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 4 equal subnets. If you borrow as few bits as possible, how many usable hosts H does each subnet have?
192.168.10.0/244 subnets
H =
2.IPv4 subnettingHard
A host has the address 10.20.215.40/30. What is the fourth octet x of its network address?
10.20.215.40/30
x =
3.Computer networksMedium
HTTP is stateless. How do websites usually keep a user logged in across requests?
- ABy giving each user a fixed IP address
- BBy keeping the TCP connection open for days
- CWith cookies holding a session identifier or token
- DBy storing the password in the URL
4.Computer networksHard
What does the Diffie-Hellman key exchange achieve on its own, and what is its weakness?
- AIt signs certificates for websites, but only works with short keys
- BIt encrypts bulk data directly over a public channel, but it is far too slow for large files and video
- CIt agrees a shared secret over a public channel, but alone it is open to a man-in-the-middle
- DIt authenticates both users with passwords, but needs a central server
5.Number systemsMedium
Set the bits so they store −71 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?
N = −718 bit
U =
6.Number systemsHard
An uncompressed image is 512 × 512 pixels with 32 bits per pixel. How much memory M does it take, in KiB (1 KiB = 1024 bytes)?
512 × 512 px32 bit/px1 KiB = 1024 B
M = KiB
3Solutions
1.IPv4 subnetting
Answer: H = 62
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
22 = 4 ≥ 4 → b = 2, /26
Hosts per subnet
26 − 2 = 62
Answer
H = 62
2.IPv4 subnetting
Answer: x = 40
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
22 = 4
Subnet of this host
network: ⌊40/4⌋·4 = 40
broadcast: 40 + 4 − 1 = 43
usable hosts: 41 … 42
Answer
x = 40
3.Computer networks
Answer: C · With cookies holding a session identifier or token
Why this is the answer
The server sets a cookie containing a session ID or token, and the browser returns it with each request so the server can recognise the user.
Answer
C · With cookies holding a session identifier or token
4.Computer networks
Answer: C · It agrees a shared secret over a public channel, but alone it is open to a man-in-the-middle
Why this is the answer
Diffie-Hellman lets two parties derive the same secret that eavesdroppers cannot compute. An attacker in the middle can run separate exchanges with each side unless the values are signed or otherwise authenticated.
Answer
C · It agrees a shared secret over a public channel, but alone it is open to a man-in-the-middle
5.Number systems
Answer: U = 185
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
71 = 0100 0111
inverted: 1011 1000
+1 → 1011 1001
Read it as unsigned
1011 1001₂ = 256 − 71 = 185
Answer
U = 185
6.Number systems
Answer: M = 1024 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
512·512·32 = 8388608 bit = 1048576 B
Convert to KiB
1048576/1024 = 1024 KiB
Answer
M = 1024 KiB
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