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
/?11111111111111111111100000000000network bitshost bits

A department needs 1432 host addresses in a single subnet. What is the longest prefix length p (smallest subnet) that is big enough?

1432 hosts
p =
2.IPv4 subnettingHard
third octet = 103012816413201608141211networkhost

A host has the address 172.20.103.178/20. What is the third octet x of its subnet's broadcast address?

172.20.103.178/20
x =
3.Computer networksMedium

How long does it take to put a 1500-byte packet onto a 12 Mbit/s link (transmission delay)?

  1. A8 ms
  2. B12 ms
  3. C0.125 ms
  4. D1 ms
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 5 Mbit/s
  2. BAbout 0.5 Mbit/s
  3. CAbout 640 Mbit/s
  4. DAbout 50 Mbit/s
5.Number systemsMedium
8-bit two's complement?−128?64?32?16?8?4?2?1

Write −38 as an 8-bit two's complement pattern. Read the pattern as an unsigned binary number: what is its value U in decimal?

N = −388 bit
U =
6.Number systemsHard
32-bit two's complement, N = ??sign???????????????????????????????

What is the most negative integer N that can be stored in 32-bit two's complement?

32 bit
N =

3Solutions

1.IPv4 subnetting

Answer: p = 21

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

210 − 2 = 1022 < 1432
211 − 2 = 2046 ≥ 1432 → h = 11

Prefix length

p = 32 − 11 = 21

Answer

p = /21
2.IPv4 subnetting

Answer: x = 111

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 in the third octet

24 = 16

Subnet of this host

network: 172.20.96.0
broadcast: 172.20.111.255

Answer

x = 111
3.Computer networks

Answer: D · 1 ms

Why this is the answer

1500 bytes = 12,000 bits. 12,000 / 12,000,000 bit/s = 0.001 s = 1 ms. Transmission delay depends on packet size and link rate, not distance.

Answer

D · 1 ms
4.Computer networks

Answer: A · 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

A · About 5 Mbit/s
5.Number systems

Answer: U = 218

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.

Invert and add one

38 = 0010 0110
inverted: 1101 1001
+1 → 1101 1010

Read it as unsigned

1101 1010₂ = 256 − 38 = 218

Answer

U = 218
6.Number systems

Answer: N = −2.147 × 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.

Most negative value

−231 = −2147483648

Answer

N = −2147483648

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