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
The network 192.168.10.0/24 must be divided into at least 8 equal subnets. If you borrow as few bits as possible, how many usable hosts H does each subnet have?
192.168.10.0/248 subnets
H =
2.IPv4 subnettingHard
A host has the address 10.20.181.47/26. What is the last number (fourth octet) x of its subnet's broadcast address?
10.20.181.47/26
x =
3.Computer networksMedium
Which subnet mask corresponds to /20?
- A255.255.224.0
- B255.255.248.0
- C255.255.255.240
- D255.255.240.0
4.Computer networksHard
An 8-port switch (no VLANs) has one host on each port. How many collision domains and broadcast domains are there?
- A8 collision domains, 1 broadcast domain
- B1 collision domain, 1 broadcast domain
- C8 collision domains, 8 broadcast domains
- D1 collision domain, 8 broadcast domains
5.Number systemsMedium
Set the bits so they store −59 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?
N = −598 bit
U =
6.Number systemsHard
What is the most negative integer N that can be stored in 20-bit two's complement?
20 bit
N =
3Solutions
1.IPv4 subnetting
Answer: H = 30
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).
Bits to borrow
23 = 8 ≥ 8 → b = 3, /27
Hosts per subnet
25 − 2 = 30
Answer
H = 30
2.IPv4 subnetting
Answer: x = 63
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).
Block size
26 = 64
Subnet of this host
network: ⌊47/64⌋·64 = 0
broadcast: 0 + 64 − 1 = 63
usable hosts: 1 … 62
Answer
x = 63
3.Computer networks
Answer: D · 255.255.240.0
Why this is the answer
/20 means 8 + 8 + 4 one-bits. Four one-bits in the third octet are 11110000 = 240, so the mask is 255.255.240.0.
Answer
D · 255.255.240.0
4.Computer networks
Answer: A · 8 collision domains, 1 broadcast domain
Why this is the answer
Each switch port is its own collision domain, but broadcasts are forwarded to all ports, so the whole switch is one broadcast domain. A hub would give one of each.
Answer
A · 8 collision domains, 1 broadcast domain
5.Number systems
Answer: U = 197
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
59 = 0011 1011
inverted: 1100 0100
+1 → 1100 0101
Read it as unsigned
1100 0101₂ = 256 − 59 = 197
Answer
U = 197
6.Number systems
Answer: N = −524288
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.
Most negative value
−219 = −524288
Answer
N = −524288
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