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 1298 host addresses in a single subnet. What is the longest prefix length p (smallest subnet) that is big enough?
1298 hosts
p =
2.IPv4 subnettingHard
A host has the address 10.20.82.51/26. What is the fourth octet x of the last usable host address in its subnet?
10.20.82.51/26
x =
3.Computer networksMedium
Which subnet mask corresponds to /20?
- A255.255.240.0
- B255.255.255.240
- C255.255.224.0
- D255.255.248.0
4.Computer networksHard
Why does IPv6 not need NAT for address shortage?
- AIts 128-bit address space is large enough to give every device a global address
- BIPv6 only works on local networks, so addresses never need to be shared globally
- CIPv6 encrypts all addresses
- DIPv6 does not use addresses
5.Number systemsMedium
Set the bits so they store −90 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?
N = −908 bit
U =
6.Number systemsHard
An uncompressed image is 256 × 256 pixels with 32 bits per pixel. How much memory M does it take, in KiB (1 KiB = 1024 bytes)?
256 × 256 px32 bit/px1 KiB = 1024 B
M = KiB
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 < 1298
211 − 2 = 2046 ≥ 1298 → h = 11
Prefix length
p = 32 − 11 = 21
Answer
p = /21
2.IPv4 subnetting
Answer: x = 62
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: ⌊51/64⌋·64 = 0
broadcast: 0 + 64 − 1 = 63
usable hosts: 1 … 62
Answer
x = 62
3.Computer networks
Answer: A · 255.255.240.0
Why this is the answer
/20 means 8 + 8 + 4 one-bits. Four one-bits in the third octet are 11110000 = 240, so the mask is 255.255.240.0.
Answer
A · 255.255.240.0
4.Computer networks
Answer: A · Its 128-bit address space is large enough to give every device a global address
Why this is the answer
IPv6 has about 3.4 × 10³⁸ addresses, so the scarcity that drove NAT in IPv4 disappears; firewalls still provide security.
Answer
A · Its 128-bit address space is large enough to give every device a global address
5.Number systems
Answer: U = 166
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
90 = 0101 1010
inverted: 1010 0101
+1 → 1010 0110
Read it as unsigned
1010 0110₂ = 256 − 90 = 166
Answer
U = 166
6.Number systems
Answer: M = 256 KiB
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.
Bits and bytes
256·256·32 = 2097152 bit = 262144 B
Convert to KiB
262144/1024 = 256 KiB
Answer
M = 256 KiB
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