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 192.168.10.0/24 must be divided into at least 3 equal subnets. If you borrow as few bits as possible, how many usable hosts H does each subnet have?
192.168.10.0/243 subnets
H =
2.IPv4 subnettingHard
The network 192.168.65.0/24 is split with VLSM for these LANs: A: 35 hosts, B: 112 hosts, C: 19 hosts, D: 13 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: 35B: 112C: 19D: 13
x =
3.Computer networksMedium
How does TLS typically use asymmetric and symmetric cryptography?
- ASymmetric methods agree keys and authenticate; asymmetric ciphers encrypt all the data
- BOnly symmetric cryptography is used, with a fixed shared password
- CAsymmetric methods agree keys and authenticate; symmetric ciphers encrypt the data
- DOnly asymmetric cryptography is used throughout
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/28
- C/25
- D/27
5.Number systemsMedium
Write −18 as an 8-bit two's complement pattern. Read the pattern as an unsigned binary number: what is its value U in decimal?
N = −188 bit
U =
6.Number systemsHard
Set the bits to the 8-bit result of 1001 1000 AND 0111 0110. What is the result's value R in decimal?
1001 1000AND0111 0110
R =
3Solutions
1.IPv4 subnetting
Answer: H = 62
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).
Bits to borrow
22 = 4 ≥ 3 → b = 2, /26
Hosts per subnet
26 − 2 = 62
Answer
H = 62
2.IPv4 subnetting
Answer: x = 224
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
B: 112 hosts → 27 = 128 (/25)
A: 35 hosts → 26 = 64 (/26)
C: 19 hosts → 25 = 32 (/27)
D: 13 hosts → 24 = 16 (/28)
Allocation, largest first
B: .0 – .127
A: .128 – .191
C: .192 – .223
D: .224 – .239
Answer
x = 224
LAN D: 192.168.65.224/28.
3.Computer networks
Answer: C · Asymmetric methods agree keys and authenticate; symmetric ciphers encrypt the data
Why this is the answer
Public-key methods and certificates set up a shared secret and prove identity. Fast symmetric ciphers such as AES then protect the bulk data.
Answer
C · Asymmetric methods agree keys and authenticate; symmetric ciphers encrypt the data
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 = 238
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
18 = 0001 0010
inverted: 1110 1101
+1 → 1110 1110
Read it as unsigned
1110 1110₂ = 256 − 18 = 238
Answer
U = 238
6.Number systems
Answer: R = 16
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.
Bit by bit
AND gives 1 only where both bits are 1.
1001 1000
AND 0111 0110
= 0001 0000
Value of the result
0001 0000₂ = 16
Answer
R = 16
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