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 department needs 1853 host addresses in a single subnet. What is the longest prefix length p (smallest subnet) that is big enough?
1853 hosts
p =
2.IPv4 subnettingHard
The network 192.168.168.0/24 is split with VLSM for these LANs: A: 22 hosts, B: 90 hosts, C: 5 hosts, D: 21 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 A?
A: 22B: 90C: 5D: 21
x =
3.Computer networksMedium
What does a subnet mask of 255.255.255.0 correspond to in CIDR notation?
- A/24
- B/16
- C/32
- D/8
4.Computer networksHard
In a SYN flood attack, what resource is the attacker trying to exhaust?
- AThe server's table of half-open connections
- BThe DNS cache of the client
- CThe server's disk space
- DThe switch's MAC address table
5.Number systemsMedium
Set the bits so they store −40 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?
N = −408 bit
U =
6.Number systemsHard
What is the most negative integer N that can be stored in 16-bit two's complement?
16 bit
N =
3Solutions
1.IPv4 subnetting
Answer: p = 21
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).
Host bits needed
210 − 2 = 1022 < 1853
211 − 2 = 2046 ≥ 1853 → h = 11
Prefix length
p = 32 − 11 = 21
Answer
p = /21
2.IPv4 subnetting
Answer: x = 128
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
B: 90 hosts → 27 = 128 (/25)
A: 22 hosts → 25 = 32 (/27)
D: 21 hosts → 25 = 32 (/27)
C: 5 hosts → 23 = 8 (/29)
Allocation, largest first
B: .0 – .127
A: .128 – .159
D: .160 – .191
C: .192 – .199
Answer
x = 128
LAN A: 192.168.168.128/27.
3.Computer networks
Answer: A · /24
Why this is the answer
Three octets of 255 are 24 one-bits, so 255.255.255.0 = /24.
Answer
A · /24
4.Computer networks
Answer: A · The server's table of half-open connections
Why this is the answer
Each SYN makes the server reserve state and wait for an ACK that never comes. SYN cookies avoid storing state until the handshake completes.
Answer
A · The server's table of half-open connections
5.Number systems
Answer: U = 216
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
40 = 0010 1000
inverted: 1101 0111
+1 → 1101 1000
Read it as unsigned
1101 1000₂ = 256 − 40 = 216
Answer
U = 216
6.Number systems
Answer: N = −32768
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.
Most negative value
−215 = −32768
Answer
N = −32768
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