Question

Use the major network address of 192.168.10.0 using a 27 bit mask. configure on subnet #2.

Find:

Find:

1) Subnet Mask

2.) First usable IP address

3.) Last usable IP address

4.) 5th usable IP address

Note: Show details and consider 27 bit mask and subnet #2.

Answer 1

Given Address = 192.168.10.0

So it is class C network (class C ranges from 192 - 223)

Default subnet mask for class C network is 255.255.255.0 (Network.Network.Network.Host)

i.e, 11111111.11111111.11111111.0000 0000

1)

Given Netmask = 27

so network address is  192.168.10.0 / 27

so, 32- 27 = 5, so there are 5 host bits and 3 subnetting bits in last octet.(3 bits are borrowed from host bits i.e, last octet).

Default mask: 255.255.255.0-------------------> 11111111.11111111.11111111. 000 00000

Our subNetmask: 255.255.255.224 = 27 ----> 11111111.11111111.11111111.111 00000 (3 bits borrowed)

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Address: 192.168.10.0 ----------------> 11000000.10101000.00001010.000 00000

Netmask: 255.255.255.224 = 27 ----> 11111111.11111111.11111111.111 00000

• ANDing Address and Netmak gives us Network Address.

Network: 192.168.10.0/27 =>11000000.10101000.00001010.000 00000 (Class C)

we will get broadcast address by making 5 host bits in last octet,

Broadcast: 192.168.10.31 => 11000000.10101000.00001010.000 11111

• First IP Address is Network ID I.e, 192.168.10.0
• Last IP Address is Broadcast Address => 192.168.10.31
• So first usable IP Addres is 192.168.10.1
• Last usable IP Address is 192.168.10.30
• 5th usable IP Address is 192.168.10.5

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Summary:

Address: 192.168.10.0 --------------> 11000000.10101000.00001010.000 00000
Netmask: 255.255.255.224 = 27 ----> 11111111.11111111.11111111.111 00000

• ANDing Address and Netmak gives us Network Address.

Network: 192.168.10.0/27 => 11000000.10101000.00001010.000 00000 (Class C)(1st IP Address)
Broadcast: 192.168.10.31 => 11000000.10101000.00001010.000 11111 (Last IP Address)
HostMin: 192.168.10.1 ===> 11000000.10101000.00001010.000 00001 (First Usable IP Address)
HostMax: 192.168.10.30 ==> 11000000.10101000.00001010.000 11110 (Last usable IP Address)
Hosts/Net: 30 => (Private Internet)

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Question 2, 3 and 4)

Subnet mask = 255.255.255.224

Network Address = 192.168.10.0

we have already calculated for subnet #0, now we will calculate for subnet #2

subnet bits used in mask = 3 bits or 2³-2= 6 subnets (subracted 2 because two of these subnets are not generally usable)

host bits are available per subnet = 5 bits or 2⁵-2=30 hosts per subnet

subnet address are : 256-224 =32, so subnets can be found by continuing to add 32 to itself ,till <224

32, 64, 96, 128, 160 and 192 (Six subnets)

subnet 1: Address  192.168.10.32

subnet 2: Address  192.168.10.64

subnet 3: Address  192.168.10.96

subnet 4: Address  192.168.10.128

subnet 5: Address  192.168.10.160

subnet 6: Address  192.168.10.192

For subnet #2:

Subnet Network address =>192.168.10.64 (First IP adress)

2) First usuable IP Address =>192.168.10.65

3) Last usuable IP Address =>192.168.10.94

4) 5th usable IP adress =>192.168.10.69

Broadcast Address =>192.168.10.95