Hi,

I met one confusion answer. My question as below:-

Practice Example #8B: 255.255.255.224 (/27)
172.16.0.0 = Network Address
255.255.255.224 = Subnet mask

Is it subnet = 2^11 = 2048 OR 2^3 = 8 ?????

* According to CCNA book Todd Lammle, Sixth Edition answer is 2048.
Which one correct ???

Currently im taking CCNA course and we are referring to this book to study in class. My letcurer also confused of this problem. He said MAYBE its wrong. Its really confusing us.

Please anyone help me on this. Your kind help will I appreciate.

Thank u.

maybe i dont know..

Since they already gave you the info that it's class B, so the third octet is the subnet number as well.
When you are calculating the number of subnets, you have to use third octect + fourth octect, which is 8+3, which will be 11.
So 2^11 is correct.

Right.

But it's more like a 27 minus 16 = 11.

kogula14: Recall that the default subnet mask for Class B addresses is a /16 (255.255.0.0).


Who's your lecturer? From which University?

He is from philiphine. So actually yours is correct. Tq very much

Ask for refund if your lecturer cannot give the correct answer.

Yup the correct answer is 2^11=2048

i agree wit d book.

class B ip got 255.255.0.0 subnet mask by default. if u make it to /27 or 255.255.255.224 subnet mask then u've made eleven(11) bits into subnet bits.
27-16 =11. so...number of subnets is 2^11=2048.

i give u a picture of that:
default classB mask: 11111111.11111111.00000000.00000000
your new mask: 11111111.11111111.11111111.11100000

see the diff?

Its good to write on a BIG paper the powers of 2



2^2=4

2^3=8

2^4=16

2^5=32

2^6=64

2^7=128

2^8=256

2^9=512

2^10=1024

2^11=2048

2^12=4096



And write on that paper the numbers



128 192 224 240 248 252 254



Cause this are the Numbers, you will allways need in calculating Subnets.


if u master this, calculator is not a issue anymore. just my thoughts anyway.





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172.16.0.0/27
subnet = 255.255.255.254
new subnet bit = 8+8+8+3 = 27
subnet bit = 27-16(class B) = 11
total subnet = 2^11-2 = 2048-2 = 2046
new host bit = 32-27 = 5 bits
total host = 2^5 -2 = 32-2=30

2^3 if correct only if this is a Class C network!! Class B is 2^11!! but dont forget to minus out the 2 for network and broadcast address

here's the list

172.16.0.0 - 172.16.0.31 (1)
172.16.0.32 - 172.16.0.63 (2)
172.16.0.64 - 172.16.0.95 (3)
172.16.0.96 - 172.16.0.127 (4)
172.16.0.128 - 172.16.0.159 (5)
172.16.0.160 - 172.16.0.191 (6)
172.16.0.192 - 172.16.0.223 (7)
172.16.0.224 - 172.16.0.256 (8)

* 8?? WRONG!! the list is not done, unless this is Class C

the list continue again

172.16.x.0 - 172.16.x.31 (x1)
172.16.x.32 - 172.16.x.63 (x2)
172.16.x.64 - 172.16.x.95 (x3)
172.16.x.96 - 172.16.x.127 (x4)
172.16.x.128 - 172.16.x.159 (x5)
172.16.x.160 - 172.16.x.191 (x6)
172.16.x.192 - 172.16.x.223 (x7)
172.16.x.224 - 172.16.x.256 (x8)

x= 0 -> 256
so, 256 x 8 = 2048

minus out the 2 = 2046!!

it's the same as u calculate it as 2^8 + 2^3 = 2^11

2^8 = 256
2^3 = 8



** please asked ur lecturer to relearn his/her subnetting first before he/she proceed to mislead all his/her student!!
what the fish, this kinda of thing he/she also dunno, how he/she get his CCNA/CCNP??

i dont have an CCNA cert also i know how to count!

[quote=ckboon,Jun 16 2008, 04:58 PM]
172.16.0.0/27
subnet = 255.255.255.254
new subnet bit = 8+8+8+3 = 27
subnet bit = 27-16(class B) = 11
total subnet = 2^11-2 = 2048-2 = 2046
new host bit = 32-27 = 5 bits
total host = 2^5 -2 = 32-2=30

Correct me if wrong.

I think it should call 30 host per subnet...

Total host should be 2046 * 30 = 61380