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 10.164.88.0/22 is divided into equal subnets with prefix /28. How many subnets N are there?
10.164.88.0/22/28
N =
2.IPv4 subnettingHard
A host has the address 10.20.45.146/27. What is the fourth octet x of the last usable host address in its subnet?
10.20.45.146/27
x =
3.Computer networksMedium
How many usable host addresses does a /24 IPv4 network have?
- A256
- B255
- C254
- D252
4.Computer networksHard
How does fragmentation differ in IPv6 compared with IPv4?
- AIPv6 fragments at the transport layer inside TCP
- BRouters never fragment IPv6 packets; only the source can, using path MTU discovery
- CIPv6 routers fragment every packet larger than 64 bytes before forwarding it on the path
- DIPv6 does not allow packets larger than 576 bytes
5.Number systemsMedium
Set the bits so they store −24 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?
N = −248 bit
U =
6.Number systemsHard
A system must give every one of 3092 items its own binary code of the same length. What is the smallest number of bits b needed?
3092
b = bit
3Solutions
1.IPv4 subnetting
Answer: N = 64
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).
Borrowed bits
28 − 22 = 6
Number of subnets
N = 26 = 64
Answer
N = 64
Each /28 subnet has 24 − 2 = 14 usable hosts.
2.IPv4 subnetting
Answer: x = 158
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
25 = 32
Subnet of this host
network: ⌊146/32⌋·32 = 128
broadcast: 128 + 32 − 1 = 159
usable hosts: 129 … 158
Answer
x = 158
3.Computer networks
Answer: C · 254
Why this is the answer
A /24 has 8 host bits, so 2⁸ = 256 addresses. The network and broadcast addresses are reserved, leaving 254 for hosts.
Answer
C · 254
4.Computer networks
Answer: B · 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
B · Routers never fragment IPv6 packets; only the source can, using path MTU discovery
5.Number systems
Answer: U = 232
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
24 = 0001 1000
inverted: 1110 0111
+1 → 1110 1000
Read it as unsigned
1110 1000₂ = 256 − 24 = 232
Answer
U = 232
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 < 3092
212 = 4096 ≥ 3092
Answer
b = 12
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