Description

NAME: FONG WEI TZE
STUDENT ID: 22030274
PART A: Packet Tracer – Subnet an IPv4 Network
Step 1: Create a subnetting scheme that meets the required number of subnets and required number of host addresses.

How many host addresses are needed in the largest required subnet?
50

What is the minimum number of subnets required?
4

The network that you are tasked to subnet is 192.168.0.0/24. What is the /24 subnet mask in binary? 11111111.11111111.11111111.00000000

In the network mask, what do the ones represent?
The network portion.

In the network mask, what do the zeros represent?
The host portion

Given each of the possible subnet masks depicted in the following binary format, how many subnets and how many hosts are created in each example?

(/25) 11111111.11111111.11111111.10000000
Dotted decimal subnet mask equivalent: 255.255.255.128
Number of subnets = 2
Number of hosts: 126 hosts in each subnet.

(/26) 11111111.11111111.11111111.11000000
Dotted decimal subnet mask equivalent: 255.255.255.192
Number of subnets: 4
Number of hosts: 62 hosts in each subnet

(/27) 11111111.11111111.11111111.11100000
Dotted decimal subnet mask equivalent: 255.255.255.224
Number of subnets: 8
Number of hosts: 30 hosts in each subnet

(/28) 11111111.11111111.11111111.11110000
Dotted decimal subnet mask equivalent: 255.255.255.240
Number of subnets: 16
Number of hosts: 14 hosts in each subnet

(/29) 11111111.11111111.11111111.11111000
Dotted decimal subnet mask equivalent: 255.255.255.248
Number of subnets: 32
Number of hosts: 6 hosts in each subnet

(/30) 11111111.11111111.11111111.11111100
Dotted decimal subnet mask equivalent: 255.255.255.252
Number of subnets: 64
Number of hosts: 2 hosts in each subnet
Considering your answers above, which subnet masks meet the required number of minimum host addresses?
(/25) 11111111.11111111.11111111.10000000
(/26) 11111111.11111111.11111111.11000000

Considering your answers above, which subnet masks meets the minimum number of subnets required?
(/26) 11111111.11111111.11111111.11000000
(/27) 11111111.11111111.11111111.11100000
(/28) 11111111.11111111.11111111.11110000
(/29) 11111111.11111111.11111111.11111000
(/30) 11111111.11111111.11111111.11111100

Considering your answers above, which subnet mask meets both the required minimum number of hosts and the minimum number of subnets required?
(/26) 11111111.11111111.11111111.11000000

When you have determined which subnet mask meets all of the stated network requirements, derive each of the subnets. List the subnets from first to last in the table. Remember that the first subnet is 192.168.0.0 with the chosen subnet mask.

Subnet Address Prefix Subnet Mask
192.168.0.0 /26blank 255.255.255.192blank
192.168.0.64 /26blank 255.255.255.192nk
192.168.0.128blank /26blank 255.255.255.192blank
192.168.0.192blank /26blank 255.255.255.192blank

Step 2: Fill in the missing IP addresses in the Addressing Table
Device Interface IP Address Subnet Mask Default Gateway
CustomerRouter G0/0 192.168.0.1blank 255.255.255.192 N/A
N/A
N/A
G0/1 192.168.0.65 255.255.255.192
S0/1/0 209.165.201.2 255.255.255.252
LAN-A Switch VLAN1 192.168.0.2 255.255.255.192blank 192.168.0.1
LAN-B Switch VLAN1 192.168.0.66 255.255.255.192blank 192.168.0.65nk
PC-A NIC l192.168.0.63ank 255.255.255.192nk 192.168.0.1lank
PC-B NIC 192.168.0.126 255.255.255.192k 192.168.0.65
ISPRouter G0/0 209.165.200.225 255.255.255.224 N/A N/A
S0/1/0 209.165.201.1 255.255.255.252
ISPSwitch VLAN1 209.165.200.226 255.255.255.224 209.165.200.225
ISP Workstation NIC 209.165.200.235 255.255.255.224 209.165.200.225
ISP Server NIC 209.165.200.240 255.255.255.224 209.165.200.225

Part 3: Test and Troubleshoot the Network
Determine if PC-A can communicate with its default gateway. Do you get a reply? Yes

Determine if PC-B can communicate with its default gateway. Do you get a reply? Yes

Determine if PC-A can communicate with PC-B. Do you get a reply?
Yes

PART B: Lab – Calculate IPv4 Subnets

Problem 1:
Given:
Host IP Address: 192.168.200.139
Original Subnet Mask 255.255.255.0
New Subnet Mask: 255.255.255.224

Find:
Number of Subnet Bits 3
Number of Subnets Created 8
Number of Host Bits per Subnet 5
Number of Hosts per Subnet 30blank
Network Address of this Subnet 192.168.200.128
IPv4 Address of First Host on this Subnet 192.168.200.129
IPv4 Address of Last Host on this Subnet 192.168.200.158
IPv4 Broadcast Address on this Subnet 192.168.200.159

Problem 2:
Given:
Host IP Address: 10.101.99.228
Original Subnet Mask 255.0.0.0
New Subnet Mask: 255.255.128.0

Find:
Number of Subnet Bits 9
Number of Subnets Created 512
Number of Host Bits per Subnet 15
Number of Hosts per Subnet 32766
Network Address of this Subnet 10.0.128.0
IPv4 Address of First Host on this Subnet 10.0.128.1
IPv4 Address of Last Host on this Subnet 10.0.128.254
IPv4 Broadcast Address on this Subnet 10.0.128.255

Problem 3:
Given:
Host IP Address: 172.22.32.12
Original Subnet Mask 255.255.0.0
New Subnet Mask: 255.255.224.0

Find:
Number of Subnet Bits 3
Number of Subnets Created 8
Number of Host Bits per Subnet 13
Number of Hosts per Subnet 8190
Network Address of this Subnet 172.22.32.0
IPv4 Address of First Host on this Subnet 172.22.32.1
IPv4 Address of Last Host on this Subnet 172.22.32.254
IPv4 Broadcast Address on this Subnet 172.22.32.255

Problem 4:
Given:
Host IP Address: 192.168.1.245
Original Subnet Mask 255.255.255.0
New Subnet Mask: 255.255.255.252

Find:
Number of Subnet Bits 6
Number of Subnets Created 64
Number of Host Bits per Subnet 2
Number of Hosts per Subnet 2
Network Address of this Subnet 192.168.1.244
IPv4 Address of First Host on this Subnet 192.168.1.245
IPv4 Address of Last Host on this Subnet 192.168.1.246
IPv4 Broadcast Address on this Subnet 192.168.1.247

Problem 5:
Given:
Host IP Address: 128.107.0.55
Original Subnet Mask 255.255.0.0
New Subnet Mask: 255.255.255.0

Find:
Number of Subnet Bits 8
Number of Subnets Created 256
Number of Host Bits per Subnet 8
Number of Hosts per Subnet 254
Network Address of this Subnet 128.107.0.0
IPv4 Address of First Host on this Subnet 128.107.0.1
IPv4 Address of Last Host on this Subnet 128.107.0.254
IPv4 Broadcast Address on this Subnet 128.107.0.255

Problem 6:
Given:
Host IP Address: 192.135.250.180
Original Subnet Mask 255.255.255.0
New Subnet Mask: 255.255.255.248

Find:
Number of Subnet Bits 5
Number of Subnets Created 32
Number of Host Bits per Subnet 3
Number of Hosts per Subnet 6
Network Address of this Subnet 192.135.250.176
IPv4 Address of First Host on this Subnet 192.135.250.177
IPv4 Address of Last Host on this Subnet 192.135.250.182
IPv4 Broadcast Address on this Subnet 192.135.250.183

Why is the subnet mask so important when analyzing an IPv4 address?
It allows us to identify the network portion and the host portion of the address as well as identifying the number of subnet bits. The number of subnet bits can then be used to find the number of subnets which can be created.

PART C: Packet Tracer – Subnetting Scenario

Step 1: Subnet the 192.168.100.0/24 network into the appropriate number of subnets.
Based on the topology, how many subnets are needed?
5

How many bits must be borrowed to support the number of subnets in the topology table? 3

How many subnets does this create?
8

How many usable hosts does this create per subnet? 30

Calculate the binary value for the first five subnets. The first two subnets have been done for you.
Subnet Network Address Bit 7 Bit 6 Bit 5 Bit 4 Bit 3 Bit 2 Bit 1 Bit 0
0 192.168.100.0 0 0 0 0 0 0 0 0
1 192.168.100.32 0 0 1 0 0 0 0 0
2 192.168.100.64 0 1ank 0Blank 0blank 0 0 n0k 0
3 192.168.100.96 Bl0ank 1lank l1ank 0 0 n0k k 0 0
4 192.168.100.128 1lank nk 0ank Bl0ank lank 0ank blank l0ank Blank l0ank lank Bl0ank Bl0ank

Calculate the binary and decimal value of the new subnet mask.
First Octet Second
Octet Third Octet Mask
Bit 7 Mask
Bit 6 Mask
Bit 5 Mask
Bit 4 Mask
Bit 3 Mask
Bit 2 Mask
Bit 1 Mask
Bit 0
11111111 11111111 11111111 Blank 1lan Blan1lan k Blank 1lan bla0ank nk Blank 0ank Blan0ank k Blank 0ank blank 0ank
First Decimal Octet Second Decimal Octet Third Decimal Octet

Fourth Decimal Octet
blank
255. 255. 255. 224

Fill in the Subnet Table, listing the decimal value of all available subnets, the first and last usable host address, and the broadcast address. Repeat until all addresses are listed.

Subnet Number Subnet Address First Usable Host Address Last Usable Host Address Broadcast Address
0 192.168.100.0 192.168.100.1Blank 192.168.100.30Blank
192.168.100.31
1 192.168.100.32 192.168.100.33nk 192.168.100.62
192.168.100.63blank
2 192.168.100.64 192.168.100.65Blank 192.168.100.94Blank
192.168.100.95blank
3 192.168.100.96k 192.168.100.97ank 192.168.100.126Blank
192.168.100.127lank
4 192.168.100.128Blank 192.168.100.129Blank 192.168.100.158Blank
192.168.100.159
5 192.168.100.160Blank 192.168.100.161Blank 192.168.100.190
192.168.100.191k
6 192.168.100.192 192.168.100.193 192.168.100.222
192.168.100.223
7 192.168.100.224 192.168.100.225 192.168.100.254
192.168.100.255

Addressing Table
Device Interface IP Address Subnet Mask Default Gateway
R1
R1
R1 G0/0
192.168.100.1Blank Blank
255.255.255.224 N/A
Blank Blank Blank Blank Blank Blank
G0/1
192.168.100.33nk ank
255.255.255.224Blank
S0/0/0 192.168.100.Blank 129Blank 255.255.255.224Blank Blank
R2
R2
R2 G0/0 192.168.100.Blank 65Blank 255.255.255.224Blank Blank
G0/1
192.168.100.97ank Blank la255.255.255.224Blank nk
S0/0/0 192.168.100.Blank 129Blank 255.255.255.224Blank Blank
S1 VLAN 1
192.168.100.2Blank Blank
a255.255.255.224 Blank
Blank 192.168.100.Blank 1Blank
S2 VLAN 1 a192.168.100.34k nk a255.255.255.224Blank 192.168.100.ank 33nk
S3 VLAN 1 192.168.100.Blank 66Blank la255.255.255.224Blank k 192.168.100.Blank 65Blank
S4 VLAN 1
192.168.100.98ank Blank la255.255.255.224Blank 192.168.100.Blank 97ank
PC1 NIC
192.168.100.3Blank lank 255.255.255.224Blank nk 192.168.100.ank 1Blank
PC2 NIC a192.168.100.35 k 255.255.255.224Blank Blank l192.168.100. 33nk
ank
PC3 NIC 192.168.100.67ank ank la255.255.255.224Blank k 192.168.100.nk 65Blank
PC4 NIC
192.168.100.99nk nk la255.255.255.224Blank 192.168.100.97ank