Question about IP Routes - Longest match rule

im quite confused on this question even though i can subnet. Which interface will a packet with a destination address of 10.10.10.14 be forwarded from?
10.10.10.16/28
10.10.10.4/30
10.10.10.8/29

So when i did the math out. /28 = 255.255.255.240. 256-240 = 16. block size 0,16,32,64. The broadcast for 0 subnet would be 15 which is minus 1 from the next subnet, and last valid host would be 14.

/30 = 255.255.255.252. 265-252 = 4. block size 0,4,8,12,16,20. The broadcast for 12 subnet is 15. Last valid host is 14.

/29 = 255.255.255.248. 256-248 = 8. block size 0,8,16. The broadcast for 8 subnet is 15. Last valid host is 14.
The book says the answer is 10.10.10.8/29. They all have the valid host as 14 so wouldnt they all be able to work?

My best guess is that 10.10.10.14 falls in between the range for 10.10.10.8/29 which is 10.10.10.9-10.10.10.14.

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Priston

They all don't have a valid host as .14.
10.10.10.16/28 = .16 - .31
10.10.10.4/30 = .4 - 7
10.10.10.8/29 = .8 - .15

10.10.10.4/30 is not the same as 10.10.10.12/30
10.10.10.16/28 is not the same as 10.10.10.0/28

alu408

Priston wrote: »

They all don't have a valid host as .14.
10.10.10.16/28 = .16 - .31
10.10.10.4/30 = .4 - 7
10.10.10.8/29 = .8 - .15

10.10.10.4/30 is not the same as 10.10.10.12/30
10.10.10.16/28 is not the same as 10.10.10.0/28

I guess what im trying to say for /28 - in the 0 subnet the range is from .1-.15 so the .14 does fall in the 0 subnet which is valid.
and /30 = in the 12 subnet. which is .13 - .14 range, the .14 falls in the 12 subnet.
The last one which is =29 in the 8th subnet is .9-.14 which has .14. So all 3 has a valid host. Why are you focusing on one subnet and not all subnets is which im confused on.. The correct answer in the book is the last one as well.

Priston

I'm only focused on the subnets that are listed as answers.

10.10.10.12/30 is NOT listed as a choice when your answering the question.
10.10.10.4/30 and 10.10.10.12/30 are two completely separate subnets.

Sy Kosys

alu408 wrote: »

Why are you focusing on one subnet and not all subnets is which im confused on.. The correct answer in the book is the last one as well.

Because of the given possibilities:
10.10.10.16/28
10.10.10.4/30
10.10.10.8/29

To quote the greatness of Highlander, there can be only one network where a .14 IP is valid. Nowhere in the initial question are we concerned about subnet X for network Y, if that is also a condition of the question then it will alter the answer for all 3 subnets.

You basically found the correct answer already, just need to disregard subnet 0,1,2, etc from any possible answers as that is not a condition to be concerned about.

alu408

Sy Kosys wrote: »

Because of the given possibilities:
10.10.10.16/28
10.10.10.4/30
10.10.10.8/29

To quote the greatness of Highlander, there can be only one network where a .14 IP is valid. Nowhere in the initial question are we concerned about subnet X for network Y, if that is also a condition of the question then it will alter the answer for all 3 subnets.

You basically found the correct answer already, just need to disregard subnet 0,1,2, etc from any possible answers as that is not a condition to be concerned about.

I get it now, thank you very much!

Mroy32

hello. I an stuck in the same problem. /29 & /30 both got .14 as valid host. But in case of /29 its the second subnet and in case of /30 its the 4th subnet. So why only /29 and not /30 as well? Please help.

Mroy32

Fortunately i got an answer for this.
/29 and /30 both are correct but when we specifically talk about longest match rule then its /29. As per this rule the subnet with maximum number of network bits is the best path. For /29 network bits=21 and for /30 network bits=18 and hence /29 is the answer. Please refer this link for more: https://www.juniper.net/documentation/en_US/junos/topics/reference/configuration-statement/longest-match-next-hop-edit-static-routing-options.html