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 192.168.66.0/24 is divided into /28 subnets, numbered 1, 2, 3, … from the lowest address (subnet 1 starts at .0). What is the fourth octet x of the broadcast address of subnet 10?
192.168.66.0/24/28subnet 10 (subnet 1 starts at .0)
x =
2.IPv4 subnettingHard
A host has the address 172.20.102.227/20. What is the third octet x of its subnet's broadcast address?
172.20.102.227/20
x =
3.Computer networksMedium
Which DNS record type maps a host name to an IPv6 address?
- AMX
- BCNAME
- CA
- DAAAA
4.Computer networksHard
In a TCP connection, what does the receive window (rwnd) control?
- AThe number of open ports
- BThe size of each IP packet header
- CThe retransmission time-out, calculated from the measured round-trip time of each segment
- DHow much unacknowledged data the sender may send, based on the receiver's buffer
5.Number systemsMedium
Write −6 as an 8-bit two's complement pattern. Read the pattern as an unsigned binary number: what is its value U in decimal?
N = −68 bit
U =
6.Number systemsHard
An uncompressed image is 1920 × 1080 pixels with 8 bits per pixel. How much memory M does it take, in KiB (1 KiB = 1024 bytes)?
1920 × 1080 px8 bit/px1 KiB = 1024 B
M = KiB
3Solutions
1.IPv4 subnetting
Answer: x = 159
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 number 10
network: (10 − 1)·16 = 144
broadcast: 144 + 16 − 1 = 159
Answer
x = 159
2.IPv4 subnetting
Answer: x = 111
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
24 = 16
Subnet of this host
network: 172.20.96.0
broadcast: 172.20.111.255
Answer
x = 111
3.Computer networks
Answer: D · AAAA
Why this is the answer
An A record holds an IPv4 address, AAAA an IPv6 address. MX names mail servers and CNAME gives an alias.
Answer
D · AAAA
4.Computer networks
Answer: D · 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
D · How much unacknowledged data the sender may send, based on the receiver's buffer
5.Number systems
Answer: U = 250
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.
Invert and add one
6 = 0000 0110
inverted: 1111 1001
+1 → 1111 1010
Read it as unsigned
1111 1010₂ = 256 − 6 = 250
Answer
U = 250
6.Number systems
Answer: M = 2025 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
1920·1080·8 = 16588800 bit = 2073600 B
Convert to KiB
2073600/1024 = 2025 KiB
Answer
M = 2025 KiB
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