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 194 host addresses in a single subnet. What is the longest prefix length p (smallest subnet) that is big enough?
194 hosts
p =
2.IPv4 subnettingHard
A host has the address 10.20.84.110/26. What is the last number (fourth octet) x of its subnet's broadcast address?
10.20.84.110/26
x =
3.Computer networksMedium
How does a stateful firewall differ from a simple packet filter?
- AIt encrypts every packet it forwards
- BIt only checks the MAC address of each frame
- CIt blocks all incoming traffic without exceptions, including replies to outgoing requests
- DIt tracks connections and allows replies that belong to established sessions
4.Computer networksHard
In a TCP connection, what does the receive window (rwnd) control?
- AHow much unacknowledged data the sender may send, based on the receiver's buffer
- BThe retransmission time-out, calculated from the measured round-trip time of each segment
- CThe number of open ports
- DThe size of each IP packet header
5.Number systemsMedium
Set the bits so they store −13 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?
N = −138 bit
U =
6.Number systemsHard
An uncompressed image is 640 × 480 pixels with 32 bits per pixel. How much memory M does it take, in KiB (1 KiB = 1024 bytes)?
640 × 480 px32 bit/px1 KiB = 1024 B
M = KiB
3Solutions
1.IPv4 subnetting
Answer: p = 24
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
27 − 2 = 126 < 194
28 − 2 = 254 ≥ 194 → h = 8
Prefix length
p = 32 − 8 = 24
Answer
p = /24
2.IPv4 subnetting
Answer: x = 127
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
26 = 64
Subnet of this host
network: ⌊110/64⌋·64 = 64
broadcast: 64 + 64 − 1 = 127
usable hosts: 65 … 126
Answer
x = 127
3.Computer networks
Answer: D · It tracks connections and allows replies that belong to established sessions
Why this is the answer
A stateful firewall keeps a table of active connections, so return traffic is allowed automatically. A stateless filter judges each packet in isolation.
Answer
D · It tracks connections and allows replies that belong to established sessions
4.Computer networks
Answer: A · How much unacknowledged data the sender may send, based on the receiver's buffer
Why this is the answer
The receiver advertises how much buffer space it has; the sender limits unacknowledged data accordingly, preventing receiver overload.
Answer
A · How much unacknowledged data the sender may send, based on the receiver's buffer
5.Number systems
Answer: U = 243
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
13 = 0000 1101
inverted: 1111 0010
+1 → 1111 0011
Read it as unsigned
1111 0011₂ = 256 − 13 = 243
Answer
U = 243
6.Number systems
Answer: M = 1200 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
640·480·32 = 9830400 bit = 1228800 B
Convert to KiB
1228800/1024 = 1200 KiB
Answer
M = 1200 KiB
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