Your company has been assigned the following IP address by IANA: 158.80.0.0. Des

Question

Your company has been assigned the following IP address by IANA: 158.80.0.0. Design a network that consists of 50 subnetworks with each subnetwork having up to 300 hosts.

a) Write the address for the 1st subnetwork as well as the 1 host, 2nd host, the last host, and the broadcast address for the 1st subnetwork. Present the addresses in the binary and decimal forms

b) Write the address for the 2nd subnetwork as well as the 1 host, 2nd host, the last host, and the broadcast address for the 2nd subnetwork. Present the addresses in the binary and decimal forms

c) Write the address for the last subnetwork as well as the 1 host, 2nd host, the last host, and the broadcast address for the last subnetwork. Present the addresses in the binary and decimal forms

d) Use the masking operation (the AND logical operator) to show explicitly that the last host residing on the 2nd subnetwork indeed belongs to this subnetwork. Show your calculations bit by bit and align bits

Explanation / Answer

A network can have max no of hosts only in the power of 2. As here max nodes can be 300, therefore we will have to keep the count as 512.

so the last 9 bits (2^9=512) will be used to determine hosts in a subnetwork.

and we have to form 50 subnetworks, so the next 6 bits(10th to 15th from the last ) will be used to divide the subnetwork. i.e.

158.80.0.0 can also be written as 10011110.00110000.00000000.00000000

and our division will work as

10011110.00110000.0000000(*)0.00000000(*)

                              (subnetwork (host

                               digits)             digits)

a) address of first subnetwork:- 10011110.00110000.0000000 0.00000000 i.e 158.80.0.0

    address of first host:- 10011110.00110000.0000000 0.00000001 i.e. 158.80.0.1

    address of second host:- 10011110.00110000.0000000 0.00000002 i.e. 158.80.0.2

     broadcast address of first subnetwork:- 10011110.00110000.0000000 1.11111111 i.e 158.80.1.255

b) address of second subnetwork:- 10011110.00110000.0000001 0.00000000 i.e 158.80.2.0

    address of first host:- 10011110.00110000.0000001 0.00000001 i.e. 158.80.2.1

    address of second host:- 10011110.00110000.0000001 0.00000002 i.e. 158.80.2.2

     broadcast address of first subnetwork:- 10011110.00110000.0000001 1.11111111 i.e 158.80.3.255

c) address of last subnetwork:- 10011110.00110000.0111111 0.00000000 i.e 158.80.126.0

    address of first host:- 10011110.00110000.0111111 0.00000001 i.e. 158.80.126.1

    address of second host:- 10011110.00110000.0111111 0.00000002 i.e. 158.80.126.2

   broadcast address of first subnetwork:- 10011110.00110000.0111111 1.11111111 i.e158.80.127.255

d) second subnetwork is 10011110.00110000.0000001 0.00000000 i.e 158.80.2.0

     so written in cider notation , it will be 158.80.2.0/9 and hence the mask will be 11111111.11111111.11111110.00000000 i.e. 255.255.254.0

the last host on second network is 10011110.00110000.00000011.11111111

masking (anding) with the mask    11111111.11111111.11111110.00000000

the result will be                              10011110.00110000.00000010.00000000 i.e 158.80.2.0

hence proved

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