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 10.190.0.0/20 is divided into equal subnets with prefix /23. How many subnets N are there?
10.190.0.0/20/23
N =
2.IPv4 subnettingHard
A router has routes to these networks: 10.133.14.0/24, 10.133.15.0/24, 10.133.16.0/24, 10.133.17.0/24, 10.133.18.0/24, 10.133.19.0/24, 10.133.20.0/24, 10.133.21.0/24. What is the longest prefix length p of a single summary route that covers all of them?
10.133.14.0/24 … 10.133.21.0/24
p =
3.Computer networksMedium
What is the main purpose of a VLAN?
- ATo increase the speed of every switch port
- BTo give each device a public IP address
- CTo split one physical switched network into separate broadcast domains
- DTo encrypt all traffic between switches so other departments cannot read it
4.Computer networksHard
What is the main benefit of persistent connections (keep-alive) in HTTP/1.1?
- AResponses can arrive in any order on one stream, so slow replies do not block others
- BThe server can push updates over UDP
- CSeveral requests reuse one TCP connection, avoiding repeated handshakes
- DThe connection stays encrypted without TLS
5.Number systemsMedium
Set the bits so they store −48 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?
N = −488 bit
U =
6.Number systemsHard
Set the bits to the 8-bit result of 0101 1110 << 2. What is the result's value R in decimal?
0101 1110<< 2
R =
3Solutions
1.IPv4 subnetting
Answer: N = 8
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).
Borrowed bits
23 − 20 = 3
Number of subnets
N = 23 = 8
Answer
N = 8
Each /23 subnet has 29 − 2 = 510 usable hosts.
2.IPv4 subnetting
Answer: p = 19
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
14 = 00001110
21 = 00010101
common leading bits: 3
Prefix length
p = 16 + 3 = 19
Answer
p = /19
Summary route: 10.133.0.0/19 (it also covers 24 network(s) that are not in the list).
3.Computer networks
Answer: C · To split one physical switched network into separate broadcast domains
Why this is the answer
VLANs group ports logically, so broadcasts stay within each VLAN. This improves security and limits broadcast traffic.
Answer
C · To split one physical switched network into separate broadcast domains
4.Computer networks
Answer: C · Several requests reuse one TCP connection, avoiding repeated handshakes
Why this is the answer
Reusing a connection avoids a new handshake and slow start for each object, so pages with many resources load faster.
Answer
C · Several requests reuse one TCP connection, avoiding repeated handshakes
5.Number systems
Answer: U = 208
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
48 = 0011 0000
inverted: 1100 1111
+1 → 1101 0000
Read it as unsigned
1101 0000₂ = 256 − 48 = 208
Answer
U = 208
6.Number systems
Answer: R = 120
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 2 moves every bit 2 places to the left; bits that fall off the left end are lost and zeros come in on the right.
0101 1110 << 2
= 0111 1000
Value of the result
0111 1000₂ = 120
Answer
R = 120
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