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 subnet uses the mask 255.255.255.252. How many usable host addresses H does it have?
255.255.255.252
H =
2.IPv4 subnettingHard
A host has the address 10.20.90.30/29. What is the fourth octet x of the last usable host address in its subnet?
10.20.90.30/29
x =
3.Computer networksMedium
What is the IPv6 loopback address?
- Afe80::1
- B::
- Cff02::1
- D::1
4.Computer networksHard
DNS queries normally use UDP. When does DNS typically use TCP instead?
- AFor every query sent from a web browser
- BOnly when the domain name is longer than ten characters
- CFor responses too large for UDP and for zone transfers
- DOnly for reverse lookups of IP addresses
5.Number systemsMedium
The 8-bit pattern 1001 0010 stores a signed integer in two's complement. What is its value N in decimal?
1001 00108 bit
N =
6.Number systemsHard
Set the bits to the 8-bit result of 1010 0100 AND 1101 0000. What is the result's value R in decimal?
1010 0100AND1101 0000
R =
3Solutions
1.IPv4 subnetting
Answer: H = 2
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).
Prefix length
255.255.255.252 = /30
Usable hosts
22 − 2 = 2
Answer
H = 2
2.IPv4 subnetting
Answer: x = 30
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
23 = 8
Subnet of this host
network: ⌊30/8⌋·8 = 24
broadcast: 24 + 8 − 1 = 31
usable hosts: 25 … 30
Answer
x = 30
3.Computer networks
Answer: D · ::1
Why this is the answer
::1 is the loopback address. :: is the unspecified address, fe80::/10 is link-local and ff02::1 is the all-nodes multicast group.
Answer
D · ::1
4.Computer networks
Answer: C · 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
C · For responses too large for UDP and for zone transfers
5.Number systems
Answer: N = −110
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 + 16 + 2 = −110
Answer
N = −110
6.Number systems
Answer: R = 128
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.
1010 0100
AND 1101 0000
= 1000 0000
Value of the result
1000 0000₂ = 128
Answer
R = 128
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