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 10.22.16.0/20 is divided into equal subnets with prefix /28. How many subnets N are there?
10.22.16.0/20/28
N =
2.IPv4 subnettingHard
A router has routes to these networks: 10.12.242.0/24, 10.12.243.0/24. What is the longest prefix length p of a single summary route that covers all of them?
10.12.242.0/24 … 10.12.243.0/24
p =
3.Computer networksMedium
What is the network address of the host 192.168.10.77/26?
- A192.168.10.64
- B192.168.10.76
- C192.168.10.0
- D192.168.10.128
4.Computer networksHard
A firewall blocks all ICMP. Small web requests work, but large downloads over some paths hang. What is the likely cause?
- ADNS needs ICMP to resolve names
- BTLS certificates are sent over ICMP
- CPath MTU discovery fails because "fragmentation needed" messages are blocked
- DTCP cannot open connections without ICMP
5.Number systemsMedium
Set the bits so they store −21 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?
N = −218 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: N = 256
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).
Borrowed bits
28 − 20 = 8
Number of subnets
N = 28 = 256
Answer
N = 256
Each /28 subnet has 24 − 2 = 14 usable hosts.
2.IPv4 subnetting
Answer: p = 23
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
242 = 11110010
243 = 11110011
common leading bits: 7
Prefix length
p = 16 + 7 = 23
Answer
p = /23
Summary route: 10.12.242.0/23
3.Computer networks
Answer: A · 192.168.10.64
Why this is the answer
/26 subnets start at 0, 64, 128 and 192. 77 lies in the block 64–127, so the network address is 192.168.10.64.
Answer
A · 192.168.10.64
4.Computer networks
Answer: C · Path MTU discovery fails because "fragmentation needed" messages are blocked
Why this is the answer
PMTUD relies on ICMP messages to learn the smaller MTU. If they are dropped, large packets vanish silently, creating a "black hole" connection.
Answer
C · Path MTU discovery fails because "fragmentation needed" messages are blocked
5.Number systems
Answer: U = 235
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
21 = 0001 0101
inverted: 1110 1010
+1 → 1110 1011
Read it as unsigned
1110 1011₂ = 256 − 21 = 235
Answer
U = 235
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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