Subnetting

Question

You need to configure a server that is on the subnet 192.168.19.24/29. The router has the first available host address. Which of the following should you assign to the server?

A.	192.168.19.0
B.	192.168.19.33
C.	192.168.19.26
D.	192.168.19.31

Comments

if first available host address gives to router ===> it should assign 192.168.19.25

but i didn't get the meaning of

Which of the following should you assign to the server? 

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A/29 is 255.255.255.248, which is a block size of 8 in the fourth octet. The subnets are 0, 8, 16, 24, 32, 40, etc. 192.168.19.24 is the 24 subnet, and since 32 is the next subnet, the broadcast address for the 24 subnet is 31. 192.168.19.26 is the only correct answer.

The solution posted is this I am not able to understand can someone help me solve this in different way!!

Answer 1

The Given Address 192.168.19.24/29  

step one convert mask to binary i.e. 29 bits are reserved. so make it binary one

 

( IP address is belong to class C so 24 bits are of class C and of last 5 bits borrowed from host so (29 mask)

11111111.11111111.11111111.11111000     

convert only last value 24 from 192.168.19.24 to binary bcz its class C now i.e. 11000

11111111.11111111.11111111.11111000

                                              00011/000     24 in binary  { first five bits dont change bcz they are net id}

                                              00011001    first address (192.168.19.25)

                                              00011110  last address   (192.168.19.30)

                                              00011111  broadcast  (192.168.19.31)

 now range is from 192.168.19.25  to  192.168.19.30  answer os  C

Caption

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tell me how to upload image as a answer. i solved neatly in notebook so