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 10.57.151.0/24 is divided into equal subnets with prefix /28. How many subnets N are there?
10.57.151.0/24/28
N =
2.IPv4 subnettingHard
The network 192.168.229.0/24 is split with VLSM for these LANs: A: 53 hosts, B: 6 hosts, C: 13 hosts, D: 8 hosts, E: 25 hosts. Subnets are allocated from .0 upwards, the LAN with the most hosts first, each with the smallest subnet that fits. What is the fourth octet x of the network address of LAN D?
A: 53B: 6C: 13D: 8E: 25
x =
3.Computer networksMedium
How many usable host addresses does a /24 IPv4 network have?
- A254
- B252
- C256
- D255
4.Computer networksHard
What does BGP mainly do on the internet?
- AEncrypts data between routers
- BResolves names to addresses
- CExchanges routing information between autonomous systems
- DAssigns IP addresses to home users through their internet providers
5.Number systemsMedium
Write −74 as an 8-bit two's complement pattern. Read the pattern as an unsigned binary number: what is its value U in decimal?
N = −748 bit
U =
6.Number systemsHard
An uncompressed image is 512 × 512 pixels with 24 bits per pixel. How much memory M does it take, in KiB (1 KiB = 1024 bytes)?
512 × 512 px24 bit/px1 KiB = 1024 B
M = KiB
3Solutions
1.IPv4 subnetting
Answer: N = 16
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 − 24 = 4
Number of subnets
N = 24 = 16
Answer
N = 16
Each /28 subnet has 24 − 2 = 14 usable hosts.
2.IPv4 subnetting
Answer: x = 112
VLSM: each LAN gets the smallest subnet with 2ʰ − 2 ≥ its host count, i.e. prefix 32 − h. Allocating the largest subnets first keeps every block aligned on a multiple of its own size, so the network addresses follow by adding up the block sizes.
Block size per LAN
A: 53 hosts → 26 = 64 (/26)
E: 25 hosts → 25 = 32 (/27)
C: 13 hosts → 24 = 16 (/28)
D: 8 hosts → 24 = 16 (/28)
B: 6 hosts → 23 = 8 (/29)
Allocation, largest first
A: .0 – .63
E: .64 – .95
C: .96 – .111
D: .112 – .127
B: .128 – .135
Answer
x = 112
LAN D: 192.168.229.112/28.
3.Computer networks
Answer: A · 254
Why this is the answer
A /24 has 8 host bits, so 2⁸ = 256 addresses. The network and broadcast addresses are reserved, leaving 254 for hosts.
Answer
A · 254
4.Computer networks
Answer: C · Exchanges routing information between autonomous systems
Why this is the answer
BGP is the inter-domain routing protocol; autonomous systems advertise reachable prefixes and select paths based on policies.
Answer
C · Exchanges routing information between autonomous systems
5.Number systems
Answer: U = 182
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
74 = 0100 1010
inverted: 1011 0101
+1 → 1011 0110
Read it as unsigned
1011 0110₂ = 256 − 74 = 182
Answer
U = 182
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
512·512·24 = 6291456 bit = 786432 B
Convert to KiB
786432/1024 = 768 KiB
Answer
M = 768 KiB
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