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 1746 host addresses in a single subnet. What is the longest prefix length p (smallest subnet) that is big enough?
1746 hosts
p =
2.IPv4 subnettingHard
A host has the address 10.20.99.203/28. What is the last number (fourth octet) x of its subnet's broadcast address?
10.20.99.203/28
x =
3.Computer networksMedium
Which DNS record type maps a host name to an IPv6 address?
- AA
- BMX
- CCNAME
- DAAAA
4.Computer networksHard
DNS queries normally use UDP. When does DNS typically use TCP instead?
- AOnly for reverse lookups of IP addresses
- BOnly when the domain name is longer than ten characters
- CFor responses too large for UDP and for zone transfers
- DFor every query sent from a web browser
5.Number systemsMedium
Write −84 as an 8-bit two's complement pattern. Read the pattern as an unsigned binary number: what is its value U in decimal?
N = −848 bit
U =
6.Number systemsHard
Set the bits to the 8-bit result of 1101 0011 >> 2. What is the result's value R in decimal?
1101 0011>> 2
R =
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 < 1746
211 − 2 = 2046 ≥ 1746 → h = 11
Prefix length
p = 32 − 11 = 21
Answer
p = /21
2.IPv4 subnetting
Answer: x = 207
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
24 = 16
Subnet of this host
network: ⌊203/16⌋·16 = 192
broadcast: 192 + 16 − 1 = 207
usable hosts: 193 … 206
Answer
x = 207
3.Computer networks
Answer: D · 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
D · AAAA
4.Computer networks
Answer: C · For responses too large for UDP and for zone transfers
Why this is the answer
If a UDP response is truncated, the resolver retries over TCP. Zone transfers between name servers also use TCP because they carry lots of data.
Answer
C · For responses too large for UDP and for zone transfers
5.Number systems
Answer: U = 172
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
84 = 0101 0100
inverted: 1010 1011
+1 → 1010 1100
Read it as unsigned
1010 1100₂ = 256 − 84 = 172
Answer
U = 172
6.Number systems
Answer: R = 52
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 right by 2 moves every bit 2 places to the right; bits that fall off the right end are lost and zeros come in on the left.
1101 0011 >> 2
= 0011 0100
Value of the result
0011 0100₂ = 52
Answer
R = 52
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