.152 110000.10101000.01001000.10011000
192.168.72.156 110000.10101000.01001000.10011100

Step 2: Document the VLSM subnets.

Complete the Subnet Table, listing the subnet descriptions (e.g. ASW-1 LAN), number of hosts needed, then network address for the subnet, the first usable host address, and the broadcast address. Repeat until all addresses are listed.

Subnet Table
Subnet Description Number of Hosts Needed Network Address/CIDR First Usable Host Address Broadcast Address
ASW-1 LAN 7 192.168.72.12828 192.168.128.129 192.168.128.143
ASW-2 LAN 15 192.168.72.6427 192.168.72.65 192.168.72.95
ASW-3 LAN 29 192.168.72.9627 192.168.72.97 192.168.72.127
ASW-4 LAN 58 192.168.72.0/26 192.168.72.1 192.168.72.63
Serial WAN 2 192.168.72.14430 192.168.72.145 192.168.72.147

Step 3: Document the addressing schstronge.

  • ASW-1 LAN: 192.168.72.129
  • ASW-2 LAN: 192.168.72.97
  • Serial WAN: 192.168.72.145
  • ASW-3 LAN: 192.168.72.65
  • ASW-4 LAN: 192.168.72.1
  • Serial WAN: 192.168.72.146
c. Assign the second usable IP addresses to the switches.
  • ASW-1: 192.168.72.130
  • ASW-2: 192.168.72.98
  • ASW-3: 192.168.72.66
  • ASW-4: 192.168.72.2
d. Assign the last usable IP addresses to the hosts.
  • Host-A: 192.168.72.142
  • Host-B: 192.168.72.94
  • Host-C: 192.168.72.126
  • Host-D: 192.168.72.62

Part 3: Assign IP Addresses to Devices and Verify Connectivity

Now it’s just a matter of plugging in values into Packet Tracer if you haven’t already.

Addressing Table

Remote-Site 1

Remote-Site 2

Device Interface IP Address Subnet Mask Default Gateway
G0/0 192.168.72.129 255.255.255.240 N/A
G0/1 192.168.72.97 255.255.255.224 N/A
S0/0/0 192.168.72.145 255.255.255.252 N/A
G0/0 192.168.72.65 255.255.255.224 N/A
G0/1 192.168.72.1 255.255.255.192 N/A
S0/0/0 192.168.72.146 255.255.255.252 N/A
SW1 VLAN 1 192.168.72.130 255.255.255.240 192.168.72.129
SW2 VLAN 1 192.168.72.98 255.255.255.224 192.168.72.97
SW3 VLAN 1 192.168.72.66 255.255.255.224 192.168.72.65
SW4 VLAN 1 192.168.72.2 255.255.255.192 192.168.72.1
User-1 NIC 192.168.72.142 255.255.255.240 192.168.72.129
User-2 NIC 192.168.72.126 255.255.255.224 192.168.72.97
User-3 NIC 192.168.72.94 255.255.255.224 192.168.72.65
User-4 NIC 192.168.72.62 255.255.255.192 192.168.72.1
Category: Technology
Tags: #CCNA

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