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
A department needs 1016 host addresses in a single subnet. What is the longest prefix length p (smallest subnet) that is big enough?
1016 hosts
p =
2.IPv4 subnettingHard
A host has the address 10.20.197.84/29. What is the last number (fourth octet) x of its subnet's broadcast address?
10.20.197.84/29
x =
3.Computer networksMedium
What is a cyclic redundancy check (CRC) used for in Ethernet frames?
- ADetecting transmission errors
- BCorrecting every bit error automatically
- CEncrypting the frame
- DChoosing the best route
4.Computer networksHard
Why does a browser warn when a site presents a self-signed certificate?
- AThe site must be using HTTP instead of HTTPS
- BSelf-signed certificates use no encryption at all, so the traffic is sent in clear text
- CThe certificate is always expired
- DNo trusted certificate authority vouches that the key belongs to that site
5.Number systemsMedium
Write −119 as an 8-bit two's complement pattern. Read the pattern as an unsigned binary number: what is its value U in decimal?
N = −1198 bit
U =
6.Number systemsHard
An uncompressed image is 1280 × 720 pixels with 24 bits per pixel. How much memory M does it take, in KiB (1 KiB = 1024 bytes)?
1280 × 720 px24 bit/px1 KiB = 1024 B
M = KiB
3Solutions
1.IPv4 subnetting
Answer: p = 22
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).
Host bits needed
29 − 2 = 510 < 1016
210 − 2 = 1022 ≥ 1016 → h = 10
Prefix length
p = 32 − 10 = 22
Answer
p = /22
2.IPv4 subnetting
Answer: x = 87
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: ⌊84/8⌋·8 = 80
broadcast: 80 + 8 − 1 = 87
usable hosts: 81 … 86
Answer
x = 87
3.Computer networks
Answer: A · Detecting transmission errors
Why this is the answer
The sender computes a CRC over the frame; the receiver recomputes it and discards the frame if they differ. CRC detects errors but does not correct them.
Answer
A · Detecting transmission errors
4.Computer networks
Answer: D · No trusted certificate authority vouches that the key belongs to that site
Why this is the answer
Browsers trust certificates signed, directly or through a chain, by CAs in their trust store. A self-signed certificate could be made by anyone, such as a man-in-the-middle.
Answer
D · No trusted certificate authority vouches that the key belongs to that site
5.Number systems
Answer: U = 137
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
119 = 0111 0111
inverted: 1000 1000
+1 → 1000 1001
Read it as unsigned
1000 1001₂ = 256 − 119 = 137
Answer
U = 137
6.Number systems
Answer: M = 2700 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
1280·720·24 = 22118400 bit = 2764800 B
Convert to KiB
2764800/1024 = 2700 KiB
Answer
M = 2700 KiB
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