3.2 IPv4 Dotted Decimal and Binary Basics
Key Takeaways
- An IPv4 address is 32 bits written as four 8-bit octets, each ranging 0–255; 256 or higher in any octet is invalid.
- Each bit position in an octet has a place value (128 64 32 16 8 4 2 1) that sums to the decimal value.
- The subnet mask's 1 bits mark network bits and 0 bits mark host bits; the address alone is meaningless without it.
- The all-zero host portion is the network address and the all-one host portion is the broadcast; hosts use neither.
Reading IPv4 addresses correctly
Cisco expects 100-150 candidates to identify IPv4 addresses and subnet formats. An IPv4 address is a 32-bit value humans write in dotted-decimal notation: four decimal numbers separated by periods, such as 192.168.10.25. Each number is an octet — 8 bits. Eight bits represent 256 distinct values, so each octet ranges 0 through 255. 192.168.10.300 is invalid because 300 cannot fit in 8 bits; 192.168.1.256 fails for the same reason (the maximum is 255).
Dotted decimal is only a display format. Devices make decisions on bits. Each octet's bits carry fixed place values:
| Bit position | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |
|---|---|---|---|---|---|---|---|---|
| Place value | 128 | 64 | 32 | 16 | 8 | 4 | 2 | 1 |
To convert, add the place values wherever a bit is 1. 192 = 11000000 (128 + 64). 168 = 10101000 (128 + 32 + 8). 10 = 00001010 (8 + 2). 25 = 00011001 (16 + 8 + 1). You will not hand-convert every address during support work, but seeing the bit basis makes masks and CIDR far less mysterious — and the exam may show a binary octet and ask for its decimal value.
A subnet mask is also 32 bits. In dotted decimal you will see masks such as 255.255.255.0, 255.255.0.0, and 255.255.255.128. The mask's binary 1 bits mark the network portion; the 0 bits mark the host portion. With address 192.168.10.25 and mask 255.255.255.0, the first 24 bits are network, so the network is 192.168.10.0, and the last octet .25 is a host on that subnet.
Why the address alone is incomplete
An address with no mask is ambiguous. 10.1.2.3 could belong to 10.1.2.0/24, 10.1.0.0/16, 10.0.0.0/8, or a smaller custom subnet — the mask decides. That is exactly why every IP configuration screen pairs four fields together:
| Field | Purpose |
|---|---|
| IP address | The host's Layer 3 identity |
| Subnet mask / prefix | Splits network bits from host bits |
| Default gateway | Router that reaches other subnets |
| DNS server | Resolves names to addresses |
If a user reports only an address, ask for the mask or prefix before deciding whether two hosts are local to each other.
Certain host values are reserved inside a subnet. The all-zero host portion is the network address, and the all-one host portion is the directed broadcast address. In 192.168.10.0/24, 192.168.10.0 is the network, 192.168.10.255 is the broadcast, and usable hosts run 192.168.10.1 through 192.168.10.254. A host configured with the network or broadcast address will not behave normally.
A practical troubleshooting habit
Compare each address against the mask. A printer at 192.168.1.50/24 and a laptop at 192.168.2.60/24 differ in the third octet, which is inside the /24 network portion, so they are on different networks — the laptop needs a router to reach the printer. Same-switch cabling cannot fix that Layer 3 separation.
Watch for subtle data-entry faults that still look valid:
- A transposed digit (
192.168.21.5instead of192.168.12.5) drops a device into the wrong subnet. - A mask typed as
255.255.0.0instead of255.255.255.0makes a host believe far more destinations are local, so it floods ARP requests instead of forwarding to the gateway. - A wrong DHCP scope can push the same bad mask to dozens of clients at once.
Visual-similarity traps to memorize: 172.15.10.5 is public but 172.16.10.5 is private; 192.169.1.10 is public while 192.168.1.10 is private; and 255.255.255.0 is a mask, never a host address. When documenting a ticket, always record the exact address, the mask or CIDR prefix, the gateway, and whether the value came from DHCP or static assignment.
Converting between decimal and binary under exam pressure
The exam may hand you a binary octet and ask for its decimal value, or vice versa, so practice the place-value method until it is automatic. Write the eight place values across the top — 128, 64, 32, 16, 8, 4, 2, 1 — then place a 1 under each value you need and add. To turn 200 into binary, subtract the largest place value that fits: 200 − 128 = 72, so bit 1 is on; 72 − 64 = 8, so bit 2 is on; 8 fits the 8-place, so that bit is on; the rest are 0. The result is 11001000. Going the other direction, 10101010 = 128 + 32 + 8 + 2 = 170.
This skill also clarifies why subnet masks look the way they do. A valid mask is always a run of contiguous 1 bits followed by contiguous 0 bits — 11111111.11111111.11111111.00000000 for /24. You will never see a legitimate mask like 255.255.255.5, because 5 is 00000101, which interleaves 1s and 0s. Spotting an illegal, non-contiguous mask is a quick way to flag a typo on a configuration screen before it causes intermittent failures.
Finally, connect the binary view back to the four-field intake habit. When a host has address 10.1.2.3 and mask 255.255.0.0, the 1 bits cover the first sixteen positions, so only the first two octets are network — the host is on 10.1.0.0, not 10.1.2.0. Reading masks in binary, even mentally, prevents the most common Layer 3 mistake a junior technician makes: assuming the boundary always falls on the dot between octets. CIDR and masks only feel arbitrary until you see the bits underneath them, which is the entire point of this objective.
Why is 192.168.1.256 not a valid IPv4 address?
With IPv4 address 192.168.10.25 and mask 255.255.255.0, what is the network address?
The binary octet 11000000 equals which decimal value?