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
A department needs 523 host addresses in a single subnet. What is the longest prefix length p (smallest subnet) that is big enough?
523 hosts
p =
2.IPv4 subnettingHard
The network 192.168.143.0/24 is split with VLSM for these LANs: A: 50 hosts, B: 28 hosts, C: 4 hosts, D: 8 hosts, E: 10 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 C?
A: 50B: 28C: 4D: 8E: 10
x =
3.Computer networksMedium
Which IPv4 header field stops packets from circulating forever in a routing loop?
- AType of service
- BHeader checksum
- CTime to live (TTL)
- DIdentification
4.Computer networksHard
A remote-terminal application feels sluggish because small keystroke packets are held back. Which TCP mechanism is often responsible?
- ANagle's algorithm
- BFast retransmit
- CThe three-way handshake
- DSlow start
5.Number systemsMedium
Set the bits so they store −118 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?
N = −1188 bit
U =
6.Number systemsHard
A system must give every one of 3656 items its own binary code of the same length. What is the smallest number of bits b needed?
3656
b = bit
3Solutions
1.IPv4 subnetting
Answer: p = 22
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).
Host bits needed
29 − 2 = 510 < 523
210 − 2 = 1022 ≥ 523 → h = 10
Prefix length
p = 32 − 10 = 22
Answer
p = /22
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
A: 50 hosts → 26 = 64 (/26)
B: 28 hosts → 25 = 32 (/27)
E: 10 hosts → 24 = 16 (/28)
D: 8 hosts → 24 = 16 (/28)
C: 4 hosts → 23 = 8 (/29)
Allocation, largest first
A: .0 – .63
B: .64 – .95
E: .96 – .111
D: .112 – .127
C: .128 – .135
Answer
x = 128
LAN C: 192.168.143.128/29.
3.Computer networks
Answer: C · Time to live (TTL)
Why this is the answer
Every router decrements the TTL; at zero the packet is discarded and an ICMP Time Exceeded message is usually sent. IPv6 calls it hop limit.
Answer
C · Time to live (TTL)
4.Computer networks
Answer: A · Nagle's algorithm
Why this is the answer
Nagle's algorithm waits to bundle small segments while earlier data is unacknowledged. This saves bandwidth but adds delay for interactive traffic.
Answer
A · Nagle's algorithm
5.Number systems
Answer: U = 138
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
118 = 0111 0110
inverted: 1000 1001
+1 → 1000 1010
Read it as unsigned
1000 1010₂ = 256 − 118 = 138
Answer
U = 138
6.Number systems
Answer: b = 12 bit
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.
Powers of two
211 = 2048 < 3656
212 = 4096 ≥ 3656
Answer
b = 12
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