### Discussion :: Networking - Section 7 (Q.No.2)

Mahesh said: (Apr 14, 2011) | |

Ans: Its a class B IP. No. of host= 2^n-1=600 (where n=numbers of zeros) So, 2^n=601. Approx. n=10 the subnet mask will be: 11111111.11111111.11111100.00000000 (255.255.252.0) |

Harsha said: (Mar 8, 2013) | |

No. of hosts=2^n-2. So, 2^n=602, Approx. n=10. The subnet mask will be: 11111111.11111111.11111100.00000000 (255.255.252.0). |

Rohit Singh said: (Oct 7, 2016) | |

Total 16 bits available for representing subnets and hosts since Class B address has 2 octets available. No. of hosts = 600. No. of bits needed to represent 600 hosts = round( log(600) to the base (2) ) = 10. Hence, no. of bits available to represent subnets = 16 - 10 = 6. From the right end, first host bits are taken then subnet bits. The subnet mask is when all subnet bits are set to 1 and remaining host bits are set to 0 (in the available octets). Hence, subnet mask = 11111100.00000000 = 252.0 (Remaining portion will remain the same, i.e., 255.255). Final mask = 255.255.252.0. |

Shruti said: (Dec 13, 2016) | |

Thanks @Rohit Singh. |

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