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 8 equal subnets. If you borrow as few bits as possible, how many usable hosts H does each subnet have?
192.168.10.0/248 subnets
H =
2.IPv4 subnettingHard
A host has the address 10.20.249.162/28. What is the last number (fourth octet) x of its subnet's broadcast address?
10.20.249.162/28
x =
3.Computer networksMedium
Why does a switch sometimes send a frame out of all ports except the one it came in on?
- AThe destination MAC address is not yet in its address table
- BThe frame has a wrong IP checksum and must be checked by every host
- CThe frame is too large for one port
- DThe switch is acting as a router
4.Computer networksHard
A DNS cache poisoning attack aims to:
- AOverload a server with traffic so that it stops answering queries
- BSteal Wi-Fi passwords
- CMake a resolver store a false IP address for a domain
- DDelete DNS servers
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
Set the bits to the 8-bit result of 1110 1111 OR 1101 0001. What is the result's value R in decimal?
1110 1111OR1101 0001
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 ≥ 8 → 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: ⌊162/16⌋·16 = 160
broadcast: 160 + 16 − 1 = 175
usable hosts: 161 … 174
Answer
x = 175
3.Computer networks
Answer: A · The destination MAC address is not yet in its address table
Why this is the answer
For an unknown unicast address the switch floods the frame. When the destination replies, the switch learns its port from the source MAC.
Answer
A · The destination MAC address is not yet in its address table
4.Computer networks
Answer: C · Make a resolver store a false IP address for a domain
Why this is the answer
Poisoning inserts forged records into a resolver's cache. DNSSEC adds signatures so resolvers can verify records.
Answer
C · Make a resolver store a false IP address for a domain
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: R = 255
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
OR gives 1 where at least one bit is 1.
1110 1111
OR 1101 0001
= 1111 1111
Value of the result
1111 1111₂ = 255
Answer
R = 255
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