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 1584 host addresses in a single subnet. What is the longest prefix length p (smallest subnet) that is big enough?
1584 hosts
p =
2.IPv4 subnettingHard
A host has the address 10.20.195.233/28. What is the fourth octet x of its network address?
10.20.195.233/28
x =
3.Computer networksMedium
What is the purpose of the TCP three-way handshake?
- ATo establish a connection and synchronise sequence numbers
- BTo close a connection quickly
- CTo encrypt the data
- DTo find the IP address of a host before data can be sent to it
4.Computer networksHard
Stations A and C both reach access point B but cannot hear each other, so their frames collide at B. How does 802.11 help?
- ABy increasing the TTL of the frames
- BBy routing A's frames through C
- CBy making A and C use collision detection so they stop sending when frames overlap
- DOptional RTS/CTS exchanges: the AP's CTS tells all nearby stations to wait
5.Number systemsMedium
The 8-bit pattern 0100 1000 stores a signed integer in two's complement. What is its value N in decimal?
0100 10008 bit
N =
6.Number systemsHard
A system must give every one of 2030 items its own binary code of the same length. What is the smallest number of bits b needed?
2030
b = bit
3Solutions
1.IPv4 subnetting
Answer: p = 21
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
210 − 2 = 1022 < 1584
211 − 2 = 2046 ≥ 1584 → h = 11
Prefix length
p = 32 − 11 = 21
Answer
p = /21
2.IPv4 subnetting
Answer: x = 224
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: ⌊233/16⌋·16 = 224
broadcast: 224 + 16 − 1 = 239
usable hosts: 225 … 238
Answer
x = 224
3.Computer networks
Answer: A · To establish a connection and synchronise sequence numbers
Why this is the answer
The handshake sets up a connection and agrees initial sequence numbers on both sides before data is sent.
Answer
A · To establish a connection and synchronise sequence numbers
4.Computer networks
Answer: D · Optional RTS/CTS exchanges: the AP's CTS tells all nearby stations to wait
Why this is the answer
A station sends Request to Send; the AP answers Clear to Send, which every station in its range hears, so hidden stations defer for the reserved time.
Answer
D · Optional RTS/CTS exchanges: the AP's CTS tells all nearby stations to wait
5.Number systems
Answer: N = 72
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 + 8 = 72
Answer
N = 72
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 < 2030
211 = 2048 ≥ 2030
Answer
b = 11
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