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 1790 host addresses in a single subnet. What is the longest prefix length p (smallest subnet) that is big enough?
1790 hosts
p =
2.IPv4 subnettingHard
The network 192.168.7.0/24 is split with VLSM for these LANs: A: 7 hosts, B: 86 hosts, C: 55 hosts, D: 3 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: 7B: 86C: 55D: 3
x =
3.Computer networksMedium
What characterises a distributed denial-of-service (DDoS) attack?
- AAn attacker reads unencrypted Wi-Fi traffic from a nearby car
- BA virus deletes files on a single PC
- CMany compromised machines flood a target with traffic
- DOne attacker guesses a user's password
4.Computer networksHard
After repeated collisions, classic Ethernet stations choose their waiting time from a range that doubles each time. Why?
- ATo synchronise all stations to one clock
- BTo give the station with the highest MAC address priority on the shared cable
- CTo make the cable carry signals further
- DTo lower the chance of colliding again when many stations compete
5.Number systemsMedium
Write −96 as an 8-bit two's complement pattern. Read the pattern as an unsigned binary number: what is its value U in decimal?
N = −968 bit
U =
6.Number systemsHard
Set the bits to the 8-bit result of 1100 0010 XOR 0001 1010. What is the result's value R in decimal?
1100 0010XOR0001 1010
R =
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 < 1790
211 − 2 = 2046 ≥ 1790 → h = 11
Prefix length
p = 32 − 11 = 21
Answer
p = /21
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: 86 hosts → 27 = 128 (/25)
C: 55 hosts → 26 = 64 (/26)
A: 7 hosts → 24 = 16 (/28)
D: 3 hosts → 23 = 8 (/29)
Allocation, largest first
B: .0 – .127
C: .128 – .191
A: .192 – .207
D: .208 – .215
Answer
x = 128
LAN C: 192.168.7.128/26.
3.Computer networks
Answer: C · Many compromised machines flood a target with traffic
Why this is the answer
In a DDoS attack many sources overwhelm a server or link so legitimate users cannot be served. The many sources make simple blocking hard.
Answer
C · Many compromised machines flood a target with traffic
4.Computer networks
Answer: D · To lower the chance of colliding again when many stations compete
Why this is the answer
Each further collision suggests more contention, so the random range grows and stations are more likely to pick different slots. It adapts to load without central control.
Answer
D · To lower the chance of colliding again when many stations compete
5.Number systems
Answer: U = 160
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
96 = 0110 0000
inverted: 1001 1111
+1 → 1010 0000
Read it as unsigned
1010 0000₂ = 256 − 96 = 160
Answer
U = 160
6.Number systems
Answer: R = 216
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
XOR gives 1 where the two bits differ.
1100 0010
XOR 0001 1010
= 1101 1000
Value of the result
1101 1000₂ = 216
Answer
R = 216
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