# Question about IP Routes - Longest match rule

alu408
Posts:

**45**Member ■■□□□□□□□□
in CCNA & CCENT

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.

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.

0

## Comments

999Member ■■■■□□□□□□don'thave 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

A+, Network+, CCNA

45Member ■■□□□□□□□□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.

999Member ■■■■□□□□□□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.

A+, Network+, CCNA

105Member ■■□□□□□□□□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 onenetwork 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.

― Ellen Johnson Sirleaf

45Member ■■□□□□□□□□I get it now, thank you very much!

2Registered Users ■□□□□□□□□□2Registered Users ■□□□□□□□□□/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