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 192.168.2.0/24 is divided into /25 subnets, numbered 1, 2, 3, … from the lowest address (subnet 1 starts at .0). What is the fourth octet x of the broadcast address of subnet 2?
192.168.2.0/24/25subnet 2 (subnet 1 starts at .0)
x =
2.IPv4 subnettingHard
The network 192.168.27.0/24 is split with VLSM for these LANs: A: 9 hosts, B: 34 hosts, C: 8 hosts, D: 5 hosts, E: 20 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 D?
A: 9B: 34C: 8D: 5E: 20
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
A path has 100 Mbit/s bandwidth and a 40 ms round-trip time. How much data must be in flight to keep it fully used?
- A2.5 MB
- B4 kB
- C50 kB
- D500 kB
5.Number systemsMedium
Set the bits so they store −56 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?
N = −568 bit
U =
6.Number systemsHard
An uncompressed image is 1920 × 1080 pixels with 8 bits per pixel. How much memory M does it take, in KiB (1 KiB = 1024 bytes)?
1920 × 1080 px8 bit/px1 KiB = 1024 B
M = KiB
3Solutions
1.IPv4 subnetting
Answer: x = 255
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
27 = 128
Subnet number 2
network: (2 − 1)·128 = 128
broadcast: 128 + 128 − 1 = 255
Answer
x = 255
2.IPv4 subnetting
Answer: x = 128
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: 34 hosts → 26 = 64 (/26)
E: 20 hosts → 25 = 32 (/27)
A: 9 hosts → 24 = 16 (/28)
C: 8 hosts → 24 = 16 (/28)
D: 5 hosts → 23 = 8 (/29)
Allocation, largest first
B: .0 – .63
E: .64 – .95
A: .96 – .111
C: .112 – .127
D: .128 – .135
Answer
x = 128
LAN D: 192.168.27.128/29.
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: D · 500 kB
Why this is the answer
The bandwidth-delay product is 100 × 10⁶ bit/s × 0.04 s = 4 × 10⁶ bits = 500 kB. The sender's window must be at least this large.
Answer
D · 500 kB
5.Number systems
Answer: U = 200
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
56 = 0011 1000
inverted: 1100 0111
+1 → 1100 1000
Read it as unsigned
1100 1000₂ = 256 − 56 = 200
Answer
U = 200
6.Number systems
Answer: M = 2025 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
1920·1080·8 = 16588800 bit = 2073600 B
Convert to KiB
2073600/1024 = 2025 KiB
Answer
M = 2025 KiB
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