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
/?11111111111111111111111000000000network bitshost bits

A department needs 281 host addresses in a single subnet. What is the longest prefix length p (smallest subnet) that is big enough?

281 hosts
p =
2.IPv4 subnettingHard
last octet = 13012806403201618140211networkhost

A host has the address 10.20.32.13/25. What is the last number (fourth octet) x of its subnet's broadcast address?

10.20.32.13/25
x =
3.Computer networksMedium

Which of these is a private IPv4 address range?

  1. A8.8.8.0/24
  2. B192.168.0.0/16
  3. C100.0.0.0/8
  4. D200.1.1.0/24
4.Computer networksHard

By the Nyquist formula, what is the maximum bit rate of a noiseless 3 kHz channel using 4 signal levels?

  1. A24 kbit/s
  2. B6 kbit/s
  3. C3 kbit/s
  4. D12 kbit/s
5.Number systemsMedium
8-bit two's complement0−12806403201608040201

Set the bits so they store −56 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?

N = −568 bit
U =
6.Number systemsHard
32 bit1 px256 px256 pxM = ? KiB

An uncompressed image is 256 × 256 pixels with 32 bits per pixel. How much memory M does it take, in KiB (1 KiB = 1024 bytes)?

256 × 256 px32 bit/px1 KiB = 1024 B
M = KiB

3Solutions

1.IPv4 subnetting

Answer: p = 23

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).

Host bits needed

28 − 2 = 254 < 281
29 − 2 = 510 ≥ 281 → h = 9

Prefix length

p = 32 − 9 = 23

Answer

p = /23
2.IPv4 subnetting

Answer: x = 127

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

27 = 128

Subnet of this host

network: ⌊13/128⌋·128 = 0
broadcast: 0 + 128 − 1 = 127
usable hosts: 1 … 126

Answer

x = 127
3.Computer networks

Answer: B · 192.168.0.0/16

Why this is the answer

Private ranges are 10.0.0.0/8, 172.16.0.0/12 and 192.168.0.0/16; they are not routed on the public internet.

Answer

B · 192.168.0.0/16
4.Computer networks

Answer: D · 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

D · 12 kbit/s
5.Number systems

Answer: U = 200

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

56 = 0011 1000
inverted: 1100 0111
+1 → 1100 1000

Read it as unsigned

1100 1000₂ = 256 − 56 = 200

Answer

U = 200
6.Number systems

Answer: M = 256 KiB

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.

Bits and bytes

256·256·32 = 2097152 bit = 262144 B

Convert to KiB

262144/1024 = 256 KiB

Answer

M = 256 KiB

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