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 subnet uses the mask 255.255.255.192. How many usable host addresses H does it have?
255.255.255.192
H =
2.IPv4 subnettingHard
A router has routes to these networks: 10.216.116.0/24, 10.216.117.0/24, 10.216.118.0/24, 10.216.119.0/24. What is the longest prefix length p of a single summary route that covers all of them?
10.216.116.0/24 … 10.216.119.0/24
p =
3.Computer networksMedium
What is the main difference between the HTTP methods GET and POST?
- AGET is encrypted, POST is not
- BGET deletes a resource, POST reads it from the server's database
- CGET retrieves a resource, POST submits data in the request body
- DGET uses UDP, POST uses TCP
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 synchronise all stations to one clock
- CTo lower the chance of colliding again when many stations compete
- DTo make the cable carry signals further
5.Number systemsMedium
Write −89 as an 8-bit two's complement pattern. Read the pattern as an unsigned binary number: what is its value U in decimal?
N = −898 bit
U =
6.Number systemsHard
A system must give every one of 4621 items its own binary code of the same length. What is the smallest number of bits b needed?
4621
b = bit
3Solutions
1.IPv4 subnetting
Answer: H = 62
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).
Prefix length
255.255.255.192 = /26
Usable hosts
26 − 2 = 62
Answer
H = 62
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
116 = 01110100
119 = 01110111
common leading bits: 6
Prefix length
p = 16 + 6 = 22
Answer
p = /22
Summary route: 10.216.116.0/22
3.Computer networks
Answer: C · 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
C · GET retrieves a resource, POST submits data in the request body
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 = 167
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
89 = 0101 1001
inverted: 1010 0110
+1 → 1010 0111
Read it as unsigned
1010 0111₂ = 256 − 89 = 167
Answer
U = 167
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 < 4621
213 = 8192 ≥ 4621
Answer
b = 13
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