Gate2020 cs Q63

  • Option : A
  • Explanation :
    No. of host = 1500
    No. of host bits = [log2 1500] = 11 bits
    ∴ Total possible hosts = 211 = 23 × 28
    n is the netmask bits,
    Range of addresses is = 232 – n
    211 = 232-n
    Available IP address ⇒ 202.61.0.0/17

    So the IP address follow the pattern
    0.0 to 7.255
    8.0 to 15.255
    16.0 to 23.255
     :
    64.0 to ...
     :
    104.0 to ...
    ∴ The possible IP addresses are 202.61.64.0/21 & 202.61.104.0/21
Cancel reply

Your email address will not be published. Required fields are marked *


Cancel reply

Your email address will not be published. Required fields are marked *