Homework Answers
Building 1:
no of PC =800
no of addresses requireed= 800 +2 (network address + broadcast address) = 802
no of bits required for host address= ceil (log2(802)) = ceil(9.64) = 10 bits
Thus, no of bits required for network address = 32 - 10 = 22 bits.
Address range = 218.165.192.0/22 to 218.165.195.255/22
Building 2:
no of PC =600
no of addresses requireed= 600 +2 (network address + broadcast address) = 602
no of bits required for host address= ceil (log2(602)) = ceil(9.23) = 10 bits
Thus, no of bits required for network address = 32 - 10 = 22 bits.
Address range = 218.165.196.0/22 to 218.165.199.255/22
Building 3:
no of PC =400
no of addresses requireed= 400 +2 (network address + broadcast address) = 402
no of bits required for host address= ceil (log2(402)) = ceil(8.65) = 9 bits
Thus, no of bits required for network address = 32 - 9 = 23 bits.
Address range = 218.165.200.0/23 to 218.165.201.255/23
Building 4:
no of PC =300
no of addresses requireed=300 +2 (network address + broadcast address) = 302
no of bits required for host address= ceil (log2(302)) = 9 bits
Thus, no of bits required for network address = 32 - 9 = 23 bits.
Address range = 218.165.202.0/23 to 218.165.203.255/23
Building 5:
no of PC =200
no of addresses requireed= 200 +2 (network address + broadcast address) = 202
no of bits required for host address= ceil (log2(202)) = 8 bits
Thus, no of bits required for network address = 32 - 8 = 24 bits.
Address range = 218.165.204.0/24 to 218.165.204.255/24
| BUILDING | NO OF PC | SUBNET ADDRESS |
| 1 | 800 | 218.165.192.0/22 |
| 2 | 600 | 218.165.196.0/22 |
| 3 | 400 | 218.165.200.0/23 |
| 4 | 300 | 218.165.202.0/23 |
| 5 | 200 | 218.165.204.0/24 |
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