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 192.168.82.0/24 is divided into /27 subnets, numbered 1, 2, 3, … from the lowest address (subnet 1 starts at .0). What is the fourth octet x of the network address of subnet 2?
192.168.82.0/24/27subnet 2 (subnet 1 starts at .0)
x =
2.IPv4 subnettingHard
A router has routes to these networks: 10.132.136.0/24, 10.132.137.0/24, 10.132.138.0/24, 10.132.139.0/24, 10.132.140.0/24, 10.132.141.0/24, 10.132.142.0/24, 10.132.143.0/24. What is the longest prefix length p of a single summary route that covers all of them?
10.132.136.0/24 … 10.132.143.0/24
p =
3.Computer networksMedium
What does an MX record in DNS specify?
- AThe name servers for a zone
- BThe IPv4 address of a web server
- CThe mail servers that accept email for a domain
- DAn alias name for another host
4.Computer networksHard
DNS queries normally use UDP. When does DNS typically use TCP instead?
- AOnly for reverse lookups of IP addresses
- BFor responses too large for UDP and for zone transfers
- CFor every query sent from a web browser
- DOnly when the domain name is longer than ten characters
5.Number systemsMedium
Write −33 as an 8-bit two's complement pattern. Read the pattern as an unsigned binary number: what is its value U in decimal?
N = −338 bit
U =
6.Number systemsHard
What is the most negative integer N that can be stored in 16-bit two's complement?
16 bit
N =
3Solutions
1.IPv4 subnetting
Answer: x = 32
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
25 = 32
Subnet number 2
network: (2 − 1)·32 = 32
broadcast: 32 + 32 − 1 = 63
Answer
x = 32
2.IPv4 subnetting
Answer: p = 21
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
136 = 10001000
143 = 10001111
common leading bits: 5
Prefix length
p = 16 + 5 = 21
Answer
p = /21
Summary route: 10.132.136.0/21
3.Computer networks
Answer: C · The mail servers that accept email for a domain
Why this is the answer
MX (mail exchange) records list mail servers with priorities, so sending servers know where to deliver mail for that domain.
Answer
C · The mail servers that accept email for a domain
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 = 223
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
33 = 0010 0001
inverted: 1101 1110
+1 → 1101 1111
Read it as unsigned
1101 1111₂ = 256 − 33 = 223
Answer
U = 223
6.Number systems
Answer: N = −32768
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
−215 = −32768
Answer
N = −32768
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