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 20 equal subnets. If you borrow as few bits as possible, how many usable hosts H does each subnet have?
192.168.10.0/2420 subnets
H =
2.IPv4 subnettingHard
A host has the address 10.20.20.58/30. What is the fourth octet x of its network address?
10.20.20.58/30
x =
3.Computer networksMedium
Why does a switch sometimes send a frame out of all ports except the one it came in on?
- AThe frame has a wrong IP checksum and must be checked by every host
- BThe frame is too large for one port
- CThe destination MAC address is not yet in its address table
- DThe switch is acting as a router
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
- BSending the certificate twice
- CUsing a longer domain name
- DForward secrecy from ephemeral Diffie-Hellman key exchange
5.Number systemsMedium
Write −84 as an 8-bit two's complement pattern. Read the pattern as an unsigned binary number: what is its value U in decimal?
N = −848 bit
U =
6.Number systemsHard
Set the bits to the 8-bit result of 0100 0101 XOR 1101 0011. What is the result's value R in decimal?
0100 0101XOR1101 0011
R =
3Solutions
1.IPv4 subnetting
Answer: H = 6
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
25 = 32 ≥ 20 → b = 5, /29
Hosts per subnet
23 − 2 = 6
Answer
H = 6
2.IPv4 subnetting
Answer: x = 56
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: ⌊58/4⌋·4 = 56
broadcast: 56 + 4 − 1 = 59
usable hosts: 57 … 58
Answer
x = 56
3.Computer networks
Answer: C · The destination MAC address is not yet in its address table
Why this is the answer
For an unknown unicast address the switch floods the frame. When the destination replies, the switch learns its port from the source MAC.
Answer
C · The destination MAC address is not yet in its address table
4.Computer networks
Answer: D · 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
D · Forward secrecy from ephemeral Diffie-Hellman key exchange
5.Number systems
Answer: U = 172
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
84 = 0101 0100
inverted: 1010 1011
+1 → 1010 1100
Read it as unsigned
1010 1100₂ = 256 − 84 = 172
Answer
U = 172
6.Number systems
Answer: R = 150
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
XOR gives 1 where the two bits differ.
0100 0101
XOR 1101 0011
= 1001 0110
Value of the result
1001 0110₂ = 150
Answer
R = 150
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