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
/?11111111111111111111100000000000network bitshost bits

A department needs 1579 host addresses in a single subnet. What is the longest prefix length p (smallest subnet) that is big enough?

1579 hosts
p =
2.IPv4 subnettingHard
third octet = 52012806413211608140201networkhost

A host has the address 172.20.52.225/21. What is the third octet x of its subnet's broadcast address?

172.20.52.225/21
x =
3.Computer networksMedium

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

  1. A1024–49151
  2. B49152–65535
  3. C0–65535
  4. D0–1023
4.Computer networksHard

Which mechanism prevents broadcast loops in a network with redundant switch links?

  1. ASpanning Tree Protocol
  2. BARP caching
  3. CNAT
  4. DDHCP snooping
5.Number systemsMedium
8-bit two's complement0−12806403201608040201

Set the bits so they store −10 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?

N = −108 bit
U =
6.Number systemsHard
click the bits to set them012806403201608040201

Set the bits to the 8-bit result of 1000 1111 >> 1. What is the result's value R in decimal?

1000 1111>> 1
R =

3Solutions

1.IPv4 subnetting

Answer: p = 21

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

210 − 2 = 1022 < 1579
211 − 2 = 2046 ≥ 1579 → h = 11

Prefix length

p = 32 − 11 = 21

Answer

p = /21
2.IPv4 subnetting

Answer: x = 55

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: 172.20.48.0
broadcast: 172.20.55.255

Answer

x = 55
3.Computer networks

Answer: D · 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

D · 0–1023
4.Computer networks

Answer: A · Spanning Tree Protocol

Why this is the answer

STP elects a root bridge and blocks redundant paths, unblocking them if an active link fails.

Answer

A · Spanning Tree Protocol
5.Number systems

Answer: U = 246

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

10 = 0000 1010
inverted: 1111 0101
+1 → 1111 0110

Read it as unsigned

1111 0110₂ = 256 − 10 = 246

Answer

U = 246
6.Number systems

Answer: R = 71

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

1000 1111 >> 1
= 0100 0111

Value of the result

0100 0111₂ = 71

Answer

R = 71

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