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 1926 host addresses in a single subnet. What is the longest prefix length p (smallest subnet) that is big enough?
1926 hosts
p =
2.IPv4 subnettingHard
A router has routes to these networks: 10.47.205.0/24, 10.47.206.0/24, 10.47.207.0/24, 10.47.208.0/24. What is the longest prefix length p of a single summary route that covers all of them?
10.47.205.0/24 … 10.47.208.0/24
p =
3.Computer networksMedium
Which host is in the same /16 subnet as 10.1.2.3?
- A11.1.2.3
- B10.2.2.3
- C10.0.2.3
- D10.1.200.5
4.Computer networksHard
Why does TCP choose an unpredictable initial sequence number for each connection?
- ATo compress the data more efficiently
- BTo select the best route to the destination for the segments of that connection
- CTo encrypt the payload of each segment
- DTo avoid confusion with old segments and make spoofed segments harder to inject
5.Number systemsMedium
Set the bits so they store −35 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?
N = −358 bit
U =
6.Number systemsHard
A system must give every one of 349 items its own binary code of the same length. What is the smallest number of bits b needed?
349
b = bit
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 < 1926
211 − 2 = 2046 ≥ 1926 → h = 11
Prefix length
p = 32 − 11 = 21
Answer
p = /21
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
205 = 11001101
208 = 11010000
common leading bits: 3
Prefix length
p = 16 + 3 = 19
Answer
p = /19
Summary route: 10.47.192.0/19 (it also covers 28 network(s) that are not in the list).
3.Computer networks
Answer: D · 10.1.200.5
Why this is the answer
With /16 the first 16 bits (10.1) are the network part. Only 10.1.200.5 shares them, so it is on 10.1.0.0/16.
Answer
D · 10.1.200.5
4.Computer networks
Answer: D · To avoid confusion with old segments and make spoofed segments harder to inject
Why this is the answer
Random initial sequence numbers stop delayed segments from old connections being accepted and make it hard for off-path attackers to forge valid segments.
Answer
D · To avoid confusion with old segments and make spoofed segments harder to inject
5.Number systems
Answer: U = 221
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
35 = 0010 0011
inverted: 1101 1100
+1 → 1101 1101
Read it as unsigned
1101 1101₂ = 256 − 35 = 221
Answer
U = 221
6.Number systems
Answer: b = 9 bit
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.
Powers of two
28 = 256 < 349
29 = 512 ≥ 349
Answer
b = 9
ExamProblems · examproblems.com/courses/computer-networks