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 255.255.252.0. How many usable host addresses H does it have?
255.255.252.0
H =
2.IPv4 subnettingHard
A host has the address 10.20.82.73/26. What is the last number (fourth octet) x of its subnet's broadcast address?
10.20.82.73/26
x =
3.Computer networksMedium
NAT (network address translation) is mainly used to:
- ALet many devices with private addresses share one public IP address
- BEncrypt traffic between two sites
- CTranslate domain names into IP addresses for devices on a local network
- DIncrease the speed of a network cable
4.Computer networksHard
How does fragmentation differ in IPv6 compared with IPv4?
- AIPv6 fragments at the transport layer inside TCP
- BIPv6 does not allow packets larger than 576 bytes
- CIPv6 routers fragment every packet larger than 64 bytes before forwarding it on the path
- DRouters never fragment IPv6 packets; only the source can, using path MTU discovery
5.Number systemsMedium
Write −13 as an 8-bit two's complement pattern. Read the pattern as an unsigned binary number: what is its value U in decimal?
N = −138 bit
U =
6.Number systemsHard
Set the bits to the 8-bit result of 1001 0001 OR 0101 0010. What is the result's value R in decimal?
1001 0001OR0101 0010
R =
3Solutions
1.IPv4 subnetting
Answer: H = 1022
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
255.255.252.0 = /22
Usable hosts
210 − 2 = 1022
Answer
H = 1022
2.IPv4 subnetting
Answer: x = 127
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: ⌊73/64⌋·64 = 64
broadcast: 64 + 64 − 1 = 127
usable hosts: 65 … 126
Answer
x = 127
3.Computer networks
Answer: A · 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
A · Let many devices with private addresses share one public IP address
4.Computer networks
Answer: D · Routers never fragment IPv6 packets; only the source can, using path MTU discovery
Why this is the answer
An IPv6 router drops an oversized packet and returns ICMPv6 Packet Too Big. The source then sends smaller packets or uses a fragment extension header.
Answer
D · Routers never fragment IPv6 packets; only the source can, using path MTU discovery
5.Number systems
Answer: U = 243
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
13 = 0000 1101
inverted: 1111 0010
+1 → 1111 0011
Read it as unsigned
1111 0011₂ = 256 − 13 = 243
Answer
U = 243
6.Number systems
Answer: R = 211
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
OR gives 1 where at least one bit is 1.
1001 0001
OR 0101 0010
= 1101 0011
Value of the result
1101 0011₂ = 211
Answer
R = 211
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