IP addressing

I took a quiz from my Cisco Academy where a question was :

[FONT=&quot]Host A has an IP address of172.16.225.93 and a mask of 255.255.248.0. Host A needs to communicate with anew host whose IP is 172.16.231.78. Host A performs the ANDing operation on thedestination address. What two things will occur? (Choose two.)

(1) Host A will change the destination IP to the IP of the nearest router andforward the packet.
(2) Host A will broadcast an ARP request for the MAC of its default gateway.
(3)A result of 172.16.225.0 will be obtained.
[/FONT][FONT=&quot](4)Host A will broadcast an ARP request for the MAC of thedestination host.
[/FONT][FONT=&quot](5)A result of 172.16.224.0 will be obtained.
[/FONT][FONT=&quot](6)A result of 172.16.225.255 will be obtained.

[/FONT]I was told that "4" and "6" are correct where as i think "4" and "5" are correct as mask of Host A shows that the Host A has network id of 172.16.224.0 and both "new host" and Host A belong to same network.So ANDing "new host" IP address wouldd give network id which is 172.16.224.0

Kindly correct me if i am wrong.[FONT=&quot][/FONT]

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Comments

zrockstar

Yeah, I don't see any reason for 6 to be correct. The network address is indeed 172.16.224.0 and the broadcast would be 172.16.231.255, so 172.16.225.255 appears to be a valid host address for this range and therefore would have nothing to do with the communication between these two hosts. Usually the academy questions are pretty good, and the ones you swear are wrong will end up being right after you stare at it for a few minutes, but with the given information, I just don't see how this one is right.

NetworkVeteran

hassantalal785 wrote: »

Host A has an IP address of172.16.225.93 and a mask of 255.255.248.0. Host A needs to communicate with anew host whose IP is 172.16.231.78. Host A performs the ANDing operation on thedestination address.

A binary AND operation can be used to determine the Subnet ID of an IP address plus subnet mask. A CCNA's expected to know decimal, binary, and hex.

172.16.231.78 &
255.255.248.0

Hint: It's easier to understand a binary operation if you work in binary.

An AND operation works like so:

If the top and bottom number both have a one in any particular position, so does the answer. If not, the answer has a zero in that position.

10101100.00010000.11100111.01001110 ( 172.16.231.78 )
11111111.11111111.11111000.00000000 (255.255.248.0)

10101100.00010000.11100000.00000000 (172.16.224.0)

If the Subnet ID is 172.16.224.0 and 172.16.231.78 is within the range, the source address must be on the same subnet. That's because 172.16.224.0 < 172.16.225.93 < 172.16.231.78 and subnet addressing is contiguous.

(1) Host A will change the destination IP to the IP of the nearest router andforward the packet.

Bzzt. This never, ever happens. The value in the IP :: Destination_Address always refers to the final destination of the IP packet.

(2) Host A will broadcast an ARP request for the MAC of its default gateway.

Negative. Since the hosts are on the same subnet, Host A will instead directly broadcast an ARP request for the MAC address of the new host.

(3)A result of 172.16.225.0 will be obtained.

No, this was not the result of our AND operation.

(4)Host A will broadcast an ARP request for the MAC of thedestination host.

Bingo!

(5)A result of 172.16.224.0 will be obtained.

Yes, this was the result of our AND operation.T

(6)A result of 172.16.225.255 will be obtained.

No, this was not the result of our AND operation.

I was told that "4" and "6" are correct where as i think "4" and "5" are correct

You are right AND your instructor's answer is completely wrong--

What I said before. wrote:

An AND operation works like so:

If the top and bottom number both have a one in any particular position, so does the answer. If not, the answer has a zero in that position.

10101100.00010000.11100111.01001110 ( 172.16.231.78 )

There is nothing you can AND the above number with and obtain--

10101100.00010000.11100001.11111111 (172.16.225.255)

There are zeroes in the final octet of the destination address, so there must be zeroes in the final octet in the solution of any possible AND operation.