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 255.255.255.240. How many usable host addresses H does it have?
255.255.255.240
H =
2.IPv4 subnettingHard
A host has the address 10.20.81.127/28. What is the fourth octet x of the last usable host address in its subnet?
10.20.81.127/28
x =
3.Computer networksMedium
What does an MX record in DNS specify?
- AThe mail servers that accept email for a domain
- BThe name servers for a zone
- CAn alias name for another host
- DThe IPv4 address of a web server
4.Computer networksHard
Which routing protocol is a link-state protocol?
- ABGP
- BOSPF
- CRIP
- DARP
5.Number systemsMedium
Write −12 as an 8-bit two's complement pattern. Read the pattern as an unsigned binary number: what is its value U in decimal?
N = −128 bit
U =
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 = 14
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
255.255.255.240 = /28
Usable hosts
24 − 2 = 14
Answer
H = 14
2.IPv4 subnetting
Answer: x = 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).
Block size
24 = 16
Subnet of this host
network: ⌊127/16⌋·16 = 112
broadcast: 112 + 16 − 1 = 127
usable hosts: 113 … 126
Answer
x = 126
3.Computer networks
Answer: A · The mail servers that accept email for a domain
Why this is the answer
MX (mail exchange) records list mail servers with priorities, so sending servers know where to deliver mail for that domain.
Answer
A · The mail servers that accept email for a domain
4.Computer networks
Answer: B · OSPF
Why this is the answer
OSPF routers build a full map of the area and compute shortest paths with Dijkstra's algorithm. RIP is distance-vector; BGP is path-vector.
Answer
B · OSPF
5.Number systems
Answer: U = 244
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
12 = 0000 1100
inverted: 1111 0011
+1 → 1111 0100
Read it as unsigned
1111 0100₂ = 256 − 12 = 244
Answer
U = 244
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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