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 172.31.0.0/16 is divided into equal subnets with prefix /19. How many subnets N are there?
172.31.0.0/16/19
N =
2.IPv4 subnettingHard
A host has the address 10.20.51.77/30. What is the last number (fourth octet) x of its subnet's broadcast address?
10.20.51.77/30
x =
3.Computer networksMedium
Which single prefix summarises both 192.168.0.0/24 and 192.168.1.0/24?
- A192.168.1.0/23
- B192.168.0.0/23
- C192.168.0.0/22
- D192.168.0.0/25
4.Computer networksHard
A firewall blocks all ICMP. Small web requests work, but large downloads over some paths hang. What is the likely cause?
- ATLS certificates are sent over ICMP
- BTCP cannot open connections without ICMP
- CDNS needs ICMP to resolve names
- DPath MTU discovery fails because "fragmentation needed" messages are blocked
5.Number systemsMedium
Write −38 as an 8-bit two's complement pattern. Read the pattern as an unsigned binary number: what is its value U in decimal?
N = −388 bit
U =
6.Number systemsHard
An uncompressed image is 640 × 480 pixels with 8 bits per pixel. How much memory M does it take, in KiB (1 KiB = 1024 bytes)?
640 × 480 px8 bit/px1 KiB = 1024 B
M = KiB
3Solutions
1.IPv4 subnetting
Answer: N = 8
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
19 − 16 = 3
Number of subnets
N = 23 = 8
Answer
N = 8
Each /19 subnet has 213 − 2 = 8190 usable hosts.
2.IPv4 subnetting
Answer: x = 79
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
22 = 4
Subnet of this host
network: ⌊77/4⌋·4 = 76
broadcast: 76 + 4 − 1 = 79
usable hosts: 77 … 78
Answer
x = 79
3.Computer networks
Answer: B · 192.168.0.0/23
Why this is the answer
0 and 1 share all bits of the third octet except the last, so one bit less of prefix (/23) starting at 192.168.0.0 covers both exactly.
Answer
B · 192.168.0.0/23
4.Computer networks
Answer: D · 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
D · Path MTU discovery fails because "fragmentation needed" messages are blocked
5.Number systems
Answer: U = 218
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
38 = 0010 0110
inverted: 1101 1001
+1 → 1101 1010
Read it as unsigned
1101 1010₂ = 256 − 38 = 218
Answer
U = 218
6.Number systems
Answer: M = 300 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
640·480·8 = 2457600 bit = 307200 B
Convert to KiB
307200/1024 = 300 KiB
Answer
M = 300 KiB
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