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 192.168.10.0/24 must be divided into at least 4 equal subnets. If you borrow as few bits as possible, how many usable hosts H does each subnet have?
192.168.10.0/244 subnets
H =
2.IPv4 subnettingHard
A host has the address 10.20.190.200/30. What is the fourth octet x of its network address?
10.20.190.200/30
x =
3.Computer networksMedium
What is a cyclic redundancy check (CRC) used for in Ethernet frames?
- AEncrypting the frame
- BChoosing the best route
- CDetecting transmission errors
- DCorrecting every bit error automatically
4.Computer networksHard
What is the main benefit of persistent connections (keep-alive) in HTTP/1.1?
- ASeveral requests reuse one TCP connection, avoiding repeated handshakes
- BResponses can arrive in any order on one stream, so slow replies do not block others
- CThe server can push updates over UDP
- DThe connection stays encrypted without TLS
5.Number systemsMedium
The 8-bit pattern 1110 0011 stores a signed integer in two's complement. What is its value N in decimal?
1110 00118 bit
N =
6.Number systemsHard
An uncompressed image is 640 × 480 pixels with 16 bits per pixel. How much memory M does it take, in KiB (1 KiB = 1024 bytes)?
640 × 480 px16 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).
Bits to borrow
22 = 4 ≥ 4 → b = 2, /26
Hosts per subnet
26 − 2 = 62
Answer
H = 62
2.IPv4 subnetting
Answer: x = 200
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: ⌊200/4⌋·4 = 200
broadcast: 200 + 4 − 1 = 203
usable hosts: 201 … 202
Answer
x = 200
3.Computer networks
Answer: C · Detecting transmission errors
Why this is the answer
The sender computes a CRC over the frame; the receiver recomputes it and discards the frame if they differ. CRC detects errors but does not correct them.
Answer
C · Detecting transmission errors
4.Computer networks
Answer: A · Several requests reuse one TCP connection, avoiding repeated handshakes
Why this is the answer
Reusing a connection avoids a new handshake and slow start for each object, so pages with many resources load faster.
Answer
A · Several requests reuse one TCP connection, avoiding repeated handshakes
5.Number systems
Answer: N = −29
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.
Add the place values
The other bits keep their normal positive values.
−128 + 64 + 32 + 2 + 1 = −29
Answer
N = −29
6.Number systems
Answer: M = 600 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
640·480·16 = 4915200 bit = 614400 B
Convert to KiB
614400/1024 = 600 KiB
Answer
M = 600 KiB
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