Help understanding this question

hellopaul2000mail

Hi all, just come accross a subnetting question i dont understand, any chance someone could explane it to me

cheers

Select all that apply. Which of the following /24 subnets can be supernetted into a larger network?

a)10.10.20.0 /24 and 10.10.21.0 /24 into 10.10.20.0 /23

b)10.10.21.0 /24 and 10.10.22.0 /24 into 10.10.21.0 /23

c)10.10.130.0 /24, 10.10.131.0 /24, 10.10.132.0 /24, and 10.10.133.0 /24 into 10.10.130.0 /22

d)10.10.164.0 /24, 10.10.165.0 /24, 10.10.166.0 /24, and 10.10.167.0 /24 into 10.10.165.0 /22

EdTheLad

Select all that apply. Which of the following /24 subnets can be supernetted into a larger network?

a)10.10.20.0 /24 and 10.10.21.0 /24 into 10.10.20.0 /23

b)10.10.21.0 /24 and 10.10.22.0 /24 into 10.10.21.0 /23

c)10.10.130.0 /24, 10.10.131.0 /24, 10.10.132.0 /24, and 10.10.133.0 /24 into 10.10.130.0 /22

d)10.10.164.0 /24, 10.10.165.0 /24, 10.10.166.0 /24, and 10.10.167.0 /24 into 10.10.165.0 /22

---

Basically what your being asked is how can you advertise these networks using a smaller number of network bits.


a) So the question is if i advertise 10.10.21.0 /23 in my routing table will
the two routes specified be seen.

10.10.20.0/23

First look at the mask /23 i.e. 255.255.254.0
this will allow subnets 2,4,6,8,10....252
the host addresses in the first subnet 2 would be as follows
10.10.2.1 -> 10.10.3.254
and so on for each subnet after
now looking at the subnet of interest in this question which is 20
host range addresses look like
10.10.20.1 -> 10.10.21.254

a) Yes your original question will allow
10.10.20.0 /24 and 10.10.21.0 /24

b) No 10.10.21.0 /24 and 10.10.22.0 /24 cannot as in a /23 they will be
considered as 2 separate advertisements
i.e. 10.10.20.0/23 10.10.22.0/23

c) /22 255.255.252.0
subnets 4,8,12,16 .....248
so subnet 128 will have range
10.10.128.1 -> 10.10.131.254
So the answer is No as 32 and 33 are outside range.

d) Yes

hellopaul2000mail

thank you very much mate, can see where i was going wrong now.

much appreciated

joyoung31

Just another 2 cents on this one.

For those who may have a problem with this still, you can always break this down into binary and see where your numbers don't match up. For example, on a:

20 = 00010100
21 = 00010101

If you notice, the 24th bits are not the same so that is where you would draw your dividing line and it would give you .20 with the 23 bit mask. I learned VLSM this way and while I can do it in my head, I always go back to this as a failsafe if I run into trouble. Hope this helps.