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
/24 → /2811111111111111111111111111110000network bitshost bits

The network 192.168.56.0/24 is divided into /28 subnets, numbered 1, 2, 3, … from the lowest address (subnet 1 starts at .0). What is the fourth octet x of the network address of subnet 12?

192.168.56.0/24/28subnet 12 (subnet 1 starts at .0)
x =
2.IPv4 subnettingHard
last octet = 195112816403201608041211networkhost

A host has the address 10.20.61.195/28. What is the fourth octet x of its network address?

10.20.61.195/28
x =
3.Computer networksMedium

NAT (network address translation) is mainly used to:

  1. ATranslate domain names into IP addresses for devices on a local network
  2. BIncrease the speed of a network cable
  3. CLet many devices with private addresses share one public IP address
  4. DEncrypt traffic between two sites
4.Computer networksHard

What is the main benefit of persistent connections (keep-alive) in HTTP/1.1?

  1. ASeveral requests reuse one TCP connection, avoiding repeated handshakes
  2. BThe server can push updates over UDP
  3. CResponses can arrive in any order on one stream, so slow replies do not block others
  4. DThe connection stays encrypted without TLS
5.Number systemsMedium
place values in two's complement1−12816413211608040201

The 8-bit pattern 1111 0000 stores a signed integer in two's complement. What is its value N in decimal?

1111 00008 bit
N =
6.Number systemsHard
click the bits to set them012806403201608040201

Set the bits to the 8-bit result of 0111 1001 << 1. What is the result's value R in decimal?

0111 1001<< 1
R =

3Solutions

1.IPv4 subnetting

Answer: x = 176

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 number 12

network: (12 − 1)·16 = 176
broadcast: 176 + 16 − 1 = 191

Answer

x = 176
2.IPv4 subnetting

Answer: x = 192

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: ⌊195/16⌋·16 = 192
broadcast: 192 + 16 − 1 = 207
usable hosts: 193 … 206

Answer

x = 192
3.Computer networks

Answer: C · Let many devices with private addresses share one public IP address

Why this is the answer

NAT rewrites addresses (and ports) so private networks can access the internet through a smaller number of public addresses.

Answer

C · Let many devices with private addresses share one public IP address
4.Computer networks

Answer: A · Several requests reuse one TCP connection, avoiding repeated handshakes

Why this is the answer

Reusing a connection avoids a new handshake and slow start for each object, so pages with many resources load faster.

Answer

A · Several requests reuse one TCP connection, avoiding repeated handshakes
5.Number systems

Answer: N = −16

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 + 64 + 32 + 16 = −16

Answer

N = −16
6.Number systems

Answer: R = 242

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 left by 1 moves every bit 1 places to the left; bits that fall off the left end are lost and zeros come in on the right.

0111 1001 << 1
= 1111 0010

Value of the result

1111 0010₂ = 242

Answer

R = 242

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