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 10.39.225.0/24 is divided into equal subnets with prefix /30. How many subnets N are there?
10.39.225.0/24/30
N =
2.IPv4 subnettingHard
A host has the address 10.20.104.170/30. What is the last number (fourth octet) x of its subnet's broadcast address?
10.20.104.170/30
x =
3.Computer networksMedium
TLS provides which security properties for HTTPS?
- AProtection against all malware and phishing on the visited site
- BConfidentiality, integrity and server authentication
- CAnonymity of the user
- DOnly faster loading
4.Computer networksHard
Head-of-line blocking in HTTP/2 over TCP occurs because:
- AA lost TCP segment delays all multiplexed streams until it is retransmitted
- BEach request needs its own new TCP connection, so later requests wait in a queue
- CServers cannot handle parallel requests
- DHTTP/2 uses UDP
5.Number systemsMedium
Set the bits so they store −58 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?
N = −588 bit
U =
6.Number systemsHard
Set the bits to the 8-bit result of 0011 1000 OR 1110 1110. What is the result's value R in decimal?
0011 1000OR1110 1110
R =
3Solutions
1.IPv4 subnetting
Answer: N = 64
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
30 − 24 = 6
Number of subnets
N = 26 = 64
Answer
N = 64
Each /30 subnet has 22 − 2 = 2 usable hosts.
2.IPv4 subnetting
Answer: x = 171
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
22 = 4
Subnet of this host
network: ⌊170/4⌋·4 = 168
broadcast: 168 + 4 − 1 = 171
usable hosts: 169 … 170
Answer
x = 171
3.Computer networks
Answer: B · Confidentiality, integrity and server authentication
Why this is the answer
TLS encrypts traffic, detects tampering and authenticates the server with certificates (optionally the client too).
Answer
B · Confidentiality, integrity and server authentication
4.Computer networks
Answer: A · A lost TCP segment delays all multiplexed streams until it is retransmitted
Why this is the answer
TCP delivers bytes in order, so one loss stalls every stream on the connection. QUIC (HTTP/3) handles streams independently over UDP.
Answer
A · A lost TCP segment delays all multiplexed streams until it is retransmitted
5.Number systems
Answer: U = 198
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
58 = 0011 1010
inverted: 1100 0101
+1 → 1100 0110
Read it as unsigned
1100 0110₂ = 256 − 58 = 198
Answer
U = 198
6.Number systems
Answer: R = 254
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.
Bit by bit
OR gives 1 where at least one bit is 1.
0011 1000
OR 1110 1110
= 1111 1110
Value of the result
1111 1110₂ = 254
Answer
R = 254
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