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 1117 host addresses in a single subnet. What is the longest prefix length p (smallest subnet) that is big enough?
1117 hosts
p =
2.IPv4 subnettingHard
A host has the address 172.20.110.83/19. What is the third octet x of its subnet's broadcast address?
172.20.110.83/19
x =
3.Computer networksMedium
What is the main purpose of a VLAN?
- ATo split one physical switched network into separate broadcast domains
- BTo encrypt all traffic between switches so other departments cannot read it
- CTo increase the speed of every switch port
- DTo give each device a public IP address
4.Computer networksHard
DNS queries normally use UDP. When does DNS typically use TCP instead?
- AOnly for reverse lookups of IP addresses
- BFor responses too large for UDP and for zone transfers
- COnly when the domain name is longer than ten characters
- DFor every query sent from a web browser
5.Number systemsMedium
The 8-bit pattern 1101 0010 stores a signed integer in two's complement. What is its value N in decimal?
1101 00108 bit
N =
6.Number systemsHard
Set the bits to the 8-bit result of 1011 0111 AND 1001 1001. What is the result's value R in decimal?
1011 0111AND1001 1001
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 < 1117
211 − 2 = 2046 ≥ 1117 → h = 11
Prefix length
p = 32 − 11 = 21
Answer
p = /21
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
25 = 32
Subnet of this host
network: 172.20.96.0
broadcast: 172.20.127.255
Answer
x = 127
3.Computer networks
Answer: A · 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
A · To split one physical switched network into separate broadcast domains
4.Computer networks
Answer: B · For responses too large for UDP and for zone transfers
Why this is the answer
If a UDP response is truncated, the resolver retries over TCP. Zone transfers between name servers also use TCP because they carry lots of data.
Answer
B · For responses too large for UDP and for zone transfers
5.Number systems
Answer: N = −46
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 + 64 + 16 + 2 = −46
Answer
N = −46
6.Number systems
Answer: R = 145
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
AND gives 1 only where both bits are 1.
1011 0111
AND 1001 1001
= 1001 0001
Value of the result
1001 0001₂ = 145
Answer
R = 145
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