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 192.168.10.0/24 must be divided into at least 5 equal subnets. If you borrow as few bits as possible, how many usable hosts H does each subnet have?
192.168.10.0/245 subnets
H =
2.IPv4 subnettingHard
A host has the address 10.20.128.93/30. What is the fourth octet x of the last usable host address in its subnet?
10.20.128.93/30
x =
3.Computer networksMedium
Which medium access method did classic shared (half-duplex) Ethernet use?
- ACSMA/CA
- BCSMA/CD
- CTDMA
- DToken passing
4.Computer networksHard
A PC can reach printers on its own subnet but nothing beyond it. Its IP address and mask are correct. What is the most likely fault?
- AA missing or wrong default gateway
- BA faulty DNS server setting
- CA duplicate MAC address on the internet
- DThe TCP window is too small
5.Number systemsMedium
Set the bits so they store −115 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?
N = −1158 bit
U =
6.Number systemsHard
An uncompressed image is 1280 × 720 pixels with 16 bits per pixel. How much memory M does it take, in KiB (1 KiB = 1024 bytes)?
1280 × 720 px16 bit/px1 KiB = 1024 B
M = KiB
3Solutions
1.IPv4 subnetting
Answer: H = 30
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).
Bits to borrow
23 = 8 ≥ 5 → b = 3, /27
Hosts per subnet
25 − 2 = 30
Answer
H = 30
2.IPv4 subnetting
Answer: x = 94
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: ⌊93/4⌋·4 = 92
broadcast: 92 + 4 − 1 = 95
usable hosts: 93 … 94
Answer
x = 94
3.Computer networks
Answer: B · CSMA/CD
Why this is the answer
Carrier sense multiple access with collision detection: stations stop when they detect a collision and retry after a random backoff. Full-duplex switched Ethernet has no collisions.
Answer
B · CSMA/CD
4.Computer networks
Answer: A · A missing or wrong default gateway
Why this is the answer
Hosts send off-subnet packets to the default gateway. If it is wrong or missing, local traffic still works but nothing outside is reachable, even by IP address.
Answer
A · A missing or wrong default gateway
5.Number systems
Answer: U = 141
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
115 = 0111 0011
inverted: 1000 1100
+1 → 1000 1101
Read it as unsigned
1000 1101₂ = 256 − 115 = 141
Answer
U = 141
6.Number systems
Answer: M = 1800 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
1280·720·16 = 14745600 bit = 1843200 B
Convert to KiB
1843200/1024 = 1800 KiB
Answer
M = 1800 KiB
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