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 3 equal subnets. If you borrow as few bits as possible, how many usable hosts H does each subnet have?
192.168.10.0/243 subnets
H =
2.IPv4 subnettingHard
A host has the address 10.20.242.193/26. What is the last number (fourth octet) x of its subnet's broadcast address?
10.20.242.193/26
x =
3.Computer networksMedium
How does a stateful firewall differ from a simple packet filter?
- AIt tracks connections and allows replies that belong to established sessions
- BIt encrypts every packet it forwards
- CIt blocks all incoming traffic without exceptions, including replies to outgoing requests
- DIt only checks the MAC address of each frame
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
- CSending the certificate twice
- DUsing a longer domain name
5.Number systemsMedium
Write −60 as an 8-bit two's complement pattern. Read the pattern as an unsigned binary number: what is its value U in decimal?
N = −608 bit
U =
6.Number systemsHard
Set the bits to the 8-bit result of 1001 1011 OR 1001 0101. What is the result's value R in decimal?
1001 1011OR1001 0101
R =
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 ≥ 3 → b = 2, /26
Hosts per subnet
26 − 2 = 62
Answer
H = 62
2.IPv4 subnetting
Answer: x = 255
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: ⌊193/64⌋·64 = 192
broadcast: 192 + 64 − 1 = 255
usable hosts: 193 … 254
Answer
x = 255
3.Computer networks
Answer: A · It tracks connections and allows replies that belong to established sessions
Why this is the answer
A stateful firewall keeps a table of active connections, so return traffic is allowed automatically. A stateless filter judges each packet in isolation.
Answer
A · It tracks connections and allows replies that belong to established sessions
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: U = 196
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.
Invert and add one
60 = 0011 1100
inverted: 1100 0011
+1 → 1100 0100
Read it as unsigned
1100 0100₂ = 256 − 60 = 196
Answer
U = 196
6.Number systems
Answer: R = 159
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.
Bit by bit
OR gives 1 where at least one bit is 1.
1001 1011
OR 1001 0101
= 1001 1111
Value of the result
1001 1111₂ = 159
Answer
R = 159
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