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 5 equal subnets. If you borrow as few bits as possible, how many usable hosts H does each subnet have?
192.168.10.0/245 subnets
H =
2.IPv4 subnettingHard
A host has the address 10.20.38.172/28. What is the last number (fourth octet) x of its subnet's broadcast address?
10.20.38.172/28
x =
3.Computer networksMedium
Which TCP flag signals that a sender has finished sending data?
- APSH
- BSYN
- CURG
- DFIN
4.Computer networksHard
What is the maximum transmission unit (MTU) of standard Ethernet payload?
- A9000 bytes in every network
- B65,535 bytes
- C1500 bytes
- D64 bytes
5.Number systemsMedium
Write −29 as an 8-bit two's complement pattern. Read the pattern as an unsigned binary number: what is its value U in decimal?
N = −298 bit
U =
6.Number systemsHard
Set the bits to the 8-bit result of 1001 1010 >> 1. What is the result's value R in decimal?
1001 1010>> 1
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 ≥ 5 → b = 3, /27
Hosts per subnet
25 − 2 = 30
Answer
H = 30
2.IPv4 subnetting
Answer: x = 175
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).
Block size
24 = 16
Subnet of this host
network: ⌊172/16⌋·16 = 160
broadcast: 160 + 16 − 1 = 175
usable hosts: 161 … 174
Answer
x = 175
3.Computer networks
Answer: D · 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
D · FIN
4.Computer networks
Answer: C · 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
C · 1500 bytes
5.Number systems
Answer: U = 227
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
29 = 0001 1101
inverted: 1110 0010
+1 → 1110 0011
Read it as unsigned
1110 0011₂ = 256 − 29 = 227
Answer
U = 227
6.Number systems
Answer: R = 77
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 1 moves every bit 1 places to the right; bits that fall off the right end are lost and zeros come in on the left.
1001 1010 >> 1
= 0100 1101
Value of the result
0100 1101₂ = 77
Answer
R = 77
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