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 201 host addresses in a single subnet. What is the longest prefix length p (smallest subnet) that is big enough?
201 hosts
p =
2.IPv4 subnettingHard
The network 192.168.133.0/24 is split with VLSM for these LANs: A: 10 hosts, B: 59 hosts, C: 68 hosts, D: 4 hosts. Subnets are allocated from .0 upwards, the LAN with the most hosts first, each with the smallest subnet that fits. What is the fourth octet x of the network address of LAN A?
A: 10B: 59C: 68D: 4
x =
3.Computer networksMedium
Which DNS record type maps a host name to an IPv6 address?
- AA
- BAAAA
- CCNAME
- DMX
4.Computer networksHard
A student runs a game server on a PC behind a home NAT router. Friends on the internet cannot connect. What is the usual fix?
- AGive the PC a different MAC address so the router can find it from outside
- BConfigure port forwarding on the router to the PC's private address
- CTurn off the PC's DNS client
- DChange the PC's subnet mask to /32
5.Number systemsMedium
Write −85 as an 8-bit two's complement pattern. Read the pattern as an unsigned binary number: what is its value U in decimal?
N = −858 bit
U =
6.Number systemsHard
An uncompressed image is 1920 × 1080 pixels with 16 bits per pixel. How much memory M does it take, in KiB (1 KiB = 1024 bytes)?
1920 × 1080 px16 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 < 201
28 − 2 = 254 ≥ 201 → h = 8
Prefix length
p = 32 − 8 = 24
Answer
p = /24
2.IPv4 subnetting
Answer: x = 192
VLSM: each LAN gets the smallest subnet with 2ʰ − 2 ≥ its host count, i.e. prefix 32 − h. Allocating the largest subnets first keeps every block aligned on a multiple of its own size, so the network addresses follow by adding up the block sizes.
Block size per LAN
C: 68 hosts → 27 = 128 (/25)
B: 59 hosts → 26 = 64 (/26)
A: 10 hosts → 24 = 16 (/28)
D: 4 hosts → 23 = 8 (/29)
Allocation, largest first
C: .0 – .127
B: .128 – .191
A: .192 – .207
D: .208 – .215
Answer
x = 192
LAN A: 192.168.133.192/28.
3.Computer networks
Answer: B · 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
B · AAAA
4.Computer networks
Answer: B · Configure port forwarding on the router to the PC's private address
Why this is the answer
NAT creates mappings only for connections started from inside. A port-forwarding rule maps a public port to the internal host so inbound connections can reach it.
Answer
B · Configure port forwarding on the router to the PC's private address
5.Number systems
Answer: U = 171
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
85 = 0101 0101
inverted: 1010 1010
+1 → 1010 1011
Read it as unsigned
1010 1011₂ = 256 − 85 = 171
Answer
U = 171
6.Number systems
Answer: M = 4050 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
1920·1080·16 = 33177600 bit = 4147200 B
Convert to KiB
4147200/1024 = 4050 KiB
Answer
M = 4050 KiB
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