Network Address and Subnetting

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Question Number Mark
allocate
d
Mark
earne
d
Question 1: (3 marks) 3
a
)
b
)
c
)
Network address for the IP address 156.120.32.3/26
156.120.32.0/26
Direct Broadcast address
156.120.32.63
Range of address available for hosts
156.120.32.1 - 156.120.32.62
Calculation:
The following table is used for the calculation of number of bits used for subnetting.
Subnet
Mask
0 128 192 224 240 248 252 254 255
No. of bits used for
subnetting (n)
0 1 2 3 4 5 6 7 8
Subnet
IP address = 156.120.32.3
Subnet mask = 255.255.255.192
1 mark
each
item,
total 3

Total bits = 8
Number of bits used for subnetting = 2
Therefore number of bits left for host = 8 – 2
= 6
Value of last bit used for subnet masking = 2^6
= 64
Number of host per subnet = 64 -2
= 62
Question 2: (8 marks) 8
a
)
Block of address granted to the company is 128.32.16.8/22
Network address for this block is 128.32.16.0/22, and the total addresses the block
contains including special addresses is 1024.
1
b
)
Subnet No. addresses Subnet address Mask /n
1 128 128.32.16.0 /24
2 64 128.32.17.0 /25
3 32 128.32.17.128 /26
4 32 128.32.17.192 /26
Calculations:
IP address = 128.32.16.8/22
Binary Representation = 10000000.00100000.000100 00.00001000
Subnet mask = 255.255.252.0
Binary Representation = 11111111.11111111.111111 00.00000000
Network: 128.32.16.0/22 10000000.00100000.000100 00.00000000 (Class B)
6
c
)
After the allocation of the subnet 518 number of addresses are left unallocated.
Total number of IP address available for the Subnet = 1022
1