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 192.168.106.0/24 is divided into /27 subnets, numbered 1, 2, 3, … from the lowest address (subnet 1 starts at .0). What is the fourth octet x of the broadcast address of subnet 8?
192.168.106.0/24/27subnet 8 (subnet 1 starts at .0)
x =
2.IPv4 subnettingHard
A host has the address 172.20.107.142/21. What is the third octet x of its network address?
172.20.107.142/21
x =
3.Computer networksMedium
TLS provides which security properties for HTTPS?
- AConfidentiality, integrity and server authentication
- BProtection against all malware and phishing on the visited site
- COnly faster loading
- DAnonymity of the user
4.Computer networksHard
An attacker on a LAN sends forged ARP replies claiming the gateway's IP address. What is the likely result?
- AThe gateway's IP address is permanently changed to the attacker's address on every router
- BThe switch shuts down all of its ports
- 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
Write −67 as an 8-bit two's complement pattern. Read the pattern as an unsigned binary number: what is its value U in decimal?
N = −678 bit
U =
6.Number systemsHard
An uncompressed image is 256 × 256 pixels with 24 bits per pixel. How much memory M does it take, in KiB (1 KiB = 1024 bytes)?
256 × 256 px24 bit/px1 KiB = 1024 B
M = KiB
3Solutions
1.IPv4 subnetting
Answer: x = 255
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
25 = 32
Subnet number 8
network: (8 − 1)·32 = 224
broadcast: 224 + 32 − 1 = 255
Answer
x = 255
2.IPv4 subnetting
Answer: x = 104
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 in the third octet
23 = 8
Subnet of this host
network: 172.20.104.0
broadcast: 172.20.111.255
Answer
x = 104
3.Computer networks
Answer: A · Confidentiality, integrity and server authentication
Why this is the answer
TLS encrypts traffic, detects tampering and authenticates the server with certificates (optionally the client too).
Answer
A · Confidentiality, integrity and server authentication
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 = 189
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
67 = 0100 0011
inverted: 1011 1100
+1 → 1011 1101
Read it as unsigned
1011 1101₂ = 256 − 67 = 189
Answer
U = 189
6.Number systems
Answer: M = 192 KiB
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.
Bits and bytes
256·256·24 = 1572864 bit = 196608 B
Convert to KiB
196608/1024 = 192 KiB
Answer
M = 192 KiB
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