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 255.255.248.0. How many usable host addresses H does it have?
255.255.248.0
H =
2.IPv4 subnettingHard
A host has the address 172.20.203.139/20. What is the third octet x of its subnet's broadcast address?
172.20.203.139/20
x =
3.Computer networksMedium
Which IPv4 header field stops packets from circulating forever in a routing loop?
- AType of service
- BTime to live (TTL)
- CIdentification
- DHeader checksum
4.Computer networksHard
An 8-port switch (no VLANs) has one host on each port. How many collision domains and broadcast domains are there?
- A8 collision domains, 8 broadcast domains
- B8 collision domains, 1 broadcast domain
- C1 collision domain, 8 broadcast domains
- D1 collision domain, 1 broadcast domain
5.Number systemsMedium
Set the bits so they store −54 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?
N = −548 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 = 2046
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
255.255.248.0 = /21
Usable hosts
211 − 2 = 2046
Answer
H = 2046
2.IPv4 subnetting
Answer: x = 207
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
24 = 16
Subnet of this host
network: 172.20.192.0
broadcast: 172.20.207.255
Answer
x = 207
3.Computer networks
Answer: B · Time to live (TTL)
Why this is the answer
Every router decrements the TTL; at zero the packet is discarded and an ICMP Time Exceeded message is usually sent. IPv6 calls it hop limit.
Answer
B · Time to live (TTL)
4.Computer networks
Answer: B · 8 collision domains, 1 broadcast domain
Why this is the answer
Each switch port is its own collision domain, but broadcasts are forwarded to all ports, so the whole switch is one broadcast domain. A hub would give one of each.
Answer
B · 8 collision domains, 1 broadcast domain
5.Number systems
Answer: U = 202
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
54 = 0011 0110
inverted: 1100 1001
+1 → 1100 1010
Read it as unsigned
1100 1010₂ = 256 − 54 = 202
Answer
U = 202
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
ExamProblems · examproblems.com/courses/computer-networks