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 192.168.91.0/24 is divided into /25 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 2?
192.168.91.0/24/25subnet 2 (subnet 1 starts at .0)
x =
2.IPv4 subnettingHard
A host has the address 10.20.168.29/25. What is the last number (fourth octet) x of its subnet's broadcast address?
10.20.168.29/25
x =
3.Computer networksMedium
NAT (network address translation) is mainly used to:
- AIncrease the speed of a network cable
- BEncrypt traffic between two sites
- CLet many devices with private addresses share one public IP address
- DTranslate domain names into IP addresses for devices on a local network
4.Computer networksHard
Stations A and C both reach access point B but cannot hear each other, so their frames collide at B. How does 802.11 help?
- AOptional RTS/CTS exchanges: the AP's CTS tells all nearby stations to wait
- BBy routing A's frames through C
- CBy making A and C use collision detection so they stop sending when frames overlap
- DBy increasing the TTL of the frames
5.Number systemsMedium
The 8-bit pattern 1011 0110 stores a signed integer in two's complement. What is its value N in decimal?
1011 01108 bit
N =
6.Number systemsHard
Set the bits to the 8-bit result of 1001 0010 XOR 1010 0101. What is the result's value R in decimal?
1001 0010XOR1010 0101
R =
3Solutions
1.IPv4 subnetting
Answer: x = 255
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
27 = 128
Subnet number 2
network: (2 − 1)·128 = 128
broadcast: 128 + 128 − 1 = 255
Answer
x = 255
2.IPv4 subnetting
Answer: x = 127
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
27 = 128
Subnet of this host
network: ⌊29/128⌋·128 = 0
broadcast: 0 + 128 − 1 = 127
usable hosts: 1 … 126
Answer
x = 127
3.Computer networks
Answer: C · Let many devices with private addresses share one public IP address
Why this is the answer
NAT rewrites addresses (and ports) so private networks can access the internet through a smaller number of public addresses.
Answer
C · Let many devices with private addresses share one public IP address
4.Computer networks
Answer: A · Optional RTS/CTS exchanges: the AP's CTS tells all nearby stations to wait
Why this is the answer
A station sends Request to Send; the AP answers Clear to Send, which every station in its range hears, so hidden stations defer for the reserved time.
Answer
A · Optional RTS/CTS exchanges: the AP's CTS tells all nearby stations to wait
5.Number systems
Answer: N = −74
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 + 32 + 16 + 4 + 2 = −74
Answer
N = −74
6.Number systems
Answer: R = 55
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
XOR gives 1 where the two bits differ.
1001 0010
XOR 1010 0101
= 0011 0111
Value of the result
0011 0111₂ = 55
Answer
R = 55
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