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 192.168.252.0/24 is divided into /28 subnets, numbered 1, 2, 3, … from the lowest address (subnet 1 starts at .0). What is the fourth octet x of the broadcast address of subnet 12?
192.168.252.0/24/28subnet 12 (subnet 1 starts at .0)
x =
2.IPv4 subnettingHard
A host has the address 10.20.217.109/28. What is the fourth octet x of the last usable host address in its subnet?
10.20.217.109/28
x =
3.Computer networksMedium
A browser receives HTTP status code 404. What does it mean?
- AThe client must log in first
- BThe requested resource was not found
- CThe request succeeded
- DThe server had an internal error
4.Computer networksHard
Hosts in VLAN 10 and VLAN 20 on the same switch cannot communicate. What is needed?
- AThe same MAC address on both hosts
- BA router or layer 3 switch to route between the VLANs
- CA longer cable between the hosts
- DA hub connected to both VLANs
5.Number systemsMedium
Set the bits so they store −33 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?
N = −338 bit
U =
6.Number systemsHard
An uncompressed image is 512 × 512 pixels with 16 bits per pixel. How much memory M does it take, in KiB (1 KiB = 1024 bytes)?
512 × 512 px16 bit/px1 KiB = 1024 B
M = KiB
3Solutions
1.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
24 = 16
Subnet number 12
network: (12 − 1)·16 = 176
broadcast: 176 + 16 − 1 = 191
Answer
x = 191
2.IPv4 subnetting
Answer: x = 110
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
24 = 16
Subnet of this host
network: ⌊109/16⌋·16 = 96
broadcast: 96 + 16 − 1 = 111
usable hosts: 97 … 110
Answer
x = 110
3.Computer networks
Answer: B · The requested resource was not found
Why this is the answer
404 Not Found means the server has no resource at that URL. 200 is OK, 401 is Unauthorized and 500 is Internal Server Error.
Answer
B · The requested resource was not found
4.Computer networks
Answer: B · 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
B · A router or layer 3 switch to route between the VLANs
5.Number systems
Answer: U = 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.
Invert and add one
33 = 0010 0001
inverted: 1101 1110
+1 → 1101 1111
Read it as unsigned
1101 1111₂ = 256 − 33 = 223
Answer
U = 223
6.Number systems
Answer: M = 512 KiB
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.
Bits and bytes
512·512·16 = 4194304 bit = 524288 B
Convert to KiB
524288/1024 = 512 KiB
Answer
M = 512 KiB
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