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
Computer Networks: cram pack
Formula sheets, the key method steps for every topic and 6 practice problems, with full solutions at the back.
1Formulas and methods
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
A department needs 96 host addresses in a single subnet. What is the longest prefix length p (smallest subnet) that is big enough?
96 hosts
p =
2.IPv4 subnettingHard
A host has the address 10.20.232.41/28. What is the last number (fourth octet) x of its subnet's broadcast address?
10.20.232.41/28
x =
3.Computer networksMedium
What is the purpose of the TCP three-way handshake?
- ATo establish a connection and synchronise sequence numbers
- BTo find the IP address of a host before data can be sent to it
- CTo encrypt the data
- DTo close a connection quickly
4.Computer networksHard
Why does the side that closes a TCP connection first stay in the TIME_WAIT state for twice the maximum segment lifetime?
- ATo keep the encryption keys for later reuse
- BSo delayed segments die out and the final ACK can be resent if it was lost
- CTo wait for the DNS cache to expire
- DTo let the routers along the path update their routing tables for the next connection
5.Number systemsMedium
The 8-bit pattern 1110 1010 stores a signed integer in two's complement. What is its value N in decimal?
1110 10108 bit
N =
6.Number systemsHard
An uncompressed image is 640 × 480 pixels with 8 bits per pixel. How much memory M does it take, in KiB (1 KiB = 1024 bytes)?
640 × 480 px8 bit/px1 KiB = 1024 B
M = KiB
3Solutions
1.IPv4 subnetting
Answer: p = 25
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
26 − 2 = 62 < 96
27 − 2 = 126 ≥ 96 → h = 7
Prefix length
p = 32 − 7 = 25
Answer
p = /25
2.IPv4 subnetting
Answer: x = 47
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: ⌊41/16⌋·16 = 32
broadcast: 32 + 16 − 1 = 47
usable hosts: 33 … 46
Answer
x = 47
3.Computer networks
Answer: A · To establish a connection and synchronise sequence numbers
Why this is the answer
The handshake sets up a connection and agrees initial sequence numbers on both sides before data is sent.
Answer
A · To establish a connection and synchronise sequence numbers
4.Computer networks
Answer: B · So delayed segments die out and the final ACK can be resent if it was lost
Why this is the answer
TIME_WAIT ensures a lost final ACK can be repeated and old duplicate segments expire before the same address and port pair is reused.
Answer
B · So delayed segments die out and the final ACK can be resent if it was lost
5.Number systems
Answer: N = −22
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 + 64 + 32 + 8 + 2 = −22
Answer
N = −22
6.Number systems
Answer: M = 300 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
640·480·8 = 2457600 bit = 307200 B
Convert to KiB
307200/1024 = 300 KiB
Answer
M = 300 KiB
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