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 772 host addresses in a single subnet. What is the longest prefix length p (smallest subnet) that is big enough?
772 hosts
p =
2.IPv4 subnettingHard
A host has the address 172.20.106.172/17. What is the third octet x of its subnet's broadcast address?
172.20.106.172/17
x =
3.Computer networksMedium
Why does a switch sometimes send a frame out of all ports except the one it came in on?
- AThe destination MAC address is not yet in its address table
- BThe switch is acting as a router
- CThe frame is too large for one port
- DThe frame has a wrong IP checksum and must be checked by every host
4.Computer networksHard
An 8-port switch (no VLANs) has one host on each port. How many collision domains and broadcast domains are there?
- A1 collision domain, 1 broadcast domain
- B8 collision domains, 1 broadcast domain
- C1 collision domain, 8 broadcast domains
- D8 collision domains, 8 broadcast domains
5.Number systemsMedium
The 8-bit pattern 1000 1010 stores a signed integer in two's complement. What is its value N in decimal?
1000 10108 bit
N =
6.Number systemsHard
What is the largest unsigned integer N that fits in 10 bits?
10 bit
N =
3Solutions
1.IPv4 subnetting
Answer: p = 22
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
29 − 2 = 510 < 772
210 − 2 = 1022 ≥ 772 → h = 10
Prefix length
p = 32 − 10 = 22
Answer
p = /22
2.IPv4 subnetting
Answer: x = 127
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
27 = 128
Subnet of this host
network: 172.20.0.0
broadcast: 172.20.127.255
Answer
x = 127
3.Computer networks
Answer: A · The destination MAC address is not yet in its address table
Why this is the answer
For an unknown unicast address the switch floods the frame. When the destination replies, the switch learns its port from the source MAC.
Answer
A · The destination MAC address is not yet in its address table
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: N = −118
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.
Add the place values
The other bits keep their normal positive values.
−128 + 8 + 2 = −118
Answer
N = −118
6.Number systems
Answer: N = 1023
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.
Largest value
210 − 1 = 1023
Answer
N = 1023
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