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
255.255.252.011111111111111111111110000000000network bitshost bits

A subnet uses the mask 255.255.252.0. How many usable host addresses H does it have?

255.255.252.0
H =
2.IPv4 subnettingHard
last octet = 28012806403211618140201networkhost

A host has the address 10.20.68.28/26. What is the fourth octet x of the last usable host address in its subnet?

10.20.68.28/26
x =
3.Computer networksMedium

What is the IPv6 loopback address?

  1. Aff02::1
  2. B::1
  3. C::
  4. Dfe80::1
4.Computer networksHard

A PC can reach printers on its own subnet but nothing beyond it. Its IP address and mask are correct. What is the most likely fault?

  1. AA missing or wrong default gateway
  2. BA duplicate MAC address on the internet
  3. CA faulty DNS server setting
  4. DThe TCP window is too small
5.Number systemsMedium
8-bit two's complement0−12806403201608040201

Set the bits so they store −66 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?

N = −668 bit
U =
6.Number systemsHard
click the bits to set them012806403201608040201

Set the bits to the 8-bit result of 1111 0110 << 2. What is the result's value R in decimal?

1111 0110<< 2
R =

3Solutions

1.IPv4 subnetting

Answer: H = 1022

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).

Prefix length

255.255.252.0 = /22

Usable hosts

210 − 2 = 1022

Answer

H = 1022
2.IPv4 subnetting

Answer: x = 62

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

26 = 64

Subnet of this host

network: ⌊28/64⌋·64 = 0
broadcast: 0 + 64 − 1 = 63
usable hosts: 1 … 62

Answer

x = 62
3.Computer networks

Answer: B · ::1

Why this is the answer

::1 is the loopback address. :: is the unspecified address, fe80::/10 is link-local and ff02::1 is the all-nodes multicast group.

Answer

B · ::1
4.Computer networks

Answer: A · A missing or wrong default gateway

Why this is the answer

Hosts send off-subnet packets to the default gateway. If it is wrong or missing, local traffic still works but nothing outside is reachable, even by IP address.

Answer

A · A missing or wrong default gateway
5.Number systems

Answer: U = 190

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

66 = 0100 0010
inverted: 1011 1101
+1 → 1011 1110

Read it as unsigned

1011 1110₂ = 256 − 66 = 190

Answer

U = 190
6.Number systems

Answer: R = 216

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 2 moves every bit 2 places to the left; bits that fall off the left end are lost and zeros come in on the right.

1111 0110 << 2
= 1101 1000

Value of the result

1101 1000₂ = 216

Answer

R = 216

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