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 30 equal subnets. If you borrow as few bits as possible, how many usable hosts H does each subnet have?
192.168.10.0/2430 subnets
H =
2.IPv4 subnettingHard
A host has the address 10.20.25.107/29. What is the fourth octet x of its network address?
10.20.25.107/29
x =
3.Computer networksMedium
Why is UDP often preferred for live video calls?
- ALow delay matters more than retransmitting every lost packet
- BUDP encrypts data automatically
- CUDP is required for all video
- DUDP guarantees every packet arrives, so the picture never freezes
4.Computer networksHard
Hosts in VLAN 10 and VLAN 20 on the same switch cannot communicate. What is needed?
- AA router or layer 3 switch to route between the VLANs
- BA hub connected to both VLANs
- CThe same MAC address on both hosts
- DA longer cable between the hosts
5.Number systemsMedium
Write −22 as an 8-bit two's complement pattern. Read the pattern as an unsigned binary number: what is its value U in decimal?
N = −228 bit
U =
6.Number systemsHard
Set the bits to the 8-bit result of 0110 0101 OR 0111 0000. What is the result's value R in decimal?
0110 0101OR0111 0000
R =
3Solutions
1.IPv4 subnetting
Answer: H = 6
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
25 = 32 ≥ 30 → b = 5, /29
Hosts per subnet
23 − 2 = 6
Answer
H = 6
2.IPv4 subnetting
Answer: x = 104
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
23 = 8
Subnet of this host
network: ⌊107/8⌋·8 = 104
broadcast: 104 + 8 − 1 = 111
usable hosts: 105 … 110
Answer
x = 104
3.Computer networks
Answer: A · Low delay matters more than retransmitting every lost packet
Why this is the answer
Retransmitting lost packets would add delay; real-time media tolerates some loss, so UDP-based protocols like RTP are used.
Answer
A · Low delay matters more than retransmitting every lost packet
4.Computer networks
Answer: A · A router or layer 3 switch to route between the VLANs
Why this is the answer
VLANs are isolated at layer 2, so traffic between them must be routed, for example by a router on a trunk ("router on a stick") or a layer 3 switch.
Answer
A · A router or layer 3 switch to route between the VLANs
5.Number systems
Answer: U = 234
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
22 = 0001 0110
inverted: 1110 1001
+1 → 1110 1010
Read it as unsigned
1110 1010₂ = 256 − 22 = 234
Answer
U = 234
6.Number systems
Answer: R = 117
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.
0110 0101
OR 0111 0000
= 0111 0101
Value of the result
0111 0101₂ = 117
Answer
R = 117
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