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 192.168.10.0/24 must be divided into at least 5 equal subnets. If you borrow as few bits as possible, how many usable hosts H does each subnet have?
192.168.10.0/245 subnets
H =
2.IPv4 subnettingHard
A host has the address 10.20.58.145/28. What is the last number (fourth octet) x of its subnet's broadcast address?
10.20.58.145/28
x =
3.Computer networksMedium
Which DNS record type maps a host name to an IPv6 address?
- ACNAME
- BA
- CAAAA
- DMX
4.Computer networksHard
A firewall blocks all ICMP. Small web requests work, but large downloads over some paths hang. What is the likely cause?
- ATCP cannot open connections without ICMP
- BDNS needs ICMP to resolve names
- CTLS certificates are sent over ICMP
- DPath MTU discovery fails because "fragmentation needed" messages are blocked
5.Number systemsMedium
Set the bits so they store −92 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?
N = −928 bit
U =
6.Number systemsHard
What is the largest unsigned integer N that fits in 16 bits?
16 bit
N =
3Solutions
1.IPv4 subnetting
Answer: H = 30
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).
Bits to borrow
23 = 8 ≥ 5 → b = 3, /27
Hosts per subnet
25 − 2 = 30
Answer
H = 30
2.IPv4 subnetting
Answer: x = 159
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: ⌊145/16⌋·16 = 144
broadcast: 144 + 16 − 1 = 159
usable hosts: 145 … 158
Answer
x = 159
3.Computer networks
Answer: C · 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
C · AAAA
4.Computer networks
Answer: D · Path MTU discovery fails because "fragmentation needed" messages are blocked
Why this is the answer
PMTUD relies on ICMP messages to learn the smaller MTU. If they are dropped, large packets vanish silently, creating a "black hole" connection.
Answer
D · Path MTU discovery fails because "fragmentation needed" messages are blocked
5.Number systems
Answer: U = 164
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
92 = 0101 1100
inverted: 1010 0011
+1 → 1010 0100
Read it as unsigned
1010 0100₂ = 256 − 92 = 164
Answer
U = 164
6.Number systems
Answer: N = 65535
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.
Largest value
216 − 1 = 65535
Answer
N = 65535
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