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 255.255.255.192. How many usable host addresses H does it have?
255.255.255.192
H =
2.IPv4 subnettingHard
The network 192.168.119.0/24 is split with VLSM for these LANs: A: 105 hosts, B: 4 hosts, C: 14 hosts, D: 52 hosts. Subnets are allocated from .0 upwards, the LAN with the most hosts first, each with the smallest subnet that fits. What is the fourth octet x of the network address of LAN C?
A: 105B: 4C: 14D: 52
x =
3.Computer networksMedium
What is the role of a default gateway on a host?
- AIt is the switch port the host is connected to
- BIt assigns MAC addresses
- CIt is the router used for traffic to other networks
- DIt is the DNS server
4.Computer networksHard
DNS queries normally use UDP. When does DNS typically use TCP instead?
- AFor responses too large for UDP and for zone transfers
- BOnly when the domain name is longer than ten characters
- COnly for reverse lookups of IP addresses
- DFor every query sent from a web browser
5.Number systemsMedium
The 8-bit pattern 1001 1110 stores a signed integer in two's complement. What is its value N in decimal?
1001 11108 bit
N =
6.Number systemsHard
Set the bits to the 8-bit result of 1101 1101 AND 1110 1110. What is the result's value R in decimal?
1101 1101AND1110 1110
R =
3Solutions
1.IPv4 subnetting
Answer: H = 62
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
255.255.255.192 = /26
Usable hosts
26 − 2 = 62
Answer
H = 62
2.IPv4 subnetting
Answer: x = 192
VLSM: each LAN gets the smallest subnet with 2ʰ − 2 ≥ its host count, i.e. prefix 32 − h. Allocating the largest subnets first keeps every block aligned on a multiple of its own size, so the network addresses follow by adding up the block sizes.
Block size per LAN
A: 105 hosts → 27 = 128 (/25)
D: 52 hosts → 26 = 64 (/26)
C: 14 hosts → 24 = 16 (/28)
B: 4 hosts → 23 = 8 (/29)
Allocation, largest first
A: .0 – .127
D: .128 – .191
C: .192 – .207
B: .208 – .215
Answer
x = 192
LAN C: 192.168.119.192/28.
3.Computer networks
Answer: C · It is the router used for traffic to other networks
Why this is the answer
The host sends packets for non-local destinations to the default gateway, usually its local router.
Answer
C · It is the router used for traffic to other networks
4.Computer networks
Answer: A · 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
A · For responses too large for UDP and for zone transfers
5.Number systems
Answer: N = −98
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 + 8 + 4 + 2 = −98
Answer
N = −98
6.Number systems
Answer: R = 204
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.
1101 1101
AND 1110 1110
= 1100 1100
Value of the result
1100 1100₂ = 204
Answer
R = 204
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