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 1964 host addresses in a single subnet. What is the longest prefix length p (smallest subnet) that is big enough?
1964 hosts
p =
2.IPv4 subnettingHard
The network 192.168.73.0/24 is split with VLSM for these LANs: A: 6 hosts, B: 15 hosts, C: 18 hosts, D: 12 hosts, E: 60 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 C?
A: 6B: 15C: 18D: 12E: 60
x =
3.Computer networksMedium
Why is UDP often preferred for live video calls?
- ALow delay matters more than retransmitting every lost packet
- BUDP encrypts data automatically
- CUDP guarantees every packet arrives, so the picture never freezes
- DUDP is required for all video
4.Computer networksHard
Which mechanism prevents broadcast loops in a network with redundant switch links?
- ASpanning Tree Protocol
- BARP caching
- CDHCP snooping
- DNAT
5.Number systemsMedium
The 8-bit pattern 1101 1111 stores a signed integer in two's complement. What is its value N in decimal?
1101 11118 bit
N =
6.Number systemsHard
What is the most negative integer N that can be stored in 10-bit two's complement?
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 < 1964
211 − 2 = 2046 ≥ 1964 → h = 11
Prefix length
p = 32 − 11 = 21
Answer
p = /21
2.IPv4 subnetting
Answer: x = 64
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
E: 60 hosts → 26 = 64 (/26)
C: 18 hosts → 25 = 32 (/27)
B: 15 hosts → 25 = 32 (/27)
D: 12 hosts → 24 = 16 (/28)
A: 6 hosts → 23 = 8 (/29)
Allocation, largest first
E: .0 – .63
C: .64 – .95
B: .96 – .127
D: .128 – .143
A: .144 – .151
Answer
x = 64
LAN C: 192.168.73.64/27.
3.Computer networks
Answer: A · 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
A · Low delay matters more than retransmitting every lost packet
4.Computer networks
Answer: A · Spanning Tree Protocol
Why this is the answer
STP elects a root bridge and blocks redundant paths, unblocking them if an active link fails.
Answer
A · Spanning Tree Protocol
5.Number systems
Answer: N = −33
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 + 16 + 8 + 4 + 2 + 1 = −33
Answer
N = −33
6.Number systems
Answer: N = −512
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.
Most negative value
−29 = −512
Answer
N = −512
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