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 department needs 814 host addresses in a single subnet. What is the longest prefix length p (smallest subnet) that is big enough?
814 hosts
p =
2.IPv4 subnettingHard
A router has routes to these networks: 10.244.229.0/24, 10.244.230.0/24. What is the longest prefix length p of a single summary route that covers all of them?
10.244.229.0/24 … 10.244.230.0/24
p =
3.Computer networksMedium
Which TCP flag signals that a sender has finished sending data?
- ASYN
- BPSH
- CFIN
- DURG
4.Computer networksHard
A DNS cache poisoning attack aims to:
- ADelete DNS servers
- BSteal Wi-Fi passwords
- COverload a server with traffic so that it stops answering queries
- DMake a resolver store a false IP address for a domain
5.Number systemsMedium
Set the bits so they store −97 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?
N = −978 bit
U =
6.Number systemsHard
An uncompressed image is 1280 × 720 pixels with 32 bits per pixel. How much memory M does it take, in KiB (1 KiB = 1024 bytes)?
1280 × 720 px32 bit/px1 KiB = 1024 B
M = KiB
3Solutions
1.IPv4 subnetting
Answer: p = 22
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).
Host bits needed
29 − 2 = 510 < 814
210 − 2 = 1022 ≥ 814 → h = 10
Prefix length
p = 32 − 10 = 22
Answer
p = /22
2.IPv4 subnetting
Answer: p = 22
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
229 = 11100101
230 = 11100110
common leading bits: 6
Prefix length
p = 16 + 6 = 22
Answer
p = /22
Summary route: 10.244.228.0/22 (it also covers 2 network(s) that are not in the list).
3.Computer networks
Answer: C · FIN
Why this is the answer
Each side sends FIN when it has no more data, and the other side acknowledges it. SYN is for opening a connection; RST aborts one.
Answer
C · FIN
4.Computer networks
Answer: D · 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
D · Make a resolver store a false IP address for a domain
5.Number systems
Answer: U = 159
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
97 = 0110 0001
inverted: 1001 1110
+1 → 1001 1111
Read it as unsigned
1001 1111₂ = 256 − 97 = 159
Answer
U = 159
6.Number systems
Answer: M = 3600 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
1280·720·32 = 29491200 bit = 3686400 B
Convert to KiB
3686400/1024 = 3600 KiB
Answer
M = 3600 KiB
ExamProblems · examproblems.com/courses/computer-networks