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
A department needs 481 host addresses in a single subnet. What is the longest prefix length p (smallest subnet) that is big enough?
481 hosts
p =
2.IPv4 subnettingHard
The network 192.168.203.0/24 is split with VLSM for these LANs: A: 48 hosts, B: 12 hosts, C: 30 hosts, D: 74 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 C?
A: 48B: 12C: 30D: 74
x =
3.Computer networksMedium
Which single prefix summarises both 192.168.0.0/24 and 192.168.1.0/24?
- A192.168.0.0/22
- B192.168.0.0/25
- C192.168.0.0/23
- D192.168.1.0/23
4.Computer networksHard
By the Nyquist formula, what is the maximum bit rate of a noiseless 3 kHz channel using 4 signal levels?
- A24 kbit/s
- B12 kbit/s
- C6 kbit/s
- D3 kbit/s
5.Number systemsMedium
The 8-bit pattern 1000 1011 stores a signed integer in two's complement. What is its value N in decimal?
1000 10118 bit
N =
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: p = 23
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).
Host bits needed
28 − 2 = 254 < 481
29 − 2 = 510 ≥ 481 → h = 9
Prefix length
p = 32 − 9 = 23
Answer
p = /23
2.IPv4 subnetting
Answer: x = 192
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
D: 74 hosts → 27 = 128 (/25)
A: 48 hosts → 26 = 64 (/26)
C: 30 hosts → 25 = 32 (/27)
B: 12 hosts → 24 = 16 (/28)
Allocation, largest first
D: .0 – .127
A: .128 – .191
C: .192 – .223
B: .224 – .239
Answer
x = 192
LAN C: 192.168.203.192/27.
3.Computer networks
Answer: C · 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
C · 192.168.0.0/23
4.Computer networks
Answer: B · 12 kbit/s
Why this is the answer
2 × 3000 × log₂ 4 = 2 × 3000 × 2 = 12,000 bit/s. More levels raise the rate, but noise limits how many can be told apart.
Answer
B · 12 kbit/s
5.Number systems
Answer: N = −117
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.
Add the place values
The other bits keep their normal positive values.
−128 + 8 + 2 + 1 = −117
Answer
N = −117
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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