Intermediate

level 05

learn to

use multiple different network addresses
use routing table

goal 01

A and R1 must be in the same network address to communicate
calculated network address from R1: 53.192.44.0/25

goal 02

same as goal 01
note that network address of A1-R1 and B1-R2 is different

goal 03

A -> B

for A and B to communicate, A must send packet to B
A knows the destination is 133.219.230.253, but does not have (or 'know') direct connection to it
therefore it would send it to others who might have the connection
first A searches for routing table
- 133.219.230.253 falls within range of 0.0.0.0/0, which means every single IPv4 address
then it should send its packet to interface R1, since it has the connection
when R1 gets the packet, R has the connecton to B, therefore would send the packet to B

B -> A

after getting packet from A, B must also send back packets to A to ensure it has received the packet
it's the same as A -> B

level 06

learn to

send to and from internet using routing table
not use private ip address

goal 01

A -> I

A needs to send packet (destination 8.8.8.8) to internet but does not know how
ideal path would be A1 -> S -> R1 -> R2 -> I
A should send the packet to R1 using its routing table
R does not know where the packet would go either, so it should send it to R2
- then the other end from R2 would handle the packet

I -> A

I needs to send return packet to A but does not know how
ideal path would be I -> R2 -> R1 -> S -> A1
provide A1's network address so internet can toss the packet to R2
- can't just enter default to routing table
- it might have to be with that internet-router relationship is 1-many
no more need to provide network address from here since R knows A

level 07

learn to

keep multiple network addresses from overlapping

goal 01

network addresses must not overlap with each other
- there are 3 of them, A <-> R1, R1 <-> R2, C <-> R2

R11 <-> A1

R11 and R12's network ips both share 109.198.14.xxx
- so CIDR can't be /24 or both will overlap
- for convenience, all networks would have /26 and share 109.198.14
- then viable network addresses are: 109.108.14.{0, 64, 128, 192}
- so for A1 and R11, 109.108.14.0/26

R12 <-> R21

since R12's last digits are 254, network address must be 109.198.14.192/26

R122 <-> C1

choose between 109.108.14.{64, 128} since others are taken