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 16 equal subnets. If you borrow as few bits as possible, how many usable hosts H does each subnet have?
192.168.10.0/2416 subnets
H =
2.IPv4 subnettingHard
A host has the address 10.20.88.80/29. What is the fourth octet x of the last usable host address in its subnet?
10.20.88.80/29
x =
3.Computer networksMedium
NAT (network address translation) is mainly used to:
- AIncrease the speed of a network cable
- BLet many devices with private addresses share one public IP address
- CEncrypt traffic between two sites
- DTranslate domain names into IP addresses for devices on a local network
4.Computer networksHard
How does fragmentation differ in IPv6 compared with IPv4?
- ARouters never fragment IPv6 packets; only the source can, using path MTU discovery
- BIPv6 fragments at the transport layer inside TCP
- CIPv6 does not allow packets larger than 576 bytes
- DIPv6 routers fragment every packet larger than 64 bytes before forwarding it on the path
5.Number systemsMedium
Set the bits so they store −29 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?
N = −298 bit
U =
6.Number systemsHard
A system must give every one of 2208 items its own binary code of the same length. What is the smallest number of bits b needed?
2208
b = bit
3Solutions
1.IPv4 subnetting
Answer: H = 14
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
24 = 16 ≥ 16 → b = 4, /28
Hosts per subnet
24 − 2 = 14
Answer
H = 14
2.IPv4 subnetting
Answer: x = 86
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: ⌊80/8⌋·8 = 80
broadcast: 80 + 8 − 1 = 87
usable hosts: 81 … 86
Answer
x = 86
3.Computer networks
Answer: B · Let many devices with private addresses share one public IP address
Why this is the answer
NAT rewrites addresses (and ports) so private networks can access the internet through a smaller number of public addresses.
Answer
B · Let many devices with private addresses share one public IP address
4.Computer networks
Answer: A · Routers never fragment IPv6 packets; only the source can, using path MTU discovery
Why this is the answer
An IPv6 router drops an oversized packet and returns ICMPv6 Packet Too Big. The source then sends smaller packets or uses a fragment extension header.
Answer
A · Routers never fragment IPv6 packets; only the source can, using path MTU discovery
5.Number systems
Answer: U = 227
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
29 = 0001 1101
inverted: 1110 0010
+1 → 1110 0011
Read it as unsigned
1110 0011₂ = 256 − 29 = 227
Answer
U = 227
6.Number systems
Answer: b = 12 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
211 = 2048 < 2208
212 = 4096 ≥ 2208
Answer
b = 12
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