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 1593 host addresses in a single subnet. What is the longest prefix length p (smallest subnet) that is big enough?
1593 hosts
p =
2.IPv4 subnettingHard
A host has the address 10.20.133.248/25. What is the fourth octet x of the last usable host address in its subnet?
10.20.133.248/25
x =
3.Computer networksMedium
Which DNS record type maps a host name to an IPv6 address?
- AA
- BMX
- CAAAA
- DCNAME
4.Computer networksHard
An 8-port switch (no VLANs) has one host on each port. How many collision domains and broadcast domains are there?
- A1 collision domain, 8 broadcast domains
- B1 collision domain, 1 broadcast domain
- C8 collision domains, 1 broadcast domain
- D8 collision domains, 8 broadcast domains
5.Number systemsMedium
Set the bits so they store −119 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?
N = −1198 bit
U =
6.Number systemsHard
Set the bits to the 8-bit result of 1100 0010 AND 0100 1010. What is the result's value R in decimal?
1100 0010AND0100 1010
R =
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 < 1593
211 − 2 = 2046 ≥ 1593 → h = 11
Prefix length
p = 32 − 11 = 21
Answer
p = /21
2.IPv4 subnetting
Answer: x = 254
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
27 = 128
Subnet of this host
network: ⌊248/128⌋·128 = 128
broadcast: 128 + 128 − 1 = 255
usable hosts: 129 … 254
Answer
x = 254
3.Computer networks
Answer: C · AAAA
Why this is the answer
An A record holds an IPv4 address, AAAA an IPv6 address. MX names mail servers and CNAME gives an alias.
Answer
C · AAAA
4.Computer networks
Answer: C · 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
C · 8 collision domains, 1 broadcast domain
5.Number systems
Answer: U = 137
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
119 = 0111 0111
inverted: 1000 1000
+1 → 1000 1001
Read it as unsigned
1000 1001₂ = 256 − 119 = 137
Answer
U = 137
6.Number systems
Answer: R = 66
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.
Bit by bit
AND gives 1 only where both bits are 1.
1100 0010
AND 0100 1010
= 0100 0010
Value of the result
0100 0010₂ = 66
Answer
R = 66
ExamProblems · examproblems.com/courses/computer-networks