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
/?11111111111111111111110000000000network bitshost bits

A department needs 913 host addresses in a single subnet. What is the longest prefix length p (smallest subnet) that is big enough?

913 hosts
p =
2.IPv4 subnettingHard
last octet = 6012806403201608141201networkhost

A host has the address 10.20.156.6/28. What is the last number (fourth octet) x of its subnet's broadcast address?

10.20.156.6/28
x =
3.Computer networksMedium

How many usable host addresses does a /26 IPv4 subnet have?

  1. A126
  2. B64
  3. C30
  4. D62
4.Computer networksHard

In a SYN flood attack, what resource is the attacker trying to exhaust?

  1. AThe server's disk space
  2. BThe switch's MAC address table
  3. CThe DNS cache of the client
  4. DThe server's table of half-open connections
5.Number systemsMedium
8-bit two's complement0−12806403201608040201

Set the bits so they store −13 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?

N = −138 bit
U =
6.Number systemsHard
32 bits, unsigned, N = ?????????????????????????????????

What is the largest unsigned integer N that fits in 32 bits?

32 bit
N =

3Solutions

1.IPv4 subnetting

Answer: p = 22

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

29 − 2 = 510 < 913
210 − 2 = 1022 ≥ 913 → h = 10

Prefix length

p = 32 − 10 = 22

Answer

p = /22
2.IPv4 subnetting

Answer: x = 15

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: ⌊6/16⌋·16 = 0
broadcast: 0 + 16 − 1 = 15
usable hosts: 1 … 14

Answer

x = 15
3.Computer networks

Answer: D · 62

Why this is the answer

A /26 has 6 host bits: 2⁶ = 64 addresses, minus the network and broadcast addresses, gives 62 usable hosts.

Answer

D · 62
4.Computer networks

Answer: D · The server's table of half-open connections

Why this is the answer

Each SYN makes the server reserve state and wait for an ACK that never comes. SYN cookies avoid storing state until the handshake completes.

Answer

D · The server's table of half-open connections
5.Number systems

Answer: U = 243

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

13 = 0000 1101
inverted: 1111 0010
+1 → 1111 0011

Read it as unsigned

1111 0011₂ = 256 − 13 = 243

Answer

U = 243
6.Number systems

Answer: N = 4.295 × 10⁹

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.

Largest value

232 − 1 = 4294967295

Answer

N = 4294967295

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