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 10.14.240.0/24 is divided into equal subnets with prefix /26. How many subnets N are there?
10.14.240.0/24/26
N =
2.IPv4 subnettingHard
A host has the address 10.20.219.221/28. What is the fourth octet x of its network address?
10.20.219.221/28
x =
3.Computer networksMedium
What is the main purpose of a VLAN?
- ATo encrypt all traffic between switches so other departments cannot read it
- BTo split one physical switched network into separate broadcast domains
- CTo increase the speed of every switch port
- DTo give each device a public IP address
4.Computer networksHard
Which mechanism prevents broadcast loops in a network with redundant switch links?
- ANAT
- BARP caching
- CDHCP snooping
- DSpanning Tree Protocol
5.Number systemsMedium
Set the bits so they store −79 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?
N = −798 bit
U =
6.Number systemsHard
A system must give every one of 3274 items its own binary code of the same length. What is the smallest number of bits b needed?
3274
b = bit
3Solutions
1.IPv4 subnetting
Answer: N = 4
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).
Borrowed bits
26 − 24 = 2
Number of subnets
N = 22 = 4
Answer
N = 4
Each /26 subnet has 26 − 2 = 62 usable hosts.
2.IPv4 subnetting
Answer: x = 208
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
24 = 16
Subnet of this host
network: ⌊221/16⌋·16 = 208
broadcast: 208 + 16 − 1 = 223
usable hosts: 209 … 222
Answer
x = 208
3.Computer networks
Answer: B · To split one physical switched network into separate broadcast domains
Why this is the answer
VLANs group ports logically, so broadcasts stay within each VLAN. This improves security and limits broadcast traffic.
Answer
B · To split one physical switched network into separate broadcast domains
4.Computer networks
Answer: D · 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
D · Spanning Tree Protocol
5.Number systems
Answer: U = 177
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
79 = 0100 1111
inverted: 1011 0000
+1 → 1011 0001
Read it as unsigned
1011 0001₂ = 256 − 79 = 177
Answer
U = 177
6.Number systems
Answer: b = 12 bit
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.
Powers of two
211 = 2048 < 3274
212 = 4096 ≥ 3274
Answer
b = 12
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