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
The network 192.168.219.0/24 is divided into /25 subnets, numbered 1, 2, 3, … from the lowest address (subnet 1 starts at .0). What is the fourth octet x of the network address of subnet 2?
192.168.219.0/24/25subnet 2 (subnet 1 starts at .0)
x =
2.IPv4 subnettingHard
A host has the address 172.20.156.37/18. What is the third octet x of its subnet's broadcast address?
172.20.156.37/18
x =
3.Computer networksMedium
What is the purpose of the TCP three-way handshake?
- ATo close a connection quickly
- BTo encrypt the data
- CTo establish a connection and synchronise sequence numbers
- DTo find the IP address of a host before data can be sent to it
4.Computer networksHard
What does TCP fast retransmit do?
- ASkips the three-way handshake when reconnecting to known hosts, so data is resent sooner
- BResends every segment twice to be safe
- CDoubles the window after every loss
- DResends a segment after three duplicate ACKs, without waiting for the timeout
5.Number systemsMedium
Write −53 as an 8-bit two's complement pattern. Read the pattern as an unsigned binary number: what is its value U in decimal?
N = −538 bit
U =
6.Number systemsHard
Set the bits to the 8-bit result of 1101 1011 >> 1. What is the result's value R in decimal?
1101 1011>> 1
R =
3Solutions
1.IPv4 subnetting
Answer: x = 128
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
27 = 128
Subnet number 2
network: (2 − 1)·128 = 128
broadcast: 128 + 128 − 1 = 255
Answer
x = 128
2.IPv4 subnetting
Answer: x = 191
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: 172.20.128.0
broadcast: 172.20.191.255
Answer
x = 191
3.Computer networks
Answer: C · 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
C · To establish a connection and synchronise sequence numbers
4.Computer networks
Answer: D · Resends a segment after three duplicate ACKs, without waiting for the timeout
Why this is the answer
Three duplicate ACKs strongly suggest a single lost segment, so TCP retransmits it at once. This recovers much faster than waiting for the retransmission timer.
Answer
D · Resends a segment after three duplicate ACKs, without waiting for the timeout
5.Number systems
Answer: U = 203
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
53 = 0011 0101
inverted: 1100 1010
+1 → 1100 1011
Read it as unsigned
1100 1011₂ = 256 − 53 = 203
Answer
U = 203
6.Number systems
Answer: R = 109
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 right by 1 moves every bit 1 places to the right; bits that fall off the right end are lost and zeros come in on the left.
1101 1011 >> 1
= 0110 1101
Value of the result
0110 1101₂ = 109
Answer
R = 109
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