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 10.250.164.0/22 is divided into equal subnets with prefix /28. How many subnets N are there?
10.250.164.0/22/28
N =
2.IPv4 subnettingHard
A host has the address 172.20.191.66/19. What is the third octet x of its network address?
172.20.191.66/19
x =
3.Computer networksMedium
A browser receives HTTP status code 404. What does it mean?
- AThe request succeeded
- BThe server had an internal error
- CThe requested resource was not found
- DThe client must log in first
4.Computer networksHard
Which mechanism prevents broadcast loops in a network with redundant switch links?
- ADHCP snooping
- BSpanning Tree Protocol
- CNAT
- DARP caching
5.Number systemsMedium
Set the bits so they store −72 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?
N = −728 bit
U =
6.Number systemsHard
An uncompressed image is 512 × 512 pixels with 16 bits per pixel. How much memory M does it take, in KiB (1 KiB = 1024 bytes)?
512 × 512 px16 bit/px1 KiB = 1024 B
M = KiB
3Solutions
1.IPv4 subnetting
Answer: N = 64
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 − 22 = 6
Number of subnets
N = 26 = 64
Answer
N = 64
Each /28 subnet has 24 − 2 = 14 usable hosts.
2.IPv4 subnetting
Answer: x = 160
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 in the third octet
25 = 32
Subnet of this host
network: 172.20.160.0
broadcast: 172.20.191.255
Answer
x = 160
3.Computer networks
Answer: C · The requested resource was not found
Why this is the answer
404 Not Found means the server has no resource at that URL. 200 is OK, 401 is Unauthorized and 500 is Internal Server Error.
Answer
C · The requested resource was not found
4.Computer networks
Answer: B · Spanning Tree Protocol
Why this is the answer
STP elects a root bridge and blocks redundant paths, unblocking them if an active link fails.
Answer
B · Spanning Tree Protocol
5.Number systems
Answer: U = 184
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
72 = 0100 1000
inverted: 1011 0111
+1 → 1011 1000
Read it as unsigned
1011 1000₂ = 256 − 72 = 184
Answer
U = 184
6.Number systems
Answer: M = 512 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
512·512·16 = 4194304 bit = 524288 B
Convert to KiB
524288/1024 = 512 KiB
Answer
M = 512 KiB
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