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 16 equal subnets. If you borrow as few bits as possible, how many usable hosts H does each subnet have?
192.168.10.0/2416 subnets
H =
2.IPv4 subnettingHard
The network 192.168.229.0/24 is split with VLSM for these LANs: A: 57 hosts, B: 48 hosts, C: 21 hosts, D: 7 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 C?
A: 57B: 48C: 21D: 7
x =
3.Computer networksMedium
Which host is in the same /16 subnet as 10.1.2.3?
- A10.2.2.3
- B10.1.200.5
- C11.1.2.3
- D10.0.2.3
4.Computer networksHard
In a SYN flood attack, what resource is the attacker trying to exhaust?
- AThe server's disk space
- BThe DNS cache of the client
- CThe switch's MAC address table
- DThe server's table of half-open connections
5.Number systemsMedium
Set the bits so they store −106 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?
N = −1068 bit
U =
6.Number systemsHard
A system must give every one of 1879 items its own binary code of the same length. What is the smallest number of bits b needed?
1879
b = bit
3Solutions
1.IPv4 subnetting
Answer: H = 14
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
24 = 16 ≥ 16 → b = 4, /28
Hosts per subnet
24 − 2 = 14
Answer
H = 14
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
A: 57 hosts → 26 = 64 (/26)
B: 48 hosts → 26 = 64 (/26)
C: 21 hosts → 25 = 32 (/27)
D: 7 hosts → 24 = 16 (/28)
Allocation, largest first
A: .0 – .63
B: .64 – .127
C: .128 – .159
D: .160 – .175
Answer
x = 128
LAN C: 192.168.229.128/27.
3.Computer networks
Answer: B · 10.1.200.5
Why this is the answer
With /16 the first 16 bits (10.1) are the network part. Only 10.1.200.5 shares them, so it is on 10.1.0.0/16.
Answer
B · 10.1.200.5
4.Computer networks
Answer: D · 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
D · The server's table of half-open connections
5.Number systems
Answer: U = 150
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
106 = 0110 1010
inverted: 1001 0101
+1 → 1001 0110
Read it as unsigned
1001 0110₂ = 256 − 106 = 150
Answer
U = 150
6.Number systems
Answer: b = 11 bit
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.
Powers of two
210 = 1024 < 1879
211 = 2048 ≥ 1879
Answer
b = 11
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