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 192.168.14.0/24 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?
192.168.14.0/24/25subnet 2 (subnet 1 starts at .0)
x =
2.IPv4 subnettingHard
The network 192.168.243.0/24 is split with VLSM for these LANs: A: 12 hosts, B: 30 hosts, C: 56 hosts, D: 43 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: 12B: 30C: 56D: 43
x =
3.Computer networksMedium
Which range of TCP/UDP port numbers is reserved as "well-known" ports?
- A1024–49151
- B49152–65535
- C0–1023
- D0–65535
4.Computer networksHard
A PC can reach printers on its own subnet but nothing beyond it. Its IP address and mask are correct. What is the most likely fault?
- AA faulty DNS server setting
- BA duplicate MAC address on the internet
- CThe TCP window is too small
- DA missing or wrong default gateway
5.Number systemsMedium
Write −43 as an 8-bit two's complement pattern. Read the pattern as an unsigned binary number: what is its value U in decimal?
N = −438 bit
U =
6.Number systemsHard
An uncompressed image is 1024 × 768 pixels with 32 bits per pixel. How much memory M does it take, in KiB (1 KiB = 1024 bytes)?
1024 × 768 px32 bit/px1 KiB = 1024 B
M = KiB
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 = 128
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
C: 56 hosts → 26 = 64 (/26)
D: 43 hosts → 26 = 64 (/26)
B: 30 hosts → 25 = 32 (/27)
A: 12 hosts → 24 = 16 (/28)
Allocation, largest first
C: .0 – .63
D: .64 – .127
B: .128 – .159
A: .160 – .175
Answer
x = 128
LAN B: 192.168.243.128/27.
3.Computer networks
Answer: C · 0–1023
Why this is the answer
Ports 0–1023 are well-known ports for standard services. 1024–49151 are registered ports and 49152–65535 are dynamic or private ports.
Answer
C · 0–1023
4.Computer networks
Answer: D · A missing or wrong default gateway
Why this is the answer
Hosts send off-subnet packets to the default gateway. If it is wrong or missing, local traffic still works but nothing outside is reachable, even by IP address.
Answer
D · A missing or wrong default gateway
5.Number systems
Answer: U = 213
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
43 = 0010 1011
inverted: 1101 0100
+1 → 1101 0101
Read it as unsigned
1101 0101₂ = 256 − 43 = 213
Answer
U = 213
6.Number systems
Answer: M = 3072 KiB
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.
Bits and bytes
1024·768·32 = 25165824 bit = 3145728 B
Convert to KiB
3145728/1024 = 3072 KiB
Answer
M = 3072 KiB
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