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 192.168.10.0/24 must be divided into at least 5 equal subnets. If you borrow as few bits as possible, how many usable hosts H does each subnet have?
192.168.10.0/245 subnets
H =
2.IPv4 subnettingHard
A router has routes to these networks: 10.67.128.0/24, 10.67.129.0/24, 10.67.130.0/24, 10.67.131.0/24, 10.67.132.0/24, 10.67.133.0/24, 10.67.134.0/24. What is the longest prefix length p of a single summary route that covers all of them?
10.67.128.0/24 … 10.67.134.0/24
p =
3.Computer networksMedium
Which medium access method did classic shared (half-duplex) Ethernet use?
- ACSMA/CA
- BToken passing
- CTDMA
- DCSMA/CD
4.Computer networksHard
What does the Diffie-Hellman key exchange achieve on its own, and what is its weakness?
- AIt signs certificates for websites, but only works with short keys
- BIt encrypts bulk data directly over a public channel, but it is far too slow for large files and video
- CIt authenticates both users with passwords, but needs a central server
- DIt agrees a shared secret over a public channel, but alone it is open to a man-in-the-middle
5.Number systemsMedium
Set the bits so they store −19 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?
N = −198 bit
U =
6.Number systemsHard
A system must give every one of 1562 items its own binary code of the same length. What is the smallest number of bits b needed?
1562
b = bit
3Solutions
1.IPv4 subnetting
Answer: H = 30
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).
Bits to borrow
23 = 8 ≥ 5 → b = 3, /27
Hosts per subnet
25 − 2 = 30
Answer
H = 30
2.IPv4 subnetting
Answer: p = 21
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
128 = 10000000
134 = 10000110
common leading bits: 5
Prefix length
p = 16 + 5 = 21
Answer
p = /21
Summary route: 10.67.128.0/21 (it also covers 1 network(s) that are not in the list).
3.Computer networks
Answer: D · CSMA/CD
Why this is the answer
Carrier sense multiple access with collision detection: stations stop when they detect a collision and retry after a random backoff. Full-duplex switched Ethernet has no collisions.
Answer
D · CSMA/CD
4.Computer networks
Answer: D · It agrees a shared secret over a public channel, but alone it is open to a man-in-the-middle
Why this is the answer
Diffie-Hellman lets two parties derive the same secret that eavesdroppers cannot compute. An attacker in the middle can run separate exchanges with each side unless the values are signed or otherwise authenticated.
Answer
D · It agrees a shared secret over a public channel, but alone it is open to a man-in-the-middle
5.Number systems
Answer: U = 237
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
19 = 0001 0011
inverted: 1110 1100
+1 → 1110 1101
Read it as unsigned
1110 1101₂ = 256 − 19 = 237
Answer
U = 237
6.Number systems
Answer: b = 11 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
210 = 1024 < 1562
211 = 2048 ≥ 1562
Answer
b = 11
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