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a) You are given the assignment of setting subnet addresses for 5 buildings of your company. The number of Internet connected

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Answer #1

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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