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 1928 host addresses in a single subnet. What is the longest prefix length p (smallest subnet) that is big enough?
1928 hosts
p =
2.IPv4 subnettingHard
A host has the address 10.20.224.65/26. What is the fourth octet x of the last usable host address in its subnet?
10.20.224.65/26
x =
3.Computer networksMedium
TLS provides which security properties for HTTPS?
- AAnonymity of the user
- BConfidentiality, integrity and server authentication
- CProtection against all malware and phishing on the visited site
- DOnly faster loading
4.Computer networksHard
Why does IPv6 not need NAT for address shortage?
- AIPv6 does not use addresses
- BIPv6 only works on local networks, so addresses never need to be shared globally
- CIts 128-bit address space is large enough to give every device a global address
- DIPv6 encrypts all addresses
5.Number systemsMedium
Set the bits so they store −101 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?
N = −1018 bit
U =
6.Number systemsHard
An uncompressed image is 256 × 256 pixels with 32 bits per pixel. How much memory M does it take, in KiB (1 KiB = 1024 bytes)?
256 × 256 px32 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 < 1928
211 − 2 = 2046 ≥ 1928 → h = 11
Prefix length
p = 32 − 11 = 21
Answer
p = /21
2.IPv4 subnetting
Answer: x = 126
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
26 = 64
Subnet of this host
network: ⌊65/64⌋·64 = 64
broadcast: 64 + 64 − 1 = 127
usable hosts: 65 … 126
Answer
x = 126
3.Computer networks
Answer: B · Confidentiality, integrity and server authentication
Why this is the answer
TLS encrypts traffic, detects tampering and authenticates the server with certificates (optionally the client too).
Answer
B · Confidentiality, integrity and server authentication
4.Computer networks
Answer: C · Its 128-bit address space is large enough to give every device a global address
Why this is the answer
IPv6 has about 3.4 × 10³⁸ addresses, so the scarcity that drove NAT in IPv4 disappears; firewalls still provide security.
Answer
C · Its 128-bit address space is large enough to give every device a global address
5.Number systems
Answer: U = 155
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
101 = 0110 0101
inverted: 1001 1010
+1 → 1001 1011
Read it as unsigned
1001 1011₂ = 256 − 101 = 155
Answer
U = 155
6.Number systems
Answer: M = 256 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
256·256·32 = 2097152 bit = 262144 B
Convert to KiB
262144/1024 = 256 KiB
Answer
M = 256 KiB
ExamProblems · examproblems.com/courses/computer-networks