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 1284 host addresses in a single subnet. What is the longest prefix length p (smallest subnet) that is big enough?
1284 hosts
p =
2.IPv4 subnettingHard
A host has the address 172.20.229.167/21. What is the third octet x of its subnet's broadcast address?
172.20.229.167/21
x =
3.Computer networksMedium
Why does Wi-Fi use collision avoidance (CSMA/CA) instead of collision detection?
- AA wireless station cannot reliably detect collisions while it is transmitting
- BRadio signals never collide
- CCollision detection only works on fibre cables, not on shared wireless radio channels
- DWi-Fi only allows one device per network
4.Computer networksHard
An 8-port switch (no VLANs) has one host on each port. How many collision domains and broadcast domains are there?
- A8 collision domains, 8 broadcast domains
- B8 collision domains, 1 broadcast domain
- C1 collision domain, 8 broadcast domains
- D1 collision domain, 1 broadcast domain
5.Number systemsMedium
Set the bits so they store −21 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?
N = −218 bit
U =
6.Number systemsHard
Set the bits to the 8-bit result of 0111 1110 OR 1010 0011. What is the result's value R in decimal?
0111 1110OR1010 0011
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 < 1284
211 − 2 = 2046 ≥ 1284 → h = 11
Prefix length
p = 32 − 11 = 21
Answer
p = /21
2.IPv4 subnetting
Answer: x = 231
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).
Block size in the third octet
23 = 8
Subnet of this host
network: 172.20.224.0
broadcast: 172.20.231.255
Answer
x = 231
3.Computer networks
Answer: A · A wireless station cannot reliably detect collisions while it is transmitting
Why this is the answer
A radio transmitter cannot hear other weak signals while sending, and some stations are hidden from each other. So 802.11 waits random backoff times and uses acknowledgements.
Answer
A · A wireless station cannot reliably detect collisions while it is transmitting
4.Computer networks
Answer: B · 8 collision domains, 1 broadcast domain
Why this is the answer
Each switch port is its own collision domain, but broadcasts are forwarded to all ports, so the whole switch is one broadcast domain. A hub would give one of each.
Answer
B · 8 collision domains, 1 broadcast domain
5.Number systems
Answer: U = 235
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
21 = 0001 0101
inverted: 1110 1010
+1 → 1110 1011
Read it as unsigned
1110 1011₂ = 256 − 21 = 235
Answer
U = 235
6.Number systems
Answer: R = 255
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
OR gives 1 where at least one bit is 1.
0111 1110
OR 1010 0011
= 1111 1111
Value of the result
1111 1111₂ = 255
Answer
R = 255
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