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 1340 host addresses in a single subnet. What is the longest prefix length p (smallest subnet) that is big enough?
1340 hosts
p =
2.IPv4 subnettingHard
A host has the address 10.20.134.191/28. What is the fourth octet x of the last usable host address in its subnet?
10.20.134.191/28
x =
3.Computer networksMedium
Which medium access method did classic shared (half-duplex) Ethernet use?
- AToken passing
- BCSMA/CD
- CTDMA
- DCSMA/CA
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
- BHTTP/2 uses UDP
- CEach request needs its own new TCP connection, so later requests wait in a queue
- DServers cannot handle parallel requests
5.Number systemsMedium
The 8-bit pattern 1111 1010 stores a signed integer in two's complement. What is its value N in decimal?
1111 10108 bit
N =
6.Number systemsHard
An uncompressed image is 640 × 480 pixels with 24 bits per pixel. How much memory M does it take, in KiB (1 KiB = 1024 bytes)?
640 × 480 px24 bit/px1 KiB = 1024 B
M = KiB
3Solutions
1.IPv4 subnetting
Answer: p = 21
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
210 − 2 = 1022 < 1340
211 − 2 = 2046 ≥ 1340 → h = 11
Prefix length
p = 32 − 11 = 21
Answer
p = /21
2.IPv4 subnetting
Answer: x = 190
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
24 = 16
Subnet of this host
network: ⌊191/16⌋·16 = 176
broadcast: 176 + 16 − 1 = 191
usable hosts: 177 … 190
Answer
x = 190
3.Computer networks
Answer: B · CSMA/CD
Why this is the answer
Carrier sense multiple access with collision detection: stations stop when they detect a collision and retry after a random backoff. Full-duplex switched Ethernet has no collisions.
Answer
B · CSMA/CD
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: N = −6
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.
Add the place values
The other bits keep their normal positive values.
−128 + 64 + 32 + 16 + 8 + 2 = −6
Answer
N = −6
6.Number systems
Answer: M = 900 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
640·480·24 = 7372800 bit = 921600 B
Convert to KiB
921600/1024 = 900 KiB
Answer
M = 900 KiB
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