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 subnet uses the mask 255.255.252.0. How many usable host addresses H does it have?
255.255.252.0
H =
2.IPv4 subnettingHard
A host has the address 10.20.129.28/29. What is the last number (fourth octet) x of its subnet's broadcast address?
10.20.129.28/29
x =
3.Computer networksMedium
At which OSI layer does a switch normally operate?
- ANetwork (layer 3)
- BTransport (layer 4)
- CData link (layer 2)
- DPhysical (layer 1)
4.Computer networksHard
Head-of-line blocking in HTTP/2 over TCP occurs because:
- AA lost TCP segment delays all multiplexed streams until it is retransmitted
- BHTTP/2 uses UDP
- CServers cannot handle parallel requests
- DEach request needs its own new TCP connection, so later requests wait in a queue
5.Number systemsMedium
Set the bits so they store −42 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?
N = −428 bit
U =
6.Number systemsHard
Set the bits to the 8-bit result of 1101 1011 AND 1000 0000. What is the result's value R in decimal?
1101 1011AND1000 0000
R =
3Solutions
1.IPv4 subnetting
Answer: H = 1022
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).
Prefix length
255.255.252.0 = /22
Usable hosts
210 − 2 = 1022
Answer
H = 1022
2.IPv4 subnetting
Answer: x = 31
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: ⌊28/8⌋·8 = 24
broadcast: 24 + 8 − 1 = 31
usable hosts: 25 … 30
Answer
x = 31
3.Computer networks
Answer: C · Data link (layer 2)
Why this is the answer
A standard switch learns MAC addresses and forwards frames at layer 2. Layer 3 switches also route.
Answer
C · Data link (layer 2)
4.Computer networks
Answer: A · 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
A · A lost TCP segment delays all multiplexed streams until it is retransmitted
5.Number systems
Answer: U = 214
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
42 = 0010 1010
inverted: 1101 0101
+1 → 1101 0110
Read it as unsigned
1101 0110₂ = 256 − 42 = 214
Answer
U = 214
6.Number systems
Answer: R = 128
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
AND gives 1 only where both bits are 1.
1101 1011
AND 1000 0000
= 1000 0000
Value of the result
1000 0000₂ = 128
Answer
R = 128
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