443 views

A class B network address 130.50.0.0 is submitted as follows. The last 10 bits of the host id are allotted for host number and the remaining 6 bits are reserved for subnet number.

1. How many subnets and number of hosts in each subnet are possible with above scheme ?
1. 62,1022
2. 30,510
3. 14,254
4. None of these
1. What is the starting addresses of 1st and 4th subnets?
1. 130.50.4.1 and 130.50.16.1
2. 130.50.1.1 and 130.50.4.1
3. 130.50.0.0 and 130.50.3.0
4. None of these
| 443 views
0
A and B?
0
My approach :-

For 1st I am getting D.

We have subnet mask of 22 bits in which 6 bit for subnet So total subnets = 64

Total host = 2^10 - 2 = 1022

For 2st I am getting C as the answer please verify
0
Why A ?
0
D should be for the first one.. I subtracted 2 from the subnet part also but I guess I am wrong .. your approach is correct.. and following that only I got B...
0
For B part shouldn't it be C

Because see 6 bits are there the first subnet will be 000000, second one = 000001 likewise 4th subnet = 000011

So 130.50.00000000.0 = 130.50.0.0

And 130.50.00001100.0 = 130.50.12.0

Sorry the answer should be D. Too

Do u agree ?
0
I think that It should be D as for the first subnet the first address will be as told by u but we cant use it as it would be the subnet Id for that subnet.. Similarly for the 4th subnet.. Correct me if I am wrong
0
For both option A..??
0

@Soumya TiwariSoumya Explain your solution how u decided

0
For part A, question say 6 bit are for subnet so total subnet =2^6-2(one left for network id and another broadcast address) Similary host=2^10-2 =1022

For part B, first subnet bit will be 000001 and host bit 0000000001 which is equal to 130.50.00000100.00000001 =130.50.4.1

Similarly for fourth subnet bits are 000100 and host bit is same as above giving address as 130.50.16.1
0

for the first one, it should be 64,1022... but one of the GATE question, they wrongly given that it is 62, 1022

But don't go with that answer.. go with concept only :)

For the second one, it is ambiguous question.. i mean

generally the subnets are marked as 0,1,....63,

1st subnet means what in the view of questioner ?

if it subnet$_0$, then it is 130.50.000000 00 . 0000 0001 ===> 130.50.0.1

if it subnet$_1$, then it is 130.50.000001 00 . 0000 0001 ===> 130.50.4.1

4th subnet means what in the view of questioner ?

if it subnet$_3$, then it is 130.50.000011 00 . 0000 0001 ===> 130.50.3.1

if it subnet$_4$, then it is 130.50.000100 00 . 0000 0001 ===> 130.50.16.1

0

@Shaik Masthan Brother here the 0th subnet is basically the first subnet

( According to standard approach to be used in GATE)

So starting address should be : 130.50.0.0

For the fourth subnet the id should be 000011 so the id should be 130.50.00001100.0 = 130.50.12.0

0

As mentioned by Shaik ,I have gone through this link.

Regarding 2nd question I think now the subnet with all 0's will be 1st subnet and the one with 011 as 4th subnet.I think count of subnet do not depend on the the value of the bits in the subnet.

0
So my comment was right Right ?

1. Subnets possible=2^6=64

Number of hosts per subnet=2^10-2

2. let the subnet count start from 0, i.e., subnet0, subnet 1.....

subnet 1 range: 135.50.00000100.00000000 to 135.50.00000111.11111111

so ip of first host in subnet 1: 135.50.00000100.00000001=135.50.4.1

subnet 4 range: 135.50.00010100.00000000 to 135.50.00010111.11111111

so ip of first host in subnet 1: 135.50.00010100.00000001=135.50.20.1

--------------------------------------------------------------

reference-

n bits for subnet id, this implies 2^n subnets possible

n bits for host id, this implies 2^n-2 hosts per subnet possible(1 ip for subnet id and 1 ip for dba excluded)
by
edited by
#Q1.

IP addresses are possible - 2^16

Subnets  are possible - 2^6

Hosts are possible - 2^6 (2^10  - 2)

#Q2.

Option A
by

+1 vote