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 192.168.216.0/24 is divided into /26 subnets, numbered 1, 2, 3, … from the lowest address (subnet 1 starts at .0). What is the fourth octet x of the broadcast address of subnet 3?
192.168.216.0/24/26subnet 3 (subnet 1 starts at .0)
x =
2.IPv4 subnettingHard
A host has the address 172.20.58.107/21. What is the third octet x of its network address?
172.20.58.107/21
x =
3.Computer networksMedium
What is the main difference between the HTTP methods GET and POST?
- AGET retrieves a resource, POST submits data in the request body
- BGET uses UDP, POST uses TCP
- CGET deletes a resource, POST reads it from the server's database
- DGET is encrypted, POST is not
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 lower the chance of colliding again when many stations compete
- CTo synchronise all stations to one clock
- DTo give the station with the highest MAC address priority on the shared cable
5.Number systemsMedium
Set the bits so they store −3 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?
N = −38 bit
U =
6.Number systemsHard
Set the bits to the 8-bit result of 1111 0110 XOR 0100 1111. What is the result's value R in decimal?
1111 0110XOR0100 1111
R =
3Solutions
1.IPv4 subnetting
Answer: x = 191
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
26 = 64
Subnet number 3
network: (3 − 1)·64 = 128
broadcast: 128 + 64 − 1 = 191
Answer
x = 191
2.IPv4 subnetting
Answer: x = 56
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 in the third octet
23 = 8
Subnet of this host
network: 172.20.56.0
broadcast: 172.20.63.255
Answer
x = 56
3.Computer networks
Answer: A · GET retrieves a resource, POST submits data in the request body
Why this is the answer
GET asks for a resource and should not change server state. POST sends data, such as a form, in the body for the server to process.
Answer
A · GET retrieves a resource, POST submits data in the request body
4.Computer networks
Answer: B · 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
B · To lower the chance of colliding again when many stations compete
5.Number systems
Answer: U = 253
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
3 = 0000 0011
inverted: 1111 1100
+1 → 1111 1101
Read it as unsigned
1111 1101₂ = 256 − 3 = 253
Answer
U = 253
6.Number systems
Answer: R = 185
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.
1111 0110
XOR 0100 1111
= 1011 1001
Value of the result
1011 1001₂ = 185
Answer
R = 185
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