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 473 host addresses in a single subnet. What is the longest prefix length p (smallest subnet) that is big enough?
473 hosts
p =
2.IPv4 subnettingHard
A router has routes to these networks: 10.112.163.0/24, 10.112.164.0/24. What is the longest prefix length p of a single summary route that covers all of them?
10.112.163.0/24 … 10.112.164.0/24
p =
3.Computer networksMedium
Which range of TCP/UDP port numbers is reserved as "well-known" ports?
- A1024–49151
- B49152–65535
- C0–1023
- D0–65535
4.Computer networksHard
After repeated collisions, classic Ethernet stations choose their waiting time from a range that doubles each time. Why?
- ATo give the station with the highest MAC address priority on the shared cable
- BTo make the cable carry signals further
- CTo synchronise all stations to one clock
- DTo lower the chance of colliding again when many stations compete
5.Number systemsMedium
Write −14 as an 8-bit two's complement pattern. Read the pattern as an unsigned binary number: what is its value U in decimal?
N = −148 bit
U =
6.Number systemsHard
A system must give every one of 4512 items its own binary code of the same length. What is the smallest number of bits b needed?
4512
b = bit
3Solutions
1.IPv4 subnetting
Answer: p = 23
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
28 − 2 = 254 < 473
29 − 2 = 510 ≥ 473 → h = 9
Prefix length
p = 32 − 9 = 23
Answer
p = /23
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
163 = 10100011
164 = 10100100
common leading bits: 5
Prefix length
p = 16 + 5 = 21
Answer
p = /21
Summary route: 10.112.160.0/21 (it also covers 6 network(s) that are not in the list).
3.Computer networks
Answer: C · 0–1023
Why this is the answer
Ports 0–1023 are well-known ports for standard services. 1024–49151 are registered ports and 49152–65535 are dynamic or private ports.
Answer
C · 0–1023
4.Computer networks
Answer: D · To lower the chance of colliding again when many stations compete
Why this is the answer
Each further collision suggests more contention, so the random range grows and stations are more likely to pick different slots. It adapts to load without central control.
Answer
D · To lower the chance of colliding again when many stations compete
5.Number systems
Answer: U = 242
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
14 = 0000 1110
inverted: 1111 0001
+1 → 1111 0010
Read it as unsigned
1111 0010₂ = 256 − 14 = 242
Answer
U = 242
6.Number systems
Answer: b = 13 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
212 = 4096 < 4512
213 = 8192 ≥ 4512
Answer
b = 13
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