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 10.44.80.0/22 is divided into equal subnets with prefix /30. How many subnets N are there?
10.44.80.0/22/30
N =
2.IPv4 subnettingHard
A router has routes to these networks: 10.23.166.0/24, 10.23.167.0/24, 10.23.168.0/24, 10.23.169.0/24, 10.23.170.0/24, 10.23.171.0/24, 10.23.172.0/24, 10.23.173.0/24. What is the longest prefix length p of a single summary route that covers all of them?
10.23.166.0/24 … 10.23.173.0/24
p =
3.Computer networksMedium
Why does Wi-Fi use collision avoidance (CSMA/CA) instead of collision detection?
- ACollision detection only works on fibre cables, not on shared wireless radio channels
- BA wireless station cannot reliably detect collisions while it is transmitting
- CRadio signals never collide
- DWi-Fi only allows one device per network
4.Computer networksHard
What is the maximum transmission unit (MTU) of standard Ethernet payload?
- A64 bytes
- B9000 bytes in every network
- C1500 bytes
- D65,535 bytes
5.Number systemsMedium
Write −6 as an 8-bit two's complement pattern. Read the pattern as an unsigned binary number: what is its value U in decimal?
N = −68 bit
U =
6.Number systemsHard
A system must give every one of 4525 items its own binary code of the same length. What is the smallest number of bits b needed?
4525
b = bit
3Solutions
1.IPv4 subnetting
Answer: N = 256
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
30 − 22 = 8
Number of subnets
N = 28 = 256
Answer
N = 256
Each /30 subnet has 22 − 2 = 2 usable hosts.
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
166 = 10100110
173 = 10101101
common leading bits: 4
Prefix length
p = 16 + 4 = 20
Answer
p = /20
Summary route: 10.23.160.0/20 (it also covers 8 network(s) that are not in the list).
3.Computer networks
Answer: B · A wireless station cannot reliably detect collisions while it is transmitting
Why this is the answer
A radio transmitter cannot hear other weak signals while sending, and some stations are hidden from each other. So 802.11 waits random backoff times and uses acknowledgements.
Answer
B · A wireless station cannot reliably detect collisions while it is transmitting
4.Computer networks
Answer: C · 1500 bytes
Why this is the answer
Standard Ethernet carries up to 1500 bytes of payload; larger IP packets must be fragmented or use path MTU discovery.
Answer
C · 1500 bytes
5.Number systems
Answer: U = 250
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
6 = 0000 0110
inverted: 1111 1001
+1 → 1111 1010
Read it as unsigned
1111 1010₂ = 256 − 6 = 250
Answer
U = 250
6.Number systems
Answer: b = 13 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
212 = 4096 < 4525
213 = 8192 ≥ 4525
Answer
b = 13
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