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 10.14.224.0/20 is divided into equal subnets with prefix /27. How many subnets N are there?
10.14.224.0/20/27
N =
2.IPv4 subnettingHard
The network 192.168.178.0/24 is split with VLSM for these LANs: A: 108 hosts, B: 23 hosts, C: 10 hosts, D: 56 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: 108B: 23C: 10D: 56
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
Why does IPv6 not need NAT for address shortage?
- AIts 128-bit address space is large enough to give every device a global address
- BIPv6 does not use addresses
- CIPv6 encrypts all addresses
- DIPv6 only works on local networks, so addresses never need to be shared globally
5.Number systemsMedium
The 8-bit pattern 1011 0101 stores a signed integer in two's complement. What is its value N in decimal?
1011 01018 bit
N =
6.Number systemsHard
What is the largest unsigned integer N that fits in 10 bits?
10 bit
N =
3Solutions
1.IPv4 subnetting
Answer: N = 128
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).
Borrowed bits
27 − 20 = 7
Number of subnets
N = 27 = 128
Answer
N = 128
Each /27 subnet has 25 − 2 = 30 usable hosts.
2.IPv4 subnetting
Answer: x = 192
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
A: 108 hosts → 27 = 128 (/25)
D: 56 hosts → 26 = 64 (/26)
B: 23 hosts → 25 = 32 (/27)
C: 10 hosts → 24 = 16 (/28)
Allocation, largest first
A: .0 – .127
D: .128 – .191
B: .192 – .223
C: .224 – .239
Answer
x = 192
LAN B: 192.168.178.192/27.
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: A · 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
A · Its 128-bit address space is large enough to give every device a global address
5.Number systems
Answer: N = −75
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 + 1 = −75
Answer
N = −75
6.Number systems
Answer: N = 1023
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.
Largest value
210 − 1 = 1023
Answer
N = 1023
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