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 172.21.0.0/16 is divided into equal subnets with prefix /18. How many subnets N are there?
172.21.0.0/16/18
N =
2.IPv4 subnettingHard
A host has the address 10.20.20.42/27. What is the fourth octet x of its network address?
10.20.20.42/27
x =
3.Computer networksMedium
Why does a switch sometimes send a frame out of all ports except the one it came in on?
- AThe frame has a wrong IP checksum and must be checked by every host
- BThe destination MAC address is not yet in its address table
- CThe frame is too large for one port
- DThe switch is acting as a router
4.Computer networksHard
By the Nyquist formula, what is the maximum bit rate of a noiseless 3 kHz channel using 4 signal levels?
- A6 kbit/s
- B12 kbit/s
- C24 kbit/s
- D3 kbit/s
5.Number systemsMedium
Write −66 as an 8-bit two's complement pattern. Read the pattern as an unsigned binary number: what is its value U in decimal?
N = −668 bit
U =
6.Number systemsHard
Set the bits to the 8-bit result of 1011 0100 >> 1. What is the result's value R in decimal?
1011 0100>> 1
R =
3Solutions
1.IPv4 subnetting
Answer: N = 4
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
18 − 16 = 2
Number of subnets
N = 22 = 4
Answer
N = 4
Each /18 subnet has 214 − 2 = 16382 usable hosts.
2.IPv4 subnetting
Answer: x = 32
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: ⌊42/32⌋·32 = 32
broadcast: 32 + 32 − 1 = 63
usable hosts: 33 … 62
Answer
x = 32
3.Computer networks
Answer: B · The destination MAC address is not yet in its address table
Why this is the answer
For an unknown unicast address the switch floods the frame. When the destination replies, the switch learns its port from the source MAC.
Answer
B · The destination MAC address is not yet in its address table
4.Computer networks
Answer: B · 12 kbit/s
Why this is the answer
2 × 3000 × log₂ 4 = 2 × 3000 × 2 = 12,000 bit/s. More levels raise the rate, but noise limits how many can be told apart.
Answer
B · 12 kbit/s
5.Number systems
Answer: U = 190
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
66 = 0100 0010
inverted: 1011 1101
+1 → 1011 1110
Read it as unsigned
1011 1110₂ = 256 − 66 = 190
Answer
U = 190
6.Number systems
Answer: R = 90
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 right by 1 moves every bit 1 places to the right; bits that fall off the right end are lost and zeros come in on the left.
1011 0100 >> 1
= 0101 1010
Value of the result
0101 1010₂ = 90
Answer
R = 90
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