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
/?11111111111111111111110000000000network bitshost bits

A department needs 617 host addresses in a single subnet. What is the longest prefix length p (smallest subnet) that is big enough?

617 hosts
p =
2.IPv4 subnettingHard
third octet = 164112806413201608140201networkhost

A host has the address 172.20.164.176/21. What is the third octet x of its network address?

172.20.164.176/21
x =
3.Computer networksMedium

Which subnet mask corresponds to /20?

  1. A255.255.248.0
  2. B255.255.255.240
  3. C255.255.224.0
  4. D255.255.240.0
4.Computer networksHard

Why does a browser warn when a site presents a self-signed certificate?

  1. AThe certificate is always expired
  2. BNo trusted certificate authority vouches that the key belongs to that site
  3. CThe site must be using HTTP instead of HTTPS
  4. DSelf-signed certificates use no encryption at all, so the traffic is sent in clear text
5.Number systemsMedium
place values in two's complement1−12806413211618141201

The 8-bit pattern 1011 1110 stores a signed integer in two's complement. What is its value N in decimal?

1011 11108 bit
N =
6.Number systemsHard
530 codes, b = ? bits?2ᵇ⁻¹…?4?2?1

A system must give every one of 530 items its own binary code of the same length. What is the smallest number of bits b needed?

530
b = bit

3Solutions

1.IPv4 subnetting

Answer: p = 22

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

29 − 2 = 510 < 617
210 − 2 = 1022 ≥ 617 → h = 10

Prefix length

p = 32 − 10 = 22

Answer

p = /22
2.IPv4 subnetting

Answer: x = 160

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 in the third octet

23 = 8

Subnet of this host

network: 172.20.160.0
broadcast: 172.20.167.255

Answer

x = 160
3.Computer networks

Answer: D · 255.255.240.0

Why this is the answer

/20 means 8 + 8 + 4 one-bits. Four one-bits in the third octet are 11110000 = 240, so the mask is 255.255.240.0.

Answer

D · 255.255.240.0
4.Computer networks

Answer: B · No trusted certificate authority vouches that the key belongs to that site

Why this is the answer

Browsers trust certificates signed, directly or through a chain, by CAs in their trust store. A self-signed certificate could be made by anyone, such as a man-in-the-middle.

Answer

B · No trusted certificate authority vouches that the key belongs to that site
5.Number systems

Answer: N = −66

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 + 32 + 16 + 8 + 4 + 2 = −66

Answer

N = −66
6.Number systems

Answer: b = 10 bit

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.

Powers of two

29 = 512 < 530
210 = 1024 ≥ 530

Answer

b = 10

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