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 1503 host addresses in a single subnet. What is the longest prefix length p (smallest subnet) that is big enough?
1503 hosts
p =
2.IPv4 subnettingHard
A host has the address 172.20.206.13/22. What is the third octet x of its subnet's broadcast address?
172.20.206.13/22
x =
3.Computer networksMedium
What characterises a distributed denial-of-service (DDoS) attack?
- AAn attacker reads unencrypted Wi-Fi traffic from a nearby car
- BOne attacker guesses a user's password
- CA virus deletes files on a single PC
- DMany compromised machines flood a target with traffic
4.Computer networksHard
By the Nyquist formula, what is the maximum bit rate of a noiseless 3 kHz channel using 4 signal levels?
- A3 kbit/s
- B24 kbit/s
- C12 kbit/s
- D6 kbit/s
5.Number systemsMedium
Set the bits so they store −59 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?
N = −598 bit
U =
6.Number systemsHard
An uncompressed image is 512 × 512 pixels with 24 bits per pixel. How much memory M does it take, in KiB (1 KiB = 1024 bytes)?
512 × 512 px24 bit/px1 KiB = 1024 B
M = KiB
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 < 1503
211 − 2 = 2046 ≥ 1503 → h = 11
Prefix length
p = 32 − 11 = 21
Answer
p = /21
2.IPv4 subnetting
Answer: x = 207
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 in the third octet
22 = 4
Subnet of this host
network: 172.20.204.0
broadcast: 172.20.207.255
Answer
x = 207
3.Computer networks
Answer: D · 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
D · Many compromised machines flood a target with traffic
4.Computer networks
Answer: C · 12 kbit/s
Why this is the answer
2 × 3000 × log₂ 4 = 2 × 3000 × 2 = 12,000 bit/s. More levels raise the rate, but noise limits how many can be told apart.
Answer
C · 12 kbit/s
5.Number systems
Answer: U = 197
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
59 = 0011 1011
inverted: 1100 0100
+1 → 1100 0101
Read it as unsigned
1100 0101₂ = 256 − 59 = 197
Answer
U = 197
6.Number systems
Answer: M = 768 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
512·512·24 = 6291456 bit = 786432 B
Convert to KiB
786432/1024 = 768 KiB
Answer
M = 768 KiB
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