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

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.

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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. 10.1.0.0/162. 192.168.1.64726.Network II FF. FF. FF. 1100 0000 - Subnet mask Destination 192.168.1.128 FF:FF:FF: l100 0000 i 192.168.1.15. 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

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.

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Answer #2
It would all depend on how the routing path looks like. You can’t just give the source and destination
source: IT Specialist
answered by: Carlos
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