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 255.255.255.252. How many usable host addresses H does it have?
255.255.255.252
H =
2.IPv4 subnettingHard
A router has routes to these networks: 10.100.49.0/24, 10.100.50.0/24, 10.100.51.0/24, 10.100.52.0/24, 10.100.53.0/24, 10.100.54.0/24, 10.100.55.0/24. What is the longest prefix length p of a single summary route that covers all of them?
10.100.49.0/24 … 10.100.55.0/24
p =
3.Computer networksMedium
Why does Wi-Fi use collision avoidance (CSMA/CA) instead of collision detection?
- ARadio signals never collide
- BA wireless station cannot reliably detect collisions while it is transmitting
- CWi-Fi only allows one device per network
- DCollision detection only works on fibre cables, not on shared wireless radio channels
4.Computer networksHard
A DNS cache poisoning attack aims to:
- ASteal Wi-Fi passwords
- BOverload a server with traffic so that it stops answering queries
- CMake a resolver store a false IP address for a domain
- DDelete DNS servers
5.Number systemsMedium
Set the bits so they store −53 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?
N = −538 bit
U =
6.Number systemsHard
An uncompressed image is 640 × 480 pixels with 8 bits per pixel. How much memory M does it take, in KiB (1 KiB = 1024 bytes)?
640 × 480 px8 bit/px1 KiB = 1024 B
M = KiB
3Solutions
1.IPv4 subnetting
Answer: H = 2
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
255.255.255.252 = /30
Usable hosts
22 − 2 = 2
Answer
H = 2
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
49 = 00110001
55 = 00110111
common leading bits: 5
Prefix length
p = 16 + 5 = 21
Answer
p = /21
Summary route: 10.100.48.0/21 (it also covers 1 network(s) that are not in the list).
3.Computer networks
Answer: B · A wireless station cannot reliably detect collisions while it is transmitting
Why this is the answer
A radio transmitter cannot hear other weak signals while sending, and some stations are hidden from each other. So 802.11 waits random backoff times and uses acknowledgements.
Answer
B · A wireless station cannot reliably detect collisions while it is transmitting
4.Computer networks
Answer: C · Make a resolver store a false IP address for a domain
Why this is the answer
Poisoning inserts forged records into a resolver's cache. DNSSEC adds signatures so resolvers can verify records.
Answer
C · Make a resolver store a false IP address for a domain
5.Number systems
Answer: U = 203
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
53 = 0011 0101
inverted: 1100 1010
+1 → 1100 1011
Read it as unsigned
1100 1011₂ = 256 − 53 = 203
Answer
U = 203
6.Number systems
Answer: M = 300 KiB
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.
Bits and bytes
640·480·8 = 2457600 bit = 307200 B
Convert to KiB
307200/1024 = 300 KiB
Answer
M = 300 KiB
ExamProblems · examproblems.com/courses/computer-networks