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 10.78.160.0/20 is divided into equal subnets with prefix /27. How many subnets N are there?
10.78.160.0/20/27
N =
2.IPv4 subnettingHard
A router has routes to these networks: 10.78.172.0/24, 10.78.173.0/24, 10.78.174.0/24. What is the longest prefix length p of a single summary route that covers all of them?
10.78.172.0/24 … 10.78.174.0/24
p =
3.Computer networksMedium
What is the purpose of the TCP three-way handshake?
- ATo close a connection quickly
- BTo find the IP address of a host before data can be sent to it
- CTo establish a connection and synchronise sequence numbers
- DTo encrypt the data
4.Computer networksHard
Why does TCP choose an unpredictable initial sequence number for each connection?
- ATo select the best route to the destination for the segments of that connection
- BTo avoid confusion with old segments and make spoofed segments harder to inject
- CTo compress the data more efficiently
- DTo encrypt the payload of each segment
5.Number systemsMedium
Set the bits so they store −55 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?
N = −558 bit
U =
6.Number systemsHard
A system must give every one of 292 items its own binary code of the same length. What is the smallest number of bits b needed?
292
b = bit
3Solutions
1.IPv4 subnetting
Answer: N = 128
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).
Borrowed bits
27 − 20 = 7
Number of subnets
N = 27 = 128
Answer
N = 128
Each /27 subnet has 25 − 2 = 30 usable hosts.
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
172 = 10101100
174 = 10101110
common leading bits: 6
Prefix length
p = 16 + 6 = 22
Answer
p = /22
Summary route: 10.78.172.0/22 (it also covers 1 network(s) that are not in the list).
3.Computer networks
Answer: C · To establish a connection and synchronise sequence numbers
Why this is the answer
The handshake sets up a connection and agrees initial sequence numbers on both sides before data is sent.
Answer
C · To establish a connection and synchronise sequence numbers
4.Computer networks
Answer: B · To avoid confusion with old segments and make spoofed segments harder to inject
Why this is the answer
Random initial sequence numbers stop delayed segments from old connections being accepted and make it hard for off-path attackers to forge valid segments.
Answer
B · To avoid confusion with old segments and make spoofed segments harder to inject
5.Number systems
Answer: U = 201
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
55 = 0011 0111
inverted: 1100 1000
+1 → 1100 1001
Read it as unsigned
1100 1001₂ = 256 − 55 = 201
Answer
U = 201
6.Number systems
Answer: b = 9 bit
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.
Powers of two
28 = 256 < 292
29 = 512 ≥ 292
Answer
b = 9
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