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 1119 host addresses in a single subnet. What is the longest prefix length p (smallest subnet) that is big enough?
1119 hosts
p =
2.IPv4 subnettingHard
A host has the address 172.20.213.33/17. What is the third octet x of its subnet's broadcast address?
172.20.213.33/17
x =
3.Computer networksMedium
Which medium access method did classic shared (half-duplex) Ethernet use?
- AToken passing
- BCSMA/CD
- CTDMA
- DCSMA/CA
4.Computer networksHard
You must divide 192.168.1.0/24 into at least 5 subnets with up to 25 hosts each. Which prefix length fits best?
- A/25
- B/26
- C/28
- D/27
5.Number systemsMedium
Write −56 as an 8-bit two's complement pattern. Read the pattern as an unsigned binary number: what is its value U in decimal?
N = −568 bit
U =
6.Number systemsHard
An uncompressed image is 640 × 480 pixels with 24 bits per pixel. How much memory M does it take, in KiB (1 KiB = 1024 bytes)?
640 × 480 px24 bit/px1 KiB = 1024 B
M = KiB
3Solutions
1.IPv4 subnetting
Answer: p = 21
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
210 − 2 = 1022 < 1119
211 − 2 = 2046 ≥ 1119 → h = 11
Prefix length
p = 32 − 11 = 21
Answer
p = /21
2.IPv4 subnetting
Answer: x = 255
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 in the third octet
27 = 128
Subnet of this host
network: 172.20.128.0
broadcast: 172.20.255.255
Answer
x = 255
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: D · /27
Why this is the answer
/27 gives 8 subnets with 30 usable hosts each. /26 gives only 4 subnets, and /28 allows only 14 hosts per subnet.
Answer
D · /27
5.Number systems
Answer: U = 200
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
56 = 0011 1000
inverted: 1100 0111
+1 → 1100 1000
Read it as unsigned
1100 1000₂ = 256 − 56 = 200
Answer
U = 200
6.Number systems
Answer: M = 900 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
640·480·24 = 7372800 bit = 921600 B
Convert to KiB
921600/1024 = 900 KiB
Answer
M = 900 KiB
ExamProblems · examproblems.com/courses/computer-networks