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 192.168.65.0/24 is divided into /27 subnets, numbered 1, 2, 3, … from the lowest address (subnet 1 starts at .0). What is the fourth octet x of the network address of subnet 7?
192.168.65.0/24/27subnet 7 (subnet 1 starts at .0)
x =
2.IPv4 subnettingHard
A host has the address 10.20.235.58/28. What is the last number (fourth octet) x of its subnet's broadcast address?
10.20.235.58/28
x =
3.Computer networksMedium
NAT (network address translation) is mainly used to:
- ATranslate domain names into IP addresses for devices on a local network
- BLet many devices with private addresses share one public IP address
- CIncrease the speed of a network cable
- DEncrypt traffic between two sites
4.Computer networksHard
After repeated collisions, classic Ethernet stations choose their waiting time from a range that doubles each time. Why?
- ATo lower the chance of colliding again when many stations compete
- BTo make the cable carry signals further
- CTo give the station with the highest MAC address priority on the shared cable
- DTo synchronise all stations to one clock
5.Number systemsMedium
Write −54 as an 8-bit two's complement pattern. Read the pattern as an unsigned binary number: what is its value U in decimal?
N = −548 bit
U =
6.Number systemsHard
Set the bits to the 8-bit result of 1011 0001 >> 2. What is the result's value R in decimal?
1011 0001>> 2
R =
3Solutions
1.IPv4 subnetting
Answer: x = 192
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
25 = 32
Subnet number 7
network: (7 − 1)·32 = 192
broadcast: 192 + 32 − 1 = 223
Answer
x = 192
2.IPv4 subnetting
Answer: x = 63
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: ⌊58/16⌋·16 = 48
broadcast: 48 + 16 − 1 = 63
usable hosts: 49 … 62
Answer
x = 63
3.Computer networks
Answer: B · Let many devices with private addresses share one public IP address
Why this is the answer
NAT rewrites addresses (and ports) so private networks can access the internet through a smaller number of public addresses.
Answer
B · Let many devices with private addresses share one public IP address
4.Computer networks
Answer: A · 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
A · To lower the chance of colliding again when many stations compete
5.Number systems
Answer: U = 202
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
54 = 0011 0110
inverted: 1100 1001
+1 → 1100 1010
Read it as unsigned
1100 1010₂ = 256 − 54 = 202
Answer
U = 202
6.Number systems
Answer: R = 44
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
Shifting right by 2 moves every bit 2 places to the right; bits that fall off the right end are lost and zeros come in on the left.
1011 0001 >> 2
= 0010 1100
Value of the result
0010 1100₂ = 44
Answer
R = 44
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