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 1551 host addresses in a single subnet. What is the longest prefix length p (smallest subnet) that is big enough?
1551 hosts
p =
2.IPv4 subnettingHard
A host has the address 10.20.169.174/28. What is the fourth octet x of the last usable host address in its subnet?
10.20.169.174/28
x =
3.Computer networksMedium
What does ARP do?
- AAssigns IP addresses automatically to hosts when they join the local network
- BResolves domain names
- CRoutes packets on the internet
- DFinds the MAC address that belongs to an IP address on the local network
4.Computer networksHard
Head-of-line blocking in HTTP/2 over TCP occurs because:
- AEach request needs its own new TCP connection, so later requests wait in a queue
- BHTTP/2 uses UDP
- CServers cannot handle parallel requests
- DA lost TCP segment delays all multiplexed streams until it is retransmitted
5.Number systemsMedium
Set the bits so they store −15 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?
N = −158 bit
U =
6.Number systemsHard
Set the bits to the 8-bit result of 0111 1100 OR 0111 1011. What is the result's value R in decimal?
0111 1100OR0111 1011
R =
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 < 1551
211 − 2 = 2046 ≥ 1551 → h = 11
Prefix length
p = 32 − 11 = 21
Answer
p = /21
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
24 = 16
Subnet of this host
network: ⌊174/16⌋·16 = 160
broadcast: 160 + 16 − 1 = 175
usable hosts: 161 … 174
Answer
x = 174
3.Computer networks
Answer: D · Finds the MAC address that belongs to an IP address on the local network
Why this is the answer
ARP broadcasts "who has this IP?" and the owner replies with its MAC address, so frames can be delivered locally.
Answer
D · Finds the MAC address that belongs to an IP address on the local network
4.Computer networks
Answer: D · A lost TCP segment delays all multiplexed streams until it is retransmitted
Why this is the answer
TCP delivers bytes in order, so one loss stalls every stream on the connection. QUIC (HTTP/3) handles streams independently over UDP.
Answer
D · A lost TCP segment delays all multiplexed streams until it is retransmitted
5.Number systems
Answer: U = 241
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
15 = 0000 1111
inverted: 1111 0000
+1 → 1111 0001
Read it as unsigned
1111 0001₂ = 256 − 15 = 241
Answer
U = 241
6.Number systems
Answer: R = 127
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.
0111 1100
OR 0111 1011
= 0111 1111
Value of the result
0111 1111₂ = 127
Answer
R = 127
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