Computer Networks: cram pack

Formula sheets, the key method steps for every topic and 6 practice problems, with full solutions at the back.

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1Formulas and methods

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

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
255.255.255.25211111111111111111111111111111100network bitshost bits

A subnet uses the mask 255.255.255.252. How many usable host addresses H does it have?

255.255.255.252
H =
2.IPv4 subnettingHard
last octet = 85012816403211608140211networkhost

A host has the address 10.20.252.85/27. What is the fourth octet x of the last usable host address in its subnet?

10.20.252.85/27
x =
3.Computer networksMedium

Which host is in the same /16 subnet as 10.1.2.3?

  1. A11.1.2.3
  2. B10.1.200.5
  3. C10.0.2.3
  4. D10.2.2.3
4.Computer networksHard

A DNS cache poisoning attack aims to:

  1. ADelete DNS servers
  2. BOverload a server with traffic so that it stops answering queries
  3. CSteal Wi-Fi passwords
  4. DMake a resolver store a false IP address for a domain
5.Number systemsMedium
8-bit two's complement0−12806403201608040201

Set the bits so they store −68 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?

N = −688 bit
U =
6.Number systemsHard
12-bit two's complement, N = ??sign???????????

What is the most negative integer N that can be stored in 12-bit two's complement?

12 bit
N =

3Solutions

1.IPv4 subnetting

Answer: H = 2

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).

Prefix length

255.255.255.252 = /30

Usable hosts

22 − 2 = 2

Answer

H = 2
2.IPv4 subnetting

Answer: x = 94

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: ⌊85/32⌋·32 = 64
broadcast: 64 + 32 − 1 = 95
usable hosts: 65 … 94

Answer

x = 94
3.Computer networks

Answer: B · 10.1.200.5

Why this is the answer

With /16 the first 16 bits (10.1) are the network part. Only 10.1.200.5 shares them, so it is on 10.1.0.0/16.

Answer

B · 10.1.200.5
4.Computer networks

Answer: D · Make a resolver store a false IP address for a domain

Why this is the answer

Poisoning inserts forged records into a resolver's cache. DNSSEC adds signatures so resolvers can verify records.

Answer

D · Make a resolver store a false IP address for a domain
5.Number systems

Answer: U = 188

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

68 = 0100 0100
inverted: 1011 1011
+1 → 1011 1100

Read it as unsigned

1011 1100₂ = 256 − 68 = 188

Answer

U = 188
6.Number systems

Answer: N = −2048

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.

Most negative value

−211 = −2048

Answer

N = −2048

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