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 192.168.10.0/24 must be divided into at least 16 equal subnets. If you borrow as few bits as possible, how many usable hosts H does each subnet have?
192.168.10.0/2416 subnets
H =
2.IPv4 subnettingHard
The network 192.168.94.0/24 is split with VLSM for these LANs: A: 52 hosts, B: 11 hosts, C: 50 hosts, D: 28 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: 52B: 11C: 50D: 28
x =
3.Computer networksMedium
Why does Wi-Fi use collision avoidance (CSMA/CA) instead of collision detection?
- ARadio signals never collide
- BWi-Fi only allows one device per network
- CCollision detection only works on fibre cables, not on shared wireless radio channels
- DA wireless station cannot reliably detect collisions while it is transmitting
4.Computer networksHard
Which mechanism prevents broadcast loops in a network with redundant switch links?
- ANAT
- BARP caching
- CSpanning Tree Protocol
- DDHCP snooping
5.Number systemsMedium
Write −68 as an 8-bit two's complement pattern. Read the pattern as an unsigned binary number: what is its value U in decimal?
N = −688 bit
U =
6.Number systemsHard
Set the bits to the 8-bit result of 1111 1000 AND 0010 1010. What is the result's value R in decimal?
1111 1000AND0010 1010
R =
3Solutions
1.IPv4 subnetting
Answer: H = 14
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).
Bits to borrow
24 = 16 ≥ 16 → b = 4, /28
Hosts per subnet
24 − 2 = 14
Answer
H = 14
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
A: 52 hosts → 26 = 64 (/26)
C: 50 hosts → 26 = 64 (/26)
D: 28 hosts → 25 = 32 (/27)
B: 11 hosts → 24 = 16 (/28)
Allocation, largest first
A: .0 – .63
C: .64 – .127
D: .128 – .159
B: .160 – .175
Answer
x = 64
LAN C: 192.168.94.64/26.
3.Computer networks
Answer: D · A wireless station cannot reliably detect collisions while it is transmitting
Why this is the answer
A radio transmitter cannot hear other weak signals while sending, and some stations are hidden from each other. So 802.11 waits random backoff times and uses acknowledgements.
Answer
D · A wireless station cannot reliably detect collisions while it is transmitting
4.Computer networks
Answer: C · 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
C · Spanning Tree Protocol
5.Number systems
Answer: U = 188
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.
Invert and add one
68 = 0100 0100
inverted: 1011 1011
+1 → 1011 1100
Read it as unsigned
1011 1100₂ = 256 − 68 = 188
Answer
U = 188
6.Number systems
Answer: R = 40
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
AND gives 1 only where both bits are 1.
1111 1000
AND 0010 1010
= 0010 1000
Value of the result
0010 1000₂ = 40
Answer
R = 40
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