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 192.168.10.0/24 must be divided into at least 16 equal subnets. If you borrow as few bits as possible, how many usable hosts H does each subnet have?

192.168.10.0/2416 subnets
H =
2.IPv4 subnettingHard
third octet 61 … 66012806413211618140211commonvaries

A router has routes to these networks: 10.76.61.0/24, 10.76.62.0/24, 10.76.63.0/24, 10.76.64.0/24, 10.76.65.0/24, 10.76.66.0/24. What is the longest prefix length p of a single summary route that covers all of them?

10.76.61.0/24 … 10.76.66.0/24
p =
3.Computer networksMedium

TLS provides which security properties for HTTPS?

  1. AOnly faster loading
  2. BProtection against all malware and phishing on the visited site
  3. CAnonymity of the user
  4. DConfidentiality, integrity and server authentication
4.Computer networksHard

A student runs a game server on a PC behind a home NAT router. Friends on the internet cannot connect. What is the usual fix?

  1. AConfigure port forwarding on the router to the PC's private address
  2. BChange the PC's subnet mask to /32
  3. CGive the PC a different MAC address so the router can find it from outside
  4. DTurn off the PC's DNS client
5.Number systemsMedium
place values in two's complement1−12806413211608141201

The 8-bit pattern 1011 0110 stores a signed integer in two's complement. What is its value N in decimal?

1011 01108 bit
N =
6.Number systemsHard
click the bits to set them012806403201608040201

Set the bits to the 8-bit result of 1100 1111 OR 0110 1110. What is the result's value R in decimal?

1100 1111OR0110 1110
R =

3Solutions

1.IPv4 subnetting

Answer: H = 14

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).

Bits to borrow

24 = 16 ≥ 16 → b = 4, /28

Hosts per subnet

24 − 2 = 14

Answer

H = 14
2.IPv4 subnetting

Answer: p = 17

Route summarisation: a summary route a.b.c.0/p covers all networks whose first p bits match. Write the differing octet of the first and last network in binary; the number of common leading bits, plus the bits of the octets before it, is the longest summary prefix.

Compare the bits

61 = 00111101
66 = 01000010
common leading bits: 1

Prefix length

p = 16 + 1 = 17

Answer

p = /17

Summary route: 10.76.0.0/17 (it also covers 122 network(s) that are not in the list).

3.Computer networks

Answer: D · Confidentiality, integrity and server authentication

Why this is the answer

TLS encrypts traffic, detects tampering and authenticates the server with certificates (optionally the client too).

Answer

D · Confidentiality, integrity and server authentication
4.Computer networks

Answer: A · Configure port forwarding on the router to the PC's private address

Why this is the answer

NAT creates mappings only for connections started from inside. A port-forwarding rule maps a public port to the internal host so inbound connections can reach it.

Answer

A · Configure port forwarding on the router to the PC's private address
5.Number systems

Answer: N = −74

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 + 32 + 16 + 4 + 2 = −74

Answer

N = −74
6.Number systems

Answer: R = 239

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

OR gives 1 where at least one bit is 1.

1100 1111
OR 0110 1110
= 1110 1111

Value of the result

1110 1111₂ = 239

Answer

R = 239

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