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 255.255.240.0. How many usable host addresses H does it have?
255.255.240.0
H =
2.IPv4 subnettingHard
A router has routes to these networks: 10.113.202.0/24, 10.113.203.0/24, 10.113.204.0/24, 10.113.205.0/24, 10.113.206.0/24, 10.113.207.0/24, 10.113.208.0/24, 10.113.209.0/24. What is the longest prefix length p of a single summary route that covers all of them?
10.113.202.0/24 … 10.113.209.0/24
p =
3.Computer networksMedium
At which OSI layer does a switch normally operate?
- ANetwork (layer 3)
- BData link (layer 2)
- CPhysical (layer 1)
- DTransport (layer 4)
4.Computer networksHard
You must divide 192.168.1.0/24 into at least 5 subnets with up to 25 hosts each. Which prefix length fits best?
- A/27
- B/26
- C/28
- D/25
5.Number systemsMedium
Set the bits so they store −53 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?
N = −538 bit
U =
6.Number systemsHard
An uncompressed image is 1024 × 768 pixels with 24 bits per pixel. How much memory M does it take, in KiB (1 KiB = 1024 bytes)?
1024 × 768 px24 bit/px1 KiB = 1024 B
M = KiB
3Solutions
1.IPv4 subnetting
Answer: H = 4094
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
255.255.240.0 = /20
Usable hosts
212 − 2 = 4094
Answer
H = 4094
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
202 = 11001010
209 = 11010001
common leading bits: 3
Prefix length
p = 16 + 3 = 19
Answer
p = /19
Summary route: 10.113.192.0/19 (it also covers 24 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: A · /27
Why this is the answer
/27 gives 8 subnets with 30 usable hosts each. /26 gives only 4 subnets, and /28 allows only 14 hosts per subnet.
Answer
A · /27
5.Number systems
Answer: U = 203
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
53 = 0011 0101
inverted: 1100 1010
+1 → 1100 1011
Read it as unsigned
1100 1011₂ = 256 − 53 = 203
Answer
U = 203
6.Number systems
Answer: M = 2304 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
1024·768·24 = 18874368 bit = 2359296 B
Convert to KiB
2359296/1024 = 2304 KiB
Answer
M = 2304 KiB
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