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 192.168.10.0/24 must be divided into at least 16 equal subnets. If you borrow as few bits as possible, how many usable hosts H does each subnet have?

192.168.10.0/2416 subnets
H =
2.IPv4 subnettingHard
last octet = 142112806403201618141201networkhost

A host has the address 10.20.65.142/25. What is the fourth octet x of its network address?

10.20.65.142/25
x =
3.Computer networksMedium

Which range of TCP/UDP port numbers is reserved as "well-known" ports?

  1. A0–1023
  2. B1024–49151
  3. C0–65535
  4. D49152–65535
4.Computer networksHard

In the TCP/IP model, which layer adds port numbers?

  1. ATransport layer
  2. BApplication layer
  3. CLink layer
  4. DInternet layer
5.Number systemsMedium
8-bit two's complement0−12806403201608040201

Set the bits so they store −36 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?

N = −368 bit
U =
6.Number systemsHard
click the bits to set them012806403201608040201

Set the bits to the 8-bit result of 1000 1101 >> 3. What is the result's value R in decimal?

1000 1101>> 3
R =

3Solutions

1.IPv4 subnetting

Answer: H = 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).

Bits to borrow

24 = 16 ≥ 16 → b = 4, /28

Hosts per subnet

24 − 2 = 14

Answer

H = 14
2.IPv4 subnetting

Answer: x = 128

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: ⌊142/128⌋·128 = 128
broadcast: 128 + 128 − 1 = 255
usable hosts: 129 … 254

Answer

x = 128
3.Computer networks

Answer: A · 0–1023

Why this is the answer

Ports 0–1023 are well-known ports for standard services. 1024–49151 are registered ports and 49152–65535 are dynamic or private ports.

Answer

A · 0–1023
4.Computer networks

Answer: A · Transport layer

Why this is the answer

TCP and UDP headers contain source and destination ports; the internet layer (IP) adds addresses.

Answer

A · Transport layer
5.Number systems

Answer: U = 220

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

36 = 0010 0100
inverted: 1101 1011
+1 → 1101 1100

Read it as unsigned

1101 1100₂ = 256 − 36 = 220

Answer

U = 220
6.Number systems

Answer: R = 17

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 3 moves every bit 3 places to the right; bits that fall off the right end are lost and zeros come in on the left.

1000 1101 >> 3
= 0001 0001

Value of the result

0001 0001₂ = 17

Answer

R = 17

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