You have the following Network ID: 131.112.0.0. You need at
least 500 hosts per network. How many networks can you
create? What subnet mask will you use
Answers were Sorted based on User's Feedback
Answer / neeta dhakad
We can create 128 subnets and subnetmask 255.255.254.0
| Is This Answer Correct ? | 92 Yes | 16 No |
Answer / suresh
subnetmask is 255.255.252.0, we can create 4 subnet and
atleast we can connect 500host per network
| Is This Answer Correct ? | 69 Yes | 31 No |
Answer / r.k.gupta
Hi Everyone
some guys give ans correctly but not in details.
so i describing in details and 100% correct.
given net id 131.112.0.0
default s/m- 255.255.0.0/16
but we need 500 hosts /network so
s/m 255.255.254.0/23
block size=2
no of networks=128
we need 500 hosts per network. so we can be on 9 bits for
host from 32 bits to fullfil our requrements(500 hosts). 9
bits on means 512. but valid host=510
1st Network is 131.112.0.0
starting ip- 131.112.0.1
Ending ip -131.112.1.254
broadcast address- 131.112.1.255
2nd Network is 131.112.2.0
starting ip - 131.112.2.1
ending ip - 131.112.3.254
broadcast add- 131.112.3.255
3rd Network - 131.112.4.0 and so on
now solved this ques.
| Is This Answer Correct ? | 37 Yes | 2 No |
Answer / suresh
we can create 64 subnet sunbet mask 255.255.252.0 for
500host per network
| Is This Answer Correct ? | 38 Yes | 29 No |
Answer / technologist
sorry typing mistake in above answer
hi guys
we can creat 128 subnet
and subnet mask will b 255.255.254.0
| Is This Answer Correct ? | 9 Yes | 0 No |
Answer / tushar
131.112.0.0 this ip add is from class B
Defualt sumbet mask is 255.255.0.0
We have 500 hosts per network
so, las 9 digit required for 512 hosts per network
Remaing 7 digit is used for subnetworks
Subnet mask will b 255.255.254.0 (Becz we have total
16+7= 23 digits)
we can creat 128 subnet (Becz we have 23 digits)
and remaing 9 digits for hosts per subnetwork.
No.of subnets Range of ip
1. 131.112.0.0 to 131.112.1.255
2. 131.112.2.0 to 131.112.3.255
3. 131.112.4.0 to 131.112.5.255
4.
--
--
128. 131.112.254.0 to 131.112.255.255
| Is This Answer Correct ? | 10 Yes | 1 No |
Answer / rames
Here, 9 bits are required to get 500 hosts. Now, there will
be 7 bits available for the subnetting( i.e in the third
byte) so, we can create 254 subnets and 500 hosts per each
subnet with the given network ID (131.112.0.0)
Here, the subnet mask will be varying from
131.112.2.0/7 to 131.112.254.0/7 according to the no. of
bits we use for the subnetting in the third byte.
| Is This Answer Correct ? | 12 Yes | 7 No |
Answer / gaurav verma
If you need to divide it up into the maximum number of
subnets containing at least 500 hosts each, you should use a
/23 subnet mask. This will provide you with 128 networks of
510 hosts each. If you used a /24 mask, you would be limited
to 254 hosts. Similarly, a /22 mask would be wasteful,
allowing you 1022 hosts.
| Is This Answer Correct ? | 5 Yes | 0 No |
Answer / sivakumar
subnet mask is 255.255.0.0
networsk =128
host per network =510+2=512 another 2 is one brodcast ID
and an other one is network ID
starting ip of the adress 131.112.0.1
last ip of the adress 131.112.254.255
| Is This Answer Correct ? | 11 Yes | 8 No |
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