Networking - Subnetting - Discussion
Discussion Forum : Subnetting - Subnetting (Q.No. 6)
6.
You have a network that needs 29 subnets while maximizing the number of host addresses available on each subnet. How many bits must you borrow from the host field to provide the correct subnet mask?
Answer: Option
Explanation:
A 240 mask is 4 subnet bits and provides 16 subnets, each with 14 hosts. We need more subnets, so let's add subnet bits. One more subnet bit would be a 248 mask. This provides 5 subnet bits (32 subnets) with 3 host bits (6 hosts per subnet). This is the best answer.
Discussion:
16 comments Page 1 of 2.
Rahul said:
1 decade ago
Please anybody clarify me. Am confused.
Rodo said:
1 decade ago
Should be B. You need 5 SUBNET bits to achieve the subnet requirement of 29.
Anand Mohan said:
1 decade ago
No of hosts: 2^n-2.
2^5 = 32.
32-2 = 30 subnets.
255.255.255.11100000.
255.255.255.248.
2^5 = 32.
32-2 = 30 subnets.
255.255.255.11100000.
255.255.255.248.
Rinju said:
1 decade ago
How we will get 240 and 248 and how it related to 29?
Noel Gates said:
10 years ago
There needs to be a better explanation to this!
Manu Ratheesh said:
10 years ago
In order to get 2 sub-networks, we have to borrow 1 bit from the host id part.
Similarly, to get 4 subnets, we have to borrow 2 bits form the host id, to get 16 subnets, borrow 4 bits from the host id part.
If we borrow 5 bits from the HID part, we will get 2^5 = 32 subnets. Hence, minimum 5 bits should be borrowed to obtain 29 subnets. Answer (D).
Similarly, to get 4 subnets, we have to borrow 2 bits form the host id, to get 16 subnets, borrow 4 bits from the host id part.
If we borrow 5 bits from the HID part, we will get 2^5 = 32 subnets. Hence, minimum 5 bits should be borrowed to obtain 29 subnets. Answer (D).
Sechonge said:
9 years ago
Yes, it is simple you know why?
Take 2 power n =29, then apply log.
nlog2 = log29, divide by log2 both you get,
n = log29/log 2.
n = 4.8 as we know that 4.8 the same as 5.
Take 2 power n =29, then apply log.
nlog2 = log29, divide by log2 both you get,
n = log29/log 2.
n = 4.8 as we know that 4.8 the same as 5.
(1)
Vick said:
8 years ago
But we start counting from 128 which equals to 1bit. So 2nd is 64 and third becomes 32.
How can we reverse the side?
How can we reverse the side?
Chaithanya katari said:
8 years ago
2 16 24 32 this is the octet order.
29 means 24+5=29 so 5 bits we need to borrow.
29 means 24+5=29 so 5 bits we need to borrow.
Manikandan said:
8 years ago
255.255.255.0
8 .16. 24. 32
If I have maximum 29 but (24+5)29 so maximum subnet.
5.
8 .16. 24. 32
If I have maximum 29 but (24+5)29 so maximum subnet.
5.
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