Question

The diagram shows a LAN with hosts connected by a switch. initially, the switch does not know which ports any of the hosts arAddress1: 204.68.130.89 netmask: 255.255.252.0 Address: 204.60.132.3 11001100.00111100.10008010.018 do 11111111.1111111111111

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Answer #1

Answer A
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The first frame sent from Host A to C will forward packets to B and D as the initially routing table is empty.

Answer B
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The next frame sent from Host D to A will forward to the A only because now the switch table is prepared on the last transaction.

Answer A
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10 bits represent the host portion in the binary form of address of 204.60.130.89.

11001100.00111100.10000010.01011001
11111111.11111111.11111100.00000000
---------------------------------------
nnnnnnnn.nnnnnnnn.nnnnnnhh.hhhhhhhh

Network Address: 204.60.128.0
Subnet mask: 255.255.252.0
CIDR: /22
Total addresses in the range: 1022
Host range: 204.60.128.1 to 204.60.131.254

Answer C
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No, as you can see in the above range 204.60.131.254 is the last usable address of that range so address 2 falls under another subnet which starts from 204.60.132.0

Thanks

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