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 subnet uses the mask 255.255.252.0. How many usable host addresses H does it have?
255.255.252.0
H =
2.IPv4 subnettingHard
The network 192.168.181.0/24 is split with VLSM for these LANs: A: 3 hosts, B: 12 hosts, C: 47 hosts, D: 21 hosts, E: 14 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 B?
A: 3B: 12C: 47D: 21E: 14
x =
3.Computer networksMedium
What is a cyclic redundancy check (CRC) used for in Ethernet frames?
- AEncrypting the frame
- BDetecting transmission errors
- CCorrecting every bit error automatically
- DChoosing the best route
4.Computer networksHard
By Shannon's theorem, what is the capacity of a 1 MHz channel with a signal-to-noise ratio of 255?
- A8 Mbit/s
- B255 Mbit/s
- C1 Mbit/s
- D2 Mbit/s
5.Number systemsMedium
Set the bits so they store −80 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?
N = −808 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: H = 1022
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).
Prefix length
255.255.252.0 = /22
Usable hosts
210 − 2 = 1022
Answer
H = 1022
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
C: 47 hosts → 26 = 64 (/26)
D: 21 hosts → 25 = 32 (/27)
E: 14 hosts → 24 = 16 (/28)
B: 12 hosts → 24 = 16 (/28)
A: 3 hosts → 23 = 8 (/29)
Allocation, largest first
C: .0 – .63
D: .64 – .95
E: .96 – .111
B: .112 – .127
A: .128 – .135
Answer
x = 112
LAN B: 192.168.181.112/28.
3.Computer networks
Answer: B · Detecting transmission errors
Why this is the answer
The sender computes a CRC over the frame; the receiver recomputes it and discards the frame if they differ. CRC detects errors but does not correct them.
Answer
B · Detecting transmission errors
4.Computer networks
Answer: A · 8 Mbit/s
Why this is the answer
C = 10⁶ × log₂(256) = 10⁶ × 8 = 8 Mbit/s. This is the upper limit for error-free data rate on a noisy channel.
Answer
A · 8 Mbit/s
5.Number systems
Answer: U = 176
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
80 = 0101 0000
inverted: 1010 1111
+1 → 1011 0000
Read it as unsigned
1011 0000₂ = 256 − 80 = 176
Answer
U = 176
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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