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 1362 host addresses in a single subnet. What is the longest prefix length p (smallest subnet) that is big enough?
1362 hosts
p =
2.IPv4 subnettingHard
A router has routes to these networks: 10.212.150.0/24, 10.212.151.0/24, 10.212.152.0/24, 10.212.153.0/24, 10.212.154.0/24, 10.212.155.0/24, 10.212.156.0/24. What is the longest prefix length p of a single summary route that covers all of them?
10.212.150.0/24 … 10.212.156.0/24
p =
3.Computer networksMedium
What does an MX record in DNS specify?
- AThe mail servers that accept email for a domain
- BThe name servers for a zone
- CAn alias name for another host
- DThe IPv4 address of a web server
4.Computer networksHard
What is the maximum transmission unit (MTU) of standard Ethernet payload?
- A9000 bytes in every network
- B65,535 bytes
- C1500 bytes
- D64 bytes
5.Number systemsMedium
Set the bits so they store −62 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?
N = −628 bit
U =
6.Number systemsHard
Set the bits to the 8-bit result of 1100 0011 << 1. What is the result's value R in decimal?
1100 0011<< 1
R =
3Solutions
1.IPv4 subnetting
Answer: p = 21
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
210 − 2 = 1022 < 1362
211 − 2 = 2046 ≥ 1362 → h = 11
Prefix length
p = 32 − 11 = 21
Answer
p = /21
2.IPv4 subnetting
Answer: p = 20
Route summarisation: a summary route a.b.c.0/p covers all networks whose first p bits match. Write the differing octet of the first and last network in binary; the number of common leading bits, plus the bits of the octets before it, is the longest summary prefix.
Compare the bits
150 = 10010110
156 = 10011100
common leading bits: 4
Prefix length
p = 16 + 4 = 20
Answer
p = /20
Summary route: 10.212.144.0/20 (it also covers 9 network(s) that are not in the list).
3.Computer networks
Answer: A · The mail servers that accept email for a domain
Why this is the answer
MX (mail exchange) records list mail servers with priorities, so sending servers know where to deliver mail for that domain.
Answer
A · The mail servers that accept email for a domain
4.Computer networks
Answer: C · 1500 bytes
Why this is the answer
Standard Ethernet carries up to 1500 bytes of payload; larger IP packets must be fragmented or use path MTU discovery.
Answer
C · 1500 bytes
5.Number systems
Answer: U = 194
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
62 = 0011 1110
inverted: 1100 0001
+1 → 1100 0010
Read it as unsigned
1100 0010₂ = 256 − 62 = 194
Answer
U = 194
6.Number systems
Answer: R = 134
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
Shifting left by 1 moves every bit 1 places to the left; bits that fall off the left end are lost and zeros come in on the right.
1100 0011 << 1
= 1000 0110
Value of the result
1000 0110₂ = 134
Answer
R = 134
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