Binary router/switch question

jekyll

If we assign the IP address 00001010.11111110.00000001.00001001 and the subnet mask 11111111.11111111.11111111.11111000 to a switch, what is the default IP address of the router?

Answer: 00001010.11111110.00000001.00001110

How to calculate this? It's not a logic operation, not AND, not OR, not XOR.

chX

Where is this question from? It seems poorly formulated and not clear about what they want from you.

10.254.1.9/29 (255.255.255.248) is part of the 10.254.1.8/29 subnet, meaning the usable host range is: 10.254.1.9 - 10.254.1.14.
Their answer is 10.254.1.14, but you could assign any unused IP address in that range to a router.

I guess convention can be to set the last usable host as the gateway (like most home routers being something similar to 192.168.1.254 by default), but I just don't like this question at all. There is no hard and fast correct answer to this, unless I'm missing something. Correct me if I'm mistaken.

jekyll

chX said:

Where is this question from? It seems poorly formulated and not clear about what they want from you.

10.254.1.9/29 (255.255.255.248) is part of the 10.254.1.8/29 subnet, meaning the usable host range is: 10.254.1.9 - 10.254.1.14.
Their answer is 10.254.1.14, but you could assign any unused IP address in that range to a router.

I guess convention can be to set the last usable host as the gateway (like most home routers being something similar to 192.168.1.254 by default), but I just don't like this question at all. There is no hard and fast correct answer to this, unless I'm missing something. Correct me if I'm mistaken.

It is a question with four different answers to pick from, so I guess 10.254.1.14 is the only one in that usable host range. Thanks.

a) 00001010.11111110.00000001.00001110

b) 00001010.11111110.00000001.01001111

c) 00001010.11111110.00000011.01010001

d) 00001010.11111110.00000001.01010011

chX

jekyll said:

chX said:

Where is this question from? It seems poorly formulated and not clear about what they want from you.

10.254.1.9/29 (255.255.255.248) is part of the 10.254.1.8/29 subnet, meaning the usable host range is: 10.254.1.9 - 10.254.1.14.
Their answer is 10.254.1.14, but you could assign any unused IP address in that range to a router.

I guess convention can be to set the last usable host as the gateway (like most home routers being something similar to 192.168.1.254 by default), but I just don't like this question at all. There is no hard and fast correct answer to this, unless I'm missing something. Correct me if I'm mistaken.

It is a question with four different answers to pick from, so I guess 10.254.1.14 is the only one in that usable host range. Thanks.

a) 00001010.11111110.00000001.00001110

b) 00001010.11111110.00000001.01001111

c) 00001010.11111110.00000011.01010001

d) 00001010.11111110.00000001.01010011

 Ah, right. That makes sense. No problem!

lwwarner

chX said:

Where is this question from? It seems poorly formulated and not clear about what they want from you.

10.254.1.9/29 (255.255.255.248) is part of the 10.254.1.8/29 subnet, meaning the usable host range is: 10.254.1.9 - 10.254.1.14.
Their answer is 10.254.1.14, but you could assign any unused IP address in that range to a router.

I agree that the question is poorly written, but I think that the point of the question is to recognize that you can answer this very quickly just by looking at the bit patterns without doing any binary-to-decimal conversion. To illustrate this I'll line everything up and draw a vertical line between the netid & the host parts of the address:

mask       11111111.11111111.11111111.11111|000

Now, hopefully it is clear that all it takes is a quick look to see that choice 'a' is the only one here that has the same bit pattern under the mask as the given IP address. Therefore, 'a' is the only one that falls in the same subnet as the given IP address and is the only possible choice for the correct answer. The "trick" here is to recognize this quickly, and to be able to answer something like this on the exam in a few seconds and move on.