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 255.255.255.252. How many usable host addresses H does it have?
255.255.255.252
H =
2.IPv4 subnettingHard
A host has the address 10.20.69.79/30. What is the fourth octet x of the last usable host address in its subnet?
10.20.69.79/30
x =
3.Computer networksMedium
How many usable host addresses does a /26 IPv4 subnet have?
- A64
- B126
- C30
- D62
4.Computer networksHard
After repeated collisions, classic Ethernet stations choose their waiting time from a range that doubles each time. Why?
- ATo make the cable carry signals further
- BTo synchronise all stations to one clock
- CTo give the station with the highest MAC address priority on the shared cable
- DTo lower the chance of colliding again when many stations compete
5.Number systemsMedium
Set the bits so they store −55 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?
N = −558 bit
U =
6.Number systemsHard
A system must give every one of 632 items its own binary code of the same length. What is the smallest number of bits b needed?
632
b = bit
3Solutions
1.IPv4 subnetting
Answer: H = 2
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
255.255.255.252 = /30
Usable hosts
22 − 2 = 2
Answer
H = 2
2.IPv4 subnetting
Answer: x = 78
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: ⌊79/4⌋·4 = 76
broadcast: 76 + 4 − 1 = 79
usable hosts: 77 … 78
Answer
x = 78
3.Computer networks
Answer: D · 62
Why this is the answer
A /26 has 6 host bits: 2⁶ = 64 addresses, minus the network and broadcast addresses, gives 62 usable hosts.
Answer
D · 62
4.Computer networks
Answer: D · To lower the chance of colliding again when many stations compete
Why this is the answer
Each further collision suggests more contention, so the random range grows and stations are more likely to pick different slots. It adapts to load without central control.
Answer
D · To lower the chance of colliding again when many stations compete
5.Number systems
Answer: U = 201
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
55 = 0011 0111
inverted: 1100 1000
+1 → 1100 1001
Read it as unsigned
1100 1001₂ = 256 − 55 = 201
Answer
U = 201
6.Number systems
Answer: b = 10 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
29 = 512 < 632
210 = 1024 ≥ 632
Answer
b = 10
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