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bighornsheep wrote: Let's say 192.168.1.35 / 27, find the range of valid hosts Figure out your subnet mask, 255.255.255.224, now figure the block size, octet of significance is the fourth one, so 224, take that away from 256, answer is 32. So block size of 32. The subnet IDs are 192.168.1.0, .32, .64 etc... .35 must be in the network ID .32, the valid IPs are from .32 through the next subnet - 1, so .32 through .63, the first and last IP can't be used, first is the subnet ID, last is the broadcast IP, so the valid hosts are .33 through .62 Is this what you mean?
shednik wrote: bighornsheep wrote: Let's say 192.168.1.35 / 27, find the range of valid hosts Figure out your subnet mask, 255.255.255.224, now figure the block size, octet of significance is the fourth one, so 224, take that away from 256, answer is 32. So block size of 32. The subnet IDs are 192.168.1.0, .32, .64 etc... .35 must be in the network ID .32, the valid IPs are from .32 through the next subnet - 1, so .32 through .63, the first and last IP can't be used, first is the subnet ID, last is the broadcast IP, so the valid hosts are .33 through .62 Is this what you mean? Another way that test out taught me was rather than take 224 - 256 to get the block size is to take the last bit in binary that's a 1....for example in this one 255.255.255.11100000 the last bit thats a 1 is 32. for me thats been my method and it has helped me figure things out a little faster. just my 2 cents
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