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 10.85.16.0/20 is divided into equal subnets with prefix /24. How many subnets N are there?
10.85.16.0/20/24
N =
2.IPv4 subnettingHard
A router has routes to these networks: 10.102.47.0/24, 10.102.48.0/24, 10.102.49.0/24, 10.102.50.0/24, 10.102.51.0/24, 10.102.52.0/24, 10.102.53.0/24, 10.102.54.0/24. What is the longest prefix length p of a single summary route that covers all of them?
10.102.47.0/24 … 10.102.54.0/24
p =
3.Computer networksMedium
How many usable host addresses does a /24 IPv4 network have?
- A255
- B252
- C256
- D254
4.Computer networksHard
A server's private key is stolen a year from now. Which TLS property keeps today's recorded sessions safe?
- AForward secrecy from ephemeral Diffie-Hellman key exchange
- BSending the certificate twice
- CUsing a longer domain name
- DCompressing the data before encryption
5.Number systemsMedium
Set the bits so they store −13 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?
N = −138 bit
U =
6.Number systemsHard
Set the bits to the 8-bit result of 1011 1001 OR 1101 0111. What is the result's value R in decimal?
1011 1001OR1101 0111
R =
3Solutions
1.IPv4 subnetting
Answer: N = 16
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
24 − 20 = 4
Number of subnets
N = 24 = 16
Answer
N = 16
Each /24 subnet has 28 − 2 = 254 usable hosts.
2.IPv4 subnetting
Answer: p = 19
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
47 = 00101111
54 = 00110110
common leading bits: 3
Prefix length
p = 16 + 3 = 19
Answer
p = /19
Summary route: 10.102.32.0/19 (it also covers 24 network(s) that are not in the list).
3.Computer networks
Answer: D · 254
Why this is the answer
A /24 has 8 host bits, so 2⁸ = 256 addresses. The network and broadcast addresses are reserved, leaving 254 for hosts.
Answer
D · 254
4.Computer networks
Answer: A · 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
A · Forward secrecy from ephemeral Diffie-Hellman key exchange
5.Number systems
Answer: U = 243
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
13 = 0000 1101
inverted: 1111 0010
+1 → 1111 0011
Read it as unsigned
1111 0011₂ = 256 − 13 = 243
Answer
U = 243
6.Number systems
Answer: R = 255
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
OR gives 1 where at least one bit is 1.
1011 1001
OR 1101 0111
= 1111 1111
Value of the result
1111 1111₂ = 255
Answer
R = 255
ExamProblems · examproblems.com/courses/computer-networks