Question

2. Given the IP address 17.119.47.72, apply Subnet Mask 1 and answer the following questions. [2 marks] My subnet mask is: a. What is the subnet address? b. What is the broadcast address for this subnet?

Answer 1

Here IP is 17.119.47.72 (in binary  00010001 01110111 00101111 01001000)

Subnet mask is 255.240.0.0 ( in binary 11111111 11110000 00000000 00000000)

a) Subnet address/ Network address :

you need to perform a bit-wise AND operation (1+1=1, 1+0 or 0+1 =0, 0+0=0) on the host IP address and subnet mask

00010001 01110111 00101111 01001000 ( IP address)

AND 11111111 11110000 00000000 00000000 ( Subnet mask)

= 00010001 01110000 00000000 00000000 (Network address)

Subnet Address = 17.112.0.0

b) Broadcast Address :

A bitwise OR between the network address and the inverted subnet mask would give us the broadcast address:

00010001 01110000 00000000 00000000 (Network address)

OR 00000000 00001111 11111111 11111111 ( Inverted Subnet mask )

= 00010001 01111111 11111111 11111111 ( Broadcast address )

BroadCast Address : 17..127.255.255

IP Address: 17.119.47.72 Network Address: 17.112.0.0 Usable Host IP Range: 17.112.0.1 - 17.127.255.254 Broadcast Address: 17.127.255.255 Total Number of Hosts: 1,048,576 Number of Usable Hosts: 1,048,574 Subnet Mask 255.240.0.0 Wildcard Mask: 0.15.255.255 Binary Subnet Mask 11111111.11110000.00000000.00000000 IP Class: CIDR Notation: /12 IP Type: Public Short: Binary ID: Integer ID: Hex ID: in-addr.arpa IPv4 Mapped Address: 6to4 Prefix: 17.119.47.72/12 00010001011101110010111101001000 293023560 Ox11772f48 72.47.119.17.in-addr.arpa --ffff: 1177 2f48 2002:1177 2f48-148

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