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 → /2511111111111111111111111110000000network bitshost bits

The network 192.168.167.0/24 is divided into /25 subnets, numbered 1, 2, 3, … from the lowest address (subnet 1 starts at .0). What is the fourth octet x of the broadcast address of subnet 2?

192.168.167.0/24/25subnet 2 (subnet 1 starts at .0)
x =
2.IPv4 subnettingHard
last octet = 200112816403201618040201networkhost

A host has the address 10.20.56.200/28. What is the fourth octet x of its network address?

10.20.56.200/28
x =
3.Computer networksMedium

What is the IPv6 loopback address?

  1. Afe80::1
  2. Bff02::1
  3. C::1
  4. D::
4.Computer networksHard

Stations A and C both reach access point B but cannot hear each other, so their frames collide at B. How does 802.11 help?

  1. AOptional RTS/CTS exchanges: the AP's CTS tells all nearby stations to wait
  2. BBy increasing the TTL of the frames
  3. CBy making A and C use collision detection so they stop sending when frames overlap
  4. DBy routing A's frames through C
5.Number systemsMedium
8-bit two's complement0−12806403201608040201

Set the bits so they store −79 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?

N = −798 bit
U =
6.Number systemsHard
click the bits to set them012806403201608040201

Set the bits to the 8-bit result of 1000 0000 XOR 0101 0100. What is the result's value R in decimal?

1000 0000XOR0101 0100
R =

3Solutions

1.IPv4 subnetting

Answer: x = 255

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 number 2

network: (2 − 1)·128 = 128
broadcast: 128 + 128 − 1 = 255

Answer

x = 255
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: ⌊200/16⌋·16 = 192
broadcast: 192 + 16 − 1 = 207
usable hosts: 193 … 206

Answer

x = 192
3.Computer networks

Answer: C · ::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

C · ::1
4.Computer networks

Answer: A · Optional RTS/CTS exchanges: the AP's CTS tells all nearby stations to wait

Why this is the answer

A station sends Request to Send; the AP answers Clear to Send, which every station in its range hears, so hidden stations defer for the reserved time.

Answer

A · Optional RTS/CTS exchanges: the AP's CTS tells all nearby stations to wait
5.Number systems

Answer: U = 177

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

79 = 0100 1111
inverted: 1011 0000
+1 → 1011 0001

Read it as unsigned

1011 0001₂ = 256 − 79 = 177

Answer

U = 177
6.Number systems

Answer: R = 212

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

XOR gives 1 where the two bits differ.

1000 0000
XOR 0101 0100
= 1101 0100

Value of the result

1101 0100₂ = 212

Answer

R = 212

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