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 192.168.47.0/24 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 broadcast address of subnet 9?

192.168.47.0/24/28subnet 9 (subnet 1 starts at .0)
x =
2.IPv4 subnettingHard
last octet = 232112816413201618040201networkhost

A host has the address 10.20.16.232/29. What is the fourth octet x of the last usable host address in its subnet?

10.20.16.232/29
x =
3.Computer networksMedium

How does TLS typically use asymmetric and symmetric cryptography?

  1. AAsymmetric methods agree keys and authenticate; symmetric ciphers encrypt the data
  2. BOnly symmetric cryptography is used, with a fixed shared password
  3. CSymmetric methods agree keys and authenticate; asymmetric ciphers encrypt all the data
  4. DOnly asymmetric cryptography is used throughout
4.Computer networksHard

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

  1. AThe connection stays encrypted without TLS
  2. BSeveral requests reuse one TCP connection, avoiding repeated handshakes
  3. CThe server can push updates over UDP
  4. DResponses can arrive in any order on one stream, so slow replies do not block others
5.Number systemsMedium
place values in two's complement1−12816413201618141201

The 8-bit pattern 1110 1110 stores a signed integer in two's complement. What is its value N in decimal?

1110 11108 bit
N =
6.Number systemsHard
click the bits to set them012806403201608040201

Set the bits to the 8-bit result of 0100 0001 XOR 0010 0101. What is the result's value R in decimal?

0100 0001XOR0010 0101
R =

3Solutions

1.IPv4 subnetting

Answer: x = 143

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 9

network: (9 − 1)·16 = 128
broadcast: 128 + 16 − 1 = 143

Answer

x = 143
2.IPv4 subnetting

Answer: x = 238

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

23 = 8

Subnet of this host

network: ⌊232/8⌋·8 = 232
broadcast: 232 + 8 − 1 = 239
usable hosts: 233 … 238

Answer

x = 238
3.Computer networks

Answer: A · Asymmetric methods agree keys and authenticate; symmetric ciphers encrypt the data

Why this is the answer

Public-key methods and certificates set up a shared secret and prove identity. Fast symmetric ciphers such as AES then protect the bulk data.

Answer

A · Asymmetric methods agree keys and authenticate; symmetric ciphers encrypt the data
4.Computer networks

Answer: B · 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

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

Answer: N = −18

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 + 8 + 4 + 2 = −18

Answer

N = −18
6.Number systems

Answer: R = 100

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.

0100 0001
XOR 0010 0101
= 0110 0100

Value of the result

0110 0100₂ = 100

Answer

R = 100

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