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 255.255.255.128. How many usable host addresses H does it have?
255.255.255.128
H =
2.IPv4 subnettingHard
A host has the address 10.20.3.116/30. What is the last number (fourth octet) x of its subnet's broadcast address?
10.20.3.116/30
x =
3.Computer networksMedium
What is the main purpose of a VLAN?
- ATo increase the speed of every switch port
- BTo split one physical switched network into separate broadcast domains
- CTo give each device a public IP address
- DTo encrypt all traffic between switches so other departments cannot read it
4.Computer networksHard
A remote-terminal application feels sluggish because small keystroke packets are held back. Which TCP mechanism is often responsible?
- ANagle's algorithm
- BFast retransmit
- CThe three-way handshake
- DSlow start
5.Number systemsMedium
Set the bits so they store −65 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?
N = −658 bit
U =
6.Number systemsHard
What is the most negative integer N that can be stored in 10-bit two's complement?
10 bit
N =
3Solutions
1.IPv4 subnetting
Answer: H = 126
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
255.255.255.128 = /25
Usable hosts
27 − 2 = 126
Answer
H = 126
2.IPv4 subnetting
Answer: x = 119
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
22 = 4
Subnet of this host
network: ⌊116/4⌋·4 = 116
broadcast: 116 + 4 − 1 = 119
usable hosts: 117 … 118
Answer
x = 119
3.Computer networks
Answer: B · To split one physical switched network into separate broadcast domains
Why this is the answer
VLANs group ports logically, so broadcasts stay within each VLAN. This improves security and limits broadcast traffic.
Answer
B · To split one physical switched network into separate broadcast domains
4.Computer networks
Answer: A · Nagle's algorithm
Why this is the answer
Nagle's algorithm waits to bundle small segments while earlier data is unacknowledged. This saves bandwidth but adds delay for interactive traffic.
Answer
A · Nagle's algorithm
5.Number systems
Answer: U = 191
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
65 = 0100 0001
inverted: 1011 1110
+1 → 1011 1111
Read it as unsigned
1011 1111₂ = 256 − 65 = 191
Answer
U = 191
6.Number systems
Answer: N = −512
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
−29 = −512
Answer
N = −512
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