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 192.168.58.0/24 is divided into /28 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 6?
192.168.58.0/24/28subnet 6 (subnet 1 starts at .0)
x =
2.IPv4 subnettingHard
A router has routes to these networks: 10.237.243.0/24, 10.237.244.0/24. What is the longest prefix length p of a single summary route that covers all of them?
10.237.243.0/24 … 10.237.244.0/24
p =
3.Computer networksMedium
What is a cyclic redundancy check (CRC) used for in Ethernet frames?
- ADetecting transmission errors
- BCorrecting every bit error automatically
- CChoosing the best route
- DEncrypting the frame
4.Computer networksHard
A remote-terminal application feels sluggish because small keystroke packets are held back. Which TCP mechanism is often responsible?
- AFast retransmit
- BThe three-way handshake
- CSlow start
- DNagle's algorithm
5.Number systemsMedium
Write −111 as an 8-bit two's complement pattern. Read the pattern as an unsigned binary number: what is its value U in decimal?
N = −1118 bit
U =
6.Number systemsHard
An uncompressed image is 1024 × 768 pixels with 8 bits per pixel. How much memory M does it take, in KiB (1 KiB = 1024 bytes)?
1024 × 768 px8 bit/px1 KiB = 1024 B
M = KiB
3Solutions
1.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 number 6
network: (6 − 1)·16 = 80
broadcast: 80 + 16 − 1 = 95
Answer
x = 95
2.IPv4 subnetting
Answer: p = 21
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
243 = 11110011
244 = 11110100
common leading bits: 5
Prefix length
p = 16 + 5 = 21
Answer
p = /21
Summary route: 10.237.240.0/21 (it also covers 6 network(s) that are not in the list).
3.Computer networks
Answer: A · 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
A · Detecting transmission errors
4.Computer networks
Answer: D · Nagle's algorithm
Why this is the answer
Nagle's algorithm waits to bundle small segments while earlier data is unacknowledged. This saves bandwidth but adds delay for interactive traffic.
Answer
D · Nagle's algorithm
5.Number systems
Answer: U = 145
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
111 = 0110 1111
inverted: 1001 0000
+1 → 1001 0001
Read it as unsigned
1001 0001₂ = 256 − 111 = 145
Answer
U = 145
6.Number systems
Answer: M = 768 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
1024·768·8 = 6291456 bit = 786432 B
Convert to KiB
786432/1024 = 768 KiB
Answer
M = 768 KiB
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