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 3 equal subnets. If you borrow as few bits as possible, how many usable hosts H does each subnet have?
192.168.10.0/243 subnets
H =
2.IPv4 subnettingHard
A router has routes to these networks: 10.54.119.0/24, 10.54.120.0/24. What is the longest prefix length p of a single summary route that covers all of them?
10.54.119.0/24 … 10.54.120.0/24
p =
3.Computer networksMedium
How many usable host addresses does a /26 IPv4 subnet have?
- A62
- B126
- C64
- D30
4.Computer networksHard
DNS queries normally use UDP. When does DNS typically use TCP instead?
- AFor every query sent from a web browser
- BFor responses too large for UDP and for zone transfers
- COnly when the domain name is longer than ten characters
- DOnly for reverse lookups of IP addresses
5.Number systemsMedium
Write −68 as an 8-bit two's complement pattern. Read the pattern as an unsigned binary number: what is its value U in decimal?
N = −688 bit
U =
6.Number systemsHard
Set the bits to the 8-bit result of 1101 0001 XOR 0011 1000. What is the result's value R in decimal?
1101 0001XOR0011 1000
R =
3Solutions
1.IPv4 subnetting
Answer: H = 62
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
22 = 4 ≥ 3 → b = 2, /26
Hosts per subnet
26 − 2 = 62
Answer
H = 62
2.IPv4 subnetting
Answer: p = 20
Route summarisation: a summary route a.b.c.0/p covers all networks whose first p bits match. Write the differing octet of the first and last network in binary; the number of common leading bits, plus the bits of the octets before it, is the longest summary prefix.
Compare the bits
119 = 01110111
120 = 01111000
common leading bits: 4
Prefix length
p = 16 + 4 = 20
Answer
p = /20
Summary route: 10.54.112.0/20 (it also covers 14 network(s) that are not in the list).
3.Computer networks
Answer: A · 62
Why this is the answer
A /26 has 6 host bits: 2⁶ = 64 addresses, minus the network and broadcast addresses, gives 62 usable hosts.
Answer
A · 62
4.Computer networks
Answer: B · For responses too large for UDP and for zone transfers
Why this is the answer
If a UDP response is truncated, the resolver retries over TCP. Zone transfers between name servers also use TCP because they carry lots of data.
Answer
B · For responses too large for UDP and for zone transfers
5.Number systems
Answer: U = 188
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.
Invert and add one
68 = 0100 0100
inverted: 1011 1011
+1 → 1011 1100
Read it as unsigned
1011 1100₂ = 256 − 68 = 188
Answer
U = 188
6.Number systems
Answer: R = 233
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.
Bit by bit
XOR gives 1 where the two bits differ.
1101 0001
XOR 0011 1000
= 1110 1001
Value of the result
1110 1001₂ = 233
Answer
R = 233
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