Based on the routing table below: A. (2 points) What layer-2 interface would a p

Question

Based on the routing table below:

A. (2 points) What layer-2 interface would a packet with a destination IP address of 192.168.1.1 be sent to?
B. (2 points) What would be the next hop of the above packet? Justify your answer for full credit.
C. (3 points) How many local subnets is this machine attached to? Justify your answer for full credit.
D. (3 points) How many IP addresses can this machine reach in 1 hop? Justify your answer for full credit.

Destination Gateway Mask Flags Interface U etho 165.230.198.64 165.230.198.119 255.255.255.192 192.168.1.0 192.168.1.1 eth1 255.255.255.0 TU 12.0.0.0 127.0.0.1 255.0.0.0 U 100 default 165.230. 198.65 255.255.255.255 UG eth0

Explanation / Answer

A) Here , it is provided the Layer-2 interface which is an indicative that helps us to reach to next hop .So we will be computing MAC address for destination IP address of 165.230.198.64 , having the subnet mask value as 255.255.255.192 ,with gateway(default) value is 165.230.198.64 that helps us to reach out to layer-2 of the other machine, we will be implementing ARP(Adrress Resolution Protocol) for the MAC address 192.168.1.1 to reach out all the ports that are present on other machine , lets say machine B, which responds by accepting the data from Layer-2 and forward the same data to Layer-3 also.So we can say that if both the machines are on the same network we can go for ARP, whereas if netwoks are entirely different then we switch up for NAT case i.e. Network Address Translation which further ,helps us out to reach the destination, because it re-writes Routing Table evrey time , by keeping source and destination same.We can use ping and arp commands on network for looking out the values.

B) Next hop of the above packet is calculated by transferring the packet form current router to adjacent router or host on which packet is transferred in this case i.e. nothing else than the machine which receives the packet with a destination IP address of 192.168.1.1, which is transferred by Interface-2 from its current destination.

Keep a note" LAYER-2 is always intended or tracked to go to next hop" as discussed in section-1.

C) Calculating the local subnets this machine is attached to:

Destination having IP address is =165.230.198.64

mask/subnet masks in bit is = 255.255.255.192/26

Bits in mask is = 64

IP address class =B

Subnet address is =165.230.198.64

Number of Hosts= 62

For example: showing table for Calculating locals subnets it can be 1,2,4,16,32,64,128,256.

Subnet

Address

First

Host

Last

Host

Subnet

Mask

And like this you can count upto 64 values in similar way that we had created till now.

D) To complete up, 1 hop we will consider each subnet mask ,among them we will be selecting the mask that will be having maximum or higher number of network address field bit , and in this scenario, we will be having the case of 255.255.255.255/32 is having major priority of subnet mask among the existing ones which acts as a loop which runs again and again and save so many Ip addresses.

But we will opt for the subnet mask i.e. 255.255.255.192/26 for the destination i.e. 165.230.198.64/26....

So the number of Ip addresses that can be addressed to this machine is 2^(32-26)-3 = 64-3 =61 IP addresses , this machine can reach in 1 Hop.

Subnet

Address

First

Host

Last

Host

Subnet

Mask

BroadCast 165.230.198.64 165.230.198.65 165.230.198.126 255.255.255.192 165.230.198.127 165.230.198.128 165.230.198.129 165.230.198.190 255.255.255.192 165.230.198.191 165.230.198.192 165.230.198.193 65.230.198.254 255.255.255.192 165.230.198.255 166.230.199.0 165.230.199.1 165.233.199.62 255.255.255.192 165.230.199.63 165.230.199.64 165.230.199.65 165.230.199.126 255.255.255.192 165.230.199.127

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