Use logical ANDing to find the network ID for the IP address

Use logical ANDing to find the network ID for the IP address and subnet mask listed below. Show all of your work. IP address: 194.56.36.127 Subnet Mask: 255.255.255.0

Solution

194₁₀

= 194 / 2 = 97 (0 remainder)

= 97 / 2 = 48 (1 remainder)

= 48 / 2 = 24 (0 remainder)

= 24 / 2 = 12 (0 remainder)

= 12 / 2 = 6 (0 remainder)

= 6 / 2 = 3 (0 remainder)

= 3 / 2 = 1 (1 remainder)

= 1 / 2 = 0 (1 remainder)

Therefore, Binary equivalent of 192 is: 1100 0010.

56₁₀

= 56 / 2 = 28 (0 remainder)

= 28 / 2 = 14 (0 remainder)

= 14 / 2 = 7 (0 remainder)

= 7 / 2 = 3 (1 remainder)

= 3 / 2 = 1 (1 remainder)

= 1 / 2 = 0 (1 remainder)

Therefore, Binary equivalent of 56 is: 0011 1000.

36₁₀

= 36 / 2 = 18 (0 remainder)

= 18 / 2 = 9 (0 remainder)

= 9 / 2 = 4 (1 remainder)

= 4 / 2 = 2 (0 remainder)

= 2 / 2 = 1 (0 remainder)

= 1 / 2 = 0 (1 remainder)

Therefore, Binary equivalent of 36 is: 0010 0100.

127₁₀

= 127 / 2 = 63 (1 remainder)

= 63 / 2 = 31 (1 remainder)

= 31 / 2 = 15 (1 remainder)

= 15 / 2 = 7 (1 remainder)

= 7 / 2 = 3 (1 remainder)

= 3 / 2 = 1 (1 remainder)

= 1 / 2 = 0 (1 remainder)

Therefore, Binary equivalent of 127 is: 0111 1111.

255₁₀

= 255 / 2 = 127 (1 remainder)

= 127 / 2 = 63 (1 remainder)

= 63 / 2 = 31 (1 remainder)

= 31 / 2 = 15 (1 remainder)

= 15 / 2 = 7 (1 remainder)

= 7 / 2 = 3 (1 remainder)

= 3 / 2 = 1 (1 remainder)

= 1 / 2 = 0 (1 remainder)

Therefore, Binary equivalent of 255 is: 1111 1111.

Finally, the network ID is: 194.56.36.0.

	Octet 1 in Binary	Octet 2 in Binary	Octet 3 in Binary	Octet 4 in Binary
IP Address	1100 0010	0011 1000	0010 0100	0111 1111	194.56.36.127
Subnet Mask	1111 1111	1111 1111	1111 1111	0000 0000	255.255.255.0
Network ID	1100 0010	0011 1000	0010 0100	0000 0000	194.56.36.0