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 192.168.172.0/24 is divided into /28 subnets, numbered 1, 2, 3, … from the lowest address (subnet 1 starts at .0). What is the fourth octet x of the broadcast address of subnet 3?
192.168.172.0/24/28subnet 3 (subnet 1 starts at .0)
x =
2.IPv4 subnettingHard
A host has the address 10.20.8.87/27. What is the fourth octet x of the last usable host address in its subnet?
10.20.8.87/27
x =
3.Computer networksMedium
Which of these is a private IPv4 address range?
- A200.1.1.0/24
- B8.8.8.0/24
- C100.0.0.0/8
- D192.168.0.0/16
4.Computer networksHard
An attacker on a LAN sends forged ARP replies claiming the gateway's IP address. What is the likely result?
- AThe switch shuts down all of its ports
- BAll DNS queries fail across the internet
- CVictims send traffic for the gateway to the attacker, enabling a man-in-the-middle attack
- DThe gateway's IP address is permanently changed to the attacker's address on every router
5.Number systemsMedium
The 8-bit pattern 1111 1111 stores a signed integer in two's complement. What is its value N in decimal?
1111 11118 bit
N =
6.Number systemsHard
Set the bits to the 8-bit result of 1100 0010 >> 2. What is the result's value R in decimal?
1100 0010>> 2
R =
3Solutions
1.IPv4 subnetting
Answer: x = 47
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 number 3
network: (3 − 1)·16 = 32
broadcast: 32 + 16 − 1 = 47
Answer
x = 47
2.IPv4 subnetting
Answer: x = 94
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
25 = 32
Subnet of this host
network: ⌊87/32⌋·32 = 64
broadcast: 64 + 32 − 1 = 95
usable hosts: 65 … 94
Answer
x = 94
3.Computer networks
Answer: D · 192.168.0.0/16
Why this is the answer
Private ranges are 10.0.0.0/8, 172.16.0.0/12 and 192.168.0.0/16; they are not routed on the public internet.
Answer
D · 192.168.0.0/16
4.Computer networks
Answer: C · Victims send traffic for the gateway to the attacker, enabling a man-in-the-middle attack
Why this is the answer
ARP spoofing poisons hosts' ARP caches with the attacker's MAC for the gateway IP. The attacker can then read, change or drop traffic; dynamic ARP inspection helps prevent it.
Answer
C · Victims send traffic for the gateway to the attacker, enabling a man-in-the-middle attack
5.Number systems
Answer: N = −1
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.
Add the place values
The other bits keep their normal positive values.
−128 + 64 + 32 + 16 + 8 + 4 + 2 + 1 = −1
Answer
N = −1
6.Number systems
Answer: R = 48
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.
1100 0010 >> 2
= 0011 0000
Value of the result
0011 0000₂ = 48
Answer
R = 48
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