IP Address allocation

Question

• A. 64,128
• B. 32,16
• C. 32,128
• D. 64,32

Answer 1

In 219.50.0.0/16 There are 2^16=65536 IP Addresses available 

In group 1 total IP address allocated is 128*64=8192 IP

in group 2 total IP address allocated is 64*256=16384 IP

In group 3 total IP address allocated is x*y=xy IP

ISP has 38K IP address left = 38912 IP

so group 3 has IP address = Total IP addresses - (group 1 IP addresses+ group 2 IP addresses)

=65536-(8192+16384)

=2048 IP

$so$ $possible$ $value$ $of$ $x$, $y$ is $64$, $32$

Comments

@prashant281188 kindly tell me here how you interpret meaning of 38K as 38×1024 and not as 38×1000.plzz reply.

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@Shreya kumari because in IP address allocation we can allocate IP addresses in power of 2's only.

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what is the use of  38K ip addresses left with isp