Addresses and Subnetting
Binary Without the Fear
9:34Coming soon
An exam question and a thirty-second clock: a host is 192.168.1.77 with mask 255.255.255.224, what is its network address? Most people freeze there, not because the maths is hard but because nobody ever showed them the lamps. This lesson is binary for networking, and only the part you need: reading and writing one octet, the nine legal subnet masks, the block size, and the AND that every computer runs before it sends.
You'll learn
- The lamp card: 1, 2, 4, 8, 16, 32, 64, 128, each lamp worth double the one to its right — the only thing to memorise
- Reading a row by adding the lit lamps (192, 10, 77, 255, 0) and writing a number by asking "can I afford it?" eight times
- Why a subnet mask can only be one of nine values, and why the prefix length is just the count of lit lamps (240 is /28, 224 is /27)
- The block size from the last lit lamp: 240 means networks of 16, 224 means networks of 32
- The bitwise AND: 77 AND 224 is 64, so the network is 192.168.1.64/27, broadcast .95, hosts .65 to .94
- The ten-second method: count the lit lamps, take the last one's value, find the multiple at or below the host
- The real thing: Windows Calculator in programmer mode as a self-check, then a real host worked through the three steps
- Exam corner: the prefix length of 255.255.255.224
Chapters
- 0:00Thirty seconds on the clock
- 0:43The lamp card
- 1:23Reading a row: add the lit lamps
- 1:53Writing a number: "can I afford it?"
- 2:53240 into lamps
- 3:29Masks are the easy case: the nine values
- 4:03The slash is the count of lit lamps
- 4:28The last lit lamp: the street width
- 4:55The street test with the lamps: AND
- 5:42The street .64 to .95
- 6:04The ten-second version
- 6:29The real names
- 7:00The three steps
- 7:29The real thing: the calculator, then a real host
- 8:43Exam corner
- 9:00Recap and next time
Exam objectives: CompTIA Network+ N10-009: 1.7 (IPv4 subnetting: subnet masks, prefix length, CIDR), 5.2 (incorrect subnet mask), 5.5 (ipconfig).
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