Nt1310 Unit 4 Research Paper

Let's take the first 22 bits as the network address of 135.46.63.10. By looking at the table, we have 135.46.60.0. Basically, the bit pattern of the 135.46.63.10 is 10000111.00101110.00111111.00001010
If we perform the operations on the 22 bits, 1s, and 0s, it is equivalent to the last 10 bit zero. We get the network address bit pattern 10000111.00101110.00111100.00000000. Here, the first two bytes does not change but the third type changes from 63 to 60 and the fourth byte become zero. It matches the network address to the routing table. It matches to the second row, the router will forward the packet to interface 1.

Let's take the 22 bits as the network address of the 135.46.57.14. Then we have 135.45.56.0.It matched to the network address of the first row and the packet will be forwarded to the interface zero 0. …show more content…
Then we have 135.45.52.0. It does not match to the network address table; the packet will be forwarded to the Default gateway that is on the router 2.

Let's take the 23 bits as the network address of the 192.53.40.7. Then we have 192.53.40.0. It matches to the third row of the table; the packet will be forwarded to the Router 1.

Let's take the 23 bits as the network address of the 192.53.56.7. Then we have 192.53.56.0. It does not match to the network address table above; the packet will be forwarded to the Default gateway on the router

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