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 30 equal subnets. If you borrow as few bits as possible, how many usable hosts H does each subnet have?
192.168.10.0/2430 subnets
H =
2.IPv4 subnettingHard
A router has routes to these networks: 10.19.83.0/24, 10.19.84.0/24, 10.19.85.0/24, 10.19.86.0/24, 10.19.87.0/24, 10.19.88.0/24. What is the longest prefix length p of a single summary route that covers all of them?
10.19.83.0/24 … 10.19.88.0/24
p =
3.Computer networksMedium
What is the IPv6 loopback address?
- Afe80::1
- B::1
- Cff02::1
- D::
4.Computer networksHard
A server's private key is stolen a year from now. Which TLS property keeps today's recorded sessions safe?
- AUsing a longer domain name
- BSending the certificate twice
- CForward secrecy from ephemeral Diffie-Hellman key exchange
- DCompressing the data before encryption
5.Number systemsMedium
Set the bits so they store −24 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?
N = −248 bit
U =
6.Number systemsHard
What is the largest unsigned integer N that fits in 32 bits?
32 bit
N =
3Solutions
1.IPv4 subnetting
Answer: H = 6
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
25 = 32 ≥ 30 → b = 5, /29
Hosts per subnet
23 − 2 = 6
Answer
H = 6
2.IPv4 subnetting
Answer: p = 20
Route summarisation: a summary route a.b.c.0/p covers all networks whose first p bits match. Write the differing octet of the first and last network in binary; the number of common leading bits, plus the bits of the octets before it, is the longest summary prefix.
Compare the bits
83 = 01010011
88 = 01011000
common leading bits: 4
Prefix length
p = 16 + 4 = 20
Answer
p = /20
Summary route: 10.19.80.0/20 (it also covers 10 network(s) that are not in the list).
3.Computer networks
Answer: B · ::1
Why this is the answer
::1 is the loopback address. :: is the unspecified address, fe80::/10 is link-local and ff02::1 is the all-nodes multicast group.
Answer
B · ::1
4.Computer networks
Answer: C · Forward secrecy from ephemeral Diffie-Hellman key exchange
Why this is the answer
With ephemeral (EC)DHE, each session key is derived from temporary values that are discarded. The long-term key only signs the exchange, so stealing it later does not decrypt old traffic.
Answer
C · Forward secrecy from ephemeral Diffie-Hellman key exchange
5.Number systems
Answer: U = 232
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
24 = 0001 1000
inverted: 1110 0111
+1 → 1110 1000
Read it as unsigned
1110 1000₂ = 256 − 24 = 232
Answer
U = 232
6.Number systems
Answer: N = 4.295 × 10⁹
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
232 − 1 = 4294967295
Answer
N = 4294967295
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