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 subnet uses the mask 255.255.255.248. How many usable host addresses H does it have?
255.255.255.248
H =
2.IPv4 subnettingHard
A router has routes to these networks: 10.43.108.0/24, 10.43.109.0/24, 10.43.110.0/24, 10.43.111.0/24, 10.43.112.0/24, 10.43.113.0/24, 10.43.114.0/24. What is the longest prefix length p of a single summary route that covers all of them?
10.43.108.0/24 … 10.43.114.0/24
p =
3.Computer networksMedium
At which OSI layer does a switch normally operate?
- APhysical (layer 1)
- BData link (layer 2)
- CTransport (layer 4)
- DNetwork (layer 3)
4.Computer networksHard
A DNS cache poisoning attack aims to:
- ASteal Wi-Fi passwords
- BMake a resolver store a false IP address for a domain
- COverload a server with traffic so that it stops answering queries
- DDelete DNS servers
5.Number systemsMedium
Write −4 as an 8-bit two's complement pattern. Read the pattern as an unsigned binary number: what is its value U in decimal?
N = −48 bit
U =
6.Number systemsHard
Set the bits to the 8-bit result of 1100 0111 >> 1. What is the result's value R in decimal?
1100 0111>> 1
R =
3Solutions
1.IPv4 subnetting
Answer: H = 6
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).
Prefix length
255.255.255.248 = /29
Usable hosts
23 − 2 = 6
Answer
H = 6
2.IPv4 subnetting
Answer: p = 19
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
108 = 01101100
114 = 01110010
common leading bits: 3
Prefix length
p = 16 + 3 = 19
Answer
p = /19
Summary route: 10.43.96.0/19 (it also covers 25 network(s) that are not in the list).
3.Computer networks
Answer: B · Data link (layer 2)
Why this is the answer
A standard switch learns MAC addresses and forwards frames at layer 2. Layer 3 switches also route.
Answer
B · Data link (layer 2)
4.Computer networks
Answer: B · Make a resolver store a false IP address for a domain
Why this is the answer
Poisoning inserts forged records into a resolver's cache. DNSSEC adds signatures so resolvers can verify records.
Answer
B · Make a resolver store a false IP address for a domain
5.Number systems
Answer: U = 252
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
4 = 0000 0100
inverted: 1111 1011
+1 → 1111 1100
Read it as unsigned
1111 1100₂ = 256 − 4 = 252
Answer
U = 252
6.Number systems
Answer: R = 99
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 right by 1 moves every bit 1 places to the right; bits that fall off the right end are lost and zeros come in on the left.
1100 0111 >> 1
= 0110 0011
Value of the result
0110 0011₂ = 99
Answer
R = 99
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