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 192.168.65.0/24 is divided into /27 subnets, numbered 1, 2, 3, … from the lowest address (subnet 1 starts at .0). What is the fourth octet x of the network address of subnet 4?
192.168.65.0/24/27subnet 4 (subnet 1 starts at .0)
x =
2.IPv4 subnettingHard
A host has the address 10.20.234.55/29. What is the fourth octet x of the last usable host address in its subnet?
10.20.234.55/29
x =
3.Computer networksMedium
What is the main purpose of a VLAN?
- ATo increase the speed of every switch port
- BTo encrypt all traffic between switches so other departments cannot read it
- CTo split one physical switched network into separate broadcast domains
- DTo give each device a public IP address
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?
- ABy increasing the TTL of the frames
- BBy making A and C use collision detection so they stop sending when frames overlap
- COptional RTS/CTS exchanges: the AP's CTS tells all nearby stations to wait
- DBy routing A's frames through C
5.Number systemsMedium
Write −66 as an 8-bit two's complement pattern. Read the pattern as an unsigned binary number: what is its value U in decimal?
N = −668 bit
U =
6.Number systemsHard
Set the bits to the 8-bit result of 0110 1101 OR 1010 0111. What is the result's value R in decimal?
0110 1101OR1010 0111
R =
3Solutions
1.IPv4 subnetting
Answer: x = 96
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 number 4
network: (4 − 1)·32 = 96
broadcast: 96 + 32 − 1 = 127
Answer
x = 96
2.IPv4 subnetting
Answer: x = 54
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
23 = 8
Subnet of this host
network: ⌊55/8⌋·8 = 48
broadcast: 48 + 8 − 1 = 55
usable hosts: 49 … 54
Answer
x = 54
3.Computer networks
Answer: C · To split one physical switched network into separate broadcast domains
Why this is the answer
VLANs group ports logically, so broadcasts stay within each VLAN. This improves security and limits broadcast traffic.
Answer
C · To split one physical switched network into separate broadcast domains
4.Computer networks
Answer: C · 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
C · Optional RTS/CTS exchanges: the AP's CTS tells all nearby stations to wait
5.Number systems
Answer: U = 190
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
66 = 0100 0010
inverted: 1011 1101
+1 → 1011 1110
Read it as unsigned
1011 1110₂ = 256 − 66 = 190
Answer
U = 190
6.Number systems
Answer: R = 239
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.
0110 1101
OR 1010 0111
= 1110 1111
Value of the result
1110 1111₂ = 239
Answer
R = 239
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