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
255.255.255.24811111111111111111111111111111000network bitshost bits

A subnet uses the mask 255.255.255.248. How many usable host addresses H does it have?

255.255.255.248
H =
2.IPv4 subnettingHard
third octet = 79012816403201618141211networkhost

A host has the address 172.20.79.27/18. What is the third octet x of its network address?

172.20.79.27/18
x =
3.Computer networksMedium

Which protocol does the "ping" command use?

  1. AHTTP
  2. BTCP
  3. CUDP
  4. DICMP
4.Computer networksHard

A path has 100 Mbit/s bandwidth and a 40 ms round-trip time. How much data must be in flight to keep it fully used?

  1. A2.5 MB
  2. B4 kB
  3. C50 kB
  4. D500 kB
5.Number systemsMedium
place values in two's complement1−12806403201618140211

The 8-bit pattern 1000 1101 stores a signed integer in two's complement. What is its value N in decimal?

1000 11018 bit
N =
6.Number systemsHard
click the bits to set them012806403201608040201

Set the bits to the 8-bit result of 1000 0100 << 3. What is the result's value R in decimal?

1000 0100<< 3
R =

3Solutions

1.IPv4 subnetting

Answer: H = 6

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

255.255.255.248 = /29

Usable hosts

23 − 2 = 6

Answer

H = 6
2.IPv4 subnetting

Answer: x = 64

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

26 = 64

Subnet of this host

network: 172.20.64.0
broadcast: 172.20.127.255

Answer

x = 64
3.Computer networks

Answer: D · ICMP

Why this is the answer

Ping sends ICMP echo requests and measures the time until echo replies return.

Answer

D · ICMP
4.Computer networks

Answer: D · 500 kB

Why this is the answer

The bandwidth-delay product is 100 × 10⁶ bit/s × 0.04 s = 4 × 10⁶ bits = 500 kB. The sender's window must be at least this large.

Answer

D · 500 kB
5.Number systems

Answer: N = −115

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 + 8 + 4 + 1 = −115

Answer

N = −115
6.Number systems

Answer: R = 32

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

Shifting left by 3 moves every bit 3 places to the left; bits that fall off the left end are lost and zeros come in on the right.

1000 0100 << 3
= 0010 0000

Value of the result

0010 0000₂ = 32

Answer

R = 32

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