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 10.64.175.0/24 is divided into equal subnets with prefix /27. How many subnets N are there?
10.64.175.0/24/27
N =
2.IPv4 subnettingHard
A host has the address 10.20.172.209/30. What is the fourth octet x of its network address?
10.20.172.209/30
x =
3.Computer networksMedium
Which medium access method did classic shared (half-duplex) Ethernet use?
- ACSMA/CA
- BToken passing
- CTDMA
- DCSMA/CD
4.Computer networksHard
DNS queries normally use UDP. When does DNS typically use TCP instead?
- AFor every query sent from a web browser
- BOnly when the domain name is longer than ten characters
- CFor responses too large for UDP and for zone transfers
- DOnly for reverse lookups of IP addresses
5.Number systemsMedium
Write −112 as an 8-bit two's complement pattern. Read the pattern as an unsigned binary number: what is its value U in decimal?
N = −1128 bit
U =
6.Number systemsHard
An uncompressed image is 256 × 256 pixels with 8 bits per pixel. How much memory M does it take, in KiB (1 KiB = 1024 bytes)?
256 × 256 px8 bit/px1 KiB = 1024 B
M = KiB
3Solutions
1.IPv4 subnetting
Answer: N = 8
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
27 − 24 = 3
Number of subnets
N = 23 = 8
Answer
N = 8
Each /27 subnet has 25 − 2 = 30 usable hosts.
2.IPv4 subnetting
Answer: x = 208
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: ⌊209/4⌋·4 = 208
broadcast: 208 + 4 − 1 = 211
usable hosts: 209 … 210
Answer
x = 208
3.Computer networks
Answer: D · 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
D · CSMA/CD
4.Computer networks
Answer: C · For responses too large for UDP and for zone transfers
Why this is the answer
If a UDP response is truncated, the resolver retries over TCP. Zone transfers between name servers also use TCP because they carry lots of data.
Answer
C · For responses too large for UDP and for zone transfers
5.Number systems
Answer: U = 144
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
112 = 0111 0000
inverted: 1000 1111
+1 → 1001 0000
Read it as unsigned
1001 0000₂ = 256 − 112 = 144
Answer
U = 144
6.Number systems
Answer: M = 64 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
256·256·8 = 524288 bit = 65536 B
Convert to KiB
65536/1024 = 64 KiB
Answer
M = 64 KiB
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