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 10 equal subnets. If you borrow as few bits as possible, how many usable hosts H does each subnet have?
192.168.10.0/2410 subnets
H =
2.IPv4 subnettingHard
The network 192.168.175.0/24 is split with VLSM for these LANs: A: 43 hosts, B: 42 hosts, C: 7 hosts, D: 22 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: 43B: 42C: 7D: 22
x =
3.Computer networksMedium
Which TCP flag signals that a sender has finished sending data?
- ASYN
- BPSH
- CFIN
- DURG
4.Computer networksHard
A server's private key is stolen a year from now. Which TLS property keeps today's recorded sessions safe?
- ACompressing the data before encryption
- BForward secrecy from ephemeral Diffie-Hellman key exchange
- CUsing a longer domain name
- DSending the certificate twice
5.Number systemsMedium
The 8-bit pattern 1100 0100 stores a signed integer in two's complement. What is its value N in decimal?
1100 01008 bit
N =
6.Number systemsHard
An uncompressed image is 1024 × 768 pixels with 8 bits per pixel. How much memory M does it take, in KiB (1 KiB = 1024 bytes)?
1024 × 768 px8 bit/px1 KiB = 1024 B
M = KiB
3Solutions
1.IPv4 subnetting
Answer: H = 14
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
24 = 16 ≥ 10 → b = 4, /28
Hosts per subnet
24 − 2 = 14
Answer
H = 14
2.IPv4 subnetting
Answer: x = 160
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
A: 43 hosts → 26 = 64 (/26)
B: 42 hosts → 26 = 64 (/26)
D: 22 hosts → 25 = 32 (/27)
C: 7 hosts → 24 = 16 (/28)
Allocation, largest first
A: .0 – .63
B: .64 – .127
D: .128 – .159
C: .160 – .175
Answer
x = 160
LAN C: 192.168.175.160/28.
3.Computer networks
Answer: C · FIN
Why this is the answer
Each side sends FIN when it has no more data, and the other side acknowledges it. SYN is for opening a connection; RST aborts one.
Answer
C · FIN
4.Computer networks
Answer: B · Forward secrecy from ephemeral Diffie-Hellman key exchange
Why this is the answer
With ephemeral (EC)DHE, each session key is derived from temporary values that are discarded. The long-term key only signs the exchange, so stealing it later does not decrypt old traffic.
Answer
B · Forward secrecy from ephemeral Diffie-Hellman key exchange
5.Number systems
Answer: N = −60
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 + 4 = −60
Answer
N = −60
6.Number systems
Answer: M = 768 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
1024·768·8 = 6291456 bit = 786432 B
Convert to KiB
786432/1024 = 768 KiB
Answer
M = 768 KiB
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