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 10.178.144.0/20 is divided into equal subnets with prefix /22. How many subnets N are there?
10.178.144.0/20/22
N =
2.IPv4 subnettingHard
A host has the address 172.20.57.251/23. What is the third octet x of its network address?
172.20.57.251/23
x =
3.Computer networksMedium
What characterises a distributed denial-of-service (DDoS) attack?
- AMany compromised machines flood a target with traffic
- BA virus deletes files on a single PC
- CAn attacker reads unencrypted Wi-Fi traffic from a nearby car
- DOne attacker guesses a user's password
4.Computer networksHard
A student runs a game server on a PC behind a home NAT router. Friends on the internet cannot connect. What is the usual fix?
- ATurn off the PC's DNS client
- BConfigure port forwarding on the router to the PC's private address
- CChange the PC's subnet mask to /32
- DGive the PC a different MAC address so the router can find it from outside
5.Number systemsMedium
Set the bits so they store −110 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?
N = −1108 bit
U =
6.Number systemsHard
An uncompressed image is 800 × 600 pixels with 16 bits per pixel. How much memory M does it take, in KiB (1 KiB = 1024 bytes)?
800 × 600 px16 bit/px1 KiB = 1024 B
M = KiB
3Solutions
1.IPv4 subnetting
Answer: N = 4
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
22 − 20 = 2
Number of subnets
N = 22 = 4
Answer
N = 4
Each /22 subnet has 210 − 2 = 1022 usable hosts.
2.IPv4 subnetting
Answer: x = 56
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 in the third octet
21 = 2
Subnet of this host
network: 172.20.56.0
broadcast: 172.20.57.255
Answer
x = 56
3.Computer networks
Answer: A · 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
A · Many compromised machines flood a target with traffic
4.Computer networks
Answer: B · Configure port forwarding on the router to the PC's private address
Why this is the answer
NAT creates mappings only for connections started from inside. A port-forwarding rule maps a public port to the internal host so inbound connections can reach it.
Answer
B · Configure port forwarding on the router to the PC's private address
5.Number systems
Answer: U = 146
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
110 = 0110 1110
inverted: 1001 0001
+1 → 1001 0010
Read it as unsigned
1001 0010₂ = 256 − 110 = 146
Answer
U = 146
6.Number systems
Answer: M = 937.5 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
800·600·16 = 7680000 bit = 960000 B
Convert to KiB
960000/1024 = 937.5 KiB
Answer
M = 937.5 KiB
ExamProblems · examproblems.com/courses/computer-networks