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 255.255.255.192. How many usable host addresses H does it have?
255.255.255.192
H =
2.IPv4 subnettingHard
A host has the address 10.20.11.175/30. What is the fourth octet x of the last usable host address in its subnet?
10.20.11.175/30
x =
3.Computer networksMedium
How long does it take to put a 1500-byte packet onto a 12 Mbit/s link (transmission delay)?
- A1 ms
- B0.125 ms
- C12 ms
- D8 ms
4.Computer networksHard
An attacker on a LAN sends forged ARP replies claiming the gateway's IP address. What is the likely result?
- AThe switch shuts down all of its ports
- BThe gateway's IP address is permanently changed to the attacker's address on every router
- CVictims send traffic for the gateway to the attacker, enabling a man-in-the-middle attack
- DAll DNS queries fail across the internet
5.Number systemsMedium
Set the bits so they store −82 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?
N = −828 bit
U =
6.Number systemsHard
What is the largest unsigned integer N that fits in 12 bits?
12 bit
N =
3Solutions
1.IPv4 subnetting
Answer: H = 62
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
255.255.255.192 = /26
Usable hosts
26 − 2 = 62
Answer
H = 62
2.IPv4 subnetting
Answer: x = 174
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: ⌊175/4⌋·4 = 172
broadcast: 172 + 4 − 1 = 175
usable hosts: 173 … 174
Answer
x = 174
3.Computer networks
Answer: A · 1 ms
Why this is the answer
1500 bytes = 12,000 bits. 12,000 / 12,000,000 bit/s = 0.001 s = 1 ms. Transmission delay depends on packet size and link rate, not distance.
Answer
A · 1 ms
4.Computer networks
Answer: C · Victims send traffic for the gateway to the attacker, enabling a man-in-the-middle attack
Why this is the answer
ARP spoofing poisons hosts' ARP caches with the attacker's MAC for the gateway IP. The attacker can then read, change or drop traffic; dynamic ARP inspection helps prevent it.
Answer
C · Victims send traffic for the gateway to the attacker, enabling a man-in-the-middle attack
5.Number systems
Answer: U = 174
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
82 = 0101 0010
inverted: 1010 1101
+1 → 1010 1110
Read it as unsigned
1010 1110₂ = 256 − 82 = 174
Answer
U = 174
6.Number systems
Answer: N = 4095
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.
Largest value
212 − 1 = 4095
Answer
N = 4095
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