Question

Consider the following IPv4 forwarding table, using CIDR. As in exercise 1, IP address bytes are in hexadecimal, and “:” is used as the separator as a reminder.

destination	next_hop
00:0:0:0/2	A
40:0:0:0/2	B
80:0:0:0/2	C
c0:0:0:0/2	D

(a). To what next_hop would each of the following be routed? 63:b1:82:15, 9e:00:15:01, de:ad:be:ef

(b). Explain why every IP address is routed somewhere, even though there is no default entry. Hint: convert the first bytes to binary

Answer 1

Answer : Given data

* Let us convert the hexadecimal IP's into binary first, it will make our work easier to decide which destination will be taken by which IP.

63:b1:82:15 ---> 01100011 : 10110001 : 10000010 : 00010101

9e:00:15:01 ---> 10011110 : 00000000 : 00010101 : 00000001

de:ad:be:ef ---> 11011110 : 10101101 : 10111110 : 11101111

* Now let us understand the representation 40:0:0:0/2 , this means that out of 32bit only first 2 bits are used to decide Network ID.

* So only bold 2 bit in the above conversion will be used to see next hop. We will match the 2bit Network ID will all possible hops (A, B, C, D) Network ID.

* Whichever match will be found, the packet will be forwarded to that route.

Network ID (first 2 bits)

00:0:0:0/2	A	00000000 :
40:0:0:0/2	B	01000000 :
80:0:0:0/2	C	10000000 :
c0:0:0:0/2	D	11000000 :

From the above discussion, it is very clear that,

63:b1:82:15 -----> B

9e:00:15:01 ------> C

de:ad:be:ef -------> D

* Every IP address is routed to some destination, the reason being the exact match of their Network ID provided in the router table.

__________THE END____________