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 16 equal subnets. If you borrow as few bits as possible, how many usable hosts H does each subnet have?
192.168.10.0/2416 subnets
H =
2.IPv4 subnettingHard
A host has the address 10.20.96.238/26. What is the fourth octet x of its network address?
10.20.96.238/26
x =
3.Computer networksMedium
What is the network address of the host 192.168.10.77/26?
- A192.168.10.0
- B192.168.10.128
- C192.168.10.76
- D192.168.10.64
4.Computer networksHard
A DNS cache poisoning attack aims to:
- AOverload a server with traffic so that it stops answering queries
- BMake a resolver store a false IP address for a domain
- CSteal Wi-Fi passwords
- DDelete DNS servers
5.Number systemsMedium
The 8-bit pattern 1011 1111 stores a signed integer in two's complement. What is its value N in decimal?
1011 11118 bit
N =
6.Number systemsHard
A system must give every one of 3290 items its own binary code of the same length. What is the smallest number of bits b needed?
3290
b = bit
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 ≥ 16 → b = 4, /28
Hosts per subnet
24 − 2 = 14
Answer
H = 14
2.IPv4 subnetting
Answer: x = 192
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: ⌊238/64⌋·64 = 192
broadcast: 192 + 64 − 1 = 255
usable hosts: 193 … 254
Answer
x = 192
3.Computer networks
Answer: D · 192.168.10.64
Why this is the answer
/26 subnets start at 0, 64, 128 and 192. 77 lies in the block 64–127, so the network address is 192.168.10.64.
Answer
D · 192.168.10.64
4.Computer networks
Answer: B · Make a resolver store a false IP address for a domain
Why this is the answer
Poisoning inserts forged records into a resolver's cache. DNSSEC adds signatures so resolvers can verify records.
Answer
B · Make a resolver store a false IP address for a domain
5.Number systems
Answer: N = −65
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 + 32 + 16 + 8 + 4 + 2 + 1 = −65
Answer
N = −65
6.Number systems
Answer: b = 12 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
211 = 2048 < 3290
212 = 4096 ≥ 3290
Answer
b = 12
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