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 30 equal subnets. If you borrow as few bits as possible, how many usable hosts H does each subnet have?
192.168.10.0/2430 subnets
H =
2.IPv4 subnettingHard
The network 192.168.95.0/24 is split with VLSM for these LANs: A: 17 hosts, B: 6 hosts, C: 5 hosts, D: 27 hosts, E: 11 hosts. Subnets are allocated from .0 upwards, the LAN with the most hosts first, each with the smallest subnet that fits. What is the fourth octet x of the network address of LAN B?
A: 17B: 6C: 5D: 27E: 11
x =
3.Computer networksMedium
NAT (network address translation) is mainly used to:
- AIncrease the speed of a network cable
- BLet many devices with private addresses share one public IP address
- CEncrypt traffic between two sites
- DTranslate domain names into IP addresses for devices on a local network
4.Computer networksHard
After repeated collisions, classic Ethernet stations choose their waiting time from a range that doubles each time. Why?
- ATo give the station with the highest MAC address priority on the shared cable
- BTo lower the chance of colliding again when many stations compete
- CTo synchronise all stations to one clock
- DTo make the cable carry signals further
5.Number systemsMedium
Set the bits so they store −54 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?
N = −548 bit
U =
6.Number systemsHard
Set the bits to the 8-bit result of 0111 0101 OR 0110 0010. What is the result's value R in decimal?
0111 0101OR0110 0010
R =
3Solutions
1.IPv4 subnetting
Answer: H = 6
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
25 = 32 ≥ 30 → b = 5, /29
Hosts per subnet
23 − 2 = 6
Answer
H = 6
2.IPv4 subnetting
Answer: x = 80
VLSM: each LAN gets the smallest subnet with 2ʰ − 2 ≥ its host count, i.e. prefix 32 − h. Allocating the largest subnets first keeps every block aligned on a multiple of its own size, so the network addresses follow by adding up the block sizes.
Block size per LAN
D: 27 hosts → 25 = 32 (/27)
A: 17 hosts → 25 = 32 (/27)
E: 11 hosts → 24 = 16 (/28)
B: 6 hosts → 23 = 8 (/29)
C: 5 hosts → 23 = 8 (/29)
Allocation, largest first
D: .0 – .31
A: .32 – .63
E: .64 – .79
B: .80 – .87
C: .88 – .95
Answer
x = 80
LAN B: 192.168.95.80/29.
3.Computer networks
Answer: B · 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
B · Let many devices with private addresses share one public IP address
4.Computer networks
Answer: B · To lower the chance of colliding again when many stations compete
Why this is the answer
Each further collision suggests more contention, so the random range grows and stations are more likely to pick different slots. It adapts to load without central control.
Answer
B · To lower the chance of colliding again when many stations compete
5.Number systems
Answer: U = 202
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
54 = 0011 0110
inverted: 1100 1001
+1 → 1100 1010
Read it as unsigned
1100 1010₂ = 256 − 54 = 202
Answer
U = 202
6.Number systems
Answer: R = 119
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.
0111 0101
OR 0110 0010
= 0111 0111
Value of the result
0111 0111₂ = 119
Answer
R = 119
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