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 10.211.112.0/20 is divided into equal subnets with prefix /25. How many subnets N are there?
10.211.112.0/20/25
N =
2.IPv4 subnettingHard
A host has the address 10.20.213.8/28. What is the fourth octet x of the last usable host address in its subnet?
10.20.213.8/28
x =
3.Computer networksMedium
What is the usual order of messages when a client obtains an address from DHCP?
- ARequest, Offer, Discover, Acknowledge
- BDiscover, Request, Offer, Acknowledge
- COffer, Discover, Acknowledge, Request
- DDiscover, Offer, Request, Acknowledge
4.Computer networksHard
By Shannon's theorem, what is the capacity of a 1 MHz channel with a signal-to-noise ratio of 255?
- A1 Mbit/s
- B2 Mbit/s
- C8 Mbit/s
- D255 Mbit/s
5.Number systemsMedium
Write −65 as an 8-bit two's complement pattern. Read the pattern as an unsigned binary number: what is its value U in decimal?
N = −658 bit
U =
6.Number systemsHard
Set the bits to the 8-bit result of 1001 0110 >> 1. What is the result's value R in decimal?
1001 0110>> 1
R =
3Solutions
1.IPv4 subnetting
Answer: N = 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).
Borrowed bits
25 − 20 = 5
Number of subnets
N = 25 = 32
Answer
N = 32
Each /25 subnet has 27 − 2 = 126 usable hosts.
2.IPv4 subnetting
Answer: x = 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).
Block size
24 = 16
Subnet of this host
network: ⌊8/16⌋·16 = 0
broadcast: 0 + 16 − 1 = 15
usable hosts: 1 … 14
Answer
x = 14
3.Computer networks
Answer: D · Discover, Offer, Request, Acknowledge
Why this is the answer
The client broadcasts Discover, servers reply with Offers, the client Requests one offer and the chosen server confirms with an Acknowledge.
Answer
D · Discover, Offer, Request, Acknowledge
4.Computer networks
Answer: C · 8 Mbit/s
Why this is the answer
C = 10⁶ × log₂(256) = 10⁶ × 8 = 8 Mbit/s. This is the upper limit for error-free data rate on a noisy channel.
Answer
C · 8 Mbit/s
5.Number systems
Answer: U = 191
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
65 = 0100 0001
inverted: 1011 1110
+1 → 1011 1111
Read it as unsigned
1011 1111₂ = 256 − 65 = 191
Answer
U = 191
6.Number systems
Answer: R = 75
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.
1001 0110 >> 1
= 0100 1011
Value of the result
0100 1011₂ = 75
Answer
R = 75
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