Question

I have to mimic how the packets travel from the source IP address to the destination IP address within the network. How do I solve this?

Routing Paths and Subnets 192.168.0.0/24 10.1.0.0/16 192.168.1.64/26 192.168.1.0/26 10.2.1.0/16 172.16.100.0/24 

Answer 1

Question 1:

Consider the source network 192.168.0.73 and the destination network 10.2.100.1 and determine the routing path as shown below:

Determine the subnet mask for the network address 192.168.0.0/24 as shown below:

Network address in binary: 1100 0000. 1010 1000. 0000 0000. 0000 0000 - 192.168.0.0

Subnet mask: 1111 1111.1111 1111.1111 1111.0000 0000 – 255.255.255.0

First usable host: 192.168.0.1

Last usable host: 192.168. 0.254

The source IP 192.168.0.73 is present is in the network 192.168.0.0/24.

Consider the network 10.2.100.1 and determine the first and last usable hosts for the network address 10.2.1.0/16 as shown below:

Network address in binary: 0000 1010. 0000 0010. 0000 0001. 0000 0000 - 10.2.1.0

Subnet mask: 1111 1111.1111 1111.0000 0000.0000 0000 – 255.255.0.0

First usable host: 10.2.1.1

Last usable host: 10.2.255.254

The destination IP 10.2.100.1 is present in the network 10.2.1.0/16.

Therefore, the packet can travel from the source 192.168.0.73 to the destination 10.2.100.1 through the network 192.168.1.64/26.

Question 2:

Consider the source network 10.3.1.32 and the destination network 10.1.32.1 and determine the routing path as shown below:

Determine the subnet mask for the network address 10.1.0.0/16 as shown below:

Network address in binary: 0000 1010. 0000 0001. 0000 0000. 0000 0000 - 10.1.0.0

Subnet mask: 1111 1111.1111 1111. 0000 0000.0000 0000 – 255.255.0.0

First usable host: 10.1.0.1                    

Last usable host: 10.1.255.254

The source IP 10.3.1.32 is present in first and last usable host range and it is not present in the network.

Question 3:

Consider the source network 172.16.100.1 and the destination network 10.2.1.1 and determine the routing path as shown below:

Determine the subnet mask for the network address 172.16.100.0/24 as shown below:

Network address in binary: 1010 1100. 0001 0000. 0110 0100. 0000 0000 – 172.16.100.0

Subnet mask: 1111 1111.1111 1111. 1111 1111.0000 0000 – 255.255.0.0

First usable host: 172.16.100.1            

Last usable host: 172.16.100.254

The destination IP 10.2.1.1 is present in the network 10.2.1.0/16.

Therefore, the packet can travel from the source IP 172.16.100.1 to the destination 10.2.1.1 directly.

Question 4:

Consider the source network 192.168.1.254 and the destination network 10.1.1.254 and determine the routing path as shown below:

Determine the subnet mask for the network address 192.168.1.0/26 as shown below:

Network address in binary: 1100 0000. 1010 1000. 0000 0001. 0000 0000 – 192.168.1.0

Subnet mask: 1111 1111.1111 1111. 1111 1111.1100 0000 – 255.255.168.0

First usable host: 192.168.1.1              

Last usable host: 192.168.63.254

The source IP is present in the network.

The destination IP 10.1.1.254 is present in the network 10.1.0.0/16.

Therefore, the packet can travel from the source IP 192.168.1.254 to the destination IP 10.1.1.254 directly.

Question 5:

Consider the source network 10.1.240.240 and the destination network 10.2.240.240 and determine the routing path as shown below:

Determine the subnet mask for the network address 10.1.0.0/16 as shown below:

Network address in binary: 0000 1010. 0000 0001. 0000 0000. 0000 0000 – 10.1.0.0

Subnet mask: 1111 1111.1111 1111. 0000 0000.0000 0000 – 255.255.0.0

First usable host: 10.1.0.1                    

Last usable host: 10.1.255.254

The source IP is present in the network.

Determine the subnet mask for the network address 10.2.1.0/16 as shown below:

Network address in binary: 0000 1010. 0000 0010. 0000 0001. 0000 0000 – 10.2.1.0

Subnet mask: 1111 1111.1111 1111. 0000 0000.0000 0000 – 255.255.0.0

First usable host: 10.2.0.1                    

Last usable host: 10.2.255.254

The destination IP 10.2.240.240 is present in the network 10.2.1.0/16.

Therefore, the packet can travel from the source IP 10.1.240.240 to the destination IP 10.2.240.240 through the network 172.16.100.0/24.

Question 6:

Consider the source network 192.168.0.128 and the destination network 192.168.1.128 and determine the routing path as shown below:

Determine the subnet mask for the network address 192.168.0.0/24 as shown below:

Network address in binary: 1100 0000. 1010 1000. 0000 0000. 0000 0000 – 192.168.0.0

Subnet mask: 1111 1111.1111 1111. 1111 1111.0000 0000 – 255.255.255.0

First usable host: 192.168.0.1              

Last usable host: 192.168.0.254

The source IP is present in the network 192.168.0.0/24.

Determine the subnet mask for the network address 192.168.1.0/26 as shown below:

Network address in binary: 1100 0000. 1010 1000. 0000 0001. 0000 0000 – 192.168.1.0

Subnet mask: 1111 1111.1111 1111. 1111 1111.1100 0000 – 255.255.0.0

First usable host: 192.168.1.1              

Last usable host: 192.168.63.254

The destination IP is present in the network 192.168.1.0/26.

Therefore, the packet can travel from the source IP 192.168.0.128 to the destination IP 192.168.1.128 directly.

Question 7:

Consider the source network 192.168.1.1 and the destination network 192.168.1.65 and determine the routing path as shown below:

The first and last usable hosts of the network 192.168.1.0/26 are 192.168.1.1 and 192.168.63.254.

Therefore, the packet can travel from source IP 192.168.1.1 to the destination IP 192.168.1.65 as they are in the same network.

Question 8:

Consider the source network 192.168.1.81 and the destination network 10.1.254.254 and determine the routing path as shown below:

The first and last usable hosts of the network 192.168.1.0/26 are 192.168.1.1 and 192.168.63.254.

The source IP is present in the network 192.168.1.0/26.

The first and last usable hosts of the network 10.1.0.0/16 are 10.1.0.1 and 10.1.255.254.

The destination IP is present in the network 10.1.0.0/16.

Therefore, the packet can travel from the source IP 192.168.1.81 to the destination IP 10.1.254.254 directly.