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 department needs 633 host addresses in a single subnet. What is the longest prefix length p (smallest subnet) that is big enough?
633 hosts
p =
2.IPv4 subnettingHard
The network 192.168.127.0/24 is split with VLSM for these LANs: A: 3 hosts, B: 70 hosts, C: 33 hosts, D: 8 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 A?
A: 3B: 70C: 33D: 8
x =
3.Computer networksMedium
What does a subnet mask of 255.255.255.0 correspond to in CIDR notation?
- A/16
- B/24
- C/8
- D/32
4.Computer networksHard
A server's private key is stolen a year from now. Which TLS property keeps today's recorded sessions safe?
- AForward secrecy from ephemeral Diffie-Hellman key exchange
- BUsing a longer domain name
- CSending the certificate twice
- DCompressing the data before encryption
5.Number systemsMedium
Set the bits so they store −13 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?
N = −138 bit
U =
6.Number systemsHard
A system must give every one of 1781 items its own binary code of the same length. What is the smallest number of bits b needed?
1781
b = bit
3Solutions
1.IPv4 subnetting
Answer: p = 22
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).
Host bits needed
29 − 2 = 510 < 633
210 − 2 = 1022 ≥ 633 → h = 10
Prefix length
p = 32 − 10 = 22
Answer
p = /22
2.IPv4 subnetting
Answer: x = 208
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
B: 70 hosts → 27 = 128 (/25)
C: 33 hosts → 26 = 64 (/26)
D: 8 hosts → 24 = 16 (/28)
A: 3 hosts → 23 = 8 (/29)
Allocation, largest first
B: .0 – .127
C: .128 – .191
D: .192 – .207
A: .208 – .215
Answer
x = 208
LAN A: 192.168.127.208/29.
3.Computer networks
Answer: B · /24
Why this is the answer
Three octets of 255 are 24 one-bits, so 255.255.255.0 = /24.
Answer
B · /24
4.Computer networks
Answer: A · 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
A · Forward secrecy from ephemeral Diffie-Hellman key exchange
5.Number systems
Answer: U = 243
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
13 = 0000 1101
inverted: 1111 0010
+1 → 1111 0011
Read it as unsigned
1111 0011₂ = 256 − 13 = 243
Answer
U = 243
6.Number systems
Answer: b = 11 bit
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.
Powers of two
210 = 1024 < 1781
211 = 2048 ≥ 1781
Answer
b = 11
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