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 479 host addresses in a single subnet. What is the longest prefix length p (smallest subnet) that is big enough?
479 hosts
p =
2.IPv4 subnettingHard
A host has the address 10.20.195.84/28. What is the last number (fourth octet) x of its subnet's broadcast address?
10.20.195.84/28
x =
3.Computer networksMedium
What is a cyclic redundancy check (CRC) used for in Ethernet frames?
- AChoosing the best route
- BEncrypting the frame
- CCorrecting every bit error automatically
- DDetecting transmission errors
4.Computer networksHard
After repeated collisions, classic Ethernet stations choose their waiting time from a range that doubles each time. Why?
- ATo make the cable carry signals further
- BTo synchronise all stations to one clock
- CTo lower the chance of colliding again when many stations compete
- DTo give the station with the highest MAC address priority on the shared cable
5.Number systemsMedium
Set the bits so they store −20 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?
N = −208 bit
U =
6.Number systemsHard
A system must give every one of 4398 items its own binary code of the same length. What is the smallest number of bits b needed?
4398
b = bit
3Solutions
1.IPv4 subnetting
Answer: p = 23
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
28 − 2 = 254 < 479
29 − 2 = 510 ≥ 479 → h = 9
Prefix length
p = 32 − 9 = 23
Answer
p = /23
2.IPv4 subnetting
Answer: x = 95
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).
Block size
24 = 16
Subnet of this host
network: ⌊84/16⌋·16 = 80
broadcast: 80 + 16 − 1 = 95
usable hosts: 81 … 94
Answer
x = 95
3.Computer networks
Answer: D · Detecting transmission errors
Why this is the answer
The sender computes a CRC over the frame; the receiver recomputes it and discards the frame if they differ. CRC detects errors but does not correct them.
Answer
D · Detecting transmission errors
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 = 236
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
20 = 0001 0100
inverted: 1110 1011
+1 → 1110 1100
Read it as unsigned
1110 1100₂ = 256 − 20 = 236
Answer
U = 236
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 < 4398
213 = 8192 ≥ 4398
Answer
b = 13
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