How many subnets can be gained by subnetting 172.17.32.0/23
into a /27 mask, and how many usable
host addresses will there be per subnet?
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Answer / metodija pankovski
Hi all i read all answers and in every answer you have some mistake.IT ask how many subnets and how many hosts have per subnet not how many host have totally!
So here is the most logical and easy understanding of sub-netting with out need to convert to binary we all know to make binary no. from the regular number.
23 subnet mask =255.255.254.0
27 subnet mask =255.255.255.224
it ask for 27 how many subnets will have and how many hosts
so we start
we all should know that totally in subnet mask have 32-bits here first 27 are taken for the subnet mask so we left 5 bits for host 32-27=5
number of the host we get from formula
2^n-2=No of host per subnet
so we get
2^5-2=32-2=30
we take 2 address for the broadcast and subnet thats why is 2^n-2
We finish with the hosts
no lets get the no. of subnets
first given was /23 bits and the second was /27-bits
we just do this
27-23=4
no. of subnet is 2^n we don`t have here -2!!! some one make mistake in the answers so remember you don`t take minus 2 when you solve no. of subnets thats for hosts!!
so
27-23=4
2^4=16
ANSWER OF THE QUESTION IS:
Subnets 16
HOST 30 per subnet
address range like this
172.17.32.0 first subnet
172.17.32.32
172.17.32.64
172.17.32.96
172.17.32.128
172.17.32.160
172.17.32.192
172.17.32.224
172.17.33.0
172.17.33.32
172.17.33.64
172.17.33.96
172.17.33.128
172.17.33.160
172.17.33.192
172.17.33.224 last one
I hope i help to some of you:)
THIS IS TE REAL ANSWER!
| Is This Answer Correct ? | 51 Yes | 3 No |
Answer / jitendera kumar sinha
172.17.32.0/23
mask of that ip is
sunated in the
172.17.32.0/27
sunating of 4 bit
total number of n/w
2^4-2=14(according to cisco)
now taol nomber of valabe bit for host is 16-11=5
so taotal nomber of avilable host is 2^5-2=30
note**(why eleven is redused from 16.ans becoz this is a
xlass b adress and it have 16 bit for n/w and 16 bit for
host but we have the mask of 27 then we have taken 11 bit
from host for n/w.so we redused that eleven bit from the 16)
now
now hat is the mask of that ip
255.255.128+64+32+16+8+4+2+1.128+64+32
i.e
mask is 255.255.255.224/27
so the block size is 256-224=32
so the n/w is
172.17.32.0---------------------.31
.32------------------------------.63 and so on
| Is This Answer Correct ? | 10 Yes | 4 No |
Answer / not_a_dum-ass
omg, you ppl just confuse the hell out of yourselves don't
you?
very simple question:
how many subnets can be GAINED / how many USABLE hosts per
subnet......
16 subnets GAINED
30 USABLE host adresses per
| Is This Answer Correct ? | 6 Yes | 1 No |
Answer / suraj kumar gupta
172.17.32.0/23 into a /27 how many subnet?
n=23-27=4
total no subnet mask =2^n - 2=2^4=16
where n is the no host
subnet =2^n=2^4=16
Total no. of usable host are..... 30
We have used the subnet bit so remainig bits are for host
means...
5 bits are available...
so 2 to the power 5
means 32 total no. of hosts..
usable hosts are..32-2 ====30 answer
| Is This Answer Correct ? | 4 Yes | 2 No |
Answer / megadeth
as explained earliar if you go by the cisco way. remember
you dont need to do any such calculations in such
problems,even vlsm and supernetting.
you generally remember /27 tends to 32 hosts per subnet
/25 tends to 128 similarly you can figure out the rest of it
| Is This Answer Correct ? | 0 Yes | 0 No |
Answer / devender chauhan
There is 16 subnet being made when we extend /23 to /27.
Because we are using 4 bits from host bits.
2*2*2*2 = 16
so the subnets are :-
172.17.32.0 to 172.17.47.0
Thanks
| Is This Answer Correct ? | 13 Yes | 14 No |
Answer / robert davenport
I have one bit of clarification on my answer. As the IP
address of, 172.17.32.0, was given that will reduce the
amount of subnets. The third octet (.32.) has restricted us
to one subnet on the /23 prefix. so that will change the
total subnets for the /27, see the following:
Prefix Subnets in Hosts per
Length Range Given Subnet
/23 1 510
/27 14 30
So, the Subnets will increase by 13 and the hosts will drop
to 420 (14x30). Hope I got it this time.
| Is This Answer Correct ? | 0 Yes | 2 No |
Answer / devender chauhan
There is 16 subnet being made when we extend /23 to /27.
Because we are using 4 bits from host bits.
2*2*2*2 = 16
so the subnets are :-
172.17.32.0 to 172.17.32.31
172.17.32.32 to 172.17.32.63
and continue...with 32 IPs per subnet...
| Is This Answer Correct ? | 4 Yes | 9 No |
Answer / bhisham
172.17.32.0/23 into a /27 how mant subnet?
n=23-27=4
total no subnet mask =2^n - 2=2^4-2=14
where n is the no host
host =2^n-2=14
subnet =2^n=2^4=16
| Is This Answer Correct ? | 3 Yes | 8 No |
Answer / devender chauhan
There is 16 subnet being made when we extend /23 to /27.
Because we are using 4 bits from host bits.
2*2*2*2 = 16
so the subnets are :-
172.17.32.0 to 172.17.47.0
and the total number of host is 16x255 =4080 host
| Is This Answer Correct ? | 7 Yes | 13 No |
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