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 6 equal subnets. If you borrow as few bits as possible, how many usable hosts H does each subnet have?
192.168.10.0/246 subnets
H =
2.IPv4 subnettingHard
The network 192.168.23.0/24 is split with VLSM for these LANs: A: 14 hosts, B: 96 hosts, C: 6 hosts, D: 31 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: 14B: 96C: 6D: 31
x =
3.Computer networksMedium
TLS provides which security properties for HTTPS?
- AProtection against all malware and phishing on the visited site
- BAnonymity of the user
- COnly faster loading
- DConfidentiality, integrity and server authentication
4.Computer networksHard
What is the maximum transmission unit (MTU) of standard Ethernet payload?
- A9000 bytes in every network
- B1500 bytes
- C65,535 bytes
- D64 bytes
5.Number systemsMedium
Set the bits so they store −113 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?
N = −1138 bit
U =
6.Number systemsHard
Set the bits to the 8-bit result of 0001 1101 >> 2. What is the result's value R in decimal?
0001 1101>> 2
R =
3Solutions
1.IPv4 subnetting
Answer: H = 30
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
23 = 8 ≥ 6 → b = 3, /27
Hosts per subnet
25 − 2 = 30
Answer
H = 30
2.IPv4 subnetting
Answer: x = 208
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: 96 hosts → 27 = 128 (/25)
D: 31 hosts → 26 = 64 (/26)
A: 14 hosts → 24 = 16 (/28)
C: 6 hosts → 23 = 8 (/29)
Allocation, largest first
B: .0 – .127
D: .128 – .191
A: .192 – .207
C: .208 – .215
Answer
x = 208
LAN C: 192.168.23.208/29.
3.Computer networks
Answer: D · Confidentiality, integrity and server authentication
Why this is the answer
TLS encrypts traffic, detects tampering and authenticates the server with certificates (optionally the client too).
Answer
D · Confidentiality, integrity and server authentication
4.Computer networks
Answer: B · 1500 bytes
Why this is the answer
Standard Ethernet carries up to 1500 bytes of payload; larger IP packets must be fragmented or use path MTU discovery.
Answer
B · 1500 bytes
5.Number systems
Answer: U = 143
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
113 = 0111 0001
inverted: 1000 1110
+1 → 1000 1111
Read it as unsigned
1000 1111₂ = 256 − 113 = 143
Answer
U = 143
6.Number systems
Answer: R = 7
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
Shifting right by 2 moves every bit 2 places to the right; bits that fall off the right end are lost and zeros come in on the left.
0001 1101 >> 2
= 0000 0111
Value of the result
0000 0111₂ = 7
Answer
R = 7
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