Question

If my source is 192.168.0.128 & destination is 192.168.1.128. How will the packet travel? Given paths are: 192.168.0

0/24, 192.168.1.64/26, 10.2.1.0/16, 10.1.0.0/16, 192.168.1.0/26 and 172.16.100.0/24.

Answer 1

Packet Forwarding at router goes like this

   --> Take destination address and do 'logical AND' with subnet mask of each outgoing network entry in routing table

--> Check with all networks that are enlisted in it's routing table

--> Take the path with which destination address is matched

--> If multiple out network id's are matched, then take longest prefix match of destination ip

Example : -- here destination is 192.168.1.128

Router has 6 entries . Take first one 192.168.0.0/24

                                   Sub net mask is 1111 1111.1111 11111.1111 1111.0000 0000

                                 logical and ing subnet mask with destination gives us ---

                   192          . 168             . 1            . 128

                   1111 1111.1111 11111.1111 1111.0000 0000

-----------------------------------------------------------------------------------------------------

                   192          . 168              . 1              . 0    is not equal to given network id 192.168.0.0

There are opaths available router 192.168.0.128 Source :..1. 192.168.0.0/24 2. 192.168.1,64126 3. 10.2.1.0116 4. 10.1.0.0/16 5. 192.168.1.0/26 6. 172.16: 100.0/24 Destination 192.168.1.128 2. Tate Frost netwerk id, de logical. And wilts Subnet mask. 192.168.1.128 TCF FF FF 00 → 24-i's 8-o's 192.168.1.0 + 192.168.0.0

2. 192.168.1.64726.Network II FF. FF. FF. 1100 0000 - Subnet mask Destination 192.168.1.128 FF:FF:FF: l100 0000 i 192.168.1.128 #192.168.1.644 3. 10.2.1.0/16 192.168.1.128 FF. FF.00.00 192.168.0.0. #10.2.1.0' 4. 10.1.0.0/16 192.168.1.128 EE. RE.00.00 192.168.0.0 #10:1.0.0

5. 192.168.1.0/26 192.168.1.128 ER. FF. FF. 000000 192.168.1.128.2192168.1.0. 6.172.16.100.0/24 192.168.1.128 ::FF. FRIFE OO 192.168.lio#172.16.100.0 F Destination address is not matetied with any of outpit vetworks. Therefore, it passes in Default interface, it present.

Here routing table should be mentioned very clearly. But in general, default entry is always present at last. Therefore output must be 172.16.100.0/24.

Answer 2

It would all depend on how the routing path looks like. You can’t just give the source and destination