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 192.168.10.0/24 must be divided into at least 8 equal subnets. If you borrow as few bits as possible, how many usable hosts H does each subnet have?
192.168.10.0/248 subnets
H =
2.IPv4 subnettingHard
A host has the address 10.20.152.173/26. What is the fourth octet x of the last usable host address in its subnet?
10.20.152.173/26
x =
3.Computer networksMedium
NAT (network address translation) is mainly used to:
- ATranslate domain names into IP addresses for devices on a local network
- BEncrypt traffic between two sites
- CIncrease the speed of a network cable
- DLet many devices with private addresses share one public IP address
4.Computer networksHard
An attacker on a LAN sends forged ARP replies claiming the gateway's IP address. What is the likely result?
- AVictims send traffic for the gateway to the attacker, enabling a man-in-the-middle attack
- BThe gateway's IP address is permanently changed to the attacker's address on every router
- CThe switch shuts down all of its ports
- DAll DNS queries fail across the internet
5.Number systemsMedium
Write −73 as an 8-bit two's complement pattern. Read the pattern as an unsigned binary number: what is its value U in decimal?
N = −738 bit
U =
6.Number systemsHard
Set the bits to the 8-bit result of 1110 0011 << 3. What is the result's value R in decimal?
1110 0011<< 3
R =
3Solutions
1.IPv4 subnetting
Answer: H = 30
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).
Bits to borrow
23 = 8 ≥ 8 → b = 3, /27
Hosts per subnet
25 − 2 = 30
Answer
H = 30
2.IPv4 subnetting
Answer: x = 190
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: ⌊173/64⌋·64 = 128
broadcast: 128 + 64 − 1 = 191
usable hosts: 129 … 190
Answer
x = 190
3.Computer networks
Answer: D · Let many devices with private addresses share one public IP address
Why this is the answer
NAT rewrites addresses (and ports) so private networks can access the internet through a smaller number of public addresses.
Answer
D · Let many devices with private addresses share one public IP address
4.Computer networks
Answer: A · 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
A · Victims send traffic for the gateway to the attacker, enabling a man-in-the-middle attack
5.Number systems
Answer: U = 183
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
73 = 0100 1001
inverted: 1011 0110
+1 → 1011 0111
Read it as unsigned
1011 0111₂ = 256 − 73 = 183
Answer
U = 183
6.Number systems
Answer: R = 24
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
Shifting left by 3 moves every bit 3 places to the left; bits that fall off the left end are lost and zeros come in on the right.
1110 0011 << 3
= 0001 1000
Value of the result
0001 1000₂ = 24
Answer
R = 24
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