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 1272 host addresses in a single subnet. What is the longest prefix length p (smallest subnet) that is big enough?
1272 hosts
p =
2.IPv4 subnettingHard
A router has routes to these networks: 10.69.185.0/24, 10.69.186.0/24, 10.69.187.0/24. What is the longest prefix length p of a single summary route that covers all of them?
10.69.185.0/24 … 10.69.187.0/24
p =
3.Computer networksMedium
TLS provides which security properties for HTTPS?
- AOnly faster loading
- BProtection against all malware and phishing on the visited site
- CAnonymity of the user
- DConfidentiality, integrity and server authentication
4.Computer networksHard
After repeated collisions, classic Ethernet stations choose their waiting time from a range that doubles each time. Why?
- ATo give the station with the highest MAC address priority on the shared cable
- BTo make the cable carry signals further
- CTo lower the chance of colliding again when many stations compete
- DTo synchronise all stations to one clock
5.Number systemsMedium
Write −10 as an 8-bit two's complement pattern. Read the pattern as an unsigned binary number: what is its value U in decimal?
N = −108 bit
U =
6.Number systemsHard
Set the bits to the 8-bit result of 0101 0101 XOR 1001 1110. What is the result's value R in decimal?
0101 0101XOR1001 1110
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 < 1272
211 − 2 = 2046 ≥ 1272 → h = 11
Prefix length
p = 32 − 11 = 21
Answer
p = /21
2.IPv4 subnetting
Answer: p = 22
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
185 = 10111001
187 = 10111011
common leading bits: 6
Prefix length
p = 16 + 6 = 22
Answer
p = /22
Summary route: 10.69.184.0/22 (it also covers 1 network(s) that are not in the list).
3.Computer networks
Answer: D · Confidentiality, integrity and server authentication
Why this is the answer
TLS encrypts traffic, detects tampering and authenticates the server with certificates (optionally the client too).
Answer
D · Confidentiality, integrity and server authentication
4.Computer networks
Answer: C · To lower the chance of colliding again when many stations compete
Why this is the answer
Each further collision suggests more contention, so the random range grows and stations are more likely to pick different slots. It adapts to load without central control.
Answer
C · To lower the chance of colliding again when many stations compete
5.Number systems
Answer: U = 246
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
10 = 0000 1010
inverted: 1111 0101
+1 → 1111 0110
Read it as unsigned
1111 0110₂ = 256 − 10 = 246
Answer
U = 246
6.Number systems
Answer: R = 203
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
XOR gives 1 where the two bits differ.
0101 0101
XOR 1001 1110
= 1100 1011
Value of the result
1100 1011₂ = 203
Answer
R = 203
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