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 255.255.255.252. How many usable host addresses H does it have?
255.255.255.252
H =
2.IPv4 subnettingHard
The network 192.168.149.0/24 is split with VLSM for these LANs: A: 27 hosts, B: 6 hosts, C: 10 hosts, D: 14 hosts, E: 53 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: 27B: 6C: 10D: 14E: 53
x =
3.Computer networksMedium
NAT (network address translation) is mainly used to:
- AIncrease the speed of a network cable
- BTranslate domain names into IP addresses for devices on a local network
- CLet many devices with private addresses share one public IP address
- DEncrypt traffic between two sites
4.Computer networksHard
A DNS cache poisoning attack aims to:
- ASteal Wi-Fi passwords
- BDelete DNS servers
- CMake a resolver store a false IP address for a domain
- DOverload a server with traffic so that it stops answering queries
5.Number systemsMedium
Write −70 as an 8-bit two's complement pattern. Read the pattern as an unsigned binary number: what is its value U in decimal?
N = −708 bit
U =
6.Number systemsHard
Set the bits to the 8-bit result of 1011 1001 << 3. What is the result's value R in decimal?
1011 1001<< 3
R =
3Solutions
1.IPv4 subnetting
Answer: H = 2
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
255.255.255.252 = /30
Usable hosts
22 − 2 = 2
Answer
H = 2
2.IPv4 subnetting
Answer: x = 64
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
E: 53 hosts → 26 = 64 (/26)
A: 27 hosts → 25 = 32 (/27)
D: 14 hosts → 24 = 16 (/28)
C: 10 hosts → 24 = 16 (/28)
B: 6 hosts → 23 = 8 (/29)
Allocation, largest first
E: .0 – .63
A: .64 – .95
D: .96 – .111
C: .112 – .127
B: .128 – .135
Answer
x = 64
LAN A: 192.168.149.64/27.
3.Computer networks
Answer: C · 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
C · Let many devices with private addresses share one public IP address
4.Computer networks
Answer: C · Make a resolver store a false IP address for a domain
Why this is the answer
Poisoning inserts forged records into a resolver's cache. DNSSEC adds signatures so resolvers can verify records.
Answer
C · Make a resolver store a false IP address for a domain
5.Number systems
Answer: U = 186
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
70 = 0100 0110
inverted: 1011 1001
+1 → 1011 1010
Read it as unsigned
1011 1010₂ = 256 − 70 = 186
Answer
U = 186
6.Number systems
Answer: R = 200
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.
1011 1001 << 3
= 1100 1000
Value of the result
1100 1000₂ = 200
Answer
R = 200
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