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 10.69.76.0/22 is divided into equal subnets with prefix /25. How many subnets N are there?
10.69.76.0/22/25
N =
2.IPv4 subnettingHard
The network 192.168.7.0/24 is split with VLSM for these LANs: A: 13 hosts, B: 20 hosts, C: 40 hosts, D: 103 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: 13B: 20C: 40D: 103
x =
3.Computer networksMedium
What does a subnet mask of 255.255.255.0 correspond to in CIDR notation?
- A/8
- B/16
- C/32
- D/24
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/26
- B/25
- C/27
- D/28
5.Number systemsMedium
Set the bits so they store −64 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?
N = −648 bit
U =
6.Number systemsHard
A system must give every one of 2910 items its own binary code of the same length. What is the smallest number of bits b needed?
2910
b = bit
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
25 − 22 = 3
Number of subnets
N = 23 = 8
Answer
N = 8
Each /25 subnet has 27 − 2 = 126 usable hosts.
2.IPv4 subnetting
Answer: x = 128
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: 103 hosts → 27 = 128 (/25)
C: 40 hosts → 26 = 64 (/26)
B: 20 hosts → 25 = 32 (/27)
A: 13 hosts → 24 = 16 (/28)
Allocation, largest first
D: .0 – .127
C: .128 – .191
B: .192 – .223
A: .224 – .239
Answer
x = 128
LAN C: 192.168.7.128/26.
3.Computer networks
Answer: D · /24
Why this is the answer
Three octets of 255 are 24 one-bits, so 255.255.255.0 = /24.
Answer
D · /24
4.Computer networks
Answer: C · /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
C · /27
5.Number systems
Answer: U = 192
Setting bits: each bit is worth a power of two (128, 64, …, 1). One hex digit is exactly four bits, and in 8-bit two's complement the leftmost bit is worth −128. Bitwise operations work on each bit position separately.
Invert and add one
64 = 0100 0000
inverted: 1011 1111
+1 → 1100 0000
Read it as unsigned
1100 0000₂ = 256 − 64 = 192
Answer
U = 192
6.Number systems
Answer: b = 12 bit
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.
Powers of two
211 = 2048 < 2910
212 = 4096 ≥ 2910
Answer
b = 12
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