If you keep 172.17.32.0/20 as is, you have just one big network (of course). If you use a /28 mask, then you have the following subnets (relative to the original address and mask): 172.17.32.0/28 .00100000.0000 0000 (last two octets, space at /28) 172.17.32.16/28 .00100000.0001…
Read the full replyCan someone please explain how the answer to this question is obtained ?
How many subnets are gained by subnetting 172.17.32.0/20 into multiple/
They answer is 256, but I can not figure out how to come up with that answer.
Thanks for the help
If you keep 172.17.32.0/20 as is, you have just one big network (of course).
If you use a /28 mask, then you have the following subnets (relative to the original address and mask):
172.17.32.0/28 .00100000.0000 0000 (last two octets, space at /28) 172.17.32.16/28 .00100000.0001 0000 172.17.32.32/28 .00100000.0010 0000 172.17.32.48/28 .00100000.0011 0000 ... 172.17.32.240/28 .00100000.1111 0000 (16 subnets so far)then:
172.17.33.0/28 .00100001.0000 0000 ... 172.17.33.240/28 .00100001.1111 0000 (other 16 subnets )and so on until
172.17.47.0/28 .00101111.0000 0000 ... 172.17.47.240/28 .00101111.1111 0000So, you have 16*16 subnets (eg, 256) if you use a /28 mask.
You can also see this very quickly if you think that you gain 8 bits for the subnet part (/20 -> /28), thus you have 2^8 = 256 subnets.
"RIckH" ha scritto nel messaggio news: snipped-for-privacy@n2g2000hse.googlegroups.com...
Hi,
2^(28-20) = 256 subnets (with ip subnet-zero)172.17.32.0/20 172.17.32.0-172.17.47.255
172.17.32.0 (16 subnets) 172.17.32.0-172.17.32.15 172.17.32.16-172.17.32.31 .... .... 172.17.32.240-172.17.32.255172.17.33.0 (16 subnets) 172.17.33.0-172.17.33.15 172.17.33.16-172.17.33.31 .... .... 172.17.33.240-172.17.33.255 ..... .....
172.17.47.0 (16 subnets) 172.17.47.0-172.17.47.15 172.17.47.16-172.17.47.31 .... .... 172.17.47.240-172.17.47.255Regards, Gabriele
If you keep 172.17.32.0/20 as is, you have just one big network (of course).
If you use a /28 mask, then you have the following subnets (relative to the original address and mask):
172.17.32.0/28 .00100000.0000 0000 (last two octets, space at /28) 172.17.32.16/28 .00100000.0001 0000 172.17.32.32/28 .00100000.0010 0000 172.17.32.48/28 .00100000.0011 0000 ... 172.17.32.240/28 .00100000.1111 0000 (16 subnets so far)then:
172.17.33.0/28 .00100001.0000 0000 ... 172.17.33.240/28 .00100001.1111 0000 (other 16 subnets )and so on until
172.17.47.0/28 .00101111.0000 0000 ... 172.17.47.240/28 .00101111.1111 0000So, you have 16*16 subnets (eg, 256) if you use a /28 mask.
You can also see this very quickly if you think that you gain 8 bits for the subnet part (/20 -> /28), thus you have 2^8 = 256 subnets.
Ok, strictly speaking, you "gain" 255 subnets (you had one, and now you have
256), but I hope you got the idea.
Thanks, That makes perfect sense now.
I found a site that does a really good job at explaining this (In My humble oppinion)
Site:
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