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 subnet uses the mask 255.255.255.224. How many usable host addresses H does it have?
255.255.255.224
H =
2.IPv4 subnettingHard
A host has the address 10.20.165.35/30. What is the last number (fourth octet) x of its subnet's broadcast address?
10.20.165.35/30
x =
3.Computer networksMedium
What is the usual order of messages when a client obtains an address from DHCP?
- ADiscover, Offer, Request, Acknowledge
- BDiscover, Request, Offer, Acknowledge
- COffer, Discover, Acknowledge, Request
- DRequest, Offer, Discover, Acknowledge
4.Computer networksHard
Which mechanism prevents broadcast loops in a network with redundant switch links?
- ASpanning Tree Protocol
- BDHCP snooping
- CARP caching
- DNAT
5.Number systemsMedium
Set the bits so they store −118 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?
N = −1188 bit
U =
6.Number systemsHard
An uncompressed image is 1920 × 1080 pixels with 24 bits per pixel. How much memory M does it take, in KiB (1 KiB = 1024 bytes)?
1920 × 1080 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).
Prefix length
255.255.255.224 = /27
Usable hosts
25 − 2 = 30
Answer
H = 30
2.IPv4 subnetting
Answer: x = 35
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: ⌊35/4⌋·4 = 32
broadcast: 32 + 4 − 1 = 35
usable hosts: 33 … 34
Answer
x = 35
3.Computer networks
Answer: A · Discover, Offer, Request, Acknowledge
Why this is the answer
The client broadcasts Discover, servers reply with Offers, the client Requests one offer and the chosen server confirms with an Acknowledge.
Answer
A · Discover, Offer, Request, Acknowledge
4.Computer networks
Answer: A · Spanning Tree Protocol
Why this is the answer
STP elects a root bridge and blocks redundant paths, unblocking them if an active link fails.
Answer
A · Spanning Tree Protocol
5.Number systems
Answer: U = 138
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
118 = 0111 0110
inverted: 1000 1001
+1 → 1000 1010
Read it as unsigned
1000 1010₂ = 256 − 118 = 138
Answer
U = 138
6.Number systems
Answer: M = 6075 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
1920·1080·24 = 49766400 bit = 6220800 B
Convert to KiB
6220800/1024 = 6075 KiB
Answer
M = 6075 KiB
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