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 255.255.254.0. How many usable host addresses H does it have?
255.255.254.0
H =
2.IPv4 subnettingHard
A host has the address 10.20.30.195/25. What is the fourth octet x of the last usable host address in its subnet?
10.20.30.195/25
x =
3.Computer networksMedium
Which IPv4 header field stops packets from circulating forever in a routing loop?
- AType of service
- BIdentification
- CTime to live (TTL)
- DHeader checksum
4.Computer networksHard
A TCP connection has a 64 KB window and a 100 ms round-trip time, and no window scaling. What is its approximate maximum throughput?
- AAbout 5 Mbit/s
- BAbout 640 Mbit/s
- CAbout 0.5 Mbit/s
- DAbout 50 Mbit/s
5.Number systemsMedium
The 8-bit pattern 1010 1101 stores a signed integer in two's complement. What is its value N in decimal?
1010 11018 bit
N =
6.Number systemsHard
An uncompressed image is 1024 × 768 pixels with 16 bits per pixel. How much memory M does it take, in KiB (1 KiB = 1024 bytes)?
1024 × 768 px16 bit/px1 KiB = 1024 B
M = KiB
3Solutions
1.IPv4 subnetting
Answer: H = 510
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
255.255.254.0 = /23
Usable hosts
29 − 2 = 510
Answer
H = 510
2.IPv4 subnetting
Answer: x = 254
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
27 = 128
Subnet of this host
network: ⌊195/128⌋·128 = 128
broadcast: 128 + 128 − 1 = 255
usable hosts: 129 … 254
Answer
x = 254
3.Computer networks
Answer: C · Time to live (TTL)
Why this is the answer
Every router decrements the TTL; at zero the packet is discarded and an ICMP Time Exceeded message is usually sent. IPv6 calls it hop limit.
Answer
C · Time to live (TTL)
4.Computer networks
Answer: A · About 5 Mbit/s
Why this is the answer
Throughput ≤ window / RTT = 65,536 × 8 bits / 0.1 s ≈ 5.2 Mbit/s, however fast the link is.
Answer
A · About 5 Mbit/s
5.Number systems
Answer: N = −83
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.
Add the place values
The other bits keep their normal positive values.
−128 + 32 + 8 + 4 + 1 = −83
Answer
N = −83
6.Number systems
Answer: M = 1536 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
1024·768·16 = 12582912 bit = 1572864 B
Convert to KiB
1572864/1024 = 1536 KiB
Answer
M = 1536 KiB
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