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 255.255.254.0. How many usable host addresses H does it have?
255.255.254.0
H =
2.IPv4 subnettingHard
The network 192.168.197.0/24 is split with VLSM for these LANs: A: 10 hosts, B: 15 hosts, C: 18 hosts, D: 4 hosts, E: 3 hosts. Subnets are allocated from .0 upwards, the LAN with the most hosts first, each with the smallest subnet that fits. What is the fourth octet x of the network address of LAN D?
A: 10B: 15C: 18D: 4E: 3
x =
3.Computer networksMedium
What characterises a distributed denial-of-service (DDoS) attack?
- AA virus deletes files on a single PC
- BAn attacker reads unencrypted Wi-Fi traffic from a nearby car
- COne attacker guesses a user's password
- DMany compromised machines flood a target with traffic
4.Computer networksHard
In the TCP/IP model, which layer adds port numbers?
- ALink layer
- BApplication layer
- CInternet layer
- DTransport layer
5.Number systemsMedium
Set the bits so they store −118 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?
N = −1188 bit
U =
6.Number systemsHard
What is the largest unsigned integer N that fits in 24 bits?
24 bit
N =
3Solutions
1.IPv4 subnetting
Answer: H = 510
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
255.255.254.0 = /23
Usable hosts
29 − 2 = 510
Answer
H = 510
2.IPv4 subnetting
Answer: x = 80
VLSM: each LAN gets the smallest subnet with 2ʰ − 2 ≥ its host count, i.e. prefix 32 − h. Allocating the largest subnets first keeps every block aligned on a multiple of its own size, so the network addresses follow by adding up the block sizes.
Block size per LAN
C: 18 hosts → 25 = 32 (/27)
B: 15 hosts → 25 = 32 (/27)
A: 10 hosts → 24 = 16 (/28)
D: 4 hosts → 23 = 8 (/29)
E: 3 hosts → 23 = 8 (/29)
Allocation, largest first
C: .0 – .31
B: .32 – .63
A: .64 – .79
D: .80 – .87
E: .88 – .95
Answer
x = 80
LAN D: 192.168.197.80/29.
3.Computer networks
Answer: D · Many compromised machines flood a target with traffic
Why this is the answer
In a DDoS attack many sources overwhelm a server or link so legitimate users cannot be served. The many sources make simple blocking hard.
Answer
D · Many compromised machines flood a target with traffic
4.Computer networks
Answer: D · Transport layer
Why this is the answer
TCP and UDP headers contain source and destination ports; the internet layer (IP) adds addresses.
Answer
D · Transport layer
5.Number systems
Answer: U = 138
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
118 = 0111 0110
inverted: 1000 1001
+1 → 1000 1010
Read it as unsigned
1000 1010₂ = 256 − 118 = 138
Answer
U = 138
6.Number systems
Answer: N = 1.678 × 10⁷
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
224 − 1 = 16777215
Answer
N = 16777215
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