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Since the organization requires 8 subnets with at least 63 hosts each, the subnet mask should allow for at least 6 bits to be used for the subnet IDs (2^6 - 2 = 62, where 2 is subtracted for the network and broadcast addresses).

  1. Class of IP address: The first octet of the given IP address is 193, which indicates a class C IP address.

  2. Default mask: The default subnet mask for a class C IP address is 255.255.255.0.

  3. Subnet mask: Since we need to use 6 bits for the subnet IDs, the subnet mask can be calculated as follows:

    • Convert the default subnet mask to binary: 11111111.11111111.11111111.00000000
    • Set the first 26 bits to 1 to allow for 6 bits for the subnet IDs: 11111111.11111111.11111111.11000000
    • Convert back to decimal: 255.255.255.192

    Therefore, the subnet mask is 255.255.255.192.

  4. Number of subnet ID bits: As mentioned earlier, we need to use 6 bits for the subnet IDs.

  5. Broadcast address of the subnets: To calculate the broadcast address of each subnet, we need to find the last address in each subnet range. Since we have 6 bits for the subnet IDs, we can create 2^6 = 64 subnets, but we only need 8 subnets. Therefore, each subnet will have a range of 32 addresses.

    The first subnet will have an ID of 193.1.1.0, and the last address in its range will be 193.1.1.31. Therefore, its broadcast address will be 193.1.1.63.

    The second subnet will have an ID of 193.1.1.64, and the last address in its range will be 193.1.1.95. Therefore, its broadcast address will be 193.1.1.127.

    The third subnet will have an ID of 193.1.1.128, and the last address in its range will be 193.1.1.159. Therefore, its broadcast address will be 193.1.1.191.

    The fourth subnet will have an ID of 193.1.1.192, and the last address in its range will be 193.1.1.223. Therefore, its broadcast address will be 193.1.1.255.

    Note: The network address (first address) and broadcast address (last address) cannot be assigned to hosts, which is why we subtracted 2 from the total number of addresses to get the maximum number of hosts per subnet (63 - 2 = 61).

 

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This is class C network. So, first 3 octet is for net ID and  last octet is host ID. for subnetting into 8 subnet 3 bits are taken from last octet (Host ID bits).

 

 

 

 

 

 

 

 

|--- subnet-----||-------- Host ID bits------|

As we have only 5 bits left in Host we cannot allot 63 hosts to each subnet.

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