IPv4 subnetting
- Hosts per subnet
- 2(32 − prefix) − 2 (/31: 2, /32: 1)
- Block size
- 256 − (mask octet)
- Network address
- IP AND mask
- Broadcast address
- IP OR (NOT mask): all host bits set to 1
- Subnets from n borrowed bits
- 2n
Computer Networks: cram pack
Formula sheets, the key method steps for every topic and 6 practice problems, with full solutions at the back.
1Formulas and methods
Number systems
- Place values
- binary 2k, octal 8k, hex 16k
- To decimal
- Σ dk·bk
- From decimal
- divide by the base, read the remainders bottom-up
- Hex ↔ binary
- one hex digit = 4 bits
- Two’s complement
- invert the bits and add 1
2Practice problems
1.IPv4 subnettingMedium
A subnet uses the mask 255.255.252.0. How many usable host addresses H does it have?
255.255.252.0
H =
2.IPv4 subnettingHard
The network 192.168.154.0/24 is split with VLSM for these LANs: A: 71 hosts, B: 11 hosts, C: 51 hosts, D: 6 hosts. Subnets are allocated from .0 upwards, the LAN with the most hosts first, each with the smallest subnet that fits. What is the fourth octet x of the network address of LAN C?
A: 71B: 11C: 51D: 6
x =
3.Computer networksMedium
How many usable host addresses does a /24 IPv4 network have?
- A255
- B256
- C254
- D252
4.Computer networksHard
Which of these addresses is a public (not RFC 1918 private) IPv4 address?
- A192.168.255.1
- B172.31.0.1
- C10.255.0.1
- D172.32.0.1
5.Number systemsMedium
Set the bits so they store −11 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?
N = −118 bit
U =
6.Number systemsHard
What is the largest unsigned integer N that fits in 24 bits?
24 bit
N =
3Solutions
1.IPv4 subnetting
Answer: H = 1022
IPv4 subnetting: an address has 32 bits. The prefix /p says how many leading bits identify the network; the remaining 32 − p bits number the hosts. A subnet has 2³²⁻ᵖ addresses, but the first (network address) and last (broadcast address) cannot be given to hosts, so 2³²⁻ᵖ − 2 hosts fit. Exceptions: a /31 (point-to-point link, RFC 3021) uses both of its 2 addresses for hosts, and a /32 is one single address (a single host or a loopback).
Prefix length
255.255.252.0 = /22
Usable hosts
210 − 2 = 1022
Answer
H = 1022
2.IPv4 subnetting
Answer: x = 128
VLSM: each LAN gets the smallest subnet with 2ʰ − 2 ≥ its host count, i.e. prefix 32 − h. Allocating the largest subnets first keeps every block aligned on a multiple of its own size, so the network addresses follow by adding up the block sizes.
Block size per LAN
A: 71 hosts → 27 = 128 (/25)
C: 51 hosts → 26 = 64 (/26)
B: 11 hosts → 24 = 16 (/28)
D: 6 hosts → 23 = 8 (/29)
Allocation, largest first
A: .0 – .127
C: .128 – .191
B: .192 – .207
D: .208 – .215
Answer
x = 128
LAN C: 192.168.154.128/26.
3.Computer networks
Answer: C · 254
Why this is the answer
A /24 has 8 host bits, so 2⁸ = 256 addresses. The network and broadcast addresses are reserved, leaving 254 for hosts.
Answer
C · 254
4.Computer networks
Answer: D · 172.32.0.1
Why this is the answer
172.16.0.0/12 runs from 172.16.0.0 to 172.31.255.255, so 172.32.0.1 lies outside it. The other three are in private ranges.
Answer
D · 172.32.0.1
5.Number systems
Answer: U = 245
Setting bits: each bit is worth a power of two (128, 64, …, 1). One hex digit is exactly four bits, and in 8-bit two's complement the leftmost bit is worth −128. Bitwise operations work on each bit position separately.
Invert and add one
11 = 0000 1011
inverted: 1111 0100
+1 → 1111 0101
Read it as unsigned
1111 0101₂ = 256 − 11 = 245
Answer
U = 245
6.Number systems
Answer: N = 1.678 × 10⁷
Positional number systems: each digit is worth its value times a power of the base. In binary the places are worth 1, 2, 4, 8, 16, …; in hexadecimal 1, 16, 256, …, with A–F standing for 10–15. In 8-bit two's complement the leftmost bit is worth −128 instead of +128, which is how negative numbers are stored.
Largest value
224 − 1 = 16777215
Answer
N = 16777215
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