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 192.168.249.0/24 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 network address of subnet 7?
192.168.249.0/24/28subnet 7 (subnet 1 starts at .0)
x =
2.IPv4 subnettingHard
A host has the address 10.20.15.71/27. What is the last number (fourth octet) x of its subnet's broadcast address?
10.20.15.71/27
x =
3.Computer networksMedium
Which range of TCP/UDP port numbers is reserved as "well-known" ports?
- A0–1023
- B49152–65535
- C1024–49151
- D0–65535
4.Computer networksHard
Why does IPv6 not need NAT for address shortage?
- AIPv6 does not use addresses
- BIts 128-bit address space is large enough to give every device a global address
- CIPv6 only works on local networks, so addresses never need to be shared globally
- DIPv6 encrypts all addresses
5.Number systemsMedium
Set the bits so they store −73 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?
N = −738 bit
U =
6.Number systemsHard
An uncompressed image is 800 × 600 pixels with 16 bits per pixel. How much memory M does it take, in KiB (1 KiB = 1024 bytes)?
800 × 600 px16 bit/px1 KiB = 1024 B
M = KiB
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
24 = 16
Subnet number 7
network: (7 − 1)·16 = 96
broadcast: 96 + 16 − 1 = 111
Answer
x = 96
2.IPv4 subnetting
Answer: x = 95
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: ⌊71/32⌋·32 = 64
broadcast: 64 + 32 − 1 = 95
usable hosts: 65 … 94
Answer
x = 95
3.Computer networks
Answer: A · 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
A · 0–1023
4.Computer networks
Answer: B · Its 128-bit address space is large enough to give every device a global address
Why this is the answer
IPv6 has about 3.4 × 10³⁸ addresses, so the scarcity that drove NAT in IPv4 disappears; firewalls still provide security.
Answer
B · Its 128-bit address space is large enough to give every device a global address
5.Number systems
Answer: U = 183
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
73 = 0100 1001
inverted: 1011 0110
+1 → 1011 0111
Read it as unsigned
1011 0111₂ = 256 − 73 = 183
Answer
U = 183
6.Number systems
Answer: M = 937.5 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
800·600·16 = 7680000 bit = 960000 B
Convert to KiB
960000/1024 = 937.5 KiB
Answer
M = 937.5 KiB
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