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 subnet uses the mask 255.255.254.0. How many usable host addresses H does it have?
255.255.254.0
H =
2.IPv4 subnettingHard
A host has the address 10.20.114.23/28. What is the fourth octet x of the last usable host address in its subnet?
10.20.114.23/28
x =
3.Computer networksMedium
What is the main purpose of a VLAN?
- ATo encrypt all traffic between switches so other departments cannot read it
- BTo split one physical switched network into separate broadcast domains
- CTo give each device a public IP address
- DTo increase the speed of every switch port
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
The 8-bit pattern 0101 0001 stores a signed integer in two's complement. What is its value N in decimal?
0101 00018 bit
N =
6.Number systemsHard
An uncompressed image is 800 × 600 pixels with 8 bits per pixel. How much memory M does it take, in KiB (1 KiB = 1024 bytes)?
800 × 600 px8 bit/px1 KiB = 1024 B
M = KiB
3Solutions
1.IPv4 subnetting
Answer: H = 510
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).
Prefix length
255.255.254.0 = /23
Usable hosts
29 − 2 = 510
Answer
H = 510
2.IPv4 subnetting
Answer: x = 30
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
24 = 16
Subnet of this host
network: ⌊23/16⌋·16 = 16
broadcast: 16 + 16 − 1 = 31
usable hosts: 17 … 30
Answer
x = 30
3.Computer networks
Answer: B · To split one physical switched network into separate broadcast domains
Why this is the answer
VLANs group ports logically, so broadcasts stay within each VLAN. This improves security and limits broadcast traffic.
Answer
B · To split one physical switched network into separate broadcast domains
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: N = 81
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.
64 + 16 + 1 = 81
Answer
N = 81
6.Number systems
Answer: M = 468.8 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
800·600·8 = 3840000 bit = 480000 B
Convert to KiB
480000/1024 = 468.75 KiB
Answer
M = 468.75 KiB
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