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 699 host addresses in a single subnet. What is the longest prefix length p (smallest subnet) that is big enough?
699 hosts
p =
2.IPv4 subnettingHard
A host has the address 10.20.251.70/25. What is the last number (fourth octet) x of its subnet's broadcast address?
10.20.251.70/25
x =
3.Computer networksMedium
What is the main purpose of a VLAN?
- ATo increase the speed of every switch port
- BTo split one physical switched network into separate broadcast domains
- CTo encrypt all traffic between switches so other departments cannot read it
- DTo give each device a public IP address
4.Computer networksHard
What is the maximum transmission unit (MTU) of standard Ethernet payload?
- A65,535 bytes
- B9000 bytes in every network
- C64 bytes
- D1500 bytes
5.Number systemsMedium
Set the bits so they store −73 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?
N = −738 bit
U =
6.Number systemsHard
A system must give every one of 1006 items its own binary code of the same length. What is the smallest number of bits b needed?
1006
b = bit
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 < 699
210 − 2 = 1022 ≥ 699 → 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
27 = 128
Subnet of this host
network: ⌊70/128⌋·128 = 0
broadcast: 0 + 128 − 1 = 127
usable hosts: 1 … 126
Answer
x = 127
3.Computer networks
Answer: B · To split one physical switched network into separate broadcast domains
Why this is the answer
VLANs group ports logically, so broadcasts stay within each VLAN. This improves security and limits broadcast traffic.
Answer
B · To split one physical switched network into separate broadcast domains
4.Computer networks
Answer: D · 1500 bytes
Why this is the answer
Standard Ethernet carries up to 1500 bytes of payload; larger IP packets must be fragmented or use path MTU discovery.
Answer
D · 1500 bytes
5.Number systems
Answer: U = 183
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
73 = 0100 1001
inverted: 1011 0110
+1 → 1011 0111
Read it as unsigned
1011 0111₂ = 256 − 73 = 183
Answer
U = 183
6.Number systems
Answer: b = 10 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
29 = 512 < 1006
210 = 1024 ≥ 1006
Answer
b = 10
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