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Address in subnet

Posted on 2010-09-20
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Last Modified: 2013-11-16
Hi,

I have IP address 10.180.10.18 and subnet 255.192.0.0.
What?
 size of the network part of the address
 size of subnet part of the address
 size of host part
 number of host per subnet
 subnet number
 broadcast address
range of valid ip address in this network

Thanks,
JT
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Question by:jtran007
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Expert Comment

by:Blaz
ID: 33714864
http://www.subnet-calculator.com/subnet.php?net_class=A

Network address is 10 bit - 10.128.0.0/10
Subnet part is 2 bit
Host part is the remaining 22 bits (32 - 10)
There are 4194302 hosts per subnet
Not sure about subnet number, but probably it is the third subnet from four (previous beeing 10.0.0.0/10, 10.64.0.0/10 and next 10.192.0.0/10)
Bradcast address is 10.191.255.255
Range of IP addresses is 10.128.0.1 - 10.191.255.254


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Expert Comment

by:k_romych
ID: 33714982
size of the network part of the address 16581375
 size of subnet part of the address 4161600
 size of host part 4161600
 number of host per subnet 4161600
 subnet number 10.128.0.0
 broadcast address 10.191.255.255
range of valid ip address in this network 10.128.0.0-10.191.255.255
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Author Comment

by:jtran007
ID: 33725471
Hi,

Thanks for giving the website to calculate, but could you
show me how you come up with this answer?

Many thanks,
JT
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Accepted Solution

by:
Blaz earned 1000 total points
ID: 33725797
255.192.0.0 is in binary form 11111111.11000000.00000000.00000000 and
10.180.10.18 is in binary form 00001010.10110100.00001010.00010010

From this two numbers most of the answers can be made

> Network address is 10 bit - 10.128.0.0/10
There are (first) 10 bits set in binary form of subnet/netmask. If you make an bitwise AND operation between your host number and the subnet mask you get 10.128.0.0 which is your network.

> Subnet part is 2 bit
First 8 bits (of the 10) denote an A class address, next two bits can be considered subnet part.

> Host part is the remaining 22 bits (32 - 10)
And the 22 zeroes in the subnet (if writing out all 8 bits of each number)

> There are 4194302 hosts per subnet
If you have 22 bits for host part, then you have 2^22 = 4194304 possible IPs
First IP is reserved for the network address and the last IP for broadcast address. So there are 4194302 IP addresses left for hosts.

> Not sure about subnet number, but probably it is the third subnet from four (previous beeing 10.0.0.0/10, 10.64.0.0/10 and next 10.192.0.0/10)
Given that the two bits in second octet denote the subnet, you can construct 4 subnets: 10.0.0.0/10, 10.64.0.0/10, 10.128.0.0/10, 10.192.0.0/10

> Bradcast address is 10.191.255.255
As said broadcast address is the last IP address in the subnet.

> Range of IP addresses is 10.128.0.1 - 10.191.255.254
10.128.0.0 is network address, 10.191.255.255 is the broadcast address, Host IP addresses are all in between.
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Author Closing Comment

by:jtran007
ID: 33733090
Thanks
JT
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