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 255.255.255.192. How many usable host addresses H does it have?
255.255.255.192
H =
2.IPv4 subnettingHard
The network 192.168.144.0/24 is split with VLSM for these LANs: A: 8 hosts, B: 16 hosts, C: 7 hosts, D: 48 hosts, E: 3 hosts. Subnets are allocated from .0 upwards, the LAN with the most hosts first, each with the smallest subnet that fits. What is the fourth octet x of the network address of LAN C?
A: 8B: 16C: 7D: 48E: 3
x =
3.Computer networksMedium
TLS provides which security properties for HTTPS?
- AProtection against all malware and phishing on the visited site
- BAnonymity of the user
- CConfidentiality, integrity and server authentication
- DOnly faster loading
4.Computer networksHard
A user can ping 8.8.8.8 but cannot open any website by name. Where is the problem most likely?
- AThe network card driver
- BThe default gateway
- CName resolution (DNS)
- DThe physical cable
5.Number systemsMedium
Set the bits so they store −58 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?
N = −588 bit
U =
6.Number systemsHard
An uncompressed image is 1280 × 720 pixels with 8 bits per pixel. How much memory M does it take, in KiB (1 KiB = 1024 bytes)?
1280 × 720 px8 bit/px1 KiB = 1024 B
M = KiB
3Solutions
1.IPv4 subnetting
Answer: H = 62
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
255.255.255.192 = /26
Usable hosts
26 − 2 = 62
Answer
H = 62
2.IPv4 subnetting
Answer: x = 112
VLSM: each LAN gets the smallest subnet with 2ʰ − 2 ≥ its host count, i.e. prefix 32 − h. Allocating the largest subnets first keeps every block aligned on a multiple of its own size, so the network addresses follow by adding up the block sizes.
Block size per LAN
D: 48 hosts → 26 = 64 (/26)
B: 16 hosts → 25 = 32 (/27)
A: 8 hosts → 24 = 16 (/28)
C: 7 hosts → 24 = 16 (/28)
E: 3 hosts → 23 = 8 (/29)
Allocation, largest first
D: .0 – .63
B: .64 – .95
A: .96 – .111
C: .112 – .127
E: .128 – .135
Answer
x = 112
LAN C: 192.168.144.112/28.
3.Computer networks
Answer: C · 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
C · Confidentiality, integrity and server authentication
4.Computer networks
Answer: C · Name resolution (DNS)
Why this is the answer
A successful ping to an external IP proves cabling, addressing and routing work. Failing names point to a DNS server or resolver setting problem.
Answer
C · Name resolution (DNS)
5.Number systems
Answer: U = 198
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
58 = 0011 1010
inverted: 1100 0101
+1 → 1100 0110
Read it as unsigned
1100 0110₂ = 256 − 58 = 198
Answer
U = 198
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
1280·720·8 = 7372800 bit = 921600 B
Convert to KiB
921600/1024 = 900 KiB
Answer
M = 900 KiB
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