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 607 host addresses in a single subnet. What is the longest prefix length p (smallest subnet) that is big enough?
607 hosts
p =
2.IPv4 subnettingHard
A router has routes to these networks: 10.163.178.0/24, 10.163.179.0/24, 10.163.180.0/24. What is the longest prefix length p of a single summary route that covers all of them?
10.163.178.0/24 … 10.163.180.0/24
p =
3.Computer networksMedium
What does ARP do?
- AAssigns IP addresses automatically to hosts when they join the local network
- BRoutes packets on the internet
- CFinds the MAC address that belongs to an IP address on the local network
- DResolves domain names
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
- B8 collision domains, 8 broadcast domains
- C1 collision domain, 1 broadcast domain
- D1 collision domain, 8 broadcast domains
5.Number systemsMedium
Write −37 as an 8-bit two's complement pattern. Read the pattern as an unsigned binary number: what is its value U in decimal?
N = −378 bit
U =
6.Number systemsHard
An uncompressed image is 1280 × 720 pixels with 16 bits per pixel. How much memory M does it take, in KiB (1 KiB = 1024 bytes)?
1280 × 720 px16 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 < 607
210 − 2 = 1022 ≥ 607 → h = 10
Prefix length
p = 32 − 10 = 22
Answer
p = /22
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
178 = 10110010
180 = 10110100
common leading bits: 5
Prefix length
p = 16 + 5 = 21
Answer
p = /21
Summary route: 10.163.176.0/21 (it also covers 5 network(s) that are not in the list).
3.Computer networks
Answer: C · Finds the MAC address that belongs to an IP address on the local network
Why this is the answer
ARP broadcasts "who has this IP?" and the owner replies with its MAC address, so frames can be delivered locally.
Answer
C · Finds the MAC address that belongs to an IP address on the local network
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 = 219
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
37 = 0010 0101
inverted: 1101 1010
+1 → 1101 1011
Read it as unsigned
1101 1011₂ = 256 − 37 = 219
Answer
U = 219
6.Number systems
Answer: M = 1800 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
1280·720·16 = 14745600 bit = 1843200 B
Convert to KiB
1843200/1024 = 1800 KiB
Answer
M = 1800 KiB
ExamProblems · examproblems.com/courses/computer-networks