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 10.1.16.0/20 is divided into equal subnets with prefix /24. How many subnets N are there?
10.1.16.0/20/24
N =
2.IPv4 subnettingHard
A host has the address 10.20.153.87/25. What is the last number (fourth octet) x of its subnet's broadcast address?
10.20.153.87/25
x =
3.Computer networksMedium
HTTP is stateless. How do websites usually keep a user logged in across requests?
- ABy storing the password in the URL
- BWith cookies holding a session identifier or token
- CBy keeping the TCP connection open for days
- DBy giving each user a fixed IP address
4.Computer networksHard
In a SYN flood attack, what resource is the attacker trying to exhaust?
- AThe switch's MAC address table
- BThe DNS cache of the client
- CThe server's disk space
- DThe server's table of half-open connections
5.Number systemsMedium
The 8-bit pattern 0110 0110 stores a signed integer in two's complement. What is its value N in decimal?
0110 01108 bit
N =
6.Number systemsHard
Set the bits to the 8-bit result of 0001 1110 OR 1111 0110. What is the result's value R in decimal?
0001 1110OR1111 0110
R =
3Solutions
1.IPv4 subnetting
Answer: N = 16
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).
Borrowed bits
24 − 20 = 4
Number of subnets
N = 24 = 16
Answer
N = 16
Each /24 subnet has 28 − 2 = 254 usable hosts.
2.IPv4 subnetting
Answer: x = 127
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
27 = 128
Subnet of this host
network: ⌊87/128⌋·128 = 0
broadcast: 0 + 128 − 1 = 127
usable hosts: 1 … 126
Answer
x = 127
3.Computer networks
Answer: B · With cookies holding a session identifier or token
Why this is the answer
The server sets a cookie containing a session ID or token, and the browser returns it with each request so the server can recognise the user.
Answer
B · With cookies holding a session identifier or token
4.Computer networks
Answer: D · The server's table of half-open connections
Why this is the answer
Each SYN makes the server reserve state and wait for an ACK that never comes. SYN cookies avoid storing state until the handshake completes.
Answer
D · The server's table of half-open connections
5.Number systems
Answer: N = 102
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 + 32 + 4 + 2 = 102
Answer
N = 102
6.Number systems
Answer: R = 254
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
OR gives 1 where at least one bit is 1.
0001 1110
OR 1111 0110
= 1111 1110
Value of the result
1111 1110₂ = 254
Answer
R = 254
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