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 10.129.32.0/20 is divided into equal subnets with prefix /24. How many subnets N are there?
10.129.32.0/20/24
N =
2.IPv4 subnettingHard
A host has the address 10.20.60.9/30. What is the fourth octet x of its network address?
10.20.60.9/30
x =
3.Computer networksMedium
Why does a switch sometimes send a frame out of all ports except the one it came in on?
- AThe switch is acting as a router
- BThe destination MAC address is not yet in its address table
- CThe frame has a wrong IP checksum and must be checked by every host
- DThe frame is too large for one port
4.Computer networksHard
A DNS cache poisoning attack aims to:
- ASteal Wi-Fi passwords
- BOverload a server with traffic so that it stops answering queries
- CMake a resolver store a false IP address for a domain
- DDelete DNS servers
5.Number systemsMedium
Set the bits so they store −111 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?
N = −1118 bit
U =
6.Number systemsHard
Set the bits to the 8-bit result of 1011 1011 << 3. What is the result's value R in decimal?
1011 1011<< 3
R =
3Solutions
1.IPv4 subnetting
Answer: N = 16
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
24 − 20 = 4
Number of subnets
N = 24 = 16
Answer
N = 16
Each /24 subnet has 28 − 2 = 254 usable hosts.
2.IPv4 subnetting
Answer: x = 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).
Block size
22 = 4
Subnet of this host
network: ⌊9/4⌋·4 = 8
broadcast: 8 + 4 − 1 = 11
usable hosts: 9 … 10
Answer
x = 8
3.Computer networks
Answer: B · 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
B · 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 = 145
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
111 = 0110 1111
inverted: 1001 0000
+1 → 1001 0001
Read it as unsigned
1001 0001₂ = 256 − 111 = 145
Answer
U = 145
6.Number systems
Answer: R = 216
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 left by 3 moves every bit 3 places to the left; bits that fall off the left end are lost and zeros come in on the right.
1011 1011 << 3
= 1101 1000
Value of the result
1101 1000₂ = 216
Answer
R = 216
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