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
255.255.240.011111111111111111111000000000000network bitshost bits

A subnet uses the mask 255.255.240.0. How many usable host addresses H does it have?

255.255.240.0
H =
2.IPv4 subnettingHard
last octet = 26012806403211618041201networkhost

A host has the address 10.20.17.26/26. What is the fourth octet x of the last usable host address in its subnet?

10.20.17.26/26
x =
3.Computer networksMedium

Which subnet mask corresponds to /20?

  1. A255.255.240.0
  2. B255.255.224.0
  3. C255.255.255.240
  4. D255.255.248.0
4.Computer networksHard

A TCP connection has a 64 KB window and a 100 ms round-trip time, and no window scaling. What is its approximate maximum throughput?

  1. AAbout 0.5 Mbit/s
  2. BAbout 5 Mbit/s
  3. CAbout 640 Mbit/s
  4. DAbout 50 Mbit/s
5.Number systemsMedium
8-bit two's complement0−12806403201608040201

Set the bits so they store −55 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?

N = −558 bit
U =
6.Number systemsHard
8 bit1 px1280 px720 pxM = ? KiB

An uncompressed image is 1280 × 720 pixels with 8 bits per pixel. How much memory M does it take, in KiB (1 KiB = 1024 bytes)?

1280 × 720 px8 bit/px1 KiB = 1024 B
M = KiB

3Solutions

1.IPv4 subnetting

Answer: H = 4094

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

255.255.240.0 = /20

Usable hosts

212 − 2 = 4094

Answer

H = 4094
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: ⌊26/64⌋·64 = 0
broadcast: 0 + 64 − 1 = 63
usable hosts: 1 … 62

Answer

x = 62
3.Computer networks

Answer: A · 255.255.240.0

Why this is the answer

/20 means 8 + 8 + 4 one-bits. Four one-bits in the third octet are 11110000 = 240, so the mask is 255.255.240.0.

Answer

A · 255.255.240.0
4.Computer networks

Answer: B · About 5 Mbit/s

Why this is the answer

Throughput ≤ window / RTT = 65,536 × 8 bits / 0.1 s ≈ 5.2 Mbit/s, however fast the link is.

Answer

B · About 5 Mbit/s
5.Number systems

Answer: U = 201

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

55 = 0011 0111
inverted: 1100 1000
+1 → 1100 1001

Read it as unsigned

1100 1001₂ = 256 − 55 = 201

Answer

U = 201
6.Number systems

Answer: M = 900 KiB

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.

Bits and bytes

1280·720·8 = 7372800 bit = 921600 B

Convert to KiB

921600/1024 = 900 KiB

Answer

M = 900 KiB

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