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 255.255.254.0. How many usable host addresses H does it have?
255.255.254.0
H =
2.IPv4 subnettingHard
A router has routes to these networks: 10.222.222.0/24, 10.222.223.0/24, 10.222.224.0/24, 10.222.225.0/24, 10.222.226.0/24, 10.222.227.0/24, 10.222.228.0/24, 10.222.229.0/24. What is the longest prefix length p of a single summary route that covers all of them?
10.222.222.0/24 … 10.222.229.0/24
p =
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
In a TCP connection, what does the receive window (rwnd) control?
- AThe size of each IP packet header
- BHow much unacknowledged data the sender may send, based on the receiver's buffer
- CThe retransmission time-out, calculated from the measured round-trip time of each segment
- DThe number of open ports
5.Number systemsMedium
Set the bits so they store −25 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?
N = −258 bit
U =
6.Number systemsHard
An uncompressed image is 512 × 512 pixels with 32 bits per pixel. How much memory M does it take, in KiB (1 KiB = 1024 bytes)?
512 × 512 px32 bit/px1 KiB = 1024 B
M = KiB
3Solutions
1.IPv4 subnetting
Answer: H = 510
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
255.255.254.0 = /23
Usable hosts
29 − 2 = 510
Answer
H = 510
2.IPv4 subnetting
Answer: p = 18
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
222 = 11011110
229 = 11100101
common leading bits: 2
Prefix length
p = 16 + 2 = 18
Answer
p = /18
Summary route: 10.222.192.0/18 (it also covers 56 network(s) that are not in the list).
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 · How much unacknowledged data the sender may send, based on the receiver's buffer
Why this is the answer
The receiver advertises how much buffer space it has; the sender limits unacknowledged data accordingly, preventing receiver overload.
Answer
B · How much unacknowledged data the sender may send, based on the receiver's buffer
5.Number systems
Answer: U = 231
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
25 = 0001 1001
inverted: 1110 0110
+1 → 1110 0111
Read it as unsigned
1110 0111₂ = 256 − 25 = 231
Answer
U = 231
6.Number systems
Answer: M = 1024 KiB
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.
Bits and bytes
512·512·32 = 8388608 bit = 1048576 B
Convert to KiB
1048576/1024 = 1024 KiB
Answer
M = 1024 KiB
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