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 10.18.180.0/24 is divided into equal subnets with prefix /26. How many subnets N are there?
10.18.180.0/24/26
N =
2.IPv4 subnettingHard
A host has the address 10.20.104.172/27. What is the fourth octet x of the last usable host address in its subnet?
10.20.104.172/27
x =
3.Computer networksMedium
Which DNS record type maps a host name to an IPv6 address?
- AA
- BAAAA
- CCNAME
- DMX
4.Computer networksHard
Which of these addresses is a public (not RFC 1918 private) IPv4 address?
- A172.32.0.1
- B172.31.0.1
- C192.168.255.1
- D10.255.0.1
5.Number systemsMedium
Write −57 as an 8-bit two's complement pattern. Read the pattern as an unsigned binary number: what is its value U in decimal?
N = −578 bit
U =
6.Number systemsHard
An uncompressed image is 512 × 512 pixels with 16 bits per pixel. How much memory M does it take, in KiB (1 KiB = 1024 bytes)?
512 × 512 px16 bit/px1 KiB = 1024 B
M = KiB
3Solutions
1.IPv4 subnetting
Answer: N = 4
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).
Borrowed bits
26 − 24 = 2
Number of subnets
N = 22 = 4
Answer
N = 4
Each /26 subnet has 26 − 2 = 62 usable hosts.
2.IPv4 subnetting
Answer: x = 190
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
25 = 32
Subnet of this host
network: ⌊172/32⌋·32 = 160
broadcast: 160 + 32 − 1 = 191
usable hosts: 161 … 190
Answer
x = 190
3.Computer networks
Answer: B · AAAA
Why this is the answer
An A record holds an IPv4 address, AAAA an IPv6 address. MX names mail servers and CNAME gives an alias.
Answer
B · AAAA
4.Computer networks
Answer: A · 172.32.0.1
Why this is the answer
172.16.0.0/12 runs from 172.16.0.0 to 172.31.255.255, so 172.32.0.1 lies outside it. The other three are in private ranges.
Answer
A · 172.32.0.1
5.Number systems
Answer: U = 199
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
57 = 0011 1001
inverted: 1100 0110
+1 → 1100 0111
Read it as unsigned
1100 0111₂ = 256 − 57 = 199
Answer
U = 199
6.Number systems
Answer: M = 512 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
512·512·16 = 4194304 bit = 524288 B
Convert to KiB
524288/1024 = 512 KiB
Answer
M = 512 KiB
ExamProblems · examproblems.com/courses/computer-networks