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 192.168.207.0/24 is divided into /27 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 7?
192.168.207.0/24/27subnet 7 (subnet 1 starts at .0)
x =
2.IPv4 subnettingHard
A host has the address 10.20.141.53/25. What is the last number (fourth octet) x of its subnet's broadcast address?
10.20.141.53/25
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
A DNS cache poisoning attack aims to:
- ADelete DNS servers
- BOverload a server with traffic so that it stops answering queries
- CMake a resolver store a false IP address for a domain
- DSteal Wi-Fi passwords
5.Number systemsMedium
Write −32 as an 8-bit two's complement pattern. Read the pattern as an unsigned binary number: what is its value U in decimal?
N = −328 bit
U =
6.Number systemsHard
An uncompressed image is 1024 × 768 pixels with 24 bits per pixel. How much memory M does it take, in KiB (1 KiB = 1024 bytes)?
1024 × 768 px24 bit/px1 KiB = 1024 B
M = KiB
3Solutions
1.IPv4 subnetting
Answer: x = 223
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
25 = 32
Subnet number 7
network: (7 − 1)·32 = 192
broadcast: 192 + 32 − 1 = 223
Answer
x = 223
2.IPv4 subnetting
Answer: x = 127
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 of this host
network: ⌊53/128⌋·128 = 0
broadcast: 0 + 128 − 1 = 127
usable hosts: 1 … 126
Answer
x = 127
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: C · Make a resolver store a false IP address for a domain
Why this is the answer
Poisoning inserts forged records into a resolver's cache. DNSSEC adds signatures so resolvers can verify records.
Answer
C · Make a resolver store a false IP address for a domain
5.Number systems
Answer: U = 224
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
32 = 0010 0000
inverted: 1101 1111
+1 → 1110 0000
Read it as unsigned
1110 0000₂ = 256 − 32 = 224
Answer
U = 224
6.Number systems
Answer: M = 2304 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
1024·768·24 = 18874368 bit = 2359296 B
Convert to KiB
2359296/1024 = 2304 KiB
Answer
M = 2304 KiB
ExamProblems · examproblems.com/courses/computer-networks