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 795 host addresses in a single subnet. What is the longest prefix length p (smallest subnet) that is big enough?
795 hosts
p =
2.IPv4 subnettingHard
The network 192.168.55.0/24 is split with VLSM for these LANs: A: 9 hosts, B: 44 hosts, C: 4 hosts, D: 15 hosts, E: 13 hosts. Subnets are allocated from .0 upwards, the LAN with the most hosts first, each with the smallest subnet that fits. What is the fourth octet x of the network address of LAN A?
A: 9B: 44C: 4D: 15E: 13
x =
3.Computer networksMedium
Why is UDP often preferred for live video calls?
- AUDP encrypts data automatically
- BLow delay matters more than retransmitting every lost packet
- CUDP is required for all video
- DUDP guarantees every packet arrives, so the picture never freezes
4.Computer networksHard
In a SYN flood attack, what resource is the attacker trying to exhaust?
- AThe server's disk space
- BThe switch's MAC address table
- CThe server's table of half-open connections
- DThe DNS cache of the client
5.Number systemsMedium
Set the bits so they store −114 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?
N = −1148 bit
U =
6.Number systemsHard
An uncompressed image is 512 × 512 pixels with 8 bits per pixel. How much memory M does it take, in KiB (1 KiB = 1024 bytes)?
512 × 512 px8 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 < 795
210 − 2 = 1022 ≥ 795 → h = 10
Prefix length
p = 32 − 10 = 22
Answer
p = /22
2.IPv4 subnetting
Answer: x = 112
VLSM: each LAN gets the smallest subnet with 2ʰ − 2 ≥ its host count, i.e. prefix 32 − h. Allocating the largest subnets first keeps every block aligned on a multiple of its own size, so the network addresses follow by adding up the block sizes.
Block size per LAN
B: 44 hosts → 26 = 64 (/26)
D: 15 hosts → 25 = 32 (/27)
E: 13 hosts → 24 = 16 (/28)
A: 9 hosts → 24 = 16 (/28)
C: 4 hosts → 23 = 8 (/29)
Allocation, largest first
B: .0 – .63
D: .64 – .95
E: .96 – .111
A: .112 – .127
C: .128 – .135
Answer
x = 112
LAN A: 192.168.55.112/28.
3.Computer networks
Answer: B · Low delay matters more than retransmitting every lost packet
Why this is the answer
Retransmitting lost packets would add delay; real-time media tolerates some loss, so UDP-based protocols like RTP are used.
Answer
B · Low delay matters more than retransmitting every lost packet
4.Computer networks
Answer: C · The server's table of half-open connections
Why this is the answer
Each SYN makes the server reserve state and wait for an ACK that never comes. SYN cookies avoid storing state until the handshake completes.
Answer
C · The server's table of half-open connections
5.Number systems
Answer: U = 142
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
114 = 0111 0010
inverted: 1000 1101
+1 → 1000 1110
Read it as unsigned
1000 1110₂ = 256 − 114 = 142
Answer
U = 142
6.Number systems
Answer: M = 256 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
512·512·8 = 2097152 bit = 262144 B
Convert to KiB
262144/1024 = 256 KiB
Answer
M = 256 KiB
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