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 172.17.0.0/16 is divided into equal subnets with prefix /18. How many subnets N are there?
172.17.0.0/16/18
N =
2.IPv4 subnettingHard
The network 192.168.168.0/24 is split with VLSM for these LANs: A: 54 hosts, B: 26 hosts, C: 8 hosts, D: 38 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 D?
A: 54B: 26C: 8D: 38
x =
3.Computer networksMedium
Which TCP flag signals that a sender has finished sending data?
- AFIN
- BPSH
- CSYN
- DURG
4.Computer networksHard
You must divide 192.168.1.0/24 into at least 5 subnets with up to 25 hosts each. Which prefix length fits best?
- A/25
- B/28
- C/26
- D/27
5.Number systemsMedium
Write −116 as an 8-bit two's complement pattern. Read the pattern as an unsigned binary number: what is its value U in decimal?
N = −1168 bit
U =
6.Number systemsHard
An uncompressed image is 800 × 600 pixels with 8 bits per pixel. How much memory M does it take, in KiB (1 KiB = 1024 bytes)?
800 × 600 px8 bit/px1 KiB = 1024 B
M = KiB
3Solutions
1.IPv4 subnetting
Answer: N = 4
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).
Borrowed bits
18 − 16 = 2
Number of subnets
N = 22 = 4
Answer
N = 4
Each /18 subnet has 214 − 2 = 16382 usable hosts.
2.IPv4 subnetting
Answer: x = 64
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
A: 54 hosts → 26 = 64 (/26)
D: 38 hosts → 26 = 64 (/26)
B: 26 hosts → 25 = 32 (/27)
C: 8 hosts → 24 = 16 (/28)
Allocation, largest first
A: .0 – .63
D: .64 – .127
B: .128 – .159
C: .160 – .175
Answer
x = 64
LAN D: 192.168.168.64/26.
3.Computer networks
Answer: A · FIN
Why this is the answer
Each side sends FIN when it has no more data, and the other side acknowledges it. SYN is for opening a connection; RST aborts one.
Answer
A · FIN
4.Computer networks
Answer: D · /27
Why this is the answer
/27 gives 8 subnets with 30 usable hosts each. /26 gives only 4 subnets, and /28 allows only 14 hosts per subnet.
Answer
D · /27
5.Number systems
Answer: U = 140
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
116 = 0111 0100
inverted: 1000 1011
+1 → 1000 1100
Read it as unsigned
1000 1100₂ = 256 − 116 = 140
Answer
U = 140
6.Number systems
Answer: M = 468.8 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
800·600·8 = 3840000 bit = 480000 B
Convert to KiB
480000/1024 = 468.75 KiB
Answer
M = 468.75 KiB
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