edited by
8,469 views

4 Answers

Best answer
29 votes
29 votes

(C) is answer since you have $6$ zeroes so you can make $64-2$ hosts

edited by
11 votes
11 votes

 

Answer is Option C 

9 votes
9 votes

Answer (C).

maximum no. of hosts = 2(no. of bits in hid) - 2

                                     = 26 - 2

                                     = 64 - 2

                                     = 62

2 votes
2 votes

Subnet Mask = 255.255.255.192

If we only write the last byte into binary, then we get,

255.255.255.11000000

∴ Number of 0's in subnet mask = Number of bits in host ID = 6

∴ Maximum number of hosts = 2^ (Number of bits in host Id) - 2 [since the 1st and last IP address is used for Network ID and Directed Broadcast Adddress]

                                     
                          = 2^6 - 2

                          = 64 - 2

                          = 62


Option (c) is the right answer

Answer:

Related questions

24 votes
24 votes
6 answers
3
Ishrat Jahan asked Nov 2, 2014
6,145 views
In the TCP/IP protocol suite, which one of the following is NOT part of the IP header?Fragment OffsetSource IP addressDestination IP addressDestination port number