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
The network 192.168.10.0/24 must be divided into at least 20 equal subnets. If you borrow as few bits as possible, how many usable hosts H does each subnet have?
192.168.10.0/2420 subnets
H =
2.IPv4 subnettingHard
A host has the address 10.20.171.182/29. What is the fourth octet x of the last usable host address in its subnet?
10.20.171.182/29
x =
3.Computer networksMedium
Which metric does RIP use to choose routes?
- ALink bandwidth
- BMeasured delay
- CBusiness policy
- DHop count
4.Computer networksHard
In a SYN flood attack, what resource is the attacker trying to exhaust?
- AThe server's table of half-open connections
- BThe switch's MAC address table
- CThe DNS cache of the client
- DThe server's disk space
5.Number systemsMedium
Set the bits so they store −36 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?
N = −368 bit
U =
6.Number systemsHard
An uncompressed image is 800 × 600 pixels with 32 bits per pixel. How much memory M does it take, in KiB (1 KiB = 1024 bytes)?
800 × 600 px32 bit/px1 KiB = 1024 B
M = KiB
3Solutions
1.IPv4 subnetting
Answer: H = 6
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).
Bits to borrow
25 = 32 ≥ 20 → b = 5, /29
Hosts per subnet
23 − 2 = 6
Answer
H = 6
2.IPv4 subnetting
Answer: x = 182
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: ⌊182/8⌋·8 = 176
broadcast: 176 + 8 − 1 = 183
usable hosts: 177 … 182
Answer
x = 182
3.Computer networks
Answer: D · Hop count
Why this is the answer
RIP is a distance-vector protocol that counts hops, with 16 meaning unreachable. This limits it to small networks.
Answer
D · Hop count
4.Computer networks
Answer: A · The server's table of half-open connections
Why this is the answer
Each SYN makes the server reserve state and wait for an ACK that never comes. SYN cookies avoid storing state until the handshake completes.
Answer
A · The server's table of half-open connections
5.Number systems
Answer: U = 220
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
36 = 0010 0100
inverted: 1101 1011
+1 → 1101 1100
Read it as unsigned
1101 1100₂ = 256 − 36 = 220
Answer
U = 220
6.Number systems
Answer: M = 1875 KiB
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.
Bits and bytes
800·600·32 = 15360000 bit = 1920000 B
Convert to KiB
1920000/1024 = 1875 KiB
Answer
M = 1875 KiB
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