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 8 equal subnets. If you borrow as few bits as possible, how many usable hosts H does each subnet have?
192.168.10.0/248 subnets
H =
2.IPv4 subnettingHard
A host has the address 10.20.131.143/25. What is the fourth octet x of its network address?
10.20.131.143/25
x =
3.Computer networksMedium
Using even parity, which parity bit is added to the data bits 1011001?
- A2
- B4
- C0
- D1
4.Computer networksHard
A server's private key is stolen a year from now. Which TLS property keeps today's recorded sessions safe?
- AForward secrecy from ephemeral Diffie-Hellman key exchange
- BSending the certificate twice
- CCompressing the data before encryption
- DUsing a longer domain name
5.Number systemsMedium
Set the bits so they store −13 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?
N = −138 bit
U =
6.Number systemsHard
An uncompressed image is 1280 × 720 pixels with 24 bits per pixel. How much memory M does it take, in KiB (1 KiB = 1024 bytes)?
1280 × 720 px24 bit/px1 KiB = 1024 B
M = KiB
3Solutions
1.IPv4 subnetting
Answer: H = 30
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
23 = 8 ≥ 8 → b = 3, /27
Hosts per subnet
25 − 2 = 30
Answer
H = 30
2.IPv4 subnetting
Answer: x = 128
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: ⌊143/128⌋·128 = 128
broadcast: 128 + 128 − 1 = 255
usable hosts: 129 … 254
Answer
x = 128
3.Computer networks
Answer: C · 0
Why this is the answer
1011001 contains four 1-bits, already an even number, so the parity bit is 0. With odd parity it would be 1.
Answer
C · 0
4.Computer networks
Answer: A · Forward secrecy from ephemeral Diffie-Hellman key exchange
Why this is the answer
With ephemeral (EC)DHE, each session key is derived from temporary values that are discarded. The long-term key only signs the exchange, so stealing it later does not decrypt old traffic.
Answer
A · Forward secrecy from ephemeral Diffie-Hellman key exchange
5.Number systems
Answer: U = 243
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
13 = 0000 1101
inverted: 1111 0010
+1 → 1111 0011
Read it as unsigned
1111 0011₂ = 256 − 13 = 243
Answer
U = 243
6.Number systems
Answer: M = 2700 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
1280·720·24 = 22118400 bit = 2764800 B
Convert to KiB
2764800/1024 = 2700 KiB
Answer
M = 2700 KiB
ExamProblems · examproblems.com/courses/computer-networks