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Need help with this calculation.

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  • Networking
Last response: in Networking
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February 17, 2014 5:51:25 PM

I went through an article at http://www.cisco.com/web/about/ac123/ac147/archived_iss... which is as follows:

The Hardest Subnetting Problem:

The hardest subnetting problem most people face is that of trying to decide what the smallest subnet is that will provide a given number of hosts on a specific segment, and yet not waste any address space. The way this sort of problem is normally phrased is something like the following:


You have 5 subnets with the following numbers of hosts on them: 58, 14, 29, 49, and 3, and you are given the address space 10.1.1.0/24. Determine how you could divide the address space given into subnets so these hosts fit into it.

This appears to be a very difficult problem to solve, but the chart we used previously to find the jump within a single octet actually makes this task quite easy. First, we run through the steps, and then we solve the example problem to see how it actually works.
•Order the networks from the largest to the smallest.
•Find the smallest number in the chart that fits the number of the largest number of hosts + 2 (you cannot, except on point-to-point links, use the address with all 0’s or all 1’s in the host address; for point-to-point links, you can use a /31, which has no broadcast addresses).
•Continue through each space needed until you either run out of space or you finish.

This process seems pretty simple, but does it work? Let’s try it with our example.
•Reorder the numbers 58, 14, 29, 49, 3 to 58, 49, 29, 14, 3.
•Start with 58. ◦The smallest number larger than (58 + 2) is 64, and 64 is 2 bits.
◦There are 24 bits of prefix length in the address space given; add 2 for 26.
◦The first network is 10.1.1.0/26.
◦The next network is 10.1.1.0 + 64, so we start the next “round” at 10.1.1.64.

now my question is " How is 58+2 = 64 and 64 is 2 bits?"

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February 17, 2014 6:05:26 PM

your subnet mask, /24 = 11111111.11111111.11111111.00000000 binary
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February 17, 2014 6:09:15 PM

so usable part of the entire host addressable network is the last octet. All "networks" have "special" address' the Network itself and a Broadcast address. These make you do the addressing as you stated above... meaning for a network requiring 58 Hosts, you add 2 for the Network and Breoadcast making the total address' needed = 60... This needs to fit in this Binary framework.
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February 17, 2014 6:13:06 PM

the Binary framework is such:

128 64 32 16 8 4 2 1
0 0 0 0 0 0 0 0 = 0 Decimal
0 1 0 0 0 0 0 0 = 64 Decimal

as you can see above, yes 60 is less than 64 making the required subnet for a 58 Host network a /26 (or in dotted Decimal 255.255.255.192
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February 17, 2014 6:13:38 PM

helpful?
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February 17, 2014 7:42:09 PM

corroded said:
basic subnetting ... it is based on Binary...

I teach basic LAN... ie subnetting...

https://learningportal.juniper.net/juniper/user_activit...

free crash course...

I can teach this to anyone... :-)



Thanks very much COrroded. How do I enrol for your online free crash course?

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February 17, 2014 11:05:03 PM

the juniper course is free... I teach in California..
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!