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 192.168.154.0/24 is divided into /28 subnets, numbered 1, 2, 3, … from the lowest address (subnet 1 starts at .0). What is the fourth octet x of the network address of subnet 7?
192.168.154.0/24/28subnet 7 (subnet 1 starts at .0)
x =
2.IPv4 subnettingHard
A router has routes to these networks: 10.91.169.0/24, 10.91.170.0/24, 10.91.171.0/24. What is the longest prefix length p of a single summary route that covers all of them?
10.91.169.0/24 … 10.91.171.0/24
p =
3.Computer networksMedium
What is the IPv6 loopback address?
- Aff02::1
- B::
- Cfe80::1
- D::1
4.Computer networksHard
What is the maximum transmission unit (MTU) of standard Ethernet payload?
- A1500 bytes
- B65,535 bytes
- C64 bytes
- D9000 bytes in every network
5.Number systemsMedium
The 8-bit pattern 1100 1101 stores a signed integer in two's complement. What is its value N in decimal?
1100 11018 bit
N =
6.Number systemsHard
Set the bits to the 8-bit result of 1110 1001 << 3. What is the result's value R in decimal?
1110 1001<< 3
R =
3Solutions
1.IPv4 subnetting
Answer: x = 96
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
24 = 16
Subnet number 7
network: (7 − 1)·16 = 96
broadcast: 96 + 16 − 1 = 111
Answer
x = 96
2.IPv4 subnetting
Answer: p = 22
Route summarisation: a summary route a.b.c.0/p covers all networks whose first p bits match. Write the differing octet of the first and last network in binary; the number of common leading bits, plus the bits of the octets before it, is the longest summary prefix.
Compare the bits
169 = 10101001
171 = 10101011
common leading bits: 6
Prefix length
p = 16 + 6 = 22
Answer
p = /22
Summary route: 10.91.168.0/22 (it also covers 1 network(s) that are not in the list).
3.Computer networks
Answer: D · ::1
Why this is the answer
::1 is the loopback address. :: is the unspecified address, fe80::/10 is link-local and ff02::1 is the all-nodes multicast group.
Answer
D · ::1
4.Computer networks
Answer: A · 1500 bytes
Why this is the answer
Standard Ethernet carries up to 1500 bytes of payload; larger IP packets must be fragmented or use path MTU discovery.
Answer
A · 1500 bytes
5.Number systems
Answer: N = −51
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 + 64 + 8 + 4 + 1 = −51
Answer
N = −51
6.Number systems
Answer: R = 72
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
Shifting left by 3 moves every bit 3 places to the left; bits that fall off the left end are lost and zeros come in on the right.
1110 1001 << 3
= 0100 1000
Value of the result
0100 1000₂ = 72
Answer
R = 72
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