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 172.29.0.0/16 is divided into equal subnets with prefix /19. How many subnets N are there?
172.29.0.0/16/19
N =
2.IPv4 subnettingHard
The network 192.168.76.0/24 is split with VLSM for these LANs: A: 45 hosts, B: 21 hosts, C: 8 hosts, D: 62 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 C?
A: 45B: 21C: 8D: 62
x =
3.Computer networksMedium
Which subnet mask corresponds to /20?
- A255.255.248.0
- B255.255.255.240
- C255.255.224.0
- D255.255.240.0
4.Computer networksHard
In a SYN flood attack, what resource is the attacker trying to exhaust?
- AThe server's table of half-open connections
- BThe server's disk space
- CThe switch's MAC address table
- DThe DNS cache of the client
5.Number systemsMedium
Write −107 as an 8-bit two's complement pattern. Read the pattern as an unsigned binary number: what is its value U in decimal?
N = −1078 bit
U =
6.Number systemsHard
An uncompressed image is 512 × 512 pixels with 32 bits per pixel. How much memory M does it take, in KiB (1 KiB = 1024 bytes)?
512 × 512 px32 bit/px1 KiB = 1024 B
M = KiB
3Solutions
1.IPv4 subnetting
Answer: N = 8
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
19 − 16 = 3
Number of subnets
N = 23 = 8
Answer
N = 8
Each /19 subnet has 213 − 2 = 8190 usable hosts.
2.IPv4 subnetting
Answer: x = 160
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
D: 62 hosts → 26 = 64 (/26)
A: 45 hosts → 26 = 64 (/26)
B: 21 hosts → 25 = 32 (/27)
C: 8 hosts → 24 = 16 (/28)
Allocation, largest first
D: .0 – .63
A: .64 – .127
B: .128 – .159
C: .160 – .175
Answer
x = 160
LAN C: 192.168.76.160/28.
3.Computer networks
Answer: D · 255.255.240.0
Why this is the answer
/20 means 8 + 8 + 4 one-bits. Four one-bits in the third octet are 11110000 = 240, so the mask is 255.255.240.0.
Answer
D · 255.255.240.0
4.Computer networks
Answer: A · The server's table of half-open connections
Why this is the answer
Each SYN makes the server reserve state and wait for an ACK that never comes. SYN cookies avoid storing state until the handshake completes.
Answer
A · The server's table of half-open connections
5.Number systems
Answer: U = 149
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
107 = 0110 1011
inverted: 1001 0100
+1 → 1001 0101
Read it as unsigned
1001 0101₂ = 256 − 107 = 149
Answer
U = 149
6.Number systems
Answer: M = 1024 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
512·512·32 = 8388608 bit = 1048576 B
Convert to KiB
1048576/1024 = 1024 KiB
Answer
M = 1024 KiB
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