We want to divide 192.168.10.0 which is a class c network
into four networks, each with unequal nmber of IP address requirements as shown
below.
Subnet A = 32 hosts
Subnet B = 8 hosts
Subnet C = 22 hosts
Subnet D = 60 hosts
Answers:
60 hosts (Subnet D)
32 hosts (Subnet A)
22 hosts (Subnet C)
8 hosts (Subnet B)
H = 2 ^6 = 64 - 2 = 62 S
= 2^2
H = 2^6 = 64 - 2 = 62 S
= 2^2
H = 2^5 = 32 -2 = 30 S
= 2^3
H = 2 ^4 = 16 - 2 = 14 S
= 2^4
Network Name
|
Network Address
|
Host Range
|
Broadcast Address
|
Submask Net
|
Host D
|
192.168.1.0
|
192.168.1.1 – 192.168.1.62
|
192.168.1.63
|
255.255.255.192
|
Host A
|
192.168.1.64
|
192.168.1.65 – 192.168.1.126
|
192.168.1.127
|
255.255.255.192
|
Host C
|
192.168.1.128
|
192.168.1.129 – 192.168.1.158
|
192.168.1.159
|
255.255.255.224
|
Host B
|
192.168.1.160
|
192.168.1.161 – 192.168.1.174
|
192.168.1.175
|
255.255.255.240
|
Given a network of 201.4.3.0/24 subnet, the network in order to create the sub network with the following requirement.
Office 1 – 14 hosts
Office 2 – 60 hosts
Office 3 – 32 hosts
Office 4 – 7 hosts
Office 5 – 15 hosts
Answers:
60 hosts (office 2)
32 hosts (office 3)
15 hosts (office 5)
14 hosts (office 1)
7 hosts (office 4)
H = 2 ^6 = 64 - 2 = 62 S
= 2^2
H = 2^6 = 64 - 2 = 62 S
= 2^2
H = 2^5 = 32 -2 = 30 S
= 2^3
H = 2^4 = 16 - 2 = 14 S
= 2^4
H = 2^4 = 16 – 2 = 14 S
= 2^4
Network Name
|
Network Address
|
Host Range
|
Broadcast Address
|
Submask Net
|
Office 2
|
201.4.3.0
|
201.4.3.1 –
201.4.3.62
|
201.4.3.63
|
255.255.255.192
|
Office 3
|
201.4.3.64
|
201.4.3.65 – 201.4.3.126
|
201.4.3.127
|
255.255.255.192
|
Office 3
|
201.4.3.128
|
201.4.3.129 –
201.4.3.158
|
201.4.3.159
|
255.255.255.224
|
Office 1
|
201.4.3.160
|
201.4.3.159 –
201.4.3.174
|
201.4.3.175
|
255.255.255.240
|
Office 4
|
201.4.3.176
|
201.4.3.177 –
201.4.3.190
|
201.4.3.191
|
255.255.255.240
|
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