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 192.168.10.0/24 must be divided into at least 12 equal subnets. If you borrow as few bits as possible, how many usable hosts H does each subnet have?
192.168.10.0/2412 subnets
H =
2.IPv4 subnettingHard
A router has routes to these networks: 10.227.226.0/24, 10.227.227.0/24, 10.227.228.0/24, 10.227.229.0/24, 10.227.230.0/24, 10.227.231.0/24, 10.227.232.0/24. What is the longest prefix length p of a single summary route that covers all of them?
10.227.226.0/24 … 10.227.232.0/24
p =
3.Computer networksMedium
What is the purpose of the TCP three-way handshake?
- ATo find the IP address of a host before data can be sent to it
- BTo close a connection quickly
- CTo encrypt the data
- DTo establish a connection and synchronise sequence numbers
4.Computer networksHard
A remote-terminal application feels sluggish because small keystroke packets are held back. Which TCP mechanism is often responsible?
- ASlow start
- BThe three-way handshake
- CFast retransmit
- DNagle's algorithm
5.Number systemsMedium
Set the bits so they store −92 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?
N = −928 bit
U =
6.Number systemsHard
Set the bits to the 8-bit result of 1000 1100 OR 1101 1011. What is the result's value R in decimal?
1000 1100OR1101 1011
R =
3Solutions
1.IPv4 subnetting
Answer: H = 14
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).
Bits to borrow
24 = 16 ≥ 12 → b = 4, /28
Hosts per subnet
24 − 2 = 14
Answer
H = 14
2.IPv4 subnetting
Answer: p = 20
Route summarisation: a summary route a.b.c.0/p covers all networks whose first p bits match. Write the differing octet of the first and last network in binary; the number of common leading bits, plus the bits of the octets before it, is the longest summary prefix.
Compare the bits
226 = 11100010
232 = 11101000
common leading bits: 4
Prefix length
p = 16 + 4 = 20
Answer
p = /20
Summary route: 10.227.224.0/20 (it also covers 9 network(s) that are not in the list).
3.Computer networks
Answer: D · 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
D · To establish a connection and synchronise sequence numbers
4.Computer networks
Answer: D · Nagle's algorithm
Why this is the answer
Nagle's algorithm waits to bundle small segments while earlier data is unacknowledged. This saves bandwidth but adds delay for interactive traffic.
Answer
D · Nagle's algorithm
5.Number systems
Answer: U = 164
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
92 = 0101 1100
inverted: 1010 0011
+1 → 1010 0100
Read it as unsigned
1010 0100₂ = 256 − 92 = 164
Answer
U = 164
6.Number systems
Answer: R = 223
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
OR gives 1 where at least one bit is 1.
1000 1100
OR 1101 1011
= 1101 1111
Value of the result
1101 1111₂ = 223
Answer
R = 223
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