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Confusion in VLSM subnetting
This is the pdf I am studying from. This is not exactly what I need for my exams but it’s pretty clearly written and concept giving thing.
192.168.10.0/24 is given IP address to subnet.
We need 4 subnets
So
192.168.10.00 00 0000
This is an ip address that has 3 octets of network address and 1 octet of host address. So a class C address.
For 4 subnets, i need 2 bits of host to be borrowed.
00 00 0000
01
10
11So I get these IP address for subnets-:
Subnet #0->192.168.10.0/26
Subnet #1->192.168.10.64/26
Subnet #2->192.168.10.128/26
Subnet #3->192.168.10.192/26A) Take subnet #0. Perth 60 hosts
The start address here is 192.168.10.0/26
192.168.10.00 000000192.168.10.00 1111111
This gives
Network address=192.168.10.0 /26
Broadcast address=192.168.10.63 /26
B) Take subnet #1-:
Kuala Lampur 28 hosts.
Next subnet IP address was 192.168.10.64/26
Question-: Why was this the next subnet IP address?
Is it becuase of this?
So I get these IP address for subnets-:
“Subnet #0->192.168.10.0/26
Subnet #1->192.168.10.64/26
Subnet #2->192.168.10.128/26
Subnet #3->192.168.10.192/26”Is it because the next consecutive ip address after the perth’s broadcast address is 192.168.10.64 /26?
28 hosts required ie 5 host bits. I.e 27 network bits.
192.168.10.010 00000
192.168.10.010 11111 -
1 Reply
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Thank you for sharing the PDF and walking us through your VLSM subnetting process. Your explanation is incredibly detailed and helpful, especially for those of us trying to grasp these concepts. It’s evident that you have a strong understanding of networking fundamentals. Exploring additional resources like https://darkvr.io/ can provide even more insights and broaden your knowledge base. Keep up the fantastic work, and don’t hesitate to reach out if you have any more questions or insights to share.
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