Homework Answers
Question 1 :
- A bitwise AND between the IP Address and the Subnet mask would give us the network ID.
IP Address : 18.132.219.175 (convert it into binary form)
: 00010010.10000100.11011011.10101111
Subnet Mask : 255.255.248.0 (convert it into binary form)
: 11111111.11111111.11111000.00000000
Bitwise AND of IP address and subnet mask
00010010.10000100.11011011.10101111 (IP address)
AND 11111111.11111111.11111000.00000000 (subnet mask)
______________________________________________
00010010.10000100.11011000.00000000 (network ID)
______________________________________________
Network ID : 00010010.10000100.11011000.00000000 (convert it into decimal)
: 18.132.216.0
Question 2:
- A bitwise OR between the IP Address and the inverted subnet mask would give us the broadcast address.
IP Address = 18.132.219.175 (convert it into binary form)
= 00010010.10000100.11011011.10101111
Subnet Mask = 255.255.248.0
= 0.248.255.255 (inverted subnet mask) (convert it into binary form)
= 00000000.11111000.11111111.11111111
Bitwise OR of IP address and subnet mask
00010010.10000100.11011011.10101111 (IP address)
OR 00000000.11111000.11111111.11111111 (Subnet mask)
____________________________________________________
000010010.11111100.11111111.11111111 (broadcast address)
_____________________________________________________
Broadcast Address : 000010010.11111100.11111111.11111111 (convert it into decimal)
: 18.252.255.255
Therefore,
Network ID : 18.132.216.0 (First part)
Broadcast Address : 18.252.255.255 (Second part)
PLEASE REVIEW AND UPVOTE!!!
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** corrections
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R GO/1
S1
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