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 192.168.215.0/24 is divided into /29 subnets, numbered 1, 2, 3, … from the lowest address (subnet 1 starts at .0). What is the fourth octet x of the network address of subnet 11?
192.168.215.0/24/29subnet 11 (subnet 1 starts at .0)
x =
2.IPv4 subnettingHard
A host has the address 10.20.107.67/25. What is the fourth octet x of the last usable host address in its subnet?
10.20.107.67/25
x =
3.Computer networksMedium
What does an MX record in DNS specify?
- AAn alias name for another host
- BThe mail servers that accept email for a domain
- CThe name servers for a zone
- DThe IPv4 address of a web server
4.Computer networksHard
Which mechanism prevents broadcast loops in a network with redundant switch links?
- ASpanning Tree Protocol
- BDHCP snooping
- CNAT
- DARP caching
5.Number systemsMedium
Set the bits so they store −38 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?
N = −388 bit
U =
6.Number systemsHard
A system must give every one of 2249 items its own binary code of the same length. What is the smallest number of bits b needed?
2249
b = bit
3Solutions
1.IPv4 subnetting
Answer: x = 80
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
23 = 8
Subnet number 11
network: (11 − 1)·8 = 80
broadcast: 80 + 8 − 1 = 87
Answer
x = 80
2.IPv4 subnetting
Answer: x = 126
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: ⌊67/128⌋·128 = 0
broadcast: 0 + 128 − 1 = 127
usable hosts: 1 … 126
Answer
x = 126
3.Computer networks
Answer: B · The mail servers that accept email for a domain
Why this is the answer
MX (mail exchange) records list mail servers with priorities, so sending servers know where to deliver mail for that domain.
Answer
B · The mail servers that accept email for a domain
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 = 218
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
38 = 0010 0110
inverted: 1101 1001
+1 → 1101 1010
Read it as unsigned
1101 1010₂ = 256 − 38 = 218
Answer
U = 218
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 < 2249
212 = 4096 ≥ 2249
Answer
b = 12
ExamProblems · examproblems.com/courses/computer-networks