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 1872 host addresses in a single subnet. What is the longest prefix length p (smallest subnet) that is big enough?
1872 hosts
p =
2.IPv4 subnettingHard
The network 192.168.226.0/24 is split with VLSM for these LANs: A: 55 hosts, B: 19 hosts, C: 14 hosts, D: 70 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 B?
A: 55B: 19C: 14D: 70
x =
3.Computer networksMedium
Why is UDP often preferred for live video calls?
- AUDP guarantees every packet arrives, so the picture never freezes
- BLow delay matters more than retransmitting every lost packet
- CUDP encrypts data automatically
- DUDP is required for all video
4.Computer networksHard
What is the main benefit of persistent connections (keep-alive) in HTTP/1.1?
- ASeveral requests reuse one TCP connection, avoiding repeated handshakes
- BThe connection stays encrypted without TLS
- CThe server can push updates over UDP
- DResponses can arrive in any order on one stream, so slow replies do not block others
5.Number systemsMedium
Set the bits so they store −83 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?
N = −838 bit
U =
6.Number systemsHard
What is the largest unsigned integer N that fits in 10 bits?
10 bit
N =
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 < 1872
211 − 2 = 2046 ≥ 1872 → h = 11
Prefix length
p = 32 − 11 = 21
Answer
p = /21
2.IPv4 subnetting
Answer: x = 192
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
D: 70 hosts → 27 = 128 (/25)
A: 55 hosts → 26 = 64 (/26)
B: 19 hosts → 25 = 32 (/27)
C: 14 hosts → 24 = 16 (/28)
Allocation, largest first
D: .0 – .127
A: .128 – .191
B: .192 – .223
C: .224 – .239
Answer
x = 192
LAN B: 192.168.226.192/27.
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: A · Several requests reuse one TCP connection, avoiding repeated handshakes
Why this is the answer
Reusing a connection avoids a new handshake and slow start for each object, so pages with many resources load faster.
Answer
A · Several requests reuse one TCP connection, avoiding repeated handshakes
5.Number systems
Answer: U = 173
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
83 = 0101 0011
inverted: 1010 1100
+1 → 1010 1101
Read it as unsigned
1010 1101₂ = 256 − 83 = 173
Answer
U = 173
6.Number systems
Answer: N = 1023
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.
Largest value
210 − 1 = 1023
Answer
N = 1023
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