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 192.168.189.0/24 is divided into /29 subnets, numbered 1, 2, 3, … from the lowest address (subnet 1 starts at .0). What is the fourth octet x of the broadcast address of subnet 6?
192.168.189.0/24/29subnet 6 (subnet 1 starts at .0)
x =
2.IPv4 subnettingHard
A host has the address 10.20.42.101/29. What is the fourth octet x of the last usable host address in its subnet?
10.20.42.101/29
x =
3.Computer networksMedium
How does a stateful firewall differ from a simple packet filter?
- AIt blocks all incoming traffic without exceptions, including replies to outgoing requests
- BIt tracks connections and allows replies that belong to established sessions
- CIt encrypts every packet it forwards
- DIt only checks the MAC address of each frame
4.Computer networksHard
DNS queries normally use UDP. When does DNS typically use TCP instead?
- AOnly for reverse lookups of IP addresses
- BFor every query sent from a web browser
- CFor responses too large for UDP and for zone transfers
- DOnly when the domain name is longer than ten characters
5.Number systemsMedium
Write −15 as an 8-bit two's complement pattern. Read the pattern as an unsigned binary number: what is its value U in decimal?
N = −158 bit
U =
6.Number systemsHard
A system must give every one of 4287 items its own binary code of the same length. What is the smallest number of bits b needed?
4287
b = bit
3Solutions
1.IPv4 subnetting
Answer: x = 47
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 number 6
network: (6 − 1)·8 = 40
broadcast: 40 + 8 − 1 = 47
Answer
x = 47
2.IPv4 subnetting
Answer: x = 102
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: ⌊101/8⌋·8 = 96
broadcast: 96 + 8 − 1 = 103
usable hosts: 97 … 102
Answer
x = 102
3.Computer networks
Answer: B · It tracks connections and allows replies that belong to established sessions
Why this is the answer
A stateful firewall keeps a table of active connections, so return traffic is allowed automatically. A stateless filter judges each packet in isolation.
Answer
B · It tracks connections and allows replies that belong to established sessions
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: U = 241
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.
Invert and add one
15 = 0000 1111
inverted: 1111 0000
+1 → 1111 0001
Read it as unsigned
1111 0001₂ = 256 − 15 = 241
Answer
U = 241
6.Number systems
Answer: b = 13 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
212 = 4096 < 4287
213 = 8192 ≥ 4287
Answer
b = 13
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