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 376 host addresses in a single subnet. What is the longest prefix length p (smallest subnet) that is big enough?
376 hosts
p =
2.IPv4 subnettingHard
A host has the address 10.20.139.172/29. What is the fourth octet x of the last usable host address in its subnet?
10.20.139.172/29
x =
3.Computer networksMedium
What characterises a distributed denial-of-service (DDoS) attack?
- AAn attacker reads unencrypted Wi-Fi traffic from a nearby car
- BMany compromised machines flood a target with traffic
- CA virus deletes files on a single PC
- DOne attacker guesses a user's password
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?
- AOptional RTS/CTS exchanges: the AP's CTS tells all nearby stations to wait
- BBy routing A's frames through C
- CBy increasing the TTL of the frames
- DBy making A and C use collision detection so they stop sending when frames overlap
5.Number systemsMedium
Set the bits so they store −93 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?
N = −938 bit
U =
6.Number systemsHard
What is the most negative integer N that can be stored in 32-bit two's complement?
32 bit
N =
3Solutions
1.IPv4 subnetting
Answer: p = 23
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
28 − 2 = 254 < 376
29 − 2 = 510 ≥ 376 → h = 9
Prefix length
p = 32 − 9 = 23
Answer
p = /23
2.IPv4 subnetting
Answer: x = 174
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
23 = 8
Subnet of this host
network: ⌊172/8⌋·8 = 168
broadcast: 168 + 8 − 1 = 175
usable hosts: 169 … 174
Answer
x = 174
3.Computer networks
Answer: B · Many compromised machines flood a target with traffic
Why this is the answer
In a DDoS attack many sources overwhelm a server or link so legitimate users cannot be served. The many sources make simple blocking hard.
Answer
B · Many compromised machines flood a target with traffic
4.Computer networks
Answer: A · 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
A · Optional RTS/CTS exchanges: the AP's CTS tells all nearby stations to wait
5.Number systems
Answer: U = 163
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
93 = 0101 1101
inverted: 1010 0010
+1 → 1010 0011
Read it as unsigned
1010 0011₂ = 256 − 93 = 163
Answer
U = 163
6.Number systems
Answer: N = −2.147 × 10⁹
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.
Most negative value
−231 = −2147483648
Answer
N = −2147483648
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