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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

4 Answers

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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)
edited by
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#Q1.

IP addresses are possible - 2^16

Subnets  are possible - 2^6

Hosts are possible - 2^6 (2^10  - 2)

#Q2.

Option A
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Here Total 62 Subnets are Possible and 1022 hosts per Subnet is possible.

62 because :

130.50.0.0 this is the whole Network Address

and 130.50.255.255 is the DBC address of the Whole N/W

hence 2 addresses will not be alloted

Hope it’s Clear..
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For 1st part option D is correct

for 2nd part option A is correct  because the subnet with all 0’s are reserved

edited by

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