Question

Material:

Router 1941

Switches 2960

Desktop PCs and Server

LAN cables

Console cable

IPV4 ADDRESSING

1. Divide the 192.168. …… .0 /24 network into equal sized subnets that supports minimum 120 hosts each. (Where the third octet of the IP address has a value of 100 + last 2 digits of your student number)

2. Use the First subnet for the Floor A network.

3. Use the 2nd subnet to create smaller subnets that support minimum 12 hosts each.

4. Use the 3rd of the smaller subnets for the Floor B network.

5. The router interfaces get the first usable address in each subnet.

6. The PCs get the 10th and 11th usable addresses in the Floor A, and the 3rd address in the Floor B network.           

7. The server gets the second address in the subnet.

8. Complete the following table for IPv4 Addresses:

DEVICE NAME	INTERFACE	IP ADDRESS/ SUBNET MASK	GATEWAY
Building	Gi 0/0		N/A
	Gi 0/1		N/A
PC1-FA
PC2-FA
PC1-FB
SRV-FB

9. Set the hostname on router BUILDING and on switches Floor-A and Floor-B, to match the device name on Topology figure

Tests:

Test connectivity between PCs (ping)

Answer 1

1) Suppose the student number ends with 23

Address I am going to use is  192.168.123.0 /24

Given that each subnet should have a minimum of 120 hosts.

In the IP address, /24 means 24 bits out of 32 bit ip addess is in network id part and rest of the 8 bits in hostid part.

To support minimum of 120 hosts, 7 bits should be retained in hostid part and we can bring one bit among the 8 bits in host id part to net id part to create 2 subnets

2^(No of bits borrowed)=no of subnets created

2^(No of bits remaining in host id part)=no of ip addres per subnet

here 8 bits was in host id part,to create 2 subnets 1 bit is borrowed from host id part to net id part and so 7 bits in host id part

2^1=2 no of subnets created

2^7=128 no of ip address per subnet (2^6=64 which will not satisfy our requirement)

subnet 1 address range :192.168.123.0 /25 to 192.168.123.127 /25

subnet 2 address range: 192.168.123.128 /25 to 192.168.123.255 /25

2)

Floor A subnet address range 192.168.123.0 /25 to 192.168.123.127 /25

3)

Using 2nd subnet to create smaller subnets each with a minimum of 12 hosts

subnet 2 address range: 192.168.123.128 /25 to 192.168.123.255 /25

subnet work address is192.168.123.128 /24 here /25 means 25 bits in netid part and 7 bits in host id part

to create subnets with minimunm requirement of 12 hosts 4 bits should be retained in host id part since(2^4=16)

2^3=8 which will not satisfy our requirement

now borrow 3 bits out of 7 to netid part and retain 4 bits in hostid part

2^3=8 no of subnets are created and 2^4=16 ip address per subnet

subnet 1: 192.168.123.128 /25 to 192.168.123.143 /25

subnet 2 :192.168.123.144 /28 to 192.168.123.159 /28

subnet 3 :192.168.123.160/28 to 192.168.123.175 /28

subnet 4 :192.168.123.176 /28 to 192.168.123.191 /28

subnet 5 :192.168.123.192 /28 to 192.168.123.207 /28

subnet 6 :192.168.123.208 /28 to 192.168.123.223 /28

subnet 7 :192.168.123.224 /28 to 192.168.123.239 /28

subnet 8 :192.168.123.240 /28 to 192.168.123.255 /28

4) using 3 rd of smaller subnet for network B

192.168.123.160/28 to 192.168.123.175 /28

5)

For floor A router interface ip address is 192.168.123.1 /25

For floor B router interface ip address is 192.168.123.161/28

6) The PCs get the 10th and 11th usable addresses in the Floor A, they are

PC1-FA 192.168.123.10 /25

PC2-FA 192.168.123.11 /25

3rd address in the Floor B network

PC1-FB 192.168.123.163/28

7)

SRV-FB 168.123.162/28

8)

х Cisco Packet Tracer File Edit Options View Tools Extensions Help 7 @ a ? IQ X Logical Physical x: 3, y: 253 [Root] 08:13:30 1941 Building 2960-4TT Floor 760-240 Floor-B PC-PT PC1-FA PC-PT PC2-FA PC-PT PC1-FB Server-PT SRV-FB MIT Time: 24:16:35 Realtime Simulation De Fire Scenario o Periodic 3o 3 2 2 2 Last Status Source Destination Successful PC1-FB PC1-FA Successful PC2-FA SRV-FB Type Color Time(sec) ICMP 0.000 ICMP 0.000 Num 0 N Edit (edit (edit New Delete N 1 MIT Toggle PDU List Window Automatically Choose Connection Type