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 10.180.188.0/24 is divided into equal subnets with prefix /28. How many subnets N are there?
10.180.188.0/24/28
N =
2.IPv4 subnettingHard
A router has routes to these networks: 10.89.39.0/24, 10.89.40.0/24, 10.89.41.0/24, 10.89.42.0/24, 10.89.43.0/24. What is the longest prefix length p of a single summary route that covers all of them?
10.89.39.0/24 … 10.89.43.0/24
p =
3.Computer networksMedium
What is the main purpose of a VLAN?
- ATo encrypt all traffic between switches so other departments cannot read it
- BTo give each device a public IP address
- CTo increase the speed of every switch port
- DTo split one physical switched network into separate broadcast domains
4.Computer networksHard
What is the maximum transmission unit (MTU) of standard Ethernet payload?
- A1500 bytes
- B65,535 bytes
- C9000 bytes in every network
- D64 bytes
5.Number systemsMedium
Set the bits so they store −81 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?
N = −818 bit
U =
6.Number systemsHard
Set the bits to the 8-bit result of 1001 0000 OR 1001 1010. What is the result's value R in decimal?
1001 0000OR1001 1010
R =
3Solutions
1.IPv4 subnetting
Answer: N = 16
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
28 − 24 = 4
Number of subnets
N = 24 = 16
Answer
N = 16
Each /28 subnet has 24 − 2 = 14 usable hosts.
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
39 = 00100111
43 = 00101011
common leading bits: 4
Prefix length
p = 16 + 4 = 20
Answer
p = /20
Summary route: 10.89.32.0/20 (it also covers 11 network(s) that are not in the list).
3.Computer networks
Answer: D · 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
D · To split one physical switched network into separate broadcast domains
4.Computer networks
Answer: A · 1500 bytes
Why this is the answer
Standard Ethernet carries up to 1500 bytes of payload; larger IP packets must be fragmented or use path MTU discovery.
Answer
A · 1500 bytes
5.Number systems
Answer: U = 175
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
81 = 0101 0001
inverted: 1010 1110
+1 → 1010 1111
Read it as unsigned
1010 1111₂ = 256 − 81 = 175
Answer
U = 175
6.Number systems
Answer: R = 154
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.
1001 0000
OR 1001 1010
= 1001 1010
Value of the result
1001 1010₂ = 154
Answer
R = 154
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