Computer Science - Networking - Discussion
Discussion Forum : Networking - Section 3 (Q.No. 8)
8.
You have a network ID of 134.57.0.0 and you need to divide it into multiple subnets in which at least 600 host IDs for each subnet are available. You desire to have the largest amount of subnets available. Which subnet mask should you assign?
Discussion:
13 comments Page 2 of 2.
Mansoor said:
1 decade ago
@Chetana Tailor.
This subnet mask is not given in the options, Why is it mistake in the given options?
This subnet mask is not given in the options, Why is it mistake in the given options?
Chetana Tailor said:
1 decade ago
You have a network ID of 134.57.0.0 and you need to divide it into multiple subnets in which at least 600 host IDs for each subnet are available. You desire to have the largest amount of subnets available. Which subnet mask should you assign?
IP belongs to class B so 255.255.0.0 mask and for that 16 bits.
remaining bits are 16 out of 16, and now we want 600 host per subnet for that we required 10 bit for host.
So, mask becomes 1111 1111.1111 1111.1111 1100.0000 0000 for subnet
134.57.252.0.
IP belongs to class B so 255.255.0.0 mask and for that 16 bits.
remaining bits are 16 out of 16, and now we want 600 host per subnet for that we required 10 bit for host.
So, mask becomes 1111 1111.1111 1111.1111 1100.0000 0000 for subnet
134.57.252.0.
Ravi kumar said:
1 decade ago
Given IP belongs to class B.
we need 600 host...means >2^9(greater than 512).
Hence we need at least 10 bit.but in option C
255.255.255.0
Binary form 11111111.11111111.11111111.00000000
we have only 8 bit(zero's which define No. of Host).so it will give 254(256-2)per host.
correct answer C.
as 255.255.248.0
Binary Form 11111111.11111111.11111000.00000000
So we have 11 bit (2^11-2=2046) or 2046 per subnet(total subnet=32=2^5).
we need 600 host...means >2^9(greater than 512).
Hence we need at least 10 bit.but in option C
255.255.255.0
Binary form 11111111.11111111.11111111.00000000
we have only 8 bit(zero's which define No. of Host).so it will give 254(256-2)per host.
correct answer C.
as 255.255.248.0
Binary Form 11111111.11111111.11111000.00000000
So we have 11 bit (2^11-2=2046) or 2046 per subnet(total subnet=32=2^5).
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