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 department needs 1312 host addresses in a single subnet. What is the longest prefix length p (smallest subnet) that is big enough?
1312 hosts
p =
2.IPv4 subnettingHard
A host has the address 10.20.98.92/26. What is the fourth octet x of its network address?
10.20.98.92/26
x =
3.Computer networksMedium
What is the usual order of messages when a client obtains an address from DHCP?
- ADiscover, Offer, Request, Acknowledge
- BRequest, Offer, Discover, Acknowledge
- CDiscover, Request, Offer, Acknowledge
- DOffer, Discover, Acknowledge, Request
4.Computer networksHard
An 8-port switch (no VLANs) has one host on each port. How many collision domains and broadcast domains are there?
- A8 collision domains, 8 broadcast domains
- B8 collision domains, 1 broadcast domain
- C1 collision domain, 1 broadcast domain
- D1 collision domain, 8 broadcast domains
5.Number systemsMedium
Write −31 as an 8-bit two's complement pattern. Read the pattern as an unsigned binary number: what is its value U in decimal?
N = −318 bit
U =
6.Number systemsHard
A system must give every one of 3773 items its own binary code of the same length. What is the smallest number of bits b needed?
3773
b = bit
3Solutions
1.IPv4 subnetting
Answer: p = 21
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).
Host bits needed
210 − 2 = 1022 < 1312
211 − 2 = 2046 ≥ 1312 → h = 11
Prefix length
p = 32 − 11 = 21
Answer
p = /21
2.IPv4 subnetting
Answer: x = 64
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
26 = 64
Subnet of this host
network: ⌊92/64⌋·64 = 64
broadcast: 64 + 64 − 1 = 127
usable hosts: 65 … 126
Answer
x = 64
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: B · 8 collision domains, 1 broadcast domain
Why this is the answer
Each switch port is its own collision domain, but broadcasts are forwarded to all ports, so the whole switch is one broadcast domain. A hub would give one of each.
Answer
B · 8 collision domains, 1 broadcast domain
5.Number systems
Answer: U = 225
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.
Invert and add one
31 = 0001 1111
inverted: 1110 0000
+1 → 1110 0001
Read it as unsigned
1110 0001₂ = 256 − 31 = 225
Answer
U = 225
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 < 3773
212 = 4096 ≥ 3773
Answer
b = 12
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