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.
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
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
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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
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
Use the major network address of 192.168.10.0 using a 27 bit mask. configure on subnet #2....
using the information given below
. what is the ip address , subnet mask , default gateway?
** corrections
# of bits in subnet for A
:6
# of bits in submet for B:
3
PC-A
R1 G0/0
R GO/1
S1
PC-B ?
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